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Problems in
Chemistry
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Problems in
Chemistry
Ranjeet Shahi
ARIHANT PRAKASHAN (Series), MEERUT
Arihant Prakashan (Series), Meerut
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PREFACE
This work is the result of my teaching the JEE aspirants over the post
decade and a half. Often I have found that the students mess up with
the chemical reactions given in the problem and get themselves
entangled into them. The reason is that they are not able to
distinguish the relationships between the various chemical principles
and concepts. This book is an effort to make them master the
fundamentals of Chemistry through problems-solving, and apply
them intelligently to arrive at a correct solution.
The problems included in the book with their solutions, aim to give
you the mastery required for solving the intricate problems asked in
the exams such as JEE.
Critical analysis of the situation is required whatever be format of the
questions. This book just gives emphasis on this, and it is hoped that it
will give a boost to all the meritorious and hard-wroking students.
I shall like to thank Mr Yogesh Chand Jain, Chairman, Arihant Group,
for bringing the book in this nice form.
Suggestions for further improvement of the book are welcome.
Ranjeet Shahi
CONTENTS
1. Mole Concept
3
2. Acid-Base Titration
7
3. Redox Titration
12
4. Gaseous State
16
5. Thermochemistry
20
6. Thermodynamics
26
7. Atomic Structure
33
8. Chemical Bonding
36
9. Chemical Equilibrium
39
10. Ionic Equilibrium
47
11. Electrochemistry
57
12. Chemical Kinetics
67
13. Colligative Properties
75
14. Solid State
78
15. Surface Chemistry
80
16. Reaction Mechanism
81
17. Stereochemistry
95
18. Hydrocarbons
103
19. Alkyl Halides
120
20. Alcohols and Ethers
133
21. Aldehydes and Ketones
161
22. Carboxylic Acids and Its Derivatives
193
23. Amines
213
24. Aromatic Compounds
216
25. Carbohydrates, Amino Acids and Polymers
264
26. Miscellaneous
265
27. Qualitative Analysis
271
28. Co-odination Compounds
278
29. Representative Elements
285
30. Metallurgy
292
Solutions
295
Problems in Chemistry
3
MOLE CONCEPT
PROBLEM 1 A crystalline hydrated salt on being rendered anhydrous, loses 45.6% of its weight. The
percentage composition of anhydrous salt is: Al = 10.5%, K = 15.1%, S = 24.8% and O = 49.6%. Find
the empirical formula of the anhydrous and crystalline salt.
PROBLEM 2 How much quantity of zinc will have to be reacted with excess of dilute HCl solution to
produce sufficient hydrogen gas for completely reacting with the oxygen obtained by decomposing
5.104 g of potassium chlorate?
PROBLEM 3 A 1.85 g sample of mixture of CuCl 2 and CuBr 2 was dissolved in water and mixed
thoroughly with 1.8 g portion of AgCl. After reaction, the solid which now contain AgCl and AgBr was
filtered, dried and weighed to be 2.052 g. What was the % by weight of CuBr 2 in the mixture?
PROBLEM 4 1.0 g of a sample containing NaCl, KCl and some inert impurity is dissolved in excess of
water and treated with excess of AgNO 3 solution. A 2.0 g precipitate of AgCl separate out. Also sample
is 23% by mass in sodium. Determine mass percentage of KCl in the sample.
PROBLEM 5 A one gram sample containing CaBr 2 , NaCl and some inert impurity was dissolved in
enough water and treated with excess of aqueous silver nitrate solution where a mixed precipitate of
AgCl and AgBr weighing 1.94 g was obtained. Precipitate was washed, dried and shaken with an
aqueous solution of NaBr where all AgCl was converted into AgBr. The new precipitate which contain
only AgBr now weighed to be 2.4 g. Determine mass percentage of CaBr 2 and NaCl in the original
sample.
PROBLEM 6 Sulphur combines with oxygen to form two oxide SO 2 and SO 3 . If 10 g of S is mixed with
12 g of O 2 , what mass of SO 2 and SO 3 will be formed, so that neither S nor oxygen will be left at the end
of reaction?
PROBLEM 7 An aqueous solution of ethanol has density 1.025 g/mL and it is 8.0 M. Determine
molality m of this solution.
PROBLEM 8 An aqueous solution of acetic acid has density 1.12 g/mL and it is 5.0 m. Determine
molarity (M).
PROBLEM 9 Octane is a component of gasoline. Incomplete combustion of octane produces some CO
along with CO 2 and H 2O, which reduces efficiency of engine. In a certain test run, 1.0 gallon of octane is
burned and total mass of CO, CO 2 and H 2O produced was found to be 11.53 kg. Calculate efficiency of
the engine, density of octane is 2.65 kg/gallon.
PROBLEM 10 The formula of a hydrated salt of barium is BaCl 2 ⋅ xH 2O. If 1.936 g of this compound
gives 1.846 g of anhydrous BaSO 4 upon treatment with H 2SO 4 , calculate x.
PROBLEM 11 A mixture of CuSO 4 ⋅ 5H 2O and MgSO 4 ⋅ 7H 2O was heated until all the water was
driven-off. If 5.0 g of mixture gave 3 g of anhydrous salts, what was the percentage by mass of
CuSO 4 ⋅ 5H 2O in the original mixture?
PROBLEM 12 A sample of clay contain 15% moisture, and rest are CaCO 3 and non-volatile SiO 2 . This
on heating loses part of its moisture, but CaCO 3 is completely converted into CaO. The partially dried
4
Problems in Chemistry
sample now contain 7.35% moisture and 51.5% SiO 2 . Determine mass percentage of CaCO 3 in the
original sample.
PROBLEM 13 Chlorine dioxide (ClO 2 ), has been used as a disinfectant in air conditioning systems. It
reacts with water according to the reaction:
ClO 2 + H 2O → HClO 3 + HCl
In an experiment, a 10.0 L sealed flask containing ClO 2 and some inert gas at 300 K and 1.0
atmosphere pressure is opened in a bath containing excess of water and all ClO 2 is reacted quantitatively.
The resulting solution required 200 mL 0.9 M NaOH solution for neutralization. Determine mole
fraction of ClO 2 in the flask.
PROBLEM 14 Potassium salt of benzoic acid (C 6 H 5COOK) can be made by the action of potassium
permanganate on toluene as follows:
C 6 H 5CH 3 + KMnO 4 → C 6 H 5COOK + MnO 2 + KOH + H 2O
If the yield of potassium benzoate can’t realistically be expected to be more than 71%, what is the
minimum number of grams of toluene needed to achieve this yield while producing 11.5 g of
C 6 H 5COOK?
PROBLEM 15 Manganese trifluoride can be prepared by the following reaction:
MnI 2 ( s) + F2 ( g ) → MnF3 + IF5
What is minimum number of grams of F2 that must be used to react with 12.0 g of MnI 2 if overall
yield of MnF3 is no more than 75%.
PROBLEM 16 A compound containing Ca, C, N and S was subjected to quantitative analysis and
formula mass determination. A 0.25 g of this compound was mixed with Na 2CO 3 to convert all Ca into
0.16 g CaCO 3 . A 0.115 g sample of compound was carried through a series of reactions until all its S was
changed into SO 2–
4 and precipitated as 0.344 g of BaSO 4 . A 0.712 g sample was processed to liberate all
of its N as NH 3 and 0.155 g NH 3 was obtained. The formula mass was found to be 156. Determine the
empirical and molecular formula of the compound.
PROBLEM 17 A 0.2 g sample, which is mixture of NaCl, NaBr and NaI was dissolved in water and
excess of AgNO 3 was added. The precipitate containing AgCl, AgBr and AgI was filtered, dried and
weighed to be 0.412 g. The solid was placed in water and treated with excess of NaBr, which converted
all AgCl into AgBr. The precipitate was then weighed to be 0.4881 g. It was then placed into water and
treated with excess of NaI, which converted all AgBr into AgI. The precipitate was then weighed to be
0.5868 g. What was the percentage of NaCl, NaBr and NaI in the original mixture.
PROBLEM 18 A mixture of NaI and NaCl when heated with H 2SO 4 produced same weight of Na 2SO 4
as that of original mixture. Calculate mass percentage of NaI in the original mixture.
PROBLEM 19 Ammonia is manufactured by the reaction of N 2 and H 2 . An equilibrium mixture
contains 5.0 g of each N 2 , H 2 and NH 3 . Calculate mass of N 2 and H 2 present initially and maximum
amount of NH 3 that can be produced.
PROBLEM 20 Consider the following reactions:
XeF2 + F2 → XeF6
and
XeF6 + —( CH 2 —CH 2—
) n → —( CF2 —CF2 —
) n + HF + XeF4
Determine mass of F2 ( g ) required for preparation of 1.0 kg fluorinated polymer.
5
Problems in Chemistry
PROBLEM 21 2.5 g of a sample containing Na 2CO 3 ; NaHCO 3 and some non-volatile impurity on
gentle heating loses 12% of its weight. Residue is dissolved in 100 mL water and its 10 mL portion
required 15 mL 0.1 M aqueous solution of BaCl 2 for complete precipitation of carbonates. Determine
mass percentage of Na 2CO 3 in the original sample.
PROBLEM 22 2.0 g of a sample containing NaCl, NaBr and some inert impurity is dissolved in enough
water and treated with excess of AgNO 3 solution. A 3.0 g of precipitate was formed. Precipitate on
shaking with aqueous NaBr gain 0.76 g of weight. Determine mass percentage of NaCl in the original
sample.
PROBLEM 23 Based on the following information, determine value of x and y:
AgNO3
(CH 3 ) x AlCl y → xCH 4 ( g ) + yCl – + Al 3+ → AgCl( s)
0.643 g
0.222 g
0.996 g
PROBLEM 24 An organic compound containing C, H, O, N and Cl was analyzed and 0.15 g of sample
on combustion produced 0.138 g of CO 2 and 0.0566 g of H 2O. All the nitrogen in different 2.0 g sample
of compound was converted into NH 3 which was found to weigh 0.238 g. Finally the chlorine in a 0.125
g sample of compound was converted to Cl – and by reacting with AgNO 3 , 0.251 g AgCl was obtained.
Deduce the empirical formula of the starting organic compound.
PROBLEM 25 A 5.0 g sample of felspar containing Na 2O, K 2O and some inert impurity is dissolved in
dilute HCl solution and NaCl and KCl formed are separated by fractional crystallization. During
crystallization some less soluble impurities also comes out. Mass of NaCl, KCl and impurity
accompanying these salts was found to be 6.47 g. Solid crystal was then re-dissolved and required 300
mL of 0.3 M AgNO 3 for complete precipitation of chlorides. The precipitate thus, obtained was found to
contain 4.23% insoluble impurity. Determine mass percentage of Na 2O and K 2O in the original sample.
PROBLEM 26 Potassium chlorate (KClO 4 ) is made in the following sequence of reactions:
Cl 2 ( g ) + KOH → KCl + KClO + H 2O
KClO → KCl + KClO 3
KClO 3 → KClO 4 + KCl
What mass of Cl 2 is needed to produce 1.0 kg of KClO 4 ?
PROBLEM 27 Titanium oxide (TiO 2 ) is heated in stream of hydrogen to give water and a new oxide
Ti x O y . If 1.598 g TiO 2 produces 1.438 g Ti x O y , what is the formula of new oxide.
PROBLEM 28 A solution of copper sulphate that contain 15% CuSO 4 by weight has a density of 1.169
g/mL. 25 mL portion of this solution was reacted with excess of ammonia solution to form a dark blue
solution. When cooled, filtered and dried, 6.127 g of dark blue solid was obtained. A 0.195g solid was
analyzed for ammonia and required 30.63 mL of 0.1036 M HCl solution to reach the equivalence point.
In a separate analysis, 0.200 g was heated at 110°C to drive off water, producing 0.185 g of anhydrous
material. Deduce formula of the compound crystallized out from blue solution assuming that it contain
only one copper atom per formula unit. Also determine the percentage yield of crystallization process.
PROBLEM 29 0.1152 g of a compound containing carbon, hydrogen, nitrogen and oxygen are burned in
oxygen. The gases produced are treated further to convert nitrogen containing product into N 2 . The
resulting mixture of CO 2 , H 2O and N 2 is passed through a CaCl 2 drying tube, which gains 0.09912 g.
The gas stream was then bubbled through water where the CO 2 forms H 2CO 3 . Titration of this solution
required 28.8 ml 0.3283 M NaOH solution to reach the phenolphthalein end point. The excess O 2 was
6
Problems in Chemistry
removed by reaction with copper metal and the N 2 was collected in a 225 mL measuring bulb where it
exerted a pressure of 65.12 mm of Hg at 25°C. In a separate analysis, the molar mass of this compound
was found to be 146 g mol –1 . Deduce molecular formula of the starting compound.
PROBLEM 30 Pb(NO 3 ) 2 and KI reacts in aqueous solution to form an yellow precipitate of PbI 2 . In
one series of experiments, the masses of two reactants varied, but the total mass of the two was held
constant at 5.0 g. What maximum mass of PbI 2 can be produced in the above experiment ?
PROBLEM 31 An element X react with hydrogen leading to formation of a class of compounds that is
analogous to hydrocarbons. 5 g of X forms 5.628 g of a mixture of two compounds of X, XH 4 and X 2 H 6
in the molar ratio of 2 : 1. Determine molar mass of X.
PROBLEM 32 The mineral Argyrodite is a stoichiometric compound that contain silver, sulphur (– 2)
and an unknown element Y ( + 4). The mass-ratio of silver and Y in the compound is,
m( Ag) : m(Y ) = 11.88
Y forms a reddish brown lower sulphide on heating the mineral in stream of H 2 ( g ), in which Y is in
+ 2 state. The residue are Ag 2S and H 2S. To convert 10 g Argyrodite completely, 0.295 L of H 2 ( g )
measured at 400K and 1.0 atmosphere is required. Determine molar mass of Y and empirical formula of
mineral.
PROBLEM 33 Uranium is isolated from its ore by dissolving it as UO 2 (NO 3 ) 2 and separating it as
solid UO 2 (C 2O 4 ) ⋅xH 2O. A 1.0 g sample of ore on treatment with nitric acid yielded 1.48 g UO 2 (NO 3 ) 2
which on further treatment with 0.4 g Na 2C 2O 4 yielded 1.23 g UO 2 (C 2O 4 ) ⋅ xH 2O. Determine weight
percentage of uranium in the original sample and x.
PROBLEM 34 When iodine was added to liquid chlorine in cold condition, orange crystal of a
compound separate out. The amount of chlorine in a sample of crystal was determined by precipitating
AgCl. A 0.467 g sample of crystal gave 0.861 g of AgCl. Deduce empirical formula of the crystal.
PROBLEM 35 Urea is manufactured on large scale by passing CO 2 (g ) through ammonia solution
followed by crystallization. CO 2 for the above reaction is prepared by combustion of hydrocarbons. If
combustion of 236 kg of a saturated hydrocarbon produces as much CO 2 as required for production of
1000 kg of urea, deduce molecular formula of hydrocarbon.
PROBLEM 36 Sodium bicarbonate can be purified by dissolving it in hot water (at 60°C), filtering to
remove insoluble impurities, cooling to 0°C to precipitate solid NaHCO 3 and the filtering to remove the
solid leaving soluble impurities in solution. Some NaHCO 3 that remain in solution is not recovered. The
solubility of NaHCO 3 in water at 60°C is 164 g/L. Its solubility in cold water at 0ºC is 69 g/L. If a 250 g
impure sample of NaHCO 3 was purified by this method by dissolving first in 250 mL water at 60°C and
then crystallizing NaHCO 3 from 100 mL water at 0°C, 150 g NaHCO 3 was recovered. Determine
percentage purity of original sample.
PROBLEM 37 A 100 g solution was prepared by dissolving 46 g CuSO 4⋅xH 2O in 54 g of water and
mole fraction of CuSO 4 in solution was found to be 0.05. Determine x.
PROBLEM 38 An ore of iron contain FeS and non-volatile impurity. Roasting of this ore converts all
FeS into Fe 2 O3 and a 4% loss in weight was observed. Determine mass percentage of FeS in ore.
PROBLEM 39 Optical measurement is a very efficient method of determining molar mass of unknown
material. In one experiment, 3.0 g of an unknown polymeric material was dissolved in 100 mL of CCl 4
and transmittance of this solution was found to be 72%. Transmittance of a 0.001 M standard solution in
7
Problems in Chemistry
the same solvent, under identical experimental condition was 60%. Determine molar mass of unknown
polymer.
PROBLEM 40
shown below,
A crystalline polymer molecule is uniform prismatic in shape with dimensions as
300 Å
100Å
If density of this polymer is 1.2 g/cm 3 , determine molar mass.
PROBLEM 41 A mother cell disintegrate into sixty identical cells and each daughter cell further
disintegrate into 24 smaller cells. The smallest cells are uniform cylindrical in shape with diameter of
120 Å and each cell is 6000 Å long. Determine molar mass of the mother cell if density of the smallest
cell is 1.12 g/cm 3 .
PROBLEM 42 A sample of rock taken for analysis weigh 1.0 g on air dried basis. After drying for one
hour at 110°C, the sample weigh 0.9437 g. The calcium is precipitated as oxalate but weighed as CaSO 4 ;
that weigh 0.5g. The magnesium is precipitated as MgNH 4 PO4 which finally ignited to 0.5 g Mg 2 P2O 7 .
Find the percentage of CaO and MgO on oven dried basis and percentage of them and H 2O on air dried
basis.
PROBLEM 43 A sample is a mixture of Mohr’s salt and (NH 4 ) 2 SO 4 . A 0.5 g sample on treatment with
excess of BaCl 2 solution gave 0.75 g BaSO 4 . Determine percentage composition of the salt mixture.
What weight of Fe 2O 3 would be obtained if 0.2 g of the sample were ignited in air?
PROBLEM 44 A chloride mixture is prepared by grinding together pure BaCl 2⋅2H 2O, KCl and NaCl.
What is the smallest and largest volume of 0.15 M AgNO 3 solution that may be used for complete
precipitation of chloride from a 0.3g sample of the mixture which may contain any one or all of the
constituents?
ACID-BASE TITRATION
PROBLEM 45 A 1.5 g sample containing oxalic acid and some inert impurity was dissolved in enough
water and volume made up to 250 mL. A 20 mL portion of this solution was then mixed with 30 mL of an
alkali solution. The resulting solution was then treated with stoichiometric amount of CaCl 2 just needed
for precipitation of oxalate as CaC 2O 4 . Solution was filtered off and filtrate was finally titrated against
0.1 M HCl solution. 8.0 mL of acid was required to reach the equivalence point. At last, the above neutral
solution was treated with excess of AgNO 3 solution and AgCl obtained was washed, dried and weighed
to be 0.4305 g. Determine mass percentage of oxalic acid in the original sample.
PROBLEM 46 A 1.5 g sample containing P2O 3 and some inert impurity was dissolved in enough water
and boiled gently where P2O 3 disproportionated quantitatively into PH 3 and H 3 PO 4 . The solution was
8
Problems in Chemistry
further boiled for some time to let-off all PH 3 ( g ) and finally cooled to room temperature and diluted to
100 mL. A 10 mL portion of this solution was then mixed with 20 mL 0.3 M NaOH solution. Excess
alkali required 11.0 mL 0.05 M H 2SO 4 solution for back titration. Determine mass percentage of P2O 3 in
the original sample.
PROBLEM 47 2.5 g of a mixture containing CaCO 3 , Ca(HCO 3 ) 2 and NaCl was dissolved in 100 mL
water and its 10 mL portion required 10 mL 0.05 M H 2SO 4 solution to reach the phenolphthalein end
point. An another 10 mL portion of the same stock solution required 32.35 mL of the same acid solution
to reach the methyl orange end point. Determine mass percentage of CaCO 3 and Ca(HCO 3 ) 2 in the
original mixture.
PROBLEM 48 A solution contain both Na 2CO 3 and NaHCO 3 . 10 mL portion of this solution is mixed
with few drops of phenolphthalein indicator and titrated against 0.08 M H 2SO 4 solution. 7.0 mL of acid
was required to reach the end point A 5.0 mL portion of this solution was then taken for further analysis
and a few drops of methyl orange was added to it and finally titrated against same acid solution. 3.53 mL
of acid was required to reach the end point. Determine mass of Na 2CO 3 and NaHCO 3 per litre of
solution. Ignore volume change due to addition of indicator.
PROBLEM 49 A mixture was known to contain both KNO 3 and K 2SO 3 . To 0.486 g of the mixture,
dissolved in enough water to give 50 mL solution, was added 50 mL of 0.15 M HCl solution. The
reaction mixture was heated to expel all SO 2 and then 25 mL of the reaction mixture was titrated with 0.1
M KOH. The titration required 13.11 mL of the base. Calculate mass percentage of K 2SO 3 in the
mixture.
PROBLEM 50 An amino acid isolated from a piece of animal tissue was believed to be glycine. A 0.05 g
sample was treated in such a way that all nitrogen in it was converted into ammonia. This ammonia was
added to 50 mL of 0.05 M HCl solution. The excess acid remaining in the solution required 30.57 mL
0.06 M NaOH solution for complete neutralization. What was the percentage by mass of nitrogen? How
does this mass compare with percentage mass of nitrogen calculated from glycine (H 2 NCH 2COOH)?
PROBLEM 51 In a reaction, calcium orthophosphate on heating with magnesium produced calcium
phosphide, magnesium metaphosphate, calcium oxide and oxygen gas. Phosphide on hydrolysis
produces PH 3 gas. The PH 3 gas is burnt completely to P2O 5 using air, which contains 21%, by volume
of oxygen. Calculate the volume of air at STP required for combustion, if 2.4 g Mg was initially reacted
with calcium orthophosphate. All volumes are measured at STP.
PROBLEM 52 9.3 g of a mixture containing Li 2CO 3 , NaHCO 3 , Na 2CO 3 on strong heating produced
7.37 g of solid residue. The residue is dissolved in 200 mL water.
A 10 mL portion of this solution is mixed with 15 mL of a normal HCl solution. The excess acid
required 12 mL 0.5 N NaOH solution to reach the equivalence point. Determine the mass percentage of
NaHCO 3 and Na 2CO 3 in the original mixture. Li = 7, Na = 23.
PROBLEM 53 4.0 g of a monobasic, saturated carboxylic acid is dissolved in 100 mL water and its 10
mL portion required 8.0 mL 0.27 M NaOH to reach the equivalence point. In an another experiment, 5.0
g of the same acid is burnt completely and CO 2 produced is absorbed completely in 500 mL of a 2.0 N
NaOH solution. A 10 mL portion of the resulting solution is treated with excess of BaCl 2 to precipitate
all carbonate and finally titrated with 0.5 N H 2SO 4 solution. Determine the volume of the acid solution
that would be required to make this solution neutral.
PROBLEM 54 5.0 g of a mixture containing NaHCO 3 , NaCl and Na 2CO 3 is dissolved in 500 mL water
and its 10 mL portion required 12.4 mL 0.1 M HCl solution to reach the equivalence point. In an another
Problems in Chemistry
9
experiment, 10 mL portion of the same stock solution is mixed with 10 mL 0.15 M NaOH solution.
Excess NaOH required 12.6 mL 0.1 M HCl solution for back titration. Determine the mass percentage of
each component in the original mixture.
PROBLEM 55 6.4 g of a pure monobasic organic acid is burnt completely in excess of oxygen and CO 2
evolved is absorbed completely in one litre of an aqueous solution of NaOH. A 10 mL portion of this
solution required 14.5 mL of a normal HCl solution to reach the phenolphthalein end point. An another
10 mL portion of the same solution required 18 mL of the same HCl solution to reach the methyl orange
end point. If the organic acid contains 25% oxygen by weight, deduce the empirical formula of this acid
and strength of original NaOH solution.
PROBLEM 56 A complex of cobalt with ammonia is analyzed for determining its formula, by titrating it
against a standardized acid as follows:
Co(NH 3 ) x Cl 3 ( aq ) + HCl → NH +4 ( aq ) + Co 3+ ( aq ) + Cl – ( aq )
A 1.58 g complex required 23.63 mL 1.5 M HCl to reach the equivalence point. Determine formula.
If the reaction mixture at equivalence point is treated with excess of AgNO 3 solution, what mass of AgCl
will precipitate out?
PROBLEM 57 One litre solution of alkali is prepared by dissolving impure solid of alkali which contain
5% Na 2CO 3 and 8% CaCO 3 and 10% NaCl. A 10 mL portion of this solution required 9.8 mL of a 0.5 M
H 2SO 4 solution for neutralization. Calculate weight of alkali dissolved initially.
PROBLEM 58 40 g of a sample of caustic soda containing NaOH, Na 2CO 3 and inert impurity is
dissolved in water to prepare 1.0 litre solution. A 25 mL portion of this solution required 23.15 mL 1.022
N HCl for complete neutralization. To 25 mL another solution, excess of BaCl 2 is added, and resulting
solution required 22.55 mL HCl of same strength to reach the end point. Calculate mass percentage of
NaOH and Na 2CO 3 in the original sample.
PROBLEM 59 1.5 g of a sample containing Na 2CO 3 and NaHCO 3 is dissolved in 100 mL of water. A
25 mL portion of this solution required 22.45 mL 0.202 N HCl using methyl orange as indicator. In a
separate analysis, 25 mL portion of the same stock solution is mixed with 30 mL 0.204 N NaOH and then
excess of BaCl 2 is added resulting in precipitation of all carbonate as BaCO 3 . Filtrate required 9.98 mL
HCl of same strength. Calculate mass percentage of Na 2CO 3 and NaHCO 3 in the mixture.
PROBLEM 60 One gram sample of a saturated hydrocarbon is burned completely and liberated CO 2
was absorbed in a 1.0 L 0.2 N NaOH solution. To the resulting solution, excess of BaCl 2 crystals was
added and the solution was filtered off to free from BaCO 3 . A 10 mL portion of the extract required 12
mL 0.025 M H 2SO 4 solution for neutralization. Determine molecular formula of the hydrocarbon.
PROBLEM 61 2.0 g of a saturated, monobasic carboxylic acid was burned and liberated CO 2 was passed
through a concentrated solution of NaOH. The resulting solution was separated into two equal half and
analyzed. One half required 71.72 mL 1.0 N HCl to reach the end point in presence of phenolphthalein
indicator. The other half required 123.44 mL 1.0 N HCl to reach the end point in presence of methyl
orange indicator. Deduce formula of acid and determine mass of NaOH present initially.
PROBLEM 62 2.5 g of a mixture containing NaHCO 3 , Na 2CO 3 and NaCl is dissolved in 100 mL water
and its 50 mL portion required 13.33 mL 1.0 N HCl solution to reach the equivalence point. On the other
hand its other 50 mL portion required 19 mL 0.25 M NaOH solution to reach the equivalence point.
Determine mass percentage of each component.
10
Problems in Chemistry
PROBLEM 63 2.0 g of a crystal of CaCO 3 is dissolved in 50 mL water and then mixed with 50 mL of a
HCl solution. The resulting solution is boiled to remove all CO 2 and its 10 mL portion required 8.0 mL of
a NaOH solution to make the solution neutral. Also 20 mL of original HCl solution is equivalent to 96
mL of NaOH solution. Determine molarity of both NaOH and HCl solution.
PROBLEM 64 2.725 g of a mixture of K 2C 2O 4 , KHC 2O 4 and H 2C 2O 4 ⋅ 2H 2O is dissolved in 100 mL
H 2O and its 10 mL portion is titrated with 0.1 N HCl solution.
20 mL acid was required to reach the equivalence point. In another experiment, 10 mL portion of the
same stock solution is titrated with 0.1 N KOH solution. 20 mL of base was required to reach the
equivalence point. Determine mass percentage of each component in the mixture.
PROBLEM 65 A 1.0 g sample containing NH 4 NO 3 , (NH 4 ) 3 PO 4 and some inert impurity was
dissolved in 100 mL water its 10 mL portion required 15 mL 0.1 M NaOH solution to reach the
equivalence point. In a separate experiment, 10 mL of the same stock solution was treated with excess of
BaCl 2 solution and 0.077 g of barium phosphate precipitate was obtained. Determine mass percentage of
ammonium nitrate in the original sample.
PROBLEM 66 10.38 mg of a diprotic acid (containing (C, H and O) is burned completely and all CO 2
was absorbed in 100 mL of alkali solution. The resulting solution is separated into two-half and one-half
required 55 mL 0.005 M H 2SO 4 solution to reach the phenolphthalein end point. Other half was titrated
in presence of methyl orange indicator and 80 mL H 2SO 4 solution of same strength was required to
reach the end point. In a separate analysis, 0.168 g of the same acid required 16.18 mL 0.125 M NaOH
solution to reach the end point. Deduce formula of the acid and determine molarity of alkali solution used
initially.
PROBLEM 67 A 3.0 g sample containing Na 2CO 3 , NaHCO 3 , NaCl and some inert impurity was
dissolved in 100 mL of water and its 10 mL portion was titrated against 0.1 M HCl solution using
phenolphthalein indicator. 11.32 mL of acid solution was required to reach the end point. The resulting
solution was then mixed with excess of AgNO 3 solution resulting in formation of 0.306 g of AgCl
precipitate. The solution was filtered-off and filtrate was again titrated, but now against 0.05 M NaOH
solution. 42.64 mL of alkali was required to reach the end point. Determine mass percentage of Na 2CO 3 ,
NaHCO 3 and NaCl in the original sample.
PROBLEM 68 In neutralization titration of Na 3 PO 4 , if phenolphthalein is used as indicator, end point is
indicated only when Na 3 PO 4 is converted into Na 2 HPO 4 while, if methyl orange is used as indicator,
end point appear only when Na 3 PO 4 is converted into H 3 PO 4 . In an experiment a 4.0 g mixture
containing Na 3 PO 4 , Na 2 HPO 4 and NaH 2 PO 4 is dissolved in 50 mL water and its 10 mL portion
required 24.4 mL 0.1 M HCl solution to reach the end point using phenolphthalein indicator. In a
separate analysis, 10 mL portion of the same stock solution required 23.572 mL 0.5 M HCl solution to
reach the end point using methyl orange as indicator. Determine mass percentage of all components in
the mixture.
PROBLEM 69 A mixture containing LiHCO 3 , NaCl and Na 2CO 3 on gentle heating loses 26.5% of its
weight. 5.0 g of this mixture was heated gently and residue was dissolved in 100 mL water. A 10 mL
portion of this solution was then treated with 20 mL 0.2 M H 2SO 4 solution. A 10 mL portion of the
resulting solution required 3.86 mL 0.1 M NaOH solution to reach the end point. Determine mass
percentage of each component in the mixture.
PROBLEM 70 A mixture containing LiHCO 3 , NaHCO 3 and CaCO 3 on gentle heating loses 48.4% of
its weight. In an experiment, 5.0 g of this mixture was dissolved in 100 mL water and its 10 mL portion
was treated with 10 mL 0.5 M NaOH solution. The resulting solution was then treated with excess of
Problems in Chemistry
11
BaCl 2 solution resulting in precipitation of all carbonates as BaCO 3 . Precipitate was separated out by
filtration and filtrate required 15.3 mL 0.1 N HCl solution to reach the end point. Determine mass
percentage of all components present in the mixture.
PROBLEM 71 5.0 g of a mixture containing NaCl, NaHCO 3 , Na 2CO 3 and CaCO 3 on gentle heating
reduces to 4.25 g of solid residue. In a separate experiment, 1.0 g of the same mixture required 10 mL 0.2
M NaOH to reach the end point. In a 3rd experiment, 1.0 g of the same mixture was dissolved in 100 mL
water and required 10 mL 1.053 M HCl solution to reach the end point. Determine mass percentage of
each component in the mixture.
PROBLEM 72 2.0 g of a sample of CaCO 3 , NaHCO 3 and some volatile, inert impurity, was heated
strongly where CaCO 3 and NaHCO 3 , were decomposed into CaO and Na 2CO 3 respectively and all CO 2
gas produced in decomposition was absorbed in a 50 mL NaOH solution. NaOH was little less than the
stoichiometric requirement therefore, CO 2 during reaction with NaOH, produced Na 2CO 3 and some
NaHCO 3 . The resulting solution was titrated first in presence of phenolphthalein indicator and 5.0 mL
1.0 M HCl was required to reach the phenolphthalein end point. Methyl orange was then added and
titration continued with HCl of same strength where 15 mL HCl was required to reach the final end point.
On the other hand, the residue obtained after heating of the original sample was dissolved in water
and treated with excess of BaCl 2 , giving 0.985 g of BaCO 3 precipitate. Determine mass percentage of
CaCO 3 and NaHCO 3 in the original sample.
PROBLEM 73 A one gram sample containing NaOH as the only basic substance and some inert
impurity was left exposed to atmosphere for a very long time so that part of NaOH got converted into
Na 2CO 3 by absorbing CO 2 from atmosphere. The resulting sample was dissolved in water and volume
made upto 100 mL. A 100 mL portion of this solution required 16 mL 0.25 M HCl solution to reach the
equivalence point when methyl orange was used as indicator. In a separate analysis, 20 mL portion of the
same solution was taken alongwith phenolphthalein indicator and mixed with 50 mL of 0.1 M HCl
solution. An additional 9.00 mL 0.1 M Ba(OH) 2 solution was required to just restore the pink colour of
solution. Determine mass percentage of NaOH in the original sample and mass percentage of Na 2CO 3 in
the sample after exposure to atmosphere.
PROBLEM 74 The monochloroacetic acid (ClCH 2COOH) preservative in a 100 mL of carbonated
beverage was extracted by shaking with dimethyl ether and then returned to aqueous solution as
ClCH 2COO – by extraction with 1.0 M NaOH. This solution was acidified and treated with 50 mL
0.0452 M AgNO 3 solution where the following reaction occurred:
ClCH 2COOH + AgNO 3 + H 2O → HOCH 2COOH + H + + NO –3 + AgCl( s)
After filtering the AgCl, titration of filtrate required 10.43 mL of an NH 4SCN solution. Titration of a
blank taken through the entire procedure used 22.98 mL of same NH 4SCN solution. Calculate weight in
mg, of ClCH 2COOH in the beverage sample.
PROBLEM 75 2.0 g of a sample containing sodium oxalate, oxalic acid dihydrate and some inert
impurity was dissolved in 100 mL water and its 20 mL portion required 23.34 mL 0.04 M acidified
permanganate solution to reach the equivalence point. In a separate analysis, 20 mL portion of the same
stock solution required 26.67 mL 0.1 N NaOH solution to reach the end point. Determine mass
percentage of Na 2C 2O 4 and H 2C 2O 4⋅2H 2O in the original sample.
PROBLEM 76 A 1.5 g sample containing (NH 4 ) 2 SO 4 , NH 4 NO 3 and some inert impurity was
dissolved in water and volume made upto 100 mL. A 20 mL portion of this solution was mixed with 50
mL 0.1 M NaOH solution. A 30 mL aliquot of this resulting solution required 9.00 mL 1/28 M H 2SO 4
solution for complete neutralization. In a separate analysis, 32 mL of the original stock solution on
12
Problems in Chemistry
treatment with excess of BaCl 2 solution produced 0.466 g BaSO 4 precipitate. Determine mass
percentage of NH 4 NO 3 and (NH 4 ) 2 SO 4 in the original sample.
PROBLEM 77 A 1.0 g impure sample containing [Zn(NH 4 ) 4 ]Cl 2 and some inert impurity was treated
with 15 mL of 1 M NaOH solution where all complex is converted into Na 2 [Zn(OH) 4 ] . The excess base
1
required 10 mL M HCl solution for back titration.
6
(a) Determine percentage purity.
(b) If the last solution obtained after neutralization was treated with excess of AgNO 3 , what weight
of AgCl would have been produced?
PROBLEM 78 1.2 g of a salt with their empirical formula K x H y (C 2O 4 ) z was dissolved in 50 mL of
water and its 10 mL portion required 11.00 mL of a 0.1 M HCl solution to reach the equivalence point. In
a separate analysis, 15 mL of the stock solution required 20 mL 0.2475 M KOH to reach the equivalence
point. Determine empirical formula of the salt.
PROBLEM 79 Impure phosphoric acid for use in the manufacture of fertilizer is produced by the
reaction of sulphuric acid on phosphate rock of which a principal component is Ca 3 (PO 4 ) 2 and rest are
silica and other inert impurity. In an analysis, 2.0 g of a sample of rock salt was dissolved in 100 mL
H 2SO 4 solution. Excess sulphuric acid left in 20 mL of this solution required 40 mL 0.02 M NaOH for
back titration. In a separate analysis 20 mL of the above solution required 50 mL 0.04 M NaOH for
complete neutralization. Determine mass percentage of Ca 3 (PO 4 ) 2 in rock-sample.
PROBLEM 80 A 10 g sample of ammonium perchlorate containing some inert impurity was mixed
with 3 g Al powder where all perchlorate reacted to produce Al 2O 3 , N 2 , HCl and H 2O. All HCl was
absorbed in 100 mL 1 M NaOH solution. Determine percentage purity of perchlorate sample and volume
of 0.5 M HCl required to neutralize the above solution.
PROBLEM 81 Potassium superoxide (KO 2 ) is utilized in closed system breathing apparatus to remove
CO 2 and water from exhaled air. The removal of H 2O generate oxygen gas and KOH and this KOH in the
subsequent step remove CO 2 as KHCO 3 . 5.0 kg of an impure sample of KO 2 is just sufficient to remove
all CO 2 and H 2O from a closed room of dimension 10 m × 5 m × 3m. Determine mass of this KO 2
required to neutralize a 100 mL 0.1 M H 2SO 4 solution in a separate analysis. Assume room conditions to
be at 1.0 atmosphere and 300 K and mole fraction of CO 2 in that room is 0.01.
PROBLEM 82 3.25 g of a saturated, tribasic carboxylic acid required 68.4 mL of a 0.750 M NaOH
solution to reach the equivalence point. Determine molecular formula of acid.
REDOX TITRATION
PROBLEM 83
A sample of chrome-vanadium steel weighing 2.0 g was dissolved in a mixture of
sulphuric acid and just sufficient oxidant was added to raise the oxidation state of iron to Fe 3+ , the
chromium to Cr 2O 72– , vanadium to VO –3 and Mn to MnO –4 . The solution was then treated with HCl and
–
resulting solution still containing Fe 3+ , Cr 2O 2–
7 and VO 3 then treated with 25 mL of 0.101 M FeSO 4 .
This resulted in reduction of dichromate and VO –3 to Cr 3+ and VO 2+ in the solution respectively. Fe 2+
and VO 2+ in the solution was then titrated with 0.02236 M KMnO 4 and required 12.6 mL to reach the
13
Problems in Chemistry
equivalence point. A small amount of Fe 2+ was then added to again reduce the VO –3 produced by
KMnO 4 back to VO 2+ and this then titrated directly with 0.02236 M KMnO 4 , a process requiring 0.86
mL to reach the equivalence point. Calculate the following quantities:
(a) Moles of Fe 2+ in 25 mL sample of standard FeSO 4 solution.
(b) Moles of Fe 2+ titrated with 12.6 mL of standard KMnO 4 .
(c) Moles of Fe 2+ consumed by Cr 2O 2–
7 .
(d) Percentage of V and Cr in the steel
[Atomic weight of V = 51, Cr = 52]
PROBLEM 84 A sample of crude uranium oxide is known to be contaminated with iron. To determine
the extent of contamination, the crude oxide were dissolved and reduced with Zn to yield a solution
containing U 4+ and Fe 2+ . A 20 mL aliquot of this solution was treated with cupferron which precipitated
all uranium and the resulting precipitate on ignition yielded 423.3 mg of U 3O 8 . A further 20 mL sample
was treated with 0.024 M KMnO 4 solution and consumed 27.23 mL. Calculate mass percentage of
contamination if the iron were present as Fe 2O 3 in a sample of crude oxide containing 100 g of U 3O 8 .
KMnO 4 solution oxidised Fe 2+ to Fe 3+ and U 4+ to UO 2+
2 . Atomic mass of U = 238.
PROBLEM 85 A 5.0 g sample containing Pb 3O 4 , PbO 2 and some inert impurity is dissolved in 250 mL
dil. HNO 3 solution and 2.7 g of Na 2C 2O 4 was added so that all lead converted into Pb 2+ . A 10 mL
portion of this solution required 8.0 mL 0.02 M KMnO 4 for titration of excess of oxalate. In an another
experiment, 25 mL of solution was taken and excess oxalate was removed by extraction, this required 10
mL of a permanganate solution for oxidation of Pb 2+ to Pb 4+ . 10 mL this permanganate solution is
equivalent to 4.48 mL 5 V H 2O 2 solution. Calculate mass percentage of PbO 2 and Pb 3O 4 in the original
sample.
[Atomic mass of Pb = 207]
PROBLEM 86 An unknown cupric salt with formula Cu x (CO 3 ) y (OH) z is analyzed to determine the
exact formula. A 1.7225 g sample of salt was dissolved in 100 mL of pure water. A 50 mL portion of this
solution required 10 mL 1.0 N H 2SO 4 solution to reach the equivalence point if phenolphthalein was
used as indicator. Another 50 mL portion was titrated using methyl orange as indicator and 15 mL acid of
same strength was required. Deduce the formula of the salt.
PROBLEM 87 Both CaCl 2 and NaCl are used to melt ice and snow on roads in winter. A certain
company was marketing a mixture of these two compounds for this purpose. A chemist, wishing to
analyze the mixture, dissolved 1 g of it in water and precipitated the calcium by adding sodium oxalate.
The calcium oxalate was then carefully filtered, dissolved in dilute sulphuric acid, and titrated with 0.1
M KMnO 4 solution. The titration required 22 mL of the KMnO 4 solution. Calculate freezing point of an
aqueous solution which is 5% (w/V) of the above mixture. K f of water is 1.86 K kg mol –1 .
PROBLEM 88 A 4.25 g sample containing CaC 2O 4 , Na 2C 2O 4 and some inert impurity is heated gently
so that CaC 2O 4 decomposed as:
CaC 2O 4 → CaO + CO( g ) + CO 2 ( g )
All gaseous products were passed through a NaOH solution where following reaction occurred
quantitatively:
2NaOH + CO 2 ( g ) → Na 2CO 3
14
Problems in Chemistry
The resulting solution is separated into two equal part (by volume) and one part required 30 mL
0.5 M HCl to reach the phenolphthalein end point while the other half solution required 50 mL 0.5 M
HCl solution to reach the methyl orange end point. In a separate experiment same mass of the same
sample is dissolved into 100 mL dilute HCl solution and its 10 mL portion required 10 mL 0.1 M
K 2Cr 2O 7 solution. Determine the mass percentage of Na 2C 2O 4 in the original sample.
PROBLEM 89 In acidic solution, 45 mL KMnO 4 solution is required to react with 50 mL 0.25 N
Na 2C 2O 4 solution. How many mL of this same KMnO 4 solution would be required to oxidise 25 mL 0.1
N K 2C 2O 4 solution in alkaline medium where KMnO 4 is reduced to MnO 2 .
PROBLEM 90 A sample weighing 0.3 g containing K 3 [Fe(C 2O 4 ) 3 ] ⋅ 3H 2O,FeCl 3 ⋅ 6H 2O and inert
impurity is dissolved in dilute sulphuric acid and volume made up to 100 mL. A 20 mL portion of this
solution required 3.75 mL of 0.005 M acidified KMnO 4 solution to reach the equivalence point. In an
another experiment, 50 mL sample of the same stock solution is treated with Zn-amalgum and the
resulting solution required 17.5 mL of permanganate solution of same strength. Determine mass
percentage of FeCl 3 ⋅ 6H 2O in the original sample.
PROBLEM 91 A 6.1 g sample containing oxalic acid dihydrate, sodium oxalate and NaHC 2O 4 and is
dissolved in 100 mL of water and its 10.0 mL portion required 16 mL 0.25 M HCl to reach the
equivalence point. In another experiment 10.0 mL portion of the same stock solution required 24 mL
0.25 M NaOH to reach the equivalence point. Determine the mass percentage of all components in the
original mixture.
PROBLEM 92 A 0.127 g of an unsaturated oil was treated with 25 mL of 0.1 M ICl solution. The
unreacted ICl was then treated with excess of KI. Liberated iodine required 40 mL 0.1 M hypo solution.
Determine mass of I 2 that would have been required with 100.0 g oil if I 2 were used in place of ICl.
PROBLEM 93 Alkali metal nitrate on heating decomposes to metal nitrite and oxygen whereas alkaline
earth metal on heating decomposes into metal oxide, NO 2 and oxygen. In an experiment 15 g mixture of
NaNO 3 and Mg(NO 3 ) 2 was heated until no more gas were evolved. The water soluble part of residue
was used for analysis and dissolved in 1.0 litre water. 10 mL portion of this solution was reacted with 20
mL 0.02 M acidified KMnO 4 solution. The excess reagent required 10.00 mL 0.05 M oxalic acid
solution. Determine mass percentage of each nitrate in the mixture. Also determine the molar ratio of
oxygen to NO 2 in the gaseous products given off.
PROBLEM 94 The mass percentage of MnO 2 in a sample of mineral is determined by reacting with
As 2O 3 in acid solution. A 0.225 g sample of mineral is ground and boiled with 75 mL 0.0125 M As 2O 3
solution. After the reaction is complete, the solution is cooled and titrated with 2.28 ×10 –3 M acidified
KMnO 4 solution. 16.34 mL of the oxidizing agent solution was required to reach the end point.
Determine mass percentage of MnO 2 in the sample.
PROBLEM 95 A driver is arrested and asked to pass “breath analyzer” test. A sample consisting 56.5
mL of exhaled air is then bubbled into a spectrometer cell containing 3 mL 0.025% (w/V) K 2Cr 2O 7
solution. The transmittance of the solution was 41.5% initially and 43.4% after bubbling the sample
through the reaction cell. It is known that the alcohol content in blood stream is 2300 times higher than in
exhaled air and that the legal limit is 80 mg of alcohol per 100 mL of blood. Determine the concentration
of alcohol in the blood and state whether or not the driver should be charged with drunk driving.
Problems in Chemistry
15
PROBLEM 96 A sample of 0.3657 g powder containing only Ba(NO 3 ) 2 and Ca(NO 3 ) 2 are dissolved in
50 mL water. Ammonia is added to the solution to raise the pH than an excess of Na 2C 2O 4 is added to
precipitate the metals. The precipitate is then filtered, washed with 1.0 L of water and transferred to a
beaker containing 50 mL H 2O. The solution is acidified to solublise the precipitate and finally titrated
with 0.05 M KMnO 4 solution. A total of 13.94 mL of oxidizing agent solution was required to reach the
end point. Find the composition of the initial mixture. K sp. of BaC 2O 4 = 1.5 × 10 –8 and of
CaC 2O 4 = 2.34 × 10 –9 .
PROBLEM 97 1.0 g sample containing KO 2 and some inert impurity is dissolved in excess of aqueous
HI solution and finally diluted to 100 mL. The solution is filtered off and 20 mL of filtrate required 15
mL 0.4 M Na 2S 2O 3 solution to reduce the liberated iodine. Determine mass % of KO 2 in the original
sample.
PROBLEM 98 Cuprous ion is known to disproportionate quantitatively in acid medium. A 3.0 g sample
of Cu 2O is dissolved in dilute H 2SO 4 solution. The solution is filtered off and 8.3 g pure KI crystal is
added to filtrate. This caused precipitation of CuI with evolution of I 2 . The solution is filtered off and
filtrate is boiled till all I 2 is expelled off. Now, excess of an oxidizing agent is added to filtrate and
liberated iodine required 10 mL 1.0 N Na 2S 2O 3 solution. Calculate mass percentage of Cu 2O in the
original sample.
PROBLEM 99 To a 10 mL 1.0 M aqueous solution of Br 2 , excess of NaOH is added so that all Br 2
disproportionated to Br – and BrO –3 . The resulting solution is freed from bromide ion by extraction and
excess of OH – neutralized by acidifying the solution. The resulting solution is just sufficient to react
with 1.5 g of an impure CaC 2O 4 sample. Calculate percentage purity of oxalate sample.
PROBLEM 100 One gram of an impure sample of NaCl was dissolved in water and treated with excess
of AgNO 3 solution. The precipitate AgCl thus, formed undergo decomposition into Ag and Cl 2 ( g ) and
latter disproportionate into chlorate (V) and chloride ions and chloride is re-precipitated due to presence
of excess of AgNO 3 . If the original precipitate was 60% decomposed and final precipitate weigh 1.5
gram, determine mass percentage of NaCl in original sample.
PROBLEM 101 0.4 g of a sample containing CuCO 3 and some inert impurity was dissolved in diute
sulphuric acid and volume made up to 50 mL. To this solution was added 50 mL 0.04 M KI solution
where copper precipitate as CuI and iodide ion is oxidized into I –3 . A 10 mL portion of this solution is
taken for analysis, filtered, made free from I –3 and treated with excess of acidic permanganate solution.
Liberated iodine required 20 mL 2.5 m M sodium thiosulphate solution to reach the end point. Determine
mass percentage of CuCO 3 in the original sample.
PROBLEM 102 One gram of an unknown sample of NaCN is dissolved in 50 mL 0.33 M alkaline
solution of KMnO 4 and refluxed so that all cyanide is converted into cyanate (OCN – ). The reaction
mixture was cooled and its 5.0 mL portion was acidified by adding excess of sulphuric acid solution and
finally titrated with 19.0 mL 0.1 M FeSO 4 solution. Determine mass percentage of NaCN in the original
sample.
PROBLEM 103 5.0 mL of a pure liquid toluene is dissolved in 100 mL of dilute alkaline KMnO 4
solution and refluxed so that all toluene is oxidized into benzoic acid and a dark brown precipitate of
MnO 2 is formed. Solution is filtered off and filtrate and precipitate were analyzed separately. Precipitate
16
Problems in Chemistry
was re-dissolved into 100 mL 1.0 M acidified solution of Na 2C 2O 4 and excess of oxalate required 50.73
mL 0.1 M acidic dichromate solution for back titration. On the other hand 10.5 mL of filtrate was
acidified by adding excess of sulphuric acid and titrated with 0.1 M acidified solution of Na 2C 2O 4 . A 38
mL of oxalate solution was required to reach the end point. Determine density of liquid toluene and
molarity of original permanganate solution.
PROBLEM 104 A 2.0 g sample containing CaOCl 2 and NaOCl is dissolved in 100 mL water and its 10
mL portion was titrated against 0.15 M acidified solution of Na 2C 2O 4 . 10 mL of oxalate solution was
required to reach the end point. Titrated solution was then treated with excess of aqueous solution of
AgNO 3 where all chloride precipitates as AgCl and weighed to be 0.287 g. Determine mass percentage
of CaOCl 2 and NaOCl in original sample.
GASEOUS STATE
PROBLEM 105 6.0 g of He having average velocity 4 × 10 2 ms –1 is mixed with 12.0 g of Ne 20 having
the same average velocity. What is the average kinetic energy per mole in the mixture?
PROBLEM 106 The valve of a commercial cylinder of N 2 gas was left slightly open so that small
amount of gas leaked into the laboratory. The leak rate was proportional to the pressure difference
(internal pressure one atm). If the initial leak rate was found to be 1 g s –1 and initial pressure inside the
7.28 m 3 tank was 17180 kPa, what would be the pressure inside the tank after 10 days assuming
temperature of the lab to be 27°C.
PROBLEM 107 Calculate pressure exerted by 22.0 g of CO 2 in 0.5 L bulb at 300 K assuming it to be real
gas with a = 363 kPaL2 mol –2 and b = 42.67 cc/ mol.
PROBLEM 108 Molar volume of He at 10.1325 Mpa and 273 K is 0.011075 times its molar volume at
101.325 kPa. Calculate radius of He atom assuming negligible ‘a’.
PROBLEM 109 A gas mixture containing 5% by mass of butane and 95% by mass of Ar (40) is to be
prepared by allowing gaseous butane to fill an evacuated 40 L cylinder at 1.0 atm and 27°C. Calculate
mass of Ar that gives the desired composition and total pressure of the final mixture.
PROBLEM 110 Cl 2O 7 gas decomposes as:
Cl 2O 7 → Cl 2 + O 2
A partially decomposed gaseous mixture is allowed to effuse through a pin-hole and the gas coming
out initially was analyzed. The mole fraction of the O 2 was found to be 0.60, determine the degree of
dissociation.
PROBLEM 111 Proportion of a lighter isotope in a gaseous mixture containing both heavier and lighter
isotopes is increased by successive effusion of the gas mixture. A sample of neon gas has
22
Ne = 90% and 20 Ne = 10% by moles. In how many stages of successive effusion, 25% enrichment of
20
Ne would be achieved?
Problems in Chemistry
17
PROBLEM 112 The density of vapour of a substance at 1.0 atm and 500 K is 0.35 k/ m 3 . The vapour
effuses through a small hole at a rate of 1.33 times faster than oxygen under similar condition.
(a) Determine (i) Molecular weight (ii) Molar volume (iii) Compression factor (Z) of the
vapour (iv) Which forces among the gas molecules are dominating, the attractive or repulsive?
(b) If the vapour behaves ideally at 1000 K, determine the average translational kinetic energy
possessed by a molecule.
PROBLEM 113 Using van der Waals’ equation of state, calculate the pressure correction factor for two
moles of a gas confined in a four litre flask that exert a pressure of 11 atmosphere at 300 K.
b = 0.05 L mol –1 .
PROBLEM 114 For a van der Waals’ gas Z (compressibility factor) was found to be 1.5 at 273 K and one
atmosphere and TB of the gas is 107 K. Determine value of a and b.
PROBLEM 115 A flask containing 2.0 moles of He gas at 1.0 atm and 300K is connected to another
flask containing N 2 ( g ) at the same temperature and pressure by a narrow tube of negligible volume.
Volume of the nitrogen flask is three times volume of He-flask. Now the He-flask is placed in a
thermostat at 200 K and N 2 -flask in another thermostat at 400 K. Determine final pressure and final
number of moles in each flask.
PROBLEM 116 In a spherical glass flask A of radius 1.0 m, containing 300 g H 2 (g ), there was a rubber
balloon B containing some N 2 ( g ). Inside B, there was another rubber balloon C containing some oxygen
gas. At 27°C, it was found that the balloon B had radius 60 cm and of C was 30 cm. Calculate the total
weight of the gas inside the flask. Now 50 g H 2 ( g ) is further added to A, what would be the volume of B
and C.
PROBLEM 117 A partially decomposed PCl 5 (g ) along with its dissociation product is subjected to
diffusion study and the gases coming out initially collected in an another flask. The rate of effusion of
collected gaseous mixture was found to be 0.45 times rate of effusion of pure oxygen gas. Determine the
degree of dissociation of PCl 5 ( g ) in the original sample.
PROBLEM 118 One mole of a monoatomic gas confined in a 22.5 litre flask at 273 K exert a pressure of
0.98 atm, whereas expected pressure was 1.0 atm has the gas behaved ideally. Determine the van der
Waal’s constants ‘a’ and ‘b’ and Boyle’s temperature (TB ).
PROBLEM 119 One litre of a gas at 300 atm and 473 K is compressed to a pressure of 600 atm and 273
K. The compressibility factors found to be 1.072 and 1.375 respectively at the initial and final states.
Calculate the final volume.
PROBLEM 120 Calculate the van der Waal’s constants for ethylene. TC = 282 K,PC = 50 atm.
PROBLEM 121 The second Virial coefficient of an imperfect gas is 2 × 10 –2 (L/ mol) 2 . Calculate the
volume of a gm mole of the gas at 27°C and 5 atmosphere pressure.
PROBLEM 122 The van der Waal’s constant ‘b’ of a gas is 4.42 centilitre/mol. How near can the centres
of the two molecules approach each other?
PROBLEM 123 For carbon dioxide, critical density is 0.45 g/cc and its TC = 300 K. Determine its van
der Waal’s constants.
18
Problems in Chemistry
PROBLEM 124 The Virial equation for ethane gas is given by PV = RT + BP . At 0°C,
B = – 0.1814 L/ mol. Calculate volume of one mole of ethane at 10 atm, and ‘a’.
PROBLEM 125 An unknown gas (X) at 2.0 atmosphere and Ar (40) at 1.0 atmosphere were injected
simultaneously from the two ends of a 1.0 metre long glass tube and the first collision between X and Ar
occurred at a distance of 38 cm from Ar-end.Determine the molar mass of X assuming that gases were
injected at same temperature and through the pin-hole of identical geometry.
PROBLEM 126 Using van der Waals’ equation of state, calculate pressure developed by 100 g of CO 2
contained in a volume of 5.0 litre at 40°C. Also compare this value with that calculated using ideal gas
law and determine the percentage deviation from ideality. a = 3.6 atm L2 mol –2 , b = 44 cm 3 mol –1 .
PROBLEM 127 An equation of state for a non-ideal gas can be written as: PVm = A + BP + CP 2 ; where
Vm is the molar volume and P is the gas pressure in atmosphere. B = – 2.879 × 10 –2 and C = 14.98 × 10 –5
in litre atmosphere unit. Under the experimental condition, determine the pressure at which PV-P curve
will attain minimum.
PROBLEM 128 A modified form of van der Waal’s equation of state for 1.0 mole of gas is given as:
α 

P +
 (V – β) = RT

TV 2 
Deduce expression for the first Virial coefficient (B) and Boyle’s temperature in term of α and β if
Virial equation of state is:
PV
B C
= 1 + + 2 +…
RT
V V
PROBLEM 129 Assuming that dry air contain 79% N 2 and 21% O 2 by volume, calculate the density of
moist air at 25°C and 1.0 atmosphere when the relative humidity is 60%. The vapour pressure of water at
25°C is 23.76 mm of Hg.
PROBLEM 130 At what temperature, three moles of SO 2 will occupy 10 litre at a pressure of 15.0 atm if
it is a van der Waal’s gas with a = 6.71 atm L2 mol –2 and b = 56.4 cm 3 mol –1 .
PROBLEM 131 Pressure of He gas confined in a steel chamber drops from 4.0 to 1.0 atmosphere in 4.0
hours due to diffusion through a pin-hole in the steel chamber. If an equimolar mixture of He and
methane gas at 20 atmosphere and the same temperature are confined in the same chamber, what will be
the partial pressure of He and methane after 1.0 hour. Assume rate of diffusion to be linear function of
gas pressure and inverse function of square root of molar masses.
PROBLEM 132 One mole of a van der Waal’s gas at 0°C and 600 atmosphere occupies 0.075 L. If
b = 0.024 L mol –1 , determine compressibility factor (Z) and predict the type of force dominating among
the gas molecule.
PROBLEM 133 A one litre flask containing NH 3 (g ) at 2.0 atmosphere and 300 K is connected to
another 800 mL flask containing HCl(g) at 8.0 atmosphere and 300 K by means of a narrow tube of
negligible volume and gases were allowed to react quantitatively as:
NH 3 ( g ) + HCl( g ) → NH 4Cl( s); ∆H = – 43kJ/ mol
19
Problems in Chemistry
If heat capacity of HCl(g) CV is 20 JK –1 mol –1 , determine final pressure inside the flask assuming
negligible heat capacity of flask and negligible volume of solid NH 4Cl.
PROBLEM 134 A long cylindrical glass tube, equipped with a porous disc at the centre, contain
methane gas at 5.0 atmosphere on one side and He gas at 2.0 atmosphere on the other side of the disc as
shown in the diagram below:
Disc is permeable to both gases and rate of diffusion is directly proportional to the gas pressure and
inversely proportional to square root of molar masses as:
CH4
5.0 atm.
–
He
2.0 atm.
dp
P
where, k is a constant.
=k
dt
M
If k for the diffusion of methane gas is 2.5 ×10 –2 second –1 , determine time after which pressure of
methane chamber will drop to 4.0 atmospheres.
PROBLEM 135 At a given condition of temperature, rate of change of r.m.s. of He gas is twice the rate
of change of absolute temperature. Determine rms of He in the given condition.
PROBLEM 136 1.6 moles of ammonia gas at 300 K is taken in a 2.0 litre flask, sealed and heated to 500
K. At this temperature, ammonia is partially decomposed into N 2 and H 2 and a pressure measurement at
this point gave 48.5 atmosphere. Determine number of moles of each component present at 500 K.
PROBLEM 137 Decomposition of KClO 3 produces oxygen gas and KCl solid. In a typical experiment,
some KClO 3 was decomposed and 36.00 mL oxygen gas was collected over water at 23°C. The
laboratory barometer reads 751 mm and vapour pressure of water at 23°C is 21.1 mm of Hg. Find the
volume of the dry oxygen at 0°C and 1.0 atmosphere.
PROBLEM 138 A narrow tube of negligible volume connects two evacuated bulb of 1.0 litre capacity
each. One bulb is placed in a 200 K thermostat bath and other in a 300 K thermostat bath and then 1.0
mole of an ideal gas is injected into the system. Find the pressure in the two flasks.
PROBLEM 139 Isothermal compressibility (κ ) of a gas is defined as:
1  ∂V 
κ =– 

V  ∂P  T , n
Determine isothermal compressibility for an ideal gas at 1.0 atmosphere.
PROBLEM 140 What will be the temperature difference needed in a hot air balloon to lift 1.0 kg weight.
Assume that the volume of balloon is100 m 3 , the temperature of atmosphere is 25°C and pressure is 1.0
atmosphere. Average molar mass of air is 29 amu.
PROBLEM 141 Using van der Waals’ equation of state, find pressure at which the PV vs P curve
acquires minima for 1.0 mole of oxygen gas at 0°C. a =1.36 L2 atm mol –2 , and b = 32 cm 3 mol –1 .
PROBLEM 142 The van der Waals’ constant ‘a’ is a correction factor to the ideal gas law for
intermolecular force of attractions within the substance. Match the following values of ‘a’ (
L2 atm mol –1 ): 0.2107, 5.464, 18.00 and 24.06 with gases benzene, toluene, Ne and steam.
20
Problems in Chemistry
PROBLEM 143 The van der Waals’ constant ‘b’ is a correction factor to the ideal gas law for the
intrinsic volume of the molecule. Match the following values of ‘b’(L mol –1 ): 0.017, 0.0305, 0.1154 and
0.1463, with the gases: toluene, benzene, Ne and steam.
THERMOCHEMISTRY
PROBLEM 144 The specific heat capacity of water is 4.18 J(° C) –1 g –1 and that of copper is
0.38 J( ° C) –1 g –1 . Calculate the heat that must be supplied to a 500 g copper kettel containing 450 g of
water to raise its temperature from 25°C to the boiling point of water. What percentage of heat is used to
raise the temperature of the water?
PROBLEM 145 How much heat can be produced from a reaction mixture of 50 g of iron (III) oxide and
25 g of aluminium in the thermite reaction:
Fe 2O 3 ( s) + 2Al( s) → Al 2O 3 ( s) + 2Fe( s); ∆H = – 851.5 kJ/ mol
PROBLEM 146 Calculate the reaction enthalpy for the hydrogenation of ethyne to ethane, given
standard enthalpy of combustion of ethyne, ethane and hydrogen; – 1300, – 1560 and – 286 kJ/mol
respectively.
PROBLEM 147 Calculate the reaction enthalpy for the synthesis of HCl(g) from the following data:
NH 3 ( g ) + HCl( g ) → NH 4Cl( s) ∆H = – 176 kJ
N 2 ( g ) + 3H 2 ( g ) → 2NH 3 ( g ) ∆H = – 92.22 kJ
N 2 ( g ) + 4H 2 ( g ) + Cl 2 ( g ) → 2NH 4Cl( s) ∆H = – 628.86 kJ
PROBLEM 148 An important reaction that occurs in the atmosphere is
NO 2 ( g ) → NO( g ) + O( g )
Which is brought about by the sunlight. How mucy energy the sun to cause it must supply? Given,
dissociation energy of oxygen = 498 kJ/ mol and
NO( g ) + O 3 ( g ) → NO 3 ( g ) + O 2 ( g ) ∆H = – 200 kJ
3O 2 ( g ) → 2O 3 ( g )
∆H = 285.4 kJ
PROBLEM 149 Using reaction a, b and c determine the enthalpy change of this reaction:
3
CH 4 ( g ) + O 2 ( g ) → CO( g ) + 2H 2O( g )
2
(a) CH 4 ( g ) + 2O 2 ( g ) → CO 2 ( g ) + 2H 2O( g )
∆H ° = – 802 kJ/ mol
(b) CH 4 ( g ) + CO 2 ( g ) → 2CO( g ) + 2H 2 ( g )
∆H ° = + 206 kJ/ mol
(c) CH 4 ( g ) + H 2O( g ) → CO( g ) + 3H 2 ( g )
∆H ° = + 247 kJ/ mol
PROBLEM 150 The bond energy of H 2 (g ) is 436 kJ/mol and that of N 2 (g ) is 941.3 kJ/mol. Calculate
the average bond energy of an N—H bond in ammonia if ∆H ° f of ammonia is – 46 kJ/mol.
21
Problems in Chemistry
PROBLEM 151 The heat of formation of PCl 3 and PH 3 are 306 kJ/mol and 8 kJ/mol respectively, and
the heats of atomization of phosphorus, chlorine and hydrogen are 314, 121 and 216.5 kJ/mol
respectively. Calculate P-Cl and P-H bond energy.
PROBLEM 152 At 25°C, the molar heat of formation of SO 2 and H 2O are – 296.81 and – 285.83 kJ
respectively. Using the information from the following reactions,
2H 2S( g ) + Fe( s) → FeS 2 + 2H 2 ( g )
3
H 2S( g ) + O 2 ( g ) → H 2O( l) + SO 2 ( g )
2
Calculate heat of formation of H 2S( g ) and FeS 2 ( s) at 25°C.
∆H ° = – 137 kJ/ mol
∆H ° = – 562 kJ/ mol
PROBLEM 153 The standard molar enthalpy of formation of cyclohexane (l) and benzene (l) at 25°C
are – 156 and + 49 kJ/ mol respectively. The standard enthalpy of hydrogenation of cyclohexene (l) at
25°C is – 119 kJ/ mol. Use this data to estimate the magnitude of the resonance energy of benzene.
PROBLEM 154 For the reaction cis-2-butene → trans-2-butene and cis-2-butene → 1-butene,
∆H = – 950 and +1771cal/ mol respectively. The heat of combustion of 1-butene is – 649.8 kcal/mol.
Determine the heat of combustion of trans-2-butene. Also calculate the bond energy of C==C bond in
trans-2-butene. Given B.E of C==O =196, O—H =110, O==O = 118, C—C = 80 and C—H = 98 kcal/mol
respectively. ∆H v (H 2O) = 11 kcal / mol.
PROBLEM 155 Using the data (all values are in kJ/mol at 25°C) given below:
(i) Enthalpy of polymerization of ethylene = – 72.
(ii) Enthalpy of formation of benzene(l) = 49
(iii) Enthalpy of vaporization of benzene(l) = 30
(iv) Resonance energy of benzene(l) = – 152
(v) Heat of formation of gaseous atoms from the elements in their standard states H = 218, C = 715.
Average bond energy of C—H = 415. Calculate the B.E. of C—C and C==C. [ A : 331 and 590 kJ/mol]
PROBLEM 156 Calculate energy of aromatization of cyclohexane according to the following reaction,
both cyclohexane and benzene are in liquid state:
Given, bond energies: C—C = 348, C—H = 415, C==C = 600, H—H = 436 kJ/mol respectively,
sublimation energy of C(gr) is 717 kJ/mol, resonance energy of C 6 H 6 ( l) = – 152 kJ/ mol, ∆H v° of
benzene = 30.8 and of cyclohexane is 33 kJ/mol.
+ 3H2(g)
PROBLEM 157 Enthalpy of polymerization of ethylene and acetylene into corresponding polymers are
– 86 kJ/ mol and – 148 kJ/mol respectively. Enthalpy of hydrogenation of ethylene is – 132 kJ/mol,
determine C==C bond energy. B.E. of H 2 ( g ) is 436 kJ/mol and of C—H = 415 kJ/ mol.
PROBLEM 158 ∆HComb of methane and ethane are – 210 kcal/mol and – 368 kcal/mol respectively.
Determine ∆HComb of decane.
PROBLEM 159 Determine resonance energy of benzene [C 6 H 6 (l)] from the following information :
∆H °f of C 6 H 6 ( l) = + 49 kJ;
∆H °f of C 2 H 2 ( g ) = + 75 kJ ∆H °v of C 6 H 6 ( l) = + 45 kJ
B.E.
C ≡≡C = 930 kJ/ mol; C==C = 615 kJ/ mol; C  C = 348 kJ/ mol
22
Problems in Chemistry
PROBLEM 160 Consider the following thermodynamic data:
Enthalpy of formation of CaC 2 ( s) = – 60 kJ/ mol;
Enthalpy of sublimation of Ca( s) = 179 kJ/ mol;
Enthalpy of sublimation of C( s) = 718 kJ/ mol;
First ionization energy of Ca(g) = 590 kJ/ mol;
Second ionization energy of Ca(g) 1143 kJ/mol;
Bond energy of C 2 ( g ) = 614 kJ/ mol;
First electron affinity of C 2 ( g ) = – 315 kJ/ mol;
Second electron affinity of C 2 ( g ) = + 410 kJ/ mol.
Draw a clear Born-Haber cycle and determine lattice energy of CaC 2 ( s).
PROBLEM 161 Normal L.P.G. contains 90% propane and 10% methane by weight. If combustion of
L.P.G. produces acetylene, CO(g) and H 2O( l), calculate the heat evolved by combustion of 100 g of
L.P.G.
Given: enthalpy of combustion of methane = – 890 kJ/ mol, C 3 H 8 = – 2220 kJ/ mol,
C 2 H 2 = – 1300 kJ/ mol and CO( g ) = – 285 kJ/ mol.
PROBLEM 162 A swimmer breaths 20 times in one minute when swimming and inhale 200 mL of air in
one breath. Inhaled air contain 20% O 2 by volume and exhaled air contain 10% O 2 by volume. If all
oxygen are consumed in combustion of glucose in the body and 25% of energy obtained from
combustion is available for muscular work. Determine the maximum distance this swimmer can swim in
one hour if 100 kJ energy is required for 1.0 km swimming. Standard molar enthalpy of combustion of
glucose is – 2880 kJ/ mol and body temperature is 37°C.
PROBLEM 163 Standard molar enthalpies of formation of H 2O(l) and H 2O 2 (l) are – 285 and – 200
kJ/mol respectively and their molar enthalpies of vaporization are 41 and 60 kJ respectively. If enthalpy
of atomization of O 2 ( g ) is 298 kJ/mol, determine bond energy of O—O bond.
PROBLEM 164 Determine resonance energy of 1,3-butadiene using the following information:
Enthalpy
of
combustion
:
1,3-butadiene
= – 2841 kJ/ mol, C(gr ) = – 394 kJ/ mol,
H 2 ( g ) = – 285 kJ/ mol
Bond enthalpy : C—C = 348 kJ/ mol, C==C = 615 kJ/ mol. Also standard enthalpy of formation of
cyclobutene =130 kJ/ mol,
PROBLEM 165 Standard molar enthalpy of formation of hydrazine liquid (N 2 H 4 ) is 50 kJ/mol,
NH 3 ( g ) = – 46 kJ/ mol. Average N—H and H—H bond energies are 393 and 436 kJ/mol respectively. If
enthalpy of vaporization of N 2 H 4 is 18 kJ/mol, determine N—N bond energy in N 2 H 4 .
PROBLEM 166 Using following standard enthalpies:
∆H °f HF( aq ) = – 329 kJ/ mol ∆H ° f H 2O( l) = – 285 kJ/ mol,
∆H ° f F – ( aq ) = – 320 kJ /mol and
H + ( aq ) + OH – ( aq ) → H 2O( l), ∆H = – 56 kJ/ mol.
Determine enthalpy of neutralization of HF against a strong base.
PROBLEM 167 From the following reactions and thermal information at 25°C:
3
2Fe( s) + O 2 → Fe 2O 3 ( s)
∆H ° = – 821.4 kJ
2
23
Problems in Chemistry
1
2FeO( s) + O 2 → Fe 2O 3 ( s)
2
∆H ° = – 284 kJ/ mol
Fe( s) + 2H + ( aq.) → Fe 2+ ( aq.) + H 2 ( g )
∆H ° = – 87.8 kJ
1
H 2 ( g ) → H + ( aq.)
2
1
H 2 ( g ) + O 2 ( g ) → H 2O( l)
2
Calculate ∆H ° for the reaction :
∆H ° = 0
∆H ° = – 285 kJ
FeO( s) + 2H + ( aq ) → H 2O( l) + Fe 2+ ( aq )
PROBLEM 168 A 150 cc portion of 0.4 N HCl is neutralized with excess of NH 4OH in a bomb
calorimeter which results in a temperature rise of 2.36°C. If the heat capacity of calorimeter content is
1316.7 J/°C, calculate heat of neutralization of HCl Vs NH 4OH.
PROBLEM 169 Determine S—S bond energy;
Given ∆H ° f of (C 2 H 5 ) 2 S( g ) = – 147 kJ/ mol,
∆H ° f of (C 2 H 5 ) 2 S 2 ( g ) = – 202 kJ/ mol and ∆H °Sublimation of S( s) = 223 kJ/ mol.
PROBLEM 170 Given the following standard molar enthalpies:
of
∆H °Sublimation
∆H ° f of
CH 3CN( g ) = 88 kJ/ mol,
∆H ° f of
C 2 H 6 = – 84 kJ/ mol,
C(gr) = 717 kJ/ mol, bond dissociation energy of N 2 ( g ) and H 2 ( g ) are 946 and 436 kJ/mol respectively,
B.E. (C—H) = 410 kJ/mol. Determine C—C and C ≡≡N bond energies.
PROBLEM 171 Determine standard state enthalpy of the following reaction:
CH 3COOH( l) → CH 4 ( g ) + CO 2 ( g )
Given ∆H °Combustion CH 4 = – 860 kJ/ mol. Bond energies in kJ/mol. C—H = 410, C—C = 348, C==O
= 728, C—O = 352, O—H = 463, O==O = 498. ∆H °Vaporization of acetic acid and water are 52 and 41
kJ/mol respectively.
PROBLEM 172 ∆H ° f of
enthalpy
of
combustion
of
C 2 H 5OH( l) = – 66 kcal/ mol,
CH 3OCH 3 ( g ) = – 348 kcal/ mol, ∆H ° f of water is – 68 kcal/mol and ∆H ° f of CO 2 ( g ) = – 94 kcal/ mol.
Determine enthalpy of the following isomerization reaction:
C 2 H 5OH( l) → CH 3OCH 3 ( g )
PROBLEM 173 The standard enthalpies of formation of BH 3 (g ) and B2 H 6 (g ) are 100 kJ and 36 kJ per
mol respectively and the enthalpies of formation of B( g ) and H( g ) are 563 kJ mol –1 and 218 kJ mol –1
respectively. Determine mean B H bond enthalpies in each case. Assume terminal B—H bonds have
same strengths, estimate enthalpies of the three centre B H  B bonds in B2 H 6 . Which bonds would
you expect to be longer-terminal or bridged one?
PROBLEM 174 Enthalpy of combustion of C 6 H 6 (l), C 6 H12 (l) (cyclohexane) H 2 (g ) are –3268, –3920
and –289 kJ/mol respectively. If enthalpy of hydrogenation of cyclohexane is –120 kJ/mol, determine
resonance energy of benzene( l).
PROBLEM 175 Assuming that mileage of an automobile gets is directly proportional to the heat of
combustion of fuel, calculate how many times farther an automobile could be expected to go on one litre
gasoline than on 1.0 litre ethanol.
24
Problems in Chemistry
Assume gasoline to be pure n-octane (ρ = 0.7025 gmL–1 ). Density of ethanol is 0.7893 gmL–1 . ∆H f°
of ethanol and octane are −278 kJ mol –1 and –208.4 kJ mol –1 . ∆H f° of CO 2 ( g ) and H 2O( l) are
−394 kJ mol –1 and –286 kJ mol –1 respectively.
PROBLEM 176 10 g of propane was burnt in air at 30°C and 1.0 atm pressure. Assume air to be 21.00%
O 2 , determine volume of air required for combustion process. If all the heat produced from combustion
of 10 g of propane was transferred to 8.00 kg of water at 30°C, determine final temperature of water. C P
of water is 418
. Jg –1 K –1 . Also, ∆H f° of propane, CO 2 and H 2O are: −104 kJ mol –1 , – 394 kJ mol –1 and
–286 kJ mol –1 respectively.
PROBLEM 177 With the following informations, determine standard state Gibb’s free energy of
formation of N 2O 4 ( g ).
∆G° = 86.6 kJ
…(i)
NO( g ) + 12 O 2 ( g ) → NO 2 ( g )
∆G° = − 34.82 kJ
…(ii)
2NO 2 ( g ) → N 2O 4 ( g )
∆G° = − 5.77 kJ
…(iii)
1
2
N 2 ( g ) + 12 O 2 ( g ) → NO( g )
Standard state enthalpy of formations of CO( g ) and CO 2 ( g ) are −111 kJ mol –1 and
–394 kJ mol –1 respectively. Bond dissociation energies of O 2 ( g ) and C==O( g ) are 498 kJ mol –1 and 743
kJ mol –1 respectively. Supposing that there is a double bond in CO as two double bond in CO 2 ,
determine the enthalpy of combustion of CO( g ) and compare it with actual value. Also, explain the
difference in calculate and observed value of enthalpies.
PROBLEM 178
PROBLEM 179 One mole of N 2 (g ) and 3.0 mole of H 2 (g ) taken in a flask at 25°C and heated to 450°C.
Now pressure was applied on the gaseous mixture which results in conversion of 0.1 mole of N 2 into
NH 3 . The gases are then cooled rapidly back to 25°C. Determine the net heat change in this process
given the following bond enthalpies:
N 2 ( g ) = 944 kJ mol –1 ; H 2 = 436 kJ mol –1 and average N  H bond energy = 388 kJ mol –1 .
PROBLEM 180 A 150 cc portion of 0.4 NHCl is neutralized with an excess of NH 4OH in a Dewar
vessel with a resulting rise in temperature of 2.36°C. If the heat capacity of Dewar and its contents after
the reaction is 1316.7 J/ ° C, calculate heat of neutralization.
PROBLEM 181 At 25°C, the heat of solution of anhydrous CuSO 4 in a large volume of water is −66.044
kJ mol –1 , while that of CuSO 4 ⋅ 5H 2O is –11.495 kJ. Determine heat of reaction:
CuSO 4 ( s) + 5H 2O → CuSO 4 ⋅ 5H 2O( s)
PROBLEM 182 The integral heats of solution at 25°C, for the various solid modification of CaCl 2 in the
indicated quantities of water are shown below:
CaCl 2 ( s) + 400 H 2O( l) → CaCl 2 (400 H 2O)
∆H1° = − 4.3 kJ
CaCl 2 ⋅ 2H 2O + 398 H 2O( l) → CaCl 2 (400 H 2O) ∆H 2° = − 41.925 kJ
…(i)
…(ii)
CaCl 2 ⋅ 4H 2O + 396 H 2O → CaCl 2 (400 H 2O)
∆H 3° = − 7.65 kJ
…(iii)
CaCl 2 ⋅ 6H 2O + 394 H 2O → CaCl 2 (400 H 2O)
∆H 4° = +19.06 kJ
…(iv)
25
Problems in Chemistry
Determine enthalpies of the following hydration reactions:
(a) CaCl 2 (s) + 2H 2O → CaCl 2 ⋅ 2H 2O
(b) CaCl 2 ⋅ 2H 2O + 2H 2O → CaCl 2 ⋅ 4H 2O
(c) CaCl 2 ( s) + 6H 2O → CaCl 2 ⋅ 6H 2O
PROBLEM 183 The enthalpy of following reactions at 25°C are:
(i) Na( s) + 12 Cl 2 ( g ) → NaCl( s)
∆H ° = − 410.6 kJ
(ii)
H 2 ( g ) + S( s) + 2O 2 ( g ) → H 2SO 4 ( l)
∆H ° = − 810.54 kJ
(iii) 2Na( s) + S( s) + 2O 2 ( g ) → Na 2SO 4 ( s) ∆H ° = −1381.5 kJ
(iv)
H 2 ( g ) + 12 Cl 2 ( g ) → HCl( g )
1
2
∆H ° = − 92.21 kJ
From the above thermal data, determine enthalpy of the following reaction:
2NaCl( s) + H 2SO 4 ( l) → Na 2SO 4 ( s) + 2HCl( g ).
PROBLEM 184 Given the following standard state enthalpies of reaction, calculate the standard molar
heat of formation of AgCl.
(i) Ag 2O ( s) + 2HCl( g ) → 2AgCl(g ) + H 2O( l)
(ii) 2Ag( s) + 12 O 2 ( g ) → Ag 2O( s)
(iii)
1
2
H 2 ( g ) + 12 Cl 2 ( g ) → HCl( g )
(iv) H 2 ( g ) + 12 O 2 ( g ) → H 2O( l)
∆H ° = − 324.4 kJ
∆H ° = − 30.56 kJ
∆H ° = − 92.21 kJ
∆H ° = − 394 kJ
PROBLEM 185 Draw Lewis structures of hypothetical molecule N 6 (g ) consisting of a six membered
ring of nitrogen atom. If its standard enthalpy of formation is 1072 kJ mol –1 , predict the most likely
structure. Given:
B.E. N 2 = 944 kJ mol –1 , N  N = 163 kJ mol –1 and N == N = 409 kJ mol –1 .
PROBLEM 186 A male burns 2000 kJ of energy while jogging for 1.0 hour. If the standard heat of
combustion of a typical fat is 38 kJ g –1 and only 70% energy is available for muscular activity. What
minimum hours would he need to jog if he wished to lose 0.5 g fat?
PROBLEM 187 Strong sunshine bombards the Earth with about 1 kJ m –2s –1 . If a beaker containing
ethanol, is placed in sunlight for 10 minutes, 3.24 g of liquid was vaporized. Assuming that all the heat is
used for vaporization, not to increase temperature, determine surface area of beaker. Enthalpy of
vaporization is 42.6 kJ mol –1 .
PROBLEM 188
“Benzene”.
From the following enthalpies values, determine resonance energy of C 6 H 6 ( l)
(i)
+ H2
∆H° = – 38 kJ
(ii)
+ 2H2
∆H° = – 170 kJ
Also, given the resonance energy of 1,3-cyclohexadiene is 70 kJ mol –1 .
26
Problems in Chemistry
THERMODYNAMICS
PROBLEM 189 Suppose that a gas obeys the modified van der Waals’ equation P × (Vm – b) = RT and
b = 0.02 L mol –1 . If 0.5 mol of the gas is reversibly compressed from an initial volume of 2 dm 3 to a final
volume of 0.5 dm 3 , how much work is done on the system at 27°C.
PROBLEM 190 One mole of a monoatomic, ideal gas confined in a 5 L piston fitted cylinder at 300 K is
heated such that its temperature increased to 400 K but at the same time gas also expanded to a volume of
8 L. Calculate change in enthalpy of the system.
PROBLEM 191 One mole of a monoatomic ideal gas confined in a 5 L, piston fitted cylinder at 300 K is
heated to 800 K as well as allowed to expand to a volume of 8 L simultaneously. Calculate change in
enthalpy of the system.
PROBLEM 192 100 g of nitrogen gas at 300 K are held by a piston under 30 atmosphere. The pressure is
suddenly released to 10 atmosphere and gas is allowed to expand adiabatically. If CV = 20.8 JK –1 mol –1 ,
calculate ∆S System .
PROBLEM 193 Calculate entropy change when 0.5 L of an ideal gas (CV =12.6 JK –1 mol –1 ) at 300 K
and one atmosphere is allowed to expand to double its volume and simultaneously heated to 373 K.
PROBLEM 194 10 g of ice at 0°C are added to 20 g water at 90°C in a thermally insulated flask of
negligible heat capacity. The heat of fusion of ice is 6 kJ/mol. What is the final temperature, ∆S System ?
C p = 75.42 JK –1 mol –1 .
PROBLEM 195 One mole of a supercooled liquid water at – 10°C and one atmosphere turns into ice at
– 10°C. Calculate entropy change for the system. C p for liquid water and ice are 75.42 and
37.2 JK –1 mol –1 respectively.
PROBLEM 196 In an open beaker at 27°C and one atm pressure, 100 g of zinc are caused to react with
dilute sulphuric acid. Calculate the work done by the liberated hydrogen gas assuming it to behave
ideally. What would be the work done if the reaction took place in a sealed vessel?
PROBLEM 197 A balloon is 0.5 m in diameter and contains air at 25°C and 1 bar pressure. It is then
filled with air isothermally and reversibly until the pressure reaches to 5 bar. Assume that pressure is
proportional to the diameter of the balloon, calculate (a) final diameter and (b) work done in the process.
PROBLEM 198 One mole of an ideal gas initially at 10 bar and 300 K is allowed to expand against a
constant external pressure of 2.00 bar to a final pressure of 2 bar. During this process, the temperature of
this gas falls to 250 K. Construct a reversible path connecting this initial and final state as a combination
of reversible isothermal expansion followed by reversible adiabatic expansion so that the final state is
attained and calculate work done by the system in attaining the final state. C vm = 3/ 2 R .
PROBLEM 199 With the temperature maintained at 0°C, 2 mole of an ideal gas are allowed to expand
against a piston that supports 2.0 bar pressure. The initial pressure of the gas is 10 bar and the final
pressure 2 bar.
(a) How much energy is transferred to the surrounding during the expansion?
(b) What is the change in internal energy and enthalpy of the gas?
(c) How much heat the gas has absorbed?
27
Problems in Chemistry
PROBLEM 200 A gas behaving ideally was allowed to expand reversibly and adiabatically to twice its
volume. Its initial temperature was 25°C and C vm = (5 / 2) R , calculate ∆E m and ∆H m .
PROBLEM 201 One mole of a gas at 300 K is compressed isothermally and reversibly from an initial
volume of10 dm 3 to a final volume of 0.2 dm 3 mol –1 . Calculate work done on the system if the equation
of state of the gas is (Vm – b) P = RT with b = 0.03 dm 3 mol –3 .
PROBLEM 202 One mole of a gas at 100 K is compressed isothermally from an initial volume of
20 dm 3 to a final volume of 5 dm 3 . Calculate the work done on the system if the equation state is:

a 
 P + 2  Vm = RT
Vm 

where a = 0.384 m 6 Pa mol –1
PROBLEM 203 Find q,W, ∆U and ∆H if 2.0 g of He undergoes a reversible isobaric expansion from 20
to 40 L at 0.8 atm pressure followed by reversible isochoric heating till pressure reaches to 1.0 atm.
Depict the change of state on a P-V diagram.
PROBLEM 204 One mole of liquid water at 30°C is adiabatically compressed, pressure increasing from
1.0 atm to 10.0 atm. Since, liquids and solids are rather incompressible, it is a fairly approximastion to
take V constant. Calculate q, ∆U and ∆H for the process. C p of H 2O( l) = 75.42 JK –1 mol –1 .
PROBLEM 205 For a perfect gas, C v = 2.5 R . 2.0 moles of this gas undergoes following change of state:
(a) A reversible isobaric expansion from 1.0 atm, 20 L to 1.0 atm, 40 L.
(b) A reversible isochoric change from 1.0 atm, 40 L to 0.5 atm, 40 L
(c) A reversible isothermal compression from 0.5 atm, 40 L to 1.0 atm, 20 L. Sketch each process on
the P-V diagram and calculate : q, W, ∆U and ∆H.
PROBLEM 206 A sample of an ideal gas underwent an adiabatic expansion from 298 K, 15 bar to 2.5
bar against a constant external pressure of 1.0 bar. What is the final temperature of the system and work
done by the system, assume C vm = 2.5 R?
PROBLEM 207 A gas behaves ideally and its C v is given by: C v = 21.52 + 8.2 × 10 –3 T (all parameters
in SI unit). A sample of this gas is initially at T1 = 300 K, P1 = 10 bar andV1 = 1 L. It is allowed to expand
until P2 = 1 atm and V2 = 10 L. What are ∆U and ∆H for this process? Could the process be carried out
adiabatically. Calculate C p at 300 K.
PROBLEM 208 The entropy change of argon is given to a good approximation by the expression:
Sm / JK –1 mol –1 = 36.36 + 20.79 ln T
Calculate change in Gibb’s free energy of one mole of argon gas if it is heated at constant pressure
from 25°C to 50°C.
PROBLEM 209 Initially at 300 K and 10 atm pressure, 1.0 mole of an ideal gas is allowed to expand
adiabatically against a constant pressure of 4.0 atm until equilibrium is established. Assume the gas to be
ideal with:
C p = 28.58 + 1.76 × 10 –2 T
Calculate ∆U , ∆H, and ∆S.
28
Problems in Chemistry
PROBLEM 210 An ideal gas expand against a constant external pressure of 2.0 atmosphere from 20 L to
50 L and absorb 20 kJ of energy from surrounding. What is the change in internal energy of the system?
10
atm, where V is volume of
V
gas at each stage of expansion. Further during expansion from volume 10 L to 100 L, the gas undergoes a
change in internal energy of 420 J. How much heat is absorbed by the gas during expansion?
PROBLEM 211 A gas expands against a variable pressure given by P =
PROBLEM 212 Three moles of an ideal gas is heated at constant pressure of one atmosphere from 27°C
to 127°C. If C v is expressed as: C v = 30 + 14 × 10 –3 T JK –1 mol –1 , determine W, ∆E and q.
PROBLEM 213 One mole of an ideal gas at state A (500 K, 5.0 atm) is cooled at constant volume to
B (300 K) and then expanded isothermally and reversibly to C and finally compressed adiabatically to A.
5
Sketch the change on a P-V diagram and determine the net work done in this cyclic process. γ = .
3
PROBLEM 214 One mole of an ideal gas is subjected the following change of state:
Reversible
Isothermal expansion
A(5.0 atm, 500 K) → B
Isochoric cooling
Reversible
Adiabatic compression
B → C (300 K) → A
Depicting the above mentioned change on a P-V diagram, determine the net work done in the cyclic
process. C v =1.5 R .
PROBLEM 215 One mole of an ideal gas initially at A (300 K and 5.0 bar) is heated at constant pressure
to double its volume (B). The gas is then expanded isothermally and reversibly to a new state C. The gas
is then cooled at constant pressure to another new state D (200 K) and finally compressed adiabatically
and reversibly to A. Depicting on a P-V diagram, determine the net work done in the above cyclic
process. C v =1.5 R .
PROBLEM 216 One mole of an ideal gas at A (500 K, 5.0 bar) is expanded isothermally and reversibly
to a new state B and then cooled at constant pressure to C (250 K) and finally compressed adiabatically
and reversibly to A. Depicting on a P-V diagram, determine the net work done in the cyclic process.
C v =1.5 R .
PROBLEM 217 One mole of an ideal gas at 500 K and 10 bar, defined by state A is allowed to expand
isothermally and does a work equal to 4200 J. Construct a combination of initial irreversible expansion
upto 2.0 bar followed by reversible expansion so that final state is reached and determine final pressure
of the gas.
PROBLEM 218 One mole of an ideal gas at 500 K and 10 bar is allowed to expand till the final pressure
falls to 1.0 atmosphere and final temperature falls to 250 K. Construct a combination of reversible path
of:
(a) initial adiabatic expansion followed by isothermal expansion so that final state is reached and
determine the total work done. C v =1.5 R .
(b) Determine the work by reversing the order of combination in (a) and compare the two work
done.
29
Problems in Chemistry
PROBLEM 219 One mole of an ideal gas at 300 K and 1.0 atmosphere is heated to 500 K and expanded
simultaneously to 36 litre. Determine ∆H assuming heat capacity to be independent of temperature and
C v =1.5 R .
PROBLEM 220 One mole of a gas initially at 300 K is heated to 500 K. Determine the Gibb’s free
energy change, ∆G if S = 1.5 + 3 × 10 –3 T [JK –1 mol –1 ] and C v =1.5 R .
PROBLEM 221 One mole of an ideal gas at 300 K and 1.0 atmosphere is heated as well as expanded
simultaneously to 500 K and 2.0 atmosphere. Determine ∆S if C v = 2.5 R .
PROBLEM 222 One mole of an ideal gas is taken in a one litre sealed flask at 300 K and heated till the
pressure becomes equal to 40 atmosphere. If C v = 12 + 28 × 10 –3 T (in SI unit), determine ∆S.
PROBLEM 223 A gaseous reactant A forms two different product in a parallel reaction B and C as
follows:
A → B ;
∆H ° = – 3 kJ, ∆S ° = 20 JK –1
A → C ;
∆H ° = – 3.6 kJ, ∆S ° = 10 JK –1
Discuss the relative stability of B and C on the basis of Gibb’s free energy change at 27°C.
PROBLEM 224 One mole of an ideal gas contained in a sealed flask at 1.0 bar is heated from 27°C to
127°C. Determine ∆G if: S (JK –1 ) = 10 +12 × 10 –3 T .
PROBLEM 225 Two moles of NO 2 is heated at constant volume from 27°C to 127°C and
C p (JK –1 mol –1 ) = 28 + 31 × 10 –3 T
Determine ∆S.
PROBLEM 226 Two moles of an ideal gas is expanded isothermally and irreversibly at 27°C from
volumeV1 to 2.5 V1 and 4.17 kJ heat is absorbed from surroundines. Determine ∆S sys , ∆S surr and ∆S univ .
PROBLEM 227 One mole of He(g) is mixed isothermally and reversibly with 2.0 mole of Ne(g).
Determine ∆S.
PROBLEM 228 C vm for an ideal gas is 2.5 R and it is independent of temperature. If 2.0 moles of this gas
is subjected to the following change of state :
Reversible
A (1.0 bar , 20 L ) → B (1.0 bar, 40 L)
Isobaric heating
Isochoric cooling
Reversible
Isothermal compression
B → C (0.5 bar, 40L) → A
Representing the above change of states on a P-V diagram, determine the net work done.
30
Problems in Chemistry
PROBLEM 229 An ideal gas has
C vm = a + bT
where a = 25 JK mol and b = 0.03 JK mol –1 . If 3.0 moles of this gas is subjected to a
thermodynamic change of state from A (300 K, 2.0 bar) to B (500 K, 3.0 bar), determine q, W, ∆E, ∆H
and ∆S.
–1
–2
–1
PROBLEM 230 One mole of an ideal gas defined by state A (400 K, 5.0 bar) is heated at constant
pressure to B (500 K) and then cooled at constant volume to C. The gas is then expanded isothermally
and reversibly to a new state D (1.0 bar) and finally compressed adibatically to A. Depicting on a P-V
diagram, determine the net work done in this cyclic process. C v =1.5 R ..
PROBLEM 231 One mole of an ideal gas is subjected to the following change of state:
Reversible
Isothermal expansion
Isochoric cooling
A(500 K, 5.0 bar ) → B → C (250K, 1.0 bar)
Reversible
C → D (3.0 bar); C v =1.5 R
Adiabatic compression
Depicting the above change on a P-V diagram, determine the net work done.
PROBLEM 232 The entropy of vaporization of benzene is 85 JK –1 mol –1 .
(a) Estimate the enthalpy of vaporization of benzene at its normal boiling point of 80°C.
(b) Determine the entropy change of surroundings when 100 g benzene vaporizes at its normal
boiling point.
PROBLEM 233 The entropy of vaporization of acetone is 85 JK –1 mol –1 .
(a) Estimate enthalpy of vaporization of acetone at its normal boiling point 56°C.
(b) Determine entropy change of surrounding if 100 g of acetone condenses at its boiling point.
With the help of following reduction reactions:
PROBLEM 234
at
1000
K,
given
TiO 2 ( s) + 2C( s) → Ti( s) + 2CO( g )
…(i)
TiO 2 ( s) + C( s) → Ti( s) + CO 2 ( g )
…(ii)
∆Gf° (CO) = − 200 kJ mol –1 ,
∆Gf° (CO 2 ) = − 396
kJ mol –1
and
∆G f° TiO 2 ( s) = − 762 kJ mol –1 , determine which will be the predominant mode of reduction of TiO 2 ( s)
at 1000 K.
PROBLEM 235
The reaction for the production of synthetic fuel ‘water gas’ from coal is:
C( gr ) + H 2O( g ) → CO( g ) + H 2 ( g )
Standard molar entropies of C( gr ), H 2O( g ), CO( g ) and H 2 ( g ) are 5.7, 70, 190 and 131 JK –1 mol –1
respectively. Also, standard enthalpy of formations of H 2O and CO are –242 and –111 kJ mol –1
respectively. Determine the standard reaction free energy of reaction at 27°C. Also, predict about the
spontaneity and effect of temperature on direction of reaction.
31
Problems in Chemistry
PROBLEM 236
follows:
The thermodynamic informations for isomerization of alkene (C 4 H 8 ) at 300 K are as
cis-2-butene
H 3C
trans-2-butene
CH 3
C==C
H
H 3C
1
H
∆G ° = 66 kJ mol –1
f
∆H ° = − 7 kJ mol –1
f
H
C==C
H
2
3
H
CH 3
∆G ° = 63 kJ mol –1
f
∆H ° = −11.2 kJ mol –1
f
CH 3
C==C
H
CH 3
2-methylpropene
∆G °f = + 58 kJ mol
If the temperature of the above system is increased to 400 K and equilibrium was allowed to
re-establish, mole percentage of trans-2-butene at new equilibrium was 18. Determine ∆H ° and ∆S ° for
the isomerization reactions below:
cis-2-butene
2-methylpropene
trans-2-butene
2-methylpropene
PROBLEM 237 At a temperature above 65 K, decarboxylation of acetic acid, (i.e., loss of CO 2 )
becomes spontaneous. If ∆H f° of CH 3COOH, CO 2 ( g ) and CH 4 are −484.5, − 394 and − 74.8 kJ/mol
respectively, determine standard state entropy change ( ∆S ° ) for the decarboxylation reaction. What is
the driving force for getting this reaction to proceed?
PROBLEM 238
For the reaction:
H 2 ( g ) + CO 2 ( g )
H 2O( g ) + CO( g )
∆G at 2000 K is 2540 J, where partial pressures of the species are PH = 0.25, PCO2 = 0.78,
2
PH O( g ) = 0.66 and PCO = 1.2 atm respectively. Determine equilibrium composition of the gaseous
2
mixture.
PROBLEM 239
Consider the thermal decomposition of solid CaCO 3 as:
CaCO 3 ( s)
CaO( s) + CO 2 ( g ).
The equilibrium vapour pressure of CO 2 at 700°C and 950°C are 22.6 and 1830 mm of Hg. Calculate
∆H ° and ∆S ° for the reaction.
PROBLEM 240 A certain reaction is spontaneous at 72°C. If the enthalpy change for the reaction is
19 kJ, what is the minimum value of ∆S for the reaction?
PROBLEM 241 The internal engine of a 1200 kg car is designed to run on octane whose enthalpy of
combustion is 5510 kJ/mol. If the car is moving up a slope, calculate the maximum height to which the
car can be driven on 2.0 gallon of the fuel. Assume the cylinder temperature is 2200°C and the exit
temperature is 760°C and ignore all form of friction. The mass of 1.0 gallon of fuel is 3.1 kg.
PROBLEM 242 One gram sample of oxygen undergoes free expansion from 0.75 L to 3.0 L at 298 K.
Calculate ∆S , q, W , ∆H and ∆E.
32
Problems in Chemistry
PROBLEM 243 A 550 ml sample of an ideal gas at 300 K exerts 3 atm. The thermodynamic state of the
system changes in a process. In the final state, P = 3.5 atm and V = 730 mL. Calculate ∆S and ∆E and
∆H, C vm = (5 / 2) R .
PROBLEM 244 A sample of 0.0133 mole of an ideal gas, initially at 5.00 atm, expands isothermally and
reversibly from 3.00 L to 10 L. Calculate ∆S , ∆G and ∆H.
PROBLEM 245 One mole of an ideal gas originally at a volume of 8.00 Lit. at 1000 K, is allowed to
expand adiabatically until final volume is 16.00 Lit. For the gas C v =1.5 R . Calculate values of ∆S for
the process when:
(a) The expansion takes place reversibly.
(b) The expansion takes place against a constant pressure of 3.00 atm.
(c) The change in volume involves a free expansion.
PROBLEM 246 One mole of an ideal gas at 0°C and 1.0 atm pressure is mixed adiabatically with one
mole of a different gas at 100°C and 1.0 atm to yield a mixture. If C P for each gas is (5 / 2) R, determine
∆S (mixing).
PROBLEM 247
For chloroform gas C PM is expressed as:
C PM = 24.9 + 14.8 × 10 −2 T − 9 × 10 −5 T 2 JK –1 mol –1 .
Assuming this gas to be ideal, determine entropy change involved in heating 2.0 mole of gas from
volume 100 L at 500 K to a volume of 70 Lit. at 700 K.
PROBLEM 248
For N 2 ( g ), entropy function as a function of temperature is expressed as:
S = 25.1 + 29.3 ln T
Determine Gibb’s free energy change ∆G of one mole of nitrogen if it is heated from 298 K to 348 K
at 2.0 atm pressure.
PROBLEM 249 One mole of an ideal gas initially at 400 K and 10 atm, is adiabatically expanded
against a constant pressure of 5.0 atm until equilibrium is attained. If CV = 18.8 + 0.021T JK –1 mol –1 ,
determine ∆E , ∆H and ∆S .
PROBLEM 250 Molar volume of C 6 H 6 (l) is 89 c.c. at 27°C and 1.0 atm pressure. Assuming the volume
to be constant, determine ∆G for compression of 5.00 moles of liquid benzene from 1.0 atm to 100 atm.
PROBLEM 251 One mole of an ideal gas at 25°C is subjected to a reversible isoentropic expansion until
final temperature reached to 75°C. If the initial pressure was 1.0 atm, determine final pressure
CV = (3 / 2) R .
PROBLEM 252 A flask containing 1.00 mol of N 2 at 4.00 atm and 298 K was connected to a flask
containing 1.00 mol of N 2 gas at 2.00 bar and 298 K. The gases were allowed to mix isothermally.
Determine the entropy change for the system.
PROBLEM 253 One mole of solid iron was vaporized in an oven at 3500 K. If iron boils at 3133 K and
enthalpy of vaporization is 349 JK –1 mol –1 , determine ∆S system , ∆S surroundings and ∆S universe .
33
Problems
ATOMIC STRUCTURE
PROBLEM 254 Determine the number of revolutions made by an electron in one second in the 2nd Bohr
orbit of H-atom.
PROBLEM 255 (a) What electronic transition in He + ion would have the same wavelength as the first
Lyman line of H-atom?
(b) Which electronic transition in Li 2+ ion would have same wavelength as 3rd transition in Lyman
series of H-atom?
PROBLEM 256 Energy required for the excitation of H-atom from its ground state to the 2nd excited
state is 2.67 times smaller than dissociation energy of H 2 ( g ). If H 2 ( g ) placed in a 1.0 litre flask at 27°C
and 1.0 bar is to be excited to their 2nd excited state, what will be the total energy consumption?
PROBLEM 257 A H-like species is in their excited state (A) and absorb a photon of 3.868 eV and get
excited into a new state B. The electron from B in returning back to lower energy orbit, can give a
maximum of 10 emission lines with different wavelengths, some lines having energy greater than 3.868
eV, some lines with energy equal to 3.868 eV but two lines with energy less than 3.868 eV. Determine
the orbit number of states A and B and ionization energy of the species.
PROBLEM 258 What is the speed of an electron whose de Broglie wavelength is 0.1 nm. By what
potential difference, must have such an electron be accelerated from an initial speed zero.
PROBLEM 259 Consider a colloidal particle of mass 6 × 10 –16 kg. Suppose that its position is measured
to 1 nm accuracy, calculate the uncertainty in its velocity.
PROBLEM 260 The wavelength of a photoelectric threshold of a metal is 230 nm. Determine the kinetic
energy of photoelectron ejected from the surface by UV radiation emitted from the second largest
wavelength of transition of electron in Lyman series of atomic spectrum of H-atom.
PROBLEM 261 A hydrogen like species with atomic number Z is in higher excited state ‘n’ and emits
photons of energy 25.7 and 8.7 eV when makes a transitions to 1st and 2nd excited state respectively.
Determine ‘n’ and ionization energy of the species.
PROBLEM 262 Light from a discharge tube containing H-atoms in some excited state, falls on the
surface of sodium metal. The kinetic energy of the fastest photoelectron was found to be 10.93 eV. If
He + ions were present in the same excited state, the kinetic energy of the fastest photoelectron would
have been 49.18 eV. Determine the excited state orbit number and work function of Na.
PROBLEM 263 Certain sun glasses have small crystals of AgCl incorporated in the lenses. When the
lenses are exposed to light of appropriate wavelength, the following reaction occurs:
AgCl → Ag + Cl
The Ag atoms formed produce a uniform gray colour that reduces the glare. If ∆H for the reacton is
248 kJ, what maximum wavelength would be required to induce this process?
PROBLEM 264 Ozone in the stratosphere absorbs the harmful radiation from the sun by undergoing
decomposition:
O 3 → O + O 2
34
Problems in Chemistry
Calculate the maximum wavelength of photons that possess energy to cause the decomposition of
O 3 . Standard enthalpy of formation of ozone is 142 kJ/mol and dissociation energy of oxygen is 498
kJ/mol.
PROBLEM 265 The retina of human eye can detect light when radiant energy incident on it is at least
4.0 × 10 –17 J. For light of 600 nm wavelength, how many photons must incident on retina for clear vision.
PROBLEM 266 An electron in the excited state in a hydrogen atom can return in its ground state in two
different ways : (a) via a direct transition in which a photon of wavelength λ 1 is emitted. (b) via an
intermediate excited state reached by emission of a photon of wavelength λ 2 . This intermediate excited
state then decays to the ground state by emitting another photon of wavelength λ 3 . Derive an equation
that relates λ 1 , λ 2 and λ 3 .
PROBLEM 267 Calculate wavelength of He atom whose speed is equal to the r.m.s. at 20°C.
PROBLEM 268 How many photons at 600 nm must be absorbed to melt 500 g of ice? On average how
many H 2O molecules does one photon convert from ice to liquid water? Standard enthalpy of fusion of
ice is 6.0 kJ mol –1 .
PROBLEM 269 The energy needed for the reaction Li(g ) → Li 3+ (g ) + 3e is 19612 kJ/mol. If the
first ionization energy of lithium is 520 kJ/mol, determine its second ionization energy.
PROBLEM 270 Alveoli are the tiny sacs of air in the lungs whose average diameter is 50 µm. Consider
an oxygen molecule trapped within a sac. Calculate the uncertainty in the velocity of oxygen molecule.
(Hint : The maximum uncertainty in the position of molecule is given by the diameter of sac.)
PROBLEM 271 The sun is surrounded by a white circle of gaseous material known as Corona, which
becomes visible during total eclipse of the sun. The temperature of Corona is in million degree Celsius
which is high enough to breakup molecules and remove some or all electrons from the atom. One line
having wavelength equal to the emission line of Si14+ has been observed in this Corona. On this
observation what could be the estimated temperature of Corona.
PROBLEM 272 A H-like neutral species is in some excited state (A) and on absorbing a photon of
energy 3.066 eV gets promoted to a new state B. When the electron from state B return back, photons of a
maximum ten different wavelengths can be observed in which some photons of energy smaller than
3.066 eV. Some of the equal energy and only four photon having energy greater than 3.066 eV.
Determine the orbit number of states A and B and ionization energy.
PROBLEM 273 Calculate average energy of an oscillator with frequency1014 Hz at 1000 K. Boltzmann
constant K = 1.38 × 10 −23 JK –1 .
PROBLEM 274 The threshold wavelength for photoelectric emission in tungsten is 2300 Å. What
wavelength of light must be used to eject electrons with a maximum energy of 1.5 eV?
PROBLEM 275 A beam of X-rays is scattered by loosely bound electrons at 45° to the direction of
beam. The wavelength of scattered beam is 0.22 Å. What is the wavelength of the incident beam?
PROBLEM 276 By what potential difference, an electron at rest must be accelerated to have a
de-Broglie wavelength of 0.38 Å?
PROBLEM 277 Determine the de-Broglie wavelength associated with an electron in the 3rd Bohr’s
orbit of He + ion.
35
Problems
PROBLEM 278 If uncertainty (maximum) in location of an electron in a hydrogen atom is 0.1 mm, what
would be the minimum uncertainty in its speed if measured simultaneously?
PROBLEM 279 An automobile of mass 500 kg is moving with speed of 50 ± 0.001km hr –1 . Determine
uncertainty in position of moving automobile and interprete the result.
PROBLEM 280 If a photon of wavelength 200 pm strikes an atom and one of the inner bound electrons
is ejected out with a velocity of 2 × 10 7 ms –1 , calculate the energy with which it is bound to nucleus?
PROBLEM 281 When radiation of wavelength 253.7 nm falls on a copper surface, electrons are
ejected. Calculate work function if the stopping potential is 0.5 V.
PROBLEM 282 The size of an atomic nucleus is 10 −14 m. Calculate uncertainty in momentum of an
electron if it were to exist inside the nucleus.
PROBLEM 283 Calculate de-Broglie wavelength of a hydrogen atom with translational energy
corresponding to a temperature of 27°C.
Calculate the electrostatic potential energy of two electrons separated by 3 Å in
PROBLEM 284
vacuum?
The wave function for electron in ground state of hydrogen atom is
PROBLEM 285
1
−
( πa 03 ) 2
−
r
a0
, where “a 0 ” is radius of Bohr’s orbit. Calculate the probability of finding the
ψ1s =
e
electron somewhere between 0 and 2a 0 .
The normalized wave function of the hydrogen atom for the 1s orbital is
PROBELM 286
ψ1s =
( πa 03 )
−
1
2
PROBLEM 287
−
1
2
−
e
r
a0
. Show that in such a state the most probable distance from nucleus to electron is a 0 .
The wave function for the electron in the ground state of H-atom is
− r/a
ψ1s = ( πa 03 ) e 0 . What is the probability of finding an electron somewhere inside a small sphere of
radius 10 −12 m centered on nucleus?
PROBLEM 288
Determine de-Broglie wavelength associated with He atom at room temperature 25°C.
PROLBEM 289 At what temperature the translational kinetic energy of atomic hydrogen equal that for
n2 = 1 to n2 = 2 transition.
PROBLEM 290 A hypothetical element “positronium” consists of an electron moving in space around a
nucleus consisting a positron. Using the Bohr’s atomic model, determine the first Bohr radius. Positron
is a subatomic particle similar to electron in all respect except possessing a positive charge.
36
Problems in Chemistry
CHEMICAL BONDING
PROBLEM 291 Discuss the bonding of the following species:
–
+
+
+
CO 2 , COS, CO 2–
3 , SiCl 4 , PCl 5 , NH 4 , PH 4 , PCl 4 , PCl 6 , SF6 , IF7 .
PROBLEM 292 Discuss the bonding of the following species:
SO 2 , NH 3 , H 2O, PCl 3 , NCl 3 .
PROBLEM 293 Discuss the relative bond angles in the following species:
(a) NH 3 , PH 3 , AsH 3 , SbH 3 and BiH 3 .
(b) H 2O, H 2S, H 2Se, H 2Te.
PROBLEM 294 Arrange the following species in the increasing order of their bond angles, explaining
the reason for your order:
H 2O, NH 3 , H 2Se, PH 3 , AsH 3 .
PROBLEM 295 Discuss the bonding of:
ClF3 , COCl 2 , Cl 2O, OF2 , ICl –4 , IF5 , IOCl –4 , IOCl +4 .
PROBLEM 296 Discuss the bonding of:
N –3 , I –3 , BrF5 , IF4– , BF4– , ICl +2 , ICl –2 and PF5 .
PROBLEM 297 The energy necessary to break similar bond is not always equal, it varies from molecule
to molecule as:
NCl 3 → NCl 2 + Cl ∆H = 375 kJ/ mol
and
NOCl → NO + Cl
Discuss the difference.
∆H =158 kJ/ mol
PROBLEM 298 Draw the shape of the following species:
–
–
–
AsF5 , AsF2+ , SnCl –3 , NOF, SO 2–
3 , TeF5 , GeF3 , SCl 2 , SbCl 6 .
PROBLEM 299 Discuss the bonding of following:
ClF4+ , FClO 3 , F2ClO + , SeO 2–
3 .
PROBLEM 300 Discuss the relative polarity of the following species:
(a) SCl 2 , BF3 , ICl 3 , POCl 3 , PCl 5 .
(b) XeF2 , SF4 , XeO 3 , SnCl 4 , PCl 4 F.
PROBLEM 301 Arrange the following compounds in the increasing order of their melting points
explaining reason for your order: Li 2O, LiF, Li 3 N.
37
Problems
PROBLEM 302 Arrange the following in the increasing order of their thermal stability with suitable
explanations:
(a) CaCO 3 , BaCO 3 , MgCO 3 , Na 2CO 3 , Al 2 (CO 3 ) 3 , BeCO 3 .
(b) CaCO 3 , CaSO 4 and CaSO 3 .
(c) MgSO 4 , BaSO 4 , Al 2 (SO 4 ) 3 , SrSO 4 .
(d) CaC 2O 4 , K 2C 2O 4 , FeC 2O 4 and CaCO 3 .
PROBLEM 303 The molecule XeF4 has two lone pairs at the central atom, instead the bond angle is 90°
as expected from its geometry, explain.
PROBLEM 304 Arrange the following species in the increasing order of their ionic character:
NaCl, CaCl 2 , AlCl 3 , BaCl 2 , MgCl 2 and GaCl 3 .
PROBLEM 305 In the following pairs of molecules, select one, which has greater bond angle, and
explain the reason for your answer.
(a) NH 3 or NF3
(b) PH 3 or PF3
(c) AsH 3 or AsF3
(d) AsF3 or AsCl 3
(e) H 2O or F2O
PROBLEM 306 Arrange the following sets of molecules in increasing order of bond angles providing
appropriate explanation for your order:
(a) H 2O, H 2S, NH 3 and PH 3
(b) CH –3 , SiH –3 and GeH –3
PROBLEM 307 Draw the shape of the following species indicating bond angles and distortion (if
present):
(e) ICl –2
(d) ICl +2
(c) PCl 2 BrF2
(b) PF2Cl 3
(a) PF3Cl 2
PROBLEM 308 Considering the molecules in question 223, arrange them in the increasing order of their
dipole moments.
PROBLEM 309 Arrange the following molecules in the increasing order of their polarity:
CH 3Cl, CH 2Cl 2 , CHCl 3 and CCl 4 .
PROBLEM 310 Discuss the bonding in the following molecules with respect to hybridisation of central
atom, shape and bond angles:
(a) ClF3O 2
(d) I +3
(b) XeOF4
(c) IOCl –4
PROBLEM 311 Discuss the bonding of PCl 5 in gas phase and in solid phase.
PROBLEM 312 Dimethyl ether has tetrahedral geometry of hybrid orbital at central atom whereas
disilyl ether has triangular planar geometry of hybrid orbital at central atom. Explain.
PROBLEM 313 Draw the shape of the following molecules:
(a) XeF2
(b) XeF4
(c) XeF6
(d) XeOF2
(e) XeOF4
(f) XeO 2 F2
(g) XeO 3 F2
(h) I 2Cl 6
(i) I 2 Br 2Cl 4
38
Problems in Chemistry
PROBLEM 314 Draw the shape of the following molecules:
(b) (CH 3 ) 3 P(CF3 ) 2
(a) (CH 3 ) 2 P(CF3 ) 3
PROBLEM 315 In sp 3 d-hybridized phosphorus atom in trigonal bipyramidal molecule, will the atom
have a greater electronegativity when bonding through equatorial or axial orbitals? Explain.
PROBLEM 316 B—F bond distance in BF3 is shorter than the same in BF4– , explain.
PROBLEM 317 In CCl 4 , C—Cl bond length is the sum of the covalent radii of carbon and chlorine
whereas in SiCl 4 , Si—Cl bond distance is smaller than the sum of the covalent radii of Si and Cl, explain.
PROBLEM 318 Arrange the following in increasing order of their Lewis acid strength with proper
reasoning. BCl 3 , BI 3 , BF3 and BBr 3 .
PROBLEM 319 In gas phase, N(CH 3 ) 3 acts as a good Lewis base but N(SiH 3 ) 3 doesn’t, explain.
PROBLEM 320 The molecule CHBrCHBr can have two different structures in which one is polar and
other is non-polar. Draw the structures labelling them as polar and non-polar.
PROBLEM 321 Arrange the following compounds in order of increasing dipole moment:
Cl
Cl
Cl
Cl
Cl
(a)
Cl
Cl
Cl
Cl
Cl
Cl
(b)
(c)
(d)
PROBLEM 322 Although both carbon and silicon are in the same group of periodic table, very few
Si==Si bonds are known. Account for the instability of Si==Si in general.
PROBLEM 323 Molecule N 2 F2 can acquire two different structures in which one is polar and other is
non-polar. Draw them labelling polar and non-polar.
PROBLEM 324 Compound 1,2-dichloro ethane is a non-polar whereas cis-1,2-dichloro ethene is polar,
explain the difference.
PROBLEM 325 Discuss the dipole moment of the following molecules in view of chemical bonding:
Cl
H
Cl
H
and
C==C==C==C
C==C==C
H
Cl
H
Cl
PROBLEM 326 Draw all possible structures for the molecule C 2 H 2Cl 2 and rank them in increasing
order of their dipole moment.
39
Problems
PROBLEM 327 Draw shape of a hypothetical molecule N 6 in which nitrogen atoms are part of a six
membered ring and has: (a) two pi-bonds, (b) three pi-bonds.
PROBLEM 328 Sodium chloride (NaCl) and sodium fluoride (NaF) both crystallizes in same type of
unit cell. Which is expected to have higher lattice energy and why?
PROBLEM 329 The bond energy in NO is 632 kJ mol –1 and that of each N  O bond in NO 2 is
469 kJ mol –1 . Explain.
PROBLEM 330
In the air, NO can react with NO 2 . What is the most likely structure of product?
PROBLEM 331 Draw the Lewis structure of the following species:
(a) SO 2Cl + , (b) S 2 F4 (contain S—S bond).
PROBLEM 332 The heteronuclear diatomic ion CN – has an orbital structure similar to that of N 2 . How
will the fact C has an electronegativity different from that of N affect the energy level diagram.
PROBLEM 333 From the following pair of molecules, select one which will be more soluble in a polar
solvent and explain the reason for your choice.
(a) SiF4 and PF3 , (b) SF6 and SF4 , (c) IF5 and AsF5
PROBLEM 334 In addition to forming σ- and π-bonds similar to those formed by p-orbitals, d-orbitals
may overlap in δ-bonds with two nodal planes cutting through the internuclear axis. Draw the overlap
diagrams showing how d-orbitals can overlap in these three ways.
CHEMICAL EQUILIBRIUM
PROBLEM 335 An equilibrium mixture of:
XeF2 ( g ) + OF2 ( g )
XeOF2 ( g ) + F2 ( g )
was found to contain 0.6 mole of XeF2 ( g ), 0.3 mole of OF2 ( g ), 0.1 mole of XeOF2 ( g ) and 0.4 mole
of F2 in a one litre container. How many moles of OF2 must be added to increase [XeOF2 ] to 0.2 M?
PROBLEM 336 A compound HB is formed from H and B according to the following reaction:
H+B
HB. A solution was prepared by dissolving 0.1 mole of H and 0.1 mole of B in enough water
to make the total volume equal to one litre. After equilibrium had been reached, it was found that 20% of
H had been reacted. What are the equilibrium concentration of H, B and HB? What is the equilibrium
constant, K c for this reaction?
PROBLEM 337 2.0 g of ammonium chloride was heated in a one litre flask to 300°C. From the
measurement of pressure, it was found that 98% of ammonium chloride was dissociated. If to this flask
2.0g of dry ammonia was added, what would be the percentage dissociation?
40
Problems in Chemistry
PROBLEM 338
H 2 (g ) + I 2 (g )
2HI( g )
If the system is equilibrated at [H 2 ] = 0.5, [I 2 ] = 0.5 and [HI] = 1.23 in a one litre flask and suddenly
0.6 mole of HI is removed, what will be the new equilibrium composition when the equilibrium is
re-established?
PROBLEM 339 The equilibrium constant K p for the gas phase decomposition of ter-butyl chloride is
3.45 at 500 K:
(CH 3 ) 3 CCl( g )
(CH 3 ) 2 C = CH 2 ( g ) + HCl( g )
Calculate molar concentration of reactants and products in the equilibrium mixture obtained by
heating 1.0 mole of ter-butyl chloride in a 5.0 L container at 500 K.
PROBLEM 340 A 79.2 g chunk of dry ice and 30 g of graphite were placed in an empty 5.0 L container,
and the mixture was heated to achieve equilibrium. The reaction is:
CO 2 ( g ) + C( s)
2CO( g )
What is the value of K p at 1000 K if the gas density at this temperature is 16.3 g/L. What is the value
of K p at 1100 K if the gas density at this temperature is 16.9 g/L. Also calculate the enthalpy of the
reaction.
PROBLEM 341 The reaction: N 2 + 3H 2
2NH 3 is started with an equal number of moles of N 2
and H 2 . Calculate mole fraction of NH 3 at 723 K and a total pressure of 0.5 atm given K p = 6 × 10 –3 .
1
H2 + O2
2
is 8.7 ×10 –11 . For a significant production of hydrogen, which may be used as fuel, at least 15% of water
should be dissociated. To what temperature should you heat the water? The enthalpy of reaction is 57.8
kcal. For a rough estimate assume that enthalpy is independent of temperature and P =1.0 atm.
PROBLEM 342 At 1000 K, the equilibrium constant for the dissociation of water, H 2O
PROBLEM 343 In a study of the equilibrium: H 2 + I 2
2HI, 1.0 mole of H 2 and 3 mole of I 2 gave
rise at equilibrium to x mole of HI. Addition of a further 2 mole of H 2 gave an additional x mole of HI.
What is x? What is K at the temperature of experiment?
PROBLEM 344 The equilibrium constant for the reaction:
H 2 (g ) + I 2 (g )
2HI ( g ) is 20 at
40°C, and vapour pressure of solid iodine is 0.1 bar at that temperature. If 12.7 g of solid iodine are
placed in a 10 L vessel at 40°C, what is the minimum amount of hydrogen gas that must be introduced in
order to remove all solid iodine?
PROBLEM 345 A standard solution of I 2 in water contains 0.33 g of I 2 in one litre flask. More than this
dissolve in a KI solution because of the following equilibrium:
I2 + I–
I –3
A 0.1 M KI solution actually dissolve 12.5 g of iodine per litre, most of which is converted to I –3 .
Assuming that concentration of I 2 in all saturated solution is same, calculate the equilibrium constant for
the above reaction. What is the effect of adding water to a clear saturated solution of I 2 in KI solution?
41
Problems
PROBLEM 346 The following gaseous equilibrium was obtained by heating 0.46 moles of A in a 5.0 L
vessel. The equilibrium pressure at 300 K was 3.0 atm. The equilibrium pressure changed to 3.6 atm
when temperature was raised to 320 K. Calculate percentage change in degree of dissociation of A at 350
K with respect to that of A at 300 K.
A(g )
B ( g ) + Cg )
PROBLEM 347 At 400°C, a 1 : 3 mixture of N 2 and H 2 reacts to form an equilibrium mixture of N 2 , H 2
and NH 3 . The total pressure at equilibrium was found to be 350 bar and mole fraction of NH 3 = 0.5.
Calculate K p for the reaction:
N 2 ( g ) + 3H 2 ( g )
2NH 3
PROBLEM 348 Show that the maximum yield of product at equilibrium occurs when stoichiometric
amount of reactants are used. Use the reaction:
Cl 2 ( g ) + Br 2 ( g )
2BrCl( g )
PROBLEM 349 A flask containing 0.06 mole of F2 (g ) is allowed to equilibrate with F(g) at 1000 K. If
the total pressure of the gases at equilibrium is 2.07 bar, calculate mole fraction of each gases at
equilibrium.
Given:
F2 ( g )
2F( g ), K p (1000 K) = 9.5 × 10 −3
PROBLEM 350 A mixture of air at 1.0 bar and 2000 K was passed through an electric arc to produce
nitric oxide as N 2 ( g ) + O 2 ( g )
2NO( g ), K 2000 = 4 × 10 –4 . What are the partial pressure of gases
once the equilibrium has been established.
PROBLEM 351 At a pressure of one bar, an equilibrium exist at 2000 K between 0.25 mole of Br 2 (g ),
0.75 mole of F2 ( g ) gas and 0.497 mole of BrF3 ( g ). What will be the amounts of each gas after the
pressure on the system has been increased to 2.0 bar and equilibrium at 2000 K re-established?
Br 2 ( g ) + 3F2 ( g )
2BrF3 ( g )
PROBLEM 352 Would 1.0% CO 2 in air be sufficient to prevent any loss in weight when Ag 2CO 3 is
dried at 120°C
Ag 2CO 3 ( s)
Ag 2O ( s) + CO 2 ( g ),
K p = 0.0095 at 120°C
PROBLEM 353 In a gas phase reaction 2 A + B
3C + 2D, it was found that when 1.0 mol of A, 2.0
mol of B, and 1.0 mol of D were mixed in one litre flask and came to equilibrium, the resulting mixture
contained 0.9 mole of C. Calculate equilibrium constant K c .
PROBLEM 354 A container whose volume is V constains an equilibrium mixture that consists of 2 mole
each of PCl 5 , PCl 3 and Cl 2 and the equilibrium pressure is 30.4 kPa at T.
A certain amount of Cl 2 ( g ) is now introduced keeping P and T constant, until equilibrium volume is
2V. Calculate amount of Cl 2 ( g ) added and K p .
42
Problems in Chemistry
PROBLEM 355 A 250 ml flask containing NO(g) at 0.46 atm is connected to a 100 mL flask containing
oxygen gas at 0.86 atm by means of a stop cock at 350 K. The gases are mixed by opening the stop cock
where the following equilibrium established: 2NO + O 2 → 2NO 2
N 2O 4 .
The first reaction is complete while the second is at equilibrium. Calculate K p if the final total
pressure is 0.37 atm.
PROBLEM 356 Equal moles of F2 (g ) and Cl 2 (g ) are introduced into a sealed container and heated to
certain temperature where following equilibria established:
Cl 2 ( g ) + F2 ( g )
2ClF( g ), K p = 3.2
Cl 2 ( g ) + 3F2 ( g )
2ClF3 ( g )
If partial pressure of ClF and ClF3 at equilibrium are 0.2 and 0.04 atm respectively, calculate K p for
the second equilibrium reaction.
PROBLEM 357 0.2 mole of each Cl 2 (g ) and F2 (g ) are introduced in a sealed flask and heated to 2000 K
where following equilibrium established. Cl 2 ( g ) + F2 ( g )
2ClF( g ); and at equilibrium, moles of
ClF = 0.267. At this stage, 0.1 mol of Br 2 is added and equilibrium is re-established as:
Cl 2 ( g ) + F2 ( g )
2ClF( g )
Cl 2 ( g ) + Br 2 ( g )
2BrCl( g )
Now moles of ClF found to be 0.25. Calculate K c for the second equilibrium reaction.
PROBLEM 358 A glass bulb initially contains mixture of N 2 and NO at a total pressure of one atm.
Some Br 2 ( g ) is added such that the pressure would have been 2.25 atm had not the equilibrium:
2NO + Br 2
2NOBr
been established, as a result, the actual pressure after adding Br 2 ( g ) was 2.12 atm. A second addition of
Br 2 was made such that total pressure would have been 22.75 atm had no reaction at all been occurred,
while the actual pressure found to be 22.5 atm. Calculate K p for the reaction assuming N 2 to be inert.
PROBLEM 359 When 0.1 mole of Cl 2 (g ) and an excess of solid iodine are placed in 5 L vessel at 300 K,
the total pressure is 0.767 atm. The vapour pressure of solid iodine is 0.03 atmosphere at this
temperature. After this 1st equilibrium, 0.1 mole of bromine vapour is introduced into the reaction
vessel. It is found that the weight of excess solid iodine diminished by an amount corresponding to 0.03
mole when new equilibrium is established. It is also found that partial pressure of bromine in new
equilibrium mixture is 0.2 atm. Calculate K p for the following equilibria at 300 K:
Cl 2 ( g ) + I 2 ( g )
2ICl( g )
Cl 2 ( g ) + Br 2 ( g )
2BrCl( g )
I 2 ( g ) + Br 2 ( g )
2IBr( g )
PROBLEM 360 The equilibrium constant of a reaction A2 (g ) + B 2 (g )
2 AB ( g ) at 100°C is 50. If a
one litre flask containing one mole of A2 ( g ) is connected to a 2.0 L flask containing two moles of B 2 ( g ),
how many moles of AB will be formed at 373 K.
PROBLEM 361 0.96 g of HI was heated till equilibrium 2HI(g )
H 2 + I 2 ( g ) is reached. The
reaction mixture was suddenly cooled and the amount of iodine produced required 15.7 mL N/10 hypo
solution. Calculate K p for the above equilibrium reaction.
43
Problems
PROBLEM 362 Hydrazine was taken in a constant volume container at 27°C and 0.3 atm and equal
moles of oxygen gas was injected, sealed and finally heated to 1000 K where the following equilibria
established:
N 2 H 4 + 3O 2
N 2 H4
2NO 2 + 2H 2O
N 2 + 2H 2
K p1 = 3
K p2 = ?
N 2 H4 + H2
2NH 3
K p3 = ?
If the gaseous mixture at equilibrium is passed through moisture absorbent, a decrease of 360 mm in
the equilibrium pressure was observed. Now if the dried gaseous mixture is passed through ammonia
absorber a further decrease of 20 mm in the equilibrium pressure was observed. Calculate K p 2 and K p 3 .
PROBLEM 363 Pure nirtosyl chloride (NOCl) gas was heated to 240°C in a 1.0 L container. At
equilibrium, the total pressure was found to be 1.00 atmosphere and partial pressure of NOCl was 0.64
atm. Now some Cl 2 ( g ) is added at constant pressure and equilibrium was allowed to re-establish. At new
equilibrium, the volume of the container was 1.5 L. Determine the moles of Cl 2 ( g ) added at first
equilibrium.
PROBLEM 364 N 2O 4 gas decomposes partially as N 2O 4
2NO 2 . In an experiment, some N 2O 4 is
taken in a flask and heated to 300 K, where the above equilibrium is established. At equilibrium pressure
of 1.0 atmosphere, density of the equilibrium mixture was found to be 2.33 g/L. Now, the gas is
compressed till the density of new equilibrium mixture reaches to 5.08 g/L. Determine the new
equilibrium pressure and the density of equilibrium mixture if the equilibrium pressure in this case is 1.5
atm.
PROBLEM 365 H 2S dissociates according to the following reaction: H 2S
H 2 + S( g ). At 1000 K
and total pressure of 1.0 atmosphere, degree of dissociation was found to be 0.3. Determine the degree of
dissociation if the gas is compressed isothermally to a new equilibrium pressure of 2.0 atmosphere.
PROBLEM 366 (a) PCl 5 dissociates as: PCl 5
PCl 3 + Cl 2 . When 0.03 mole of PCl 5 was brought
to equilibrium at 500 K and 1.0 atmosphere, the equilibrium volume was 2.09 L. Calculate degree of
dissociation.
(b) What will be the degree of dissociation when 0.2 mole of PCl 5 is brought to equilibrium in a 3 L
flask at 500 K.
PROBLEM 367 0.1 mole of hydrogen gas and 0.2 mole of CO 2 (g ) are introduced in an evacuated flask
at 723 K and the following reaction occurs to give an equilibrium pressure of 50.67 kPa.
H 2 ( g ) + CO 2 ( g )
H 2O( g ) + CO( g )
Analysis of the mixture shows that it contains 10 mol % of H 2O( g ). A mixture of CoO(s) and Co(s)
is then introduced such that the additional equilibria:
CoO( s) + H 2 ( g )
Co( s) + H 2O( g )
CoO( s) + CO( g )
Co( s) + CO 2 ( g )
are established. Analysis of the new equilibrium mixture thus, obtained is found to contain 30% (mol) of
H 2O( g ) are present. Determine K c for the three reactions.
PROBLEM 368 At 25°C, the equilibrium : 2NOBr
2NO + Br 2 is readily established. When 1.1 g
of NOBr is present in 1.0 litre flask at 25°C, total pressure in the flask was found to be 0.335 atmosphere.
44
Problems in Chemistry
Determine the pressure if 5.0 g of NOBr is placed in a two litre evacuated flask at 25°C, sealed and
allowed to attain the decomposition equilibrium to establish.
PROBLEM 369 A gaseous mixture containing equimolar amount of HCl and O 2 was taken in a flask,
sealed and heated to 500°C where the following equilibrium is established:
4HCl + O 2 ( g )
2Cl 2 + 2H 2O( g )
If the initial gas pressure was 1.0 atmosphere and mole fraction of HCl reacted before the
equilibrium was established, was 0.76. Determine K p .
PROBLEM 370 When N 2O 5 is heated, it dissociates as: N 2O 5
N 2O 3 + O 2 K c = 4.5. At the
same time, N 2O 3 also decomposes as: N 2O 3
N 2O + O 2 . If initially 4.0 moles of N 2O 5 are taken
in a 1.0 litre flask and allowed to attain equilibrium, concentration of O 2 was found to be 4.5 M.
Determine equilibrium concentration of other species and K c for the second equilibrium.
PROBLEM 371 At 25°C, a mixture of NO 2 and N 2O 4 are in equilibrium in a cylinder fitted with a
movable piston. The concentration of species present at equilibrium are as follows:
[ N 2O 4 ] = 0.487, [NO 2 ] = 0.0475. Now, the piston is pushed to half the volume where the equilibrium
was re-establish. Determine concentration of gases present at new equilibrium.
PROBLEM 372 Gaseous nitrosyl chloride (NOCl) and N 2 are taken in a flask, sealed and heated to
some temperature where the total pressure would have been 1.0 bar had not the following equilibrium
been established:
2NOCl
2NO + Cl 2
The actual pressure was found to be 1.2 bar. Now into the equilibrium mixture, some Cl 2 gas was
introduced so that the total pressure would have been 9.0 bar had no further reaction occurred but the
actual pressure was found to be 8.9 bar. Determine the equilibrium constant for the decomposition
equilibrium under the given experimental condition.
PROBLEM 373 20.85 g of PCl 5 (g ) is introduced in a vessel washed with a nonvolatile solvent (B.Pt.
= 350 K, M. Pt. =154). The equilibrium is established at 523 K when PCl 5 ( g ) is 52% dissociated and a
total pressure was found to be 5.5 bar. If K p for the decomposition reaction:
PCl 5
PCl 3 + Cl 2 is 1.78.
Calculate the weight of solvent left in the vessel during washing.
PROBLEM 374 Cl 2 (g ) and O 2 (g ) are taken in the molar ratio of 2 : 7 where the following equilibrium
was established:
2Cl 2 + 7O 2
2Cl 2O 7
At equilibrium, mole fraction of Cl 2O 7 was found to be 0.1 when the total pressure was 100 bar. In
an another experiment, two gases were taken in equimolar amount under identifical condition of
temperature and mole fraction of Cl 2O 7 at equilibrium was found to be 0.06. Determine the equilibrium
pressure in the new flask.
PROBLEM 375 PCl 5 (g ) is taken in a flask at 1.0 atmosphere, sealed and allowed to attain the following
equilibrium:
PCl 5
PCl 3 + Cl 2
The equilibrium mixture was then allowed to pass through a pin hole and the gases coming out of
pin-hole initially, was collected, analyzed and mole fraction of Cl 2 was found to be 0.53. Determine
equilibrium constant ( K p ) for the decomposition reaction.
45
Problems
PROBLEM 376 A solid substance A decomposes into two gaseous products B and C as:
A ( s)
2B ( g ) + C ( g )
If at equilibrium, some C(g) at 1.0 atmosphere is added in constant volume condition, 10% of B(g)
solidified before the equilibrium was re-established. Determine total pressure at final equilibrium.
PROBLEM 377 Toxic level of Pb 2+ in human blood is reduced by forming a stable Pb 2+ EDTA complex
which is excrete able through kidneys. Formation constant ( K f ) for this complex is 1018 . The ligand is
administered by infusion of a solution of Na 2 [Ca-EDTA] (K f = 5 × 1010 ). In the blood stream, exchange
of Ca 2+ for Pb 2+ occurs. The level of Pb 2+ in a patient blood was found to be 4µΜ. To this patient a
mixed solution containing Ca(NO 3 ) 2 and Na 2 [Ca-EDTA] was administered so that their initial
concentration in their blood were 2.5 µM and 1.0 µM respectively. Determine the ratio [Pb-EDTA
]2– /[Pb 2+ ] in the patient blood at equilibrium.
PROBLEM 378 NH 4 HS(s) is an unstable solid, decomposes as:NH 4 HS
NH 3 + H 2S and the
following thermodynamic informations are available:
∆H ° f ( kJ/ mol) :
NH 4 HS = −157,
NH 3 = – 46 and H 2S = – 20.5 ∆S f° (JK –1 mol –1 ) : NH 4 HS = 113.5, NH 3 = 193 and H 2S = 206.
Suppose 1.0 mole of solid NH 4 HS is introduced into an empty 25 L flask, calculate the equilibrium
pressure at 27°C.
PROBLEM 379 N 2O 4 decomposes as N 2O 4
2NO 2 and at 300 K, ∆G ° f of N 2O 4 ( g ) and NO 2 ( g )
are 98 and 52 kJ/mol respectively. Starting with one mole of N 2O 4 at one bar and 300 K, calculate the
fraction of N 2O 4 decomposed when equilibrium is established at 1.0 bar and 300 K. Also determine
percentage volume change if decomposition is carried out in a cylinder fitted with a mass-less,
frictionless piston.
PROBLEM 380 A gaseous substance AB 2 (g ) convert to AB(g) in presence of solid A(s) as:
AB 2 ( g ) + A ( s)
2 AB ( g )
The initial pressure and equilibrium pressure are 0.7 and 0.95 bar. Now the equilibrium mixture is
expanded reversibly and isothermally till the gas pressure falls to 0.4 bar. Determine volume percentage
of AB(g) and AB 2 ( g ) at the final equilibrium.
PROBLEM 381 One mole of N 2 (g ), three moles of H 2 (g ) and one mole of H 2S(g ) were taken in a one
litre flask, sealed and heated to 700 K where the following equilibria were established:
N 2 ( g ) + 3H 2 ( g )
H 2S( g ) + NH 3 ( g )
2NH 3 ( g )
NH 4 HS( g )
K p = 8 × 10 –3 atm –1
At equilibrium, concentration of ammonia gas was found to be 0.9 M. Determine K p for the first
equilibrium and equilibrium concentrations of H 2S.
PROBLEM 382 N 2O 4 (g ) is taken in a cylinder equipped with movable piston and heated first at
constant volume where the following equilibrium is established and the gas was 30% dissociated :
N 2O 4 ( g )
2NO 2 ( g )
Now the gaseous mixture was expanded isothermally till volume was doubled. Determine the
percentage of N 2O 4 dissociated at this stage.
46
Problems in Chemistry
PROBLEM 383
Consider the following reaction:
2NO 2 + O 3
N 2O 5 ( g ) + O 2 ( g )
∆H °f (O 3 ) = 143 kJ mol , ∆H °f (N 2O 5 ) = 11 kJ mol –1 and ∆H °f (NO 2 ) = 33 kJ mol –1 .
–1
The
above reaction is spontaneous at lower temperature but turned non-spontaneous as temperature
approaches to 1175 K. Assuming ∆H ° and ∆S ° to be independent of temperature, determine K p at
500 K.
PROBLEM 384
Consider formation of N 2O 5 ( g ) according to the reaction below :
2NO 2 + 12 O 2
N 2O 5
∆H ° = − 55 kJ; ∆S ° = − 227 JK –1
Also, ∆H °f (NO 2 ) = + 33.2 kJ mol –1 , S ° ( NO 2 ) = 240 JK –1 , S ° (O 2 ) = 205 JK –1 .
(a) Determine ∆H °f N 2O 5 , S ° (N 2O 5 ), ∆G ° at 25°C.
(b) State and explain, whether this reaction is spontaneous at 25°C.
(c) How the relative amounts of reactants and products would be affected at equilibrium?
PROBLEM 385 Cis-2-butene when heated to 500 K, it isomerizes into trans-2-butene and
2-methylpropene. If standard state Gibb’s free energy of formations of cis-2-butene, trans-2-butene and
2-methyl-propene are 65.85 kJ mol –1 , 62.77 kJ mol –1 and 58.07 kJ mol –1 respectively, determine the
equilibrium composition.
PROBLEM 386 A vessel containing 0.015 mol N 2 (g ) and 0.02 mol PCl 5 (g ) is heated to 227°C where
the pressure was found to be 1.843 atm`. Also at 227°C, K p for : PCl 5 ( g )
PCl 3 ( g ) + Cl 2 ( g ) = 0.4
atm. Assuming N 2 to be inert gas, determine, volume of the vessel.
PROBLEM 387
The ∆G° for the reaction:
SO 2 + 12 O 2
SO 3 is −22600 + 21T
Determine the temperature at which mixture is 80% (by mole) in SO 3 , if initial mixture contain 15%
SO 2 and 20% O 2 at one atmosphere total pressure. Assume total pressure is maintained constant at one
atmosphere throughout.
PROBLEM 388 A 2.0 L flask, initially containing one mol of each CO and H 2O, was sealed and heated
to 700 K, where the following equilibrium was established :
CO( g ) + H 2O( g )
CO 2 ( g ) + H 2 ( g );
K ( 700) = 9
Now the flask was connected to another flask containing some pure CO 2 ( g ) at same temperature and
pressure, by means of a narrow tube of negligible volume. When the equilibrium was restored, moles of
CO was found to be double of its mole at first equilibrium. Determine volume of CO 2 ( g ) flask.
PROBLEM 389
Consider the reaction:
H 2 (g ) + I 2 (g )
2HI ( g )
If we start with 0.5 moles of each H 2 ( g ) and I 2 ( g ) at 700 K in a flask, equilibrium concentration of
HI was observed to be 0.15 M. What would have been the concentration of HI at equilibrium had the
reaction been started with 0.8 mole of each H 2 and I 2 ( g ) at same temperature and in the same vessel?
PROBLEM 390 For the reaction C(s) + CO 2 (g )
2CO( g ), K p = 63 atm at 1000 K. What will be the
total pressure of the gases above an equilibrium mixture if PCO = 10 PCO2 .
Problems
47
IONIC EQUILIBRIUM
PROBLEM 391 Addition of 100 mL 0.1 M HCl to certain volume of a slightly weak monobasic (0.1 M)
acid solution decreases its pH from 2.0 to 1.7. Determine ionization constant of the weak acid and its
volume.
PROBLEM 392 Calculate pH of 0.02 M succinnic acid solution taking into account both ionization.
K a 1 = 7 × 10 –5 , K a 2 = 3 × 10 –6 .
PROBLEM 393 K a of formic acid at 27°C is1.7 ×10 –4 . What will be the pH of a 0.1 M aqueous solution
of formic acid at 47°C. Standard enthalpy of neutralization of formic acid and HCl versus NaOH are – 42
kJ and – 57 kJ respectively.
PROBLEM 394 Determine exact pD of a 10 –7 M DCl solution is D 2O. Ionic product of D 2O is
1.35 ×10 –15 .
PROBLEM 395 Calcium hypochlorite [Ca(OCl) 2 ] is used as a disinfectant for swimming pools. The
recommended pH of a swimming pool is 7.8. Calculate the percentage of HClO and ClO – in the
swimming pool. K a for HOCl is 3 × 10 –8 .
PROBLEM 396 In the vapour phase acetic acid molecule associate to the some extent to form dimmers.
At 50°C, the pressure of a certain acetic acid vapour is 0.0342 atm in a 360 mL flask. The vapour is
condensed and neutralized with 13.8 mL 0.0568 M NaOH. Calculate the degree of dissociation of the
dimmer.
PROBLEM 397 Henery’s law constant for CO 2 at 38°C is 2.28 ×10 –3 mol/ L. Determine pH of a
solution of CO 2 at 38°C in equilibrium with the gas at a partial pressure of 3.2.
For CO 2 : K1 = 4.2 × 10 –7 , K 2 = 4.8 × 10 –11 .
PROBLEM 398 Determine pH of a 5% aqueous solution of NaOCl (by weight). Density of the solution
is 1.0 g/cc. K a = 3 × 10 –8 .
PROBLEM 399 An ammonia-ammonium chloride buffer has a pH value of 9 with [NH 3 ] = 0.25. By
how much the pH will change if 75 mL of 0.1 M KOH be added to 200 mL buffer solution. K b = 2 × 10 –5 .
PROBLEM 400 A solution is prepared by dissolving 15 g of acetic acid and 25 g sodium acetate in
750 mL of water.
(a) What is the pH of this solution?
(b) What would the pH of the solution be after 25.00 mL 0.25 M NaOH is added?
(c) What would be the pH is 25 mL 0.4 M HCl is added into (b)?
K a = 2 × 10 –5 .
48
Problems in Chemistry
PROBLEM 401 Determine freezing point of a 0.5 M solution of dichloroacetic acid (K a = 0.05).
Assume density of the solution to be 1 g/cc and K f for water =1.86.
PROBLEM 402 pK w of heavy water at 20°C and 30°C are 15.05 and 14.7 respectively. Determine pD of
a pure heavy water at 50°C.
PROBLEM 403 Suppose that two hydroxides MOH and M ′ (OH) 2 , both have K sp =10 –12 and that
initially both cations are present in a solution at concentrations of 0.001 M. Which hydroxide precipitate
first and at what pH when a NaOH solution is added dropwise?
PROBLEM 404 Determine molar solubility of Fe(OH) 2 at pH − 8 and at pH = 6. K sp = 1.6 × 10 –14 .
PROBLEM 405 Determine solubility of CaF2 (K sp = 4 × 10 –11 )
(a) at pH = 7,
(b) at pH = 5.
K a = 3.45 × 10 –4 .
PROBLEM 406 Determine the molar solubility of BaF2 (K sp = 1.7 × 10 –6 ) at
(a) pH = 7,
(b) pH = 4.
PROBLEM 407 When 10 mg sodium barbiturate are dissolved in 250 mL of water to form a solution,
the resulting pH was found to be 7.71. The molar mass of the salt is 150. Determine percentage
protonation of barbiturate ion and K a of the acid.
PROBLEM 408 Determine pH of a 0.024 M hydroxyl amine hydrochloride solution. K b of hydroxyl
amine =10 –8 .
PROBLEM 409 A 25 mL 0.2 M oxalic acid is titrated with 0.2 M NaOH solution. Determine pH at the
first and second equivalence point. K a 1 = 6 × 10 –2 and K a 2 = 6.5 × 10 –5 .
PROBLEM 410 Determine molar solubility of ZnS (K sp = 1.6 × 10 –24 ) in a saturated solution of H 2S
(0.1 M) buffered at pH = 7. What would be the solubility in a saturated H 2S solution buffered at pH =10.
For H 2S K1 = 10 –7 and K 2 = 10 –14 .
PROBLEM 411 Determine simultaneous solubility of Ag 2CO 3 and Ag 2CrO 4 . K sp (Ag 2CO 3 )
= 8.1 × 10 –12 and K sp ( Ag 2CrO 4 ) = 1.2 × 10 –12 .
PROBLEM 412 To a certain volume of a weak monobasic acid, when 20 mL of NaOH solution is added,
pH of the resulting solution was found to be 3.7 whereas when 30 mL of the same NaOH is added to the
same volume of the acid from same stock, pH was found to be 4.18. Determine K a of the acid.
PROBLEM 413 Determine mass of sodium dihydrogen phosphate and volume of 1.5 M HCl solution
required for preparation of a 100 mL buffer solution of pH 2.42 with concentration of H 2 PO 4– be 0.15 M.
K1 = 7.6 × 10 –3 , K 2 = 6.2 × 10 –8 and K 3 = 2 × 10 –13 .
PROBLEM 414 Carbondioxide gas from a steel cylinder is bubbled for some time through pure water
placed in a jar. When the supply of gas is terminated, pH of the solution was found to be 3.83. What is the
pressure of gas in the cylinder if Henery’s constant of CO 2 is 2.3 ×10 –2 . For CO 2 , K1 = 4.2 × 10 −7 and
K 2 = 4.8 × 10 −11 .
49
Problems
PROBLEM 415 Determine pH of a 0.1 M Fe(NO 3 ) 2 solution. Given
Fe(OH) 2
Fe(OH) + + OH –
K1 = 10 –4
Fe(OH) +
Fe 2+ + OH –
K 2 = 2.5 × 10 –6
PROBLEM 416 H 2S is bubbled into a 0.2 M NaCN solution which is 0.02 M in each Ag(CN) –2 and
–50
, K sp of CdS = 7.1 ×10 –28 . K d
Cd(CN) 2–
4 . Determine which sulphide precipitate first. K sp ( Ag 2S) =10
–18
.
Ag(CN) –2 = 10 –20 and Kd(Cd(CN) 2–
4 ) = 7.8 × 10
PROBLEM 417 Over what range of concentration of hydrogen ion concentration (pH) is it possible to
separate Cu 2+ from Ni 2+ when both metal ions are present at 0.01 M concentration and solution is made
saturated with 0.1 M H 2S? K sp of CuS and NiS are 6 × 10 –37 and 4 × 10 –20 respectively.
PROBLEM 418 A solution contains calcium nitrate and nickel nitrate, each at concentration of 0.1 M.
CO 2 is bubbled into make its concentration equal to 0.03 M. What pH range would make possible for the
selective precipitation of metal carbonates? K sp of CaCO 3 and NiCO 3 are 4.5 ×10 –9 and 1.3 ×10 –7
respectively.
PROBLEM 419 How many moles AgBr can dissolve in 1.0 L of 1.0 M NH 3 solution. K sp of AgBr is
5 × 10 –13 and K f for Ag(NH 3 ) +2 is 1.5 ×10 7 .
PROBLEM 420 A salt whose formula is of the form MX has a value of K sp equal to 3.2 ×10 –10 . Another
sparingly soluble salt MX 3 must have what value of K sp if the molar solubilities of the two salts are to be
identical?
PROBLEM 421 A salt having formula of the type M 2 X 3 has K sp = 2.2 × 10 –20 . Another salt, M 2 X , has
to have what K sp value if M 2 X has twice the molar solubility of M 2 X 3 ?
PROBLEM 422 Suppose Na 2SO 4 is gradually added to 100 mL of a solution that contains both Ca 2+
and Sr 2+ ions each at 0.15 M concentrations.
(a) What will the Sr 2+ ion concentration be when CaSO 4 just begin to precipitate.
(b) What percentage of Sr 2+ ion has precipitated when CaSO 4 just begin to precipitate?
K sp (CaSO 4 ) = 2.4 × 10 –5 , K sp (SrSO 4 ) = 3.2 × 10 –7 .
PROBLEM 423 How many grams of solid NaCN have to be added to 1.2 L of water to dissolve 0.11
mole of Fe(OH) 3 in the form of [ Fe(CN) 6 ]3– ?
[K sp of Fe(OH) 3 = 1.6 × 10 –39 , K f =10 31 ]
PROBLEM 424 Silver ion forms a complex with thiosulphate ion with their formula [Ag(S 2O 3 ) 2 ]3– .
K f for this complex is 2 × 1013 . How many grams of AgBr ( K sp = 5 × 10 –13 ) will dissolve in 125 mL of a
1.20 M Na 2S 2O 3 solution.
PROBLEM 425 A sample of hard water was found to contain 278 ppm of calcium ion. Into 1.00 L of this
water, 1.06 g of Na 2CO 3 was added. What is the new concentration of Ca 2+ ion in ppm. Density of all
solution be 1.0 g/cc and K sp of CaCO 3 is 4.5 ×10 –9 .
50
Problems in Chemistry
PROBLEM 426 What are the concentration of Pb 2+ , Br – and I – in an aqueous solution that is in contact
with both PbI 2 and PbBr 2 ? K sp (PbBr 2 ) = 2 × 10 –6 , K sp (PbI 2 ) = 8 × 10 –9 .
PROBLEM 427 An indicator has its standard ionization constant 9 × 10 –9 . The acid colour of the
indicator is yellow while its basic colour is red. The yellow colour is visible when the ratio yellow/red is
30 and red colour becomes visible when the ratio red/yellow is 2. What is the working pH range of this
indicator?
PROBLEM 428 (a) At what maximum concentration of hydroxide ion, 1.0 m mole Zn(OH) 2 will go
–16
into 1.0 L solution as Zn(OH) 2–
and K f = 2 × 10 20 .
4 ? K sp = 3 × 10
(b) At what maximum concentration of hydroxide ion will 1.0 m mole Zn(OH) 2 will go into 1.0 L
solution as Zn 2+ ?
PROBLEM 429 Determine
the
simultaneous
solubility
of
AgCN
( K sp = 2.2 × 10 –16 )
and
AgCl( K sp = 1.6 ×10 –10 ) in 1.0 M ammonia solution. K f [ Ag(NH 3 ) +2 ] = 1.5 × 10 7 .
PROBLEM 430 What must be the minimum concentration of acetic acid in a one litre buffer solution of
acetic acid acetate ( pH = 4.74) if the pH changes by not more than 0.1 unit on addition of 50 mL 1 M HCl.
K a = 1.8 × 10 –5 .
PROBLEM 431 How much 1.0 M HCl must be added to 1.00 L of 0.10 M sodium lactate (NaC 3 H 5O 3 )
to make a buffer of pH = 3.77. K a = 1.4 × 10 –4 .
PROBLEM 432 Potas alum is KAl(SO 4 ) 2 ⋅ 12 ⋅ H 2O. As a strong electrolyte, it is considered to be 100%
3+
ionized into K + , Al 3+ and SO 2–
4 . The solution is acidic because of the hydrolysis of Al , but not so
acidic as might be expected, because the sulphate ion can spong-up some of the H + by forming HSO –4 .
Given a solution is made by dissolving 11.4 g of alum in enough water to make 100 mL of the solution.
Calculate pH considering the following hydrolysis reactions:
Al 3+ + H 2O
HSO –4
Al(OH) 2+ + H + ;
H + + SO 2–
4 ;
K h = 1.4 × 10 –5
K a 2 = 1.26 × 10 –2
PROBLEM 433 Determine molar solubility of AgCN in water considering hydrolysis of CN – ion.
Given K sp (AgCN) = 2.3 ×10 –16 and K a ( HCN ) = 5 × 10 –10 .
PROBLEM 434 To what volume a 50 mL 6.0 M ammonia solution be diluted to double the degree of
protonation? K b = 2 × 10 –5 .
PROBLEM 435 Molar solubility of PbS is 6.7 ×10 –10 . Sulphide ion hydrolysis completely into HS – but
the further hydrolysis of HS – is insignificant. Determine K sp of PbS. K a 1 (H 2S) = 10 –7 and
K a 2 (H 2S) = 10 –14 .
PROBLEM 436 Lactic acid is formed in the muscles during intense activity (anaerobic metabolism). In
Problems
51
the blood, lactic acid is neutralized by reaction with hydrogen carbonate. Lactic acid written HL is
monoprotic, and the acid dissociation constant is K (HL) = 1.4 ×10 –4 .
The acid dissociation constants for carbonic acid are: K a 1 = 4.5 × 10 –7 and K a 2 = 4.7 × 10 –11 . All
carbon dioxide remains dissolved during the reactions.
(a) Calculate pH in a 3.00 ×10 –3 M solution of HL.
(b) Calculate the value of the equilibrium constant for the reaction between lactic acid and
hydrogen carbonate.
(c) 3.00 ×10 –3 mol lactic acid (HL) is added to 1.00 L of 0.024 M solution of NaHCO 3 (no change
in volume, HL completely neutralized).
(i) Calculate the value of pH in the solution of NaHCO 3 before HL is added.
(ii) Calculate the value of pH in the solution after the addition of HL.
(d) pH in the blood of a person changed from 7.40 to 7.00 due to lactic acid formed during physical
activity. Let an aqueous solution having pH = 7.40 and [ HCO –3 ] = 0.022 M represent blood in the
following calculation. How many moles of lactic acid have been added to 1.00 L of this solution when its
pH has become 7.00?
(e) In a saturated aqueous solution of CaCO 3 ( s), pH is measured to 9.95. Calculate the solubility of
calcium carbonate in water and show that the calculated value for the solubility product constant K sp is
5 × 10 –9 .
(f) Blood contains calcium. Determine the maximum concentration of “free” calcium ions in the
solution (pH = 7.40, [HCO –3 ] = 0.022 M) given in d).
PROBLEM 437 The pH recorded when 1.7 g of a monoprotic acid (K a = 2 × 10 –10 ) was added to 100
mL of NaOH of 0.1 M strength was 10.2. Determine molar mass of the acid.
PROBLEM 438 A saturated solution of Mg(OH) 2 has pH value 10.5. A mixture of 10.0 g of Mg(OH) 2
and 100 mL 0.1 M HCl is stirred magnetically for some time at 25°C. Calculate pH of the liquid phase
when equilibrium is reached.
PROBLEM 439 The concentration of CO 2 is expected to rise 440 ppm in year 2020. What would be the
pH of rainwater in the year 2020 if molar solubility of CO 2 is 0.0343 L–1 bar –1 . K a 1 of CO 2 = 4.2 × 10 –7 .
PROBLEM 440 Nitrogen in agricultural materials is often determined by the Kjeldahl method. The
method involves a treatment of the sample with hot concentrated sulphuric acid, to convert organically
bound nitrogen to ammonium ion. Concentrated sodium hydroxide is then added, and the ammonia
formed is distilled into hydrochloric acid of known volume and concentration. The excess hydrochloric
acid is then back-titrated with a standard solution of sodium hydroxide, to determine nitrogen in the
sample.
(a) 0.2515 g of a grain sample was treated with sulphuric acid, sodium hydroxide was then added
and the ammonia distilled into 50.00 mL of 0.1010 M hydrochloric acid. The excess acid was
back-titrated with 19.30 mL of 0.1050 M sodium hydroxide.
Calculate the concentration of nitrogen in the sample, in per cent by mass.
(b) Calculate the pH of the solution which is titrated in (a), when 0 mL, 9.65 mL, 19.30 mL and
28.95 mL of sodium hydroxide has been added. Disregard any volume change during the reaction of
ammonia gas with hydrochloric acid. K a for ammonium ion is 5.7 ×10 –10 .
(c) Draw the titration curve based on the calculations in (b).
52
Problems in Chemistry
(d) What is the pH transition range of the indicator which could be used for the back titration.
(e) The Kjeldahl method can also be used to determine the molecular weight of amino acids. In a
given experiment, the molecular weight of a naturally occuring amino acid was determined by digesting.
0.2345 g of the pure acid, and distilling the ammonia into 50.00 mL of 0.1010 M hydrochloric acid. A
titration volume of 17.50 mL was obtained for the back titration with 0.1050 M sodium hydroxide.
Calculate the molecular weight of the amino acid based on one and two nitrogen groups in the
molecule, respectively.
PROBLEM 441 In the precipitation titration of KCl against AgNO 3 , K 2CrO 4 is used as an indicator
since, AgCl is white coloured. End point is detected by appearance of deep yellow coloured precipitate
of Ag 2CrO 4 . Determine the minimum concentration of chromate ion required for detection of end point.
K sp of AgCl = 2.5 × 10 –10 and K sp of Ag 2CrO 4 = 1.8 × 10 –12 .
PROBLEM 442 50 mL of a10 –4 M aqueous solution of La(NO 3 ) 3 is mixed with 50 mL of an NH +4 /NH 3
buffer that is originally 0.2 M in NH +4 and 0.4 M in NH 3 . What percentage of La 3+ has been precipitated
at equilibrium. K sp of La(OH) 3 = 10 –19 and K b of NH 3 = 2 × 10 –5 .
PROBLEM 443 Aluminium phosphate is more soluble than its expected solubility due to hydrolysis of
both Al 3+ and PO 3–
4 as:
Al 3+ + H 2O
PO 3–
4 + H 2O
Al(OH) 2+ + H +
K a =10 –5
–
HPO 2–
4 + OH
K b = 5 × 10 –2
Determine the actual solubility of AlPO 4 . K sp of AlPO 4 = 10 –20 .
PROBLEM 444 0.2 moles of MgCl 2 is mixed with 0.2 moles of NaOH in a total volume of 1.0 litre.
(a) Calculate pH of this solution. K sp of Mg(OH) 2 is 1.6 ×10 –12 .
(b) Also determine pH when 0.04 moles of NaOH is added in the solution (a).
(c) Also determine pH of solution obtained after adding 0.04 mole of HCl to solution (a).
PROBLEM 445 How many grams of PbS will precipitate from a 1.0 L saturated PbSO 4 solution if the
solution is made saturated with H 2S (0.1 M) and pH adjusted to 0. K sp of PbSO 4 = 1.6 × 10 –8 and K sp of
PbS = 2.5 × 10 –27 . Also for H 2S; K a 1 = 10 −7 and K a 2 = 10 −14 .
PROBLEM 446 A typical wine sample contain 2.3% by mass of the base urea. K b for urea is1.5 ×10 –14 .
If the density of urea solution is 1.06 g/mL and it is buffered at pH 6.35, determine the equilibrium
concentration of protonated urea.
PROBLEM 447 At 25°C K sp of AgCl is 2 × 10 –10 . Using the following standard molar Gibb’s free
energies of formations: AgCl( s) = – 110 kJ/ mol, Ag 2SO 4 ( s) = – 618.5 kJ/ mol, Cl − ( aq ) = − 131
kJ/mol and SO 24− ( aq ) = − 742 kJ/mol determine solubility of Ag 2SO 4 in a 1.0 m M Na 2SO 4 solution at
25°C.
PROBLEM 448 pH of a 100 mL HOCN solution is 2.75. Addition of 100 mL 0.05 M HOCN to the
above solution lowers the pH to 2.5. Determine concentration of the original acid solution (assume
degree of ionization to be negligible in comparison to one).
Problems
53
PROBLEM 449 1.0 metric ton coal containing 2.5% sulphur is burned and SO 2 produced in the
combustion was dissolved in rainwater of volume equivalent to 2.0 cm rain fall over 2.6 km 2 area.
Determine pH of rainwater if K a of H 2SO 3 = 1.5 × 10 –2 .
PROBLEM 450 1.0 g impure sample of Mg is burnt completely in N 2 atmosphere and nitride is
dissolved in 1.0 litre of water. pH of the resulting solution was found to be 10.75. Determine percentage
purity of the sample ignoring solubility of Mg(OH) 2 . K b of NH 3 = 2 × 10 –5 .
PROBLEM 451 Bromine in excess is added drop wise to a 0.01 M solution of SO 2 . All SO 2 is oxidized
to sulphuric acid. Excess of bromine is removed by flushing with gaseous N 2 . Determine pH of the
resulting solution assuming K a 1 of H 2SO 4 to be very large and K a 2 = 10 –2 .
PROBLEM 452 In laboratory class, a student is given three flasks that are labeled Q, R and S . Each
flask contains one of the following solutions : 1.0 M Pb(NO 3 ) 2 , 1.0 M NaCl and 1.0 M K 2CO 3 . The
student is also given two flasks that are labeled X and Y . One of these flasks contains 1.0 M AgNO 3 and
other contains 1.0 M BaCl 2 .
(a) When the student combined a sample of solution Q with a sample of solution X , a precipitate
formed. A precipitate also formed when samples of solution Q and Y were combined. Identify Q.
(b) When solution Q is mixed with solution R, a precipitate forms. However, no precipitate forms
when solution Q is mixed with solution S. Identify R and S .
(c) After knowing the identity of Q, R and S , employ them to identify X and Y .
PROBLEM 453 You are given two burettes. One is filled with 0.15 M acetic acid and other is filled with
0.10 M sodium acetate. How much of each would you mix together to produce a total of 20 mL of
solution with a pH of 4.7. You may not add any water to solution. K a of acetic acid is 2 × 10 −5 .
PROBLEM 454 You are given two burettes. One is filled with 0.15 M acetic acid and other with 0.1 M
NaOH solution. How much of each would you mix together to produce a total volume of 50 mL with a
pH = 4.7 K a of acetic acid = 2 × 10 −5 .
PROBLEM 455 1.6 g solid Ba(NO 3 ) 2 and 0.3 g H 3 PO 4 are taken in a flask and volume made up to 2.0
litre by adding sufficient water. Determine pH of the final solution assuming Ba 3 (PO 4 ) 2 to be insoluble
in water.
PROBLEM 456 A solution contains the cations Mn 2+ , Co 2+ and Ag + at an original concentration
of 0.01 M each. Determine the pH range of this precipitation when the solution is saturated in
H 2S(0.1 M). K a 1 and K a 2 of H 2S are 10 −7 and 10 −14 respectively. MnS K sp = 2.5 × 10 –10 ,
CoS : K sp = 4 × 10 −21 , Ag 2S : K sp = 6.3 × 10 –50 .
PROBLEM 457 A saturated hydrogen electrode (SHE) with [H + ] = 1.0 M is connected to a silver
electrode placed in a saturated solution of silver oxalate Ag 2C 2O 4 via a salt bridge and e.m.f was found
°+
to be 0.589 V. Calculate solubility product constant K sp for given oxalate. E Ag
= 0.80 V.
/Ag
PROBLEM 458 If 150 mg of zinc hydroxide crystals are mixed with 250 mL of a 0.1 M NaOH solution,
what mass of Zn(OH) 2 crystals will remain undissolved? K sp = 2.1 × 10 −16 and K f for
15
Zn(OH) 2–
4 = 2.8 ×10 .
54
Problems in Chemistry
PROBLEM 459
K sp for AgCl = 2 × 10 –10 and K f for Ag(NH 3 ) +2 = 2 × 10 7 . Calculate molar solubility of
AgCl in an aqueous solution that is 0.1 M in KCl and 0.1 M in NH 3 .
PROBLEM 460
values
K sp
of
silver
halides
are :
AgCl = 1.8 ×10 –10 ,
AgBr = 5 × 10 –13 ,
AgI = 8.4 ×10 –17 . State the order in which halide ions should be added to a concentration of 0.1 M so that
each precipitate will form from the [Ag + ] in equilibrium with the previous precipitate. Support your
answer with appropriate calculations.
PROBLEM 461
Phosphoric acid ionizes according to the equations :
H 3 PO 4 ( aq )
H + ( aq ) + H 2 PO 4– ( aq ) K1 = 7 × 10 –3
H 2 PO 4– ( aq )
–8
H + ( aq ) + HPO 2–
4 ( aq ) K 2 = 6 × 10
HPO 2–
4 ( aq )
H + ( aq ) + PO 3–
4 ( aq )
K 3 = 4.5 × 10 –13
(a) If you are asked to prepare a buffer with a pH = 7.00, identify the species that should be used in
the solution and calculate their ratio.
(b) Assume 50 mL of the buffer prepared in part ‘a’ is available in which more abundant species has
a concentration of 0.10 M. If to this solution 20 mL 0.1 M NaOH is added further, what will be the new
pH?
PROBLEM 462 A 0.50 gram mixture containing NaHCO 3 , Na 2CO 3 and KCl was dissolved in 50.00
mL water and titrated with 0.115 M HCl resulting in the following titration curve.
14
12
10
(11.00 mL)
8
6
4
(35 mL)
2
0
10
20
30
40
Determine weight percentage of NaHCO 3 , Na 2CO 3 and KCl in the mixture.
PROBLEM 463
K sp of Ca(H 2 PO 4 ) 2 = 10 –3 . What minimum pH must be maintained in a 0.25 M
H 3 PO 4 solution to prevent precipitation of Ca(H 2 PO 4 ) 2 if the solution is 0.15 M in calcium ion? K a 1 of
H 3 PO 4 is 7 × 10 –3 .
PROBLEM 464 Fe 3+ ion forms a blood red complexes with thiocyanate ion, SCN – . Following
equilibria exist on adding SCN – to an aqueous solution of Fe 3+ ion :
(i) Fe(OH) 3 ( s)
Fe 3+ ( aq ) + 3OH – K sp = 4 × 10 –38
(ii) Fe 3+ + 2SCN –
Fe(SCN) +2
(iii) SCN – + H 2O
HSCN + OH –
K f = 2.3 × 10 3
K a = 71
55
Problems
Discuss the effect of followings on concentration of Fe 3+ ion in solution :
(a) Addition of HNO 3 , (b) Addition of NaOH,
(c) Addition of NaSCN, (d) Addition of Fe(OH) 3 .
PROBLEM 465 What will be the effect of adding a strong acid to an aqueous solution of Cu(CN) 2–
4 ,
–10
HCN is a weak acid with K a = 6 × 10 ?
PROBLEM 466 What concentration of NH 3 must be present in a 0.1 M AgNO 3 solution to prevent
AgCl from precipitating when 4.0 g of NaCl are added to a 250 mL of this solution. K sp = 2 × 10 –10 ,
K f [Ag(NH 3 ) +2 ] = 10 7 .
PROBLEM 467 It is possible to keep Co(OH) 3 from precipitating from a 0.01 M CoCl 3 solution by
buffering the solution at pH 9.1 with a buffer that contain NH 3 and NH +4 ion. How much of 6 M NH 3 and
6 M HCl must be added per litre of this solution to prevent Co(OH) 3 from precipitating?
35
[NH 3 : K b = 2 × 10 –5 , Co(OH) 3 ( s) : K sp =10 –43 , Co(NH 3 ) 3+
6 : K f = 2 × 10 ]
PROBLEM 468 Will Co(OH) 3 precipitate from a solution that is initially 0.1 M in Co 3+ and 1M in
SCN – if this solution is buffered at pH = 7. K sp of Co(OH) 3 = 1.6 × 10 –44 and K f [Co(SCN) –4 ] = 10 3 ?
–
PROBLEM 469 Calculate the concentration of CO 2–
3 in a 0.1 M HCO 3 solution buffered with equal
2–
+
number of moles of NH 3 and NH 4 . Is this CO 3 concentration large enough to precipitate BaCO 3 when
the solution is mixed with equal volume of 0.1 M Ba 2+ ion. [K sp = 5 × 10 –9 , H 2CO 3 : K a 1 = 4.5 × 10 –7 ,
K a 2 = 4.7 × 10 –11 , NH 3 : K b = 2 × 10 –5 ]
PROBLEM 470 Enough carbonate ion CO 2–
3 can be leached out of clay to buffer ground water at
pH = 8. Assume that total concentration of HCO –3 and CO 2–
3 ions in this solution is 0.10 M. Calculate the
60
maximum concentration of Co-isotope that could leach into ground water if clay were used as a barrier
to store this radioactive isotope. K sp of CoCO 3 and Co(OH) 3 are1.4 ×10 –13 and1.6 ×10 –44 respectively.
For H 2CO 3 K a 1 = 4.5 × 10 –7 and K a 2 = 4.7 × 10 –11 .
PROBLEM 471 Determine pH and degree of hydrolysis of a 10 –3 M C 6 H 5 NH 3Cl solution.
K b = 5 × 10 –10 .
PROBLEM 472 A solution is made by dissolving 0.001 mol Ca(OI) 2 in 100 mL of water and to this
solution 100 mL 0.01 M acetic acid solution was added. Determine pH of this solution. K a (acetic acid)
= 2 × 10 –5 , K a ( HOI) = 2 × 10 –11 .
PROBLEM 473
Determine pH of a 0.01 M ClC 6 H 4 NH 3Cl, K b = 4 × 10 –12 .
PROBLEM 474 The indicator dinitrophenol has K a = 1.1 × 10 –4 . In a10 –4 M solution, it is colourless in
acid medium and yellow in basic medium. Calculate pH range over which it goes from 25% (ionized) to
75% (ionized).
PROBLEM 475
Determine concentration of ammonium ion that is required to prevent the precipitation
56
Problems in Chemistry
of Mg(OH) 2 in a solution with [Mg 2+ ] = 0.1 M and [NH 3 ] = 0.1 M. K sp [ Mg(OH) 2 ] = 1.5 ×10 –11 and
K b ( NH 3 ) = 2 × 10 –5 .
PROBLEM 476 Determine molar solubility of BaSO 4 in a 0.25 M solution of NaHSO 4 . K a for
HSO –4 = 1.2 × 10 –2 . Also determine the factor by which solubility is decreased as compared to solubility
in pure water K sp =10 –10 .
1
M HCl will have to be added to
3
500 c.c. of 0.1 M Na 2CO 3 solution in order to adjust the pH to 10? K a 1 of CO 2 = 4 × 10 –7 and
PROBLEM 477
If the final volume is to be 1 litre, how many mL of
K a 2 = 5 × 10 –11 .
PROBLEM 478 Formation constant of [Ag(CN) 2 ]– is 2.5 ×1018 . Determine concentration of Ag + (aq )
in a solution which was originally 0.1 M in KCN and 0.03 M in AgNO 3 .
PROBLEM 479 The solubility product of CaF2 at 18°C is 3.4 ×10 –11 while that of CaCO 3 is 95
. × 10 –9 .
(a) What will be the nature of 1st precipitate when a solution of CaCl 2 is added to a solution which is
0.05 M in NaF and 0.02 M in Na 2CO 3 . (b) In a 0.02 M solution of Na 2CO 3 , what is the minimum
concentration of NaF at which both CaF2 and CaCO 3 will precipitate simultaneously?
PROBLEM 480 The solubility product of PbI 2 is 7.47 ×10 –9 at 17°C and1.4 ×10 –8 at 27°C. Calculate
(a) the molar heat of solution of PbI 2 (b) the solubility of PbI 2 at 77°C.
PROBLEM 481 The equivalence conductivity of a solution containing 2.54 g CuSO 4 /L is 91
cm 2 Ω –1 equ –1 . Calculate the specific conductivity of a solution. Also determine the resistance of a cm 3
of this solution when placed between two electrodes 1.00 cm apart, each having an area of 1.00 cm 2 ?
PROBLEM 482 The equivalent conductance of a 0.01 M ammonia solution is 10 cm 2 Ω –1equ –1 . The
equivalent conductance of ammonia at infinite dilution is 238 cm 2 Ω –1equ –1 . Determine ionization
constant K b of ammonia.
PROBLEM 483 Equivalent conductance at infinite dilution for acetic acid is 390 cm 2 Ω –1 at 25°C.
Determine specific conductance and equivalent conductance of a 0.01 M acetic acid solution if K a of
acetic acid = 2 × 10 –5 .
PROBLEM 484 A conductivity cell is filled with 0.1 M KCl solution at 27°C. The measured resistance
R = 25 Ω. Calculate cell constant if specific conductance of 0.1 M KCl solution is 0.0116 Ω –1cm –1 . If
the same cell is filled with acetic acid, the cell resistance is 1982 Ω. Determine molar conductance of 0.1
M acetic acid solution.
PROBLEM 485 The conductivity of a 0.1 M NaOH solution is 0.022 Ω –1cm –1 . When an equal volume
of 0.1 M HCl solution is added, the conductivity decreases to 0.0056 Ω –1cm –1 . A further addition of HCl
solution, the volume of which is equal to that of first portion added, conductivity increases to
0.017 Ω –1cm –1 . Calculate molar conductance of NaOH, NaCl and HCl solutions in the given conditions.
PROBLEM 486 A sample of water from a large swimming pool has a resistance of 9200 Ω at 25°C
when placed in a certain conductance cell. When filled with 0.02 M KCl solution, the cell has a
Problems
57
resistance of 85 Ω at 25°C. Five hundred grams of NaCl was then dissolved in swimming pool, stirred
throughly and a sample of this solution gave a resistance of 7600 Ω. Calculate the volume of water in the
pool. Given molar conductivity of NaCl solution at that concentration is126.5 Ω –1cm 2 mol –1 and molar
conductivity of KCl solution at 0.02 M is 138.3 Ω –1cm 2 mol –1 .
PROBLEM 487 The molar conductivity of a 0.05 M aqueous solution of MgCl 2 is 195 Ω –1cm 2 mol –1 at
27°C. A cell whose electrodes are of1.5 cm 2 surface area and placed at a distance of 0.5 cm, is filled with
0.05 M MgCl 2 solution. How much current will flow if the potential difference across the electrodes
is 5 V?
PROBLEM 488 A conductivity cell when filled with 0.02 M KCl gives resistance of 85 Ω. The same
conductivity cell, when filled with a saturated solution of PbCl 2 , gives resistance of 70 Ω. If the
equivalent conductance of 0.02 N KCl solution and saturated solution of PbCl 2 , under identical
conditions are 138.3 and 106 cm 2 Ω –1eq –1 respectively, determine solubility product of PbCl 2 .
ELECTROCHEMISTRY
PROBLEM 489 (a) When a current of 150 mA is used for 8.0 hr, what volume of fluorine gas at N.T.P.
can be produced from a molten mixture of potassium and hydrogen fluoride?
(b) If the same current is passed for same interval of time, how many litre of oxygen gas could be
produced from electrolysis of water?
PROBLEM 490 An aqueous solution of Na 2SO 4 was electrolysed for 30 min; 25 mL of O 2 was
produced at the anode over water at 22°C at a total pressure of 722 torr. Determine the strength of current
that was used to produce the oxygen gas. Vapour pressure of water at 22°C is 19.83 torr.
PROBLEM 491 A piece of copper metal is to be electroplated on all side with silver to a thickness of 1
micrometre. If the metal strip measures 50 mm × 10 mm × 1 mm, how long must the solution, which
consists the Ag(CN) –2 ion, be electrolysed, using a current of 100 mA? The density of silver metal is 10.5
g/cc.
PROBLEM 492 19 g of fused SnCl 2 was electrolysed using inert electrode when 0.119 g of Sn was
deposited at cathode. If nothing was given out during electrolysis, calculate the ratio of weight of SnCl 2
and SnCl 4 in fused state after electrolysis.
PROBLEM 493 Electrolysis of molten KBr generates bromine gas, which can be used in industrial
bromination process. How long will it take to convert a 500 kg batch of phenol to monobromo phenol
using a current of 20 kA?
PROBLEM 494 The cell Pt | H 2 (1 bar), H + || KCl (1.0 M saturated) | Hg 2Cl 2 | Hg was used to measure
the pH of a solution of 0.01 M acetic acid in 0.0358 M sodium acetate. Calculate the cell potential
expected at 25° C [K a = 1.81 × 10 –5 for acetic acid; E ° (Hg 2Cl 2 / Hg, Cl – ) = 0.28 V]
58
Problems in Chemistry
PROBLEM 495 The voltage required to electrolyse certain solutions changes as the electrolysis
proceeds because the concentration in the solution is changing. In an experiment, 500 mL of a 0.1 M
solution of copper (II) bromide was electrolysed until 2.827 g of Cu was deposited. Calculated the
theoretical minimum voltage required to sustain the electrolysis reaction at the beginning and at the end
of experiment. E °Cu 2+ /Cu = 0.34 V and E ° Br / Br – = 1.07 V.
2
PROBLEM 496 Calculate the concentration of I −3 in a standard solution of iodine in 0.5 M KI, making
use of the following standard electrode potentials: E ° (I 2 /I – = 0.5355 V) and E ° (I –3 /I – = 0.5365 V)..
The molarity of I 2 in the standard solution can be assumed to be 0.5 M.
PROBLEM 497 Write both electrode reaction and overall cell reaction for the cell
Tl | TlCl( s) || CdCl 2 (0.01 M) | Cd
and calculate EMF of the cell if E ° (Tl + /Tl = – 0.34 V), E ° (Cd 2+ /Cd = – 0.40 V) and the solubility
product of TlCl is 1.6 × 10 −3 .
PROBLEM 498
The meal M forms a soluble nitrate and a very slightly soluble chloride. The cell
M | M + (0.1 M ), HNO 3 (0.2 M ) || H 2 (1 bar), Pt
Has a measured EMF = – 0.4 V. When sufficient solid KCl is added to make the solution of the cell
0.2 M in K + , the EMF changes to – 0.05 V. Calculate the solubility product of MCl.
PROBLEM 499 The voltage of the following cell is 0.987 V
Pt(H 2 , 1.0 atm ) | HOCN(1.4 × 10 –3 M ) || Ag + (0.8 M ) | Ag ( s)
Calculate K a of HOCN, E ° Ag + / Ag = 0.8 V
PROBLEM 500 For the galvanic cell:
Ag | AgCl( s), KCl(0.2 M ) || KBr(0.001 M ), AgBr ( s) | Ag
Calculate the e.m.f. generated and assign correct polarity to each electrode for a spontaneous
process after taking account of the cell reaction at 25°C. Given K sp of AgCl = 2.8 ×10 –10 and of
AgBr = 3.3 × 10 –13 .
PROBLEM 501 Zinc granules are added in excess to 750 mL of 1.5 M Ni(NO 3 ) 2 solution at 30°C untill
the equilibrium is reached. If emf of the cell
Zn ( s) | Zn 2+ (1 M ) || H + (1 M ) | H 2 ( g ) 1.0 atm, | Pt is 0.76 V and that of cell
Ni | Ni 2+ (1 M ) || H + ( pH = 0) | H 2 ( g, 1 atm ) | Pt is 0.24 V, then calculate [ Ni 2+ ] at equilibrium. Also
find the amount of zinc consumed.
PROBLEM 502 An electrochemical cell consist of a silver electrode in contact with 346 mL of 0.10 M
AgNO 3 solution and a magnesium electrode in contact with 288 mL 0.1 M Mg(NO 3 ) 2
solution. (a) Calculate E for the cell at 25°C. (b) A current is drawn from the cell until 1.2 g of silver is
deposited at the silver electrode. Calculate E for the cell at this stage of operation. E °(Ag + /Ag) = 0.8 V
and E °( Mg 2+ /Mg) = – 2.37 V.
59
Problems
PROBLEM 503 An electrochemical cell is constructed by immersing a piece of copper wire in 25 mL of
a 0.2 M CuSO 4 solution and a zinc strip in 25.0 mL of a 0.2 M ZnSO 4 solution. (a) Calculate the e.m.f.
of the cell at 25°C and predict what would happen if a small amount of concentrated NH 3 solution were
added to (i) the CuSO 4 solution and (ii) to ZnSO 4 solution? (b) In a separate experiment 25 mL of
3.0 M NH 3 are added to the CuSO 4 solution. If the e.m.f. of the cell is 0.68 V, calculate formation
constant of Cu(NH 3 ) 24+ . E °Cu 2+ /Cu = 0.34 V, E ° Zn 2+ / Zn = − 0.76 V
PROBLEM 504 The voltage of a certain cell at 25°C and 20°C are 0.3525 V and 0.3533 V respectively.
If the number of electrons involved in the overall reactions are two, calculate ∆G°, ∆S ° and ∆H ° at 25°C.
PROBLEM 505 For the cell:
Pt | H 2 ( g ) | sol X || KCl ( sat ) | Hg 2Cl 2 ( s) | Hg | Pt
the observed e.m.f. at 25°C was 612 mV. When solution X was replaced by a standard phosphate buffer
whose assigned pH is 6.86, the e.m.f. was 741 mV. Find pH of the solution X.
PROBLEM 506 Use the following data to calculate solubility product of PbF2 . A galvanic cell consists
of a hydrogen gas electrode ( p(H 2 ) = 1 bar , [ H + ] = 1 M ) and a Pb electrode in a 1.0 M KF solution in
equilibrium with PbF2 ( s). The cell potential is 0.348 V and E ° (Pb2+ / Pb) = − 0.13 V.
PROBLEM 507 A concentration cell has Zn electrodes. The electrolyte in each of the half-cells is a
solution of ZnCl 2 dissolved in water. One has a freezing point of – 2.0°C and the other has a freezing
point of – 0.9°C. What is the potential of the cell at 25°C?
PROBLEM 508 A lead storage battery is allowed to discharge until 23.92 g of PbO 2 have been reduced.
Determine the mass of PbSO 4 formed and time required to recharge the battery to its original state using
a current of strength 3.0 A.
PROBLEM 509 The chlorate ion ClO –3 can disproportionate in basic solution according to the reaction
2ClO –3 ( aq )
ClO –2 ( aq ) + ClO –4 ( aq )
What is the equilibrium concentration of the ions resulting from a solution initially at 0.1 M in
chlorate ion? E ° (ClO –4 /ClO –3 ) = 0.36 V and E ° (ClO –3 /ClO –2 ) = 0.33 V.
PROBLEM 510 A platinum electrode is in contact with a solution containing Fe 2+ and Fe 3+ ions. The
total concentration of iron ions is 0.033 M. What is the ratio of the concentrations if the potential
difference between solution and metal surface is 35 mV and E ° (Fe 3+ / Fe 2+ ) = 0.77 V ?
PROBLEM 511 Calculate Gibb’s free energy change and equilibrium constant for the reaction
Ni 2+ + 2H 2O → Ni(OH) 2 + 2H +
Given that
and
NiO 2 + 4H + + 2e → Ni 2+ + 2H 2O
NiO 2 + 2H 2O + 2e → Ni(OH) 2 + 2OH
E ° =1.678 V
–
E ° = – 0.49 V
PROBLEM 512 The standard electromotive force of the cell:
Fe | Fe 2+ ( aq ) || Cd 2+ | Cd is 0.0372 V
The temperature coefficient of e.m.f. is – 0.125 V K –1 . Calculate the quantities ∆G°, ∆H ° and ∆S ° at
25°C.
60
Problems in Chemistry
PROBLEM 513 The concentration of Ca 2+ in sea water is determined using a Ca 2+ selective electrode.
A 10 mL sample of sea water is diluted to 100 mL and its 50 mL aliquot is placed in a beaker with Ca 2+
ion selective electrode and a reference electrode. The e.m.f. was found to be – 0.053 V. A 1.0 mL aliquot
of 0.05 M Ca 2+ ion is added and emf was found to be – 0.0422 V. What is the concentration of Ca 2+ in
sea water? Ignore volume change due to addition of second aliquot.
PROBLEM 514 Use the van der Waals’ equation of state to calculate the pressure at 35°C that would
result if the chlorine produced from electrolysis of 10 kg molten NaCl is compressed into a tank with a
volume of 75 L. Assume 96% efficiency of the electrolysis. a = 6.5 atm L2 mol –2 , b = 0.0562 L mol −1 .
PROBLEM 515 The pH of the solution in the cell
Pt | H 2 ( g ) | HCl(aq.) || AgCl(s) | KCl(1M) | Ag is 0.65 and E ° (Ag + | Ag) = 0.8 V. Calculate e.m.f.
K sp AgCl = 1.6 × 10 −10 .
PROBLEM 516 The e.m.f. of the following cell:
Ag(s) | AgCl(saturated), KCl( M = 0.05) | KNO 3 | AgNO 3 ( M = 0.1) | Ag ( s) is 0.4312 V.
Calculate solubility product of AgCl.
PROBLEM 517 The e.m.f. of cell: H 2 (g ) | Buffer || Normal calomal electrode is 0.6885 V at 40°C when
the barometric pressure is 725 mm of Hg. What is the pH of the solution. E ° calomal = 0.28.
PROBLEM 518 The cell :
Ag ( s) | AgCl( s), HCl(0.1 N ) | Glass | Buffer || Saturated calomal electrode gave an e.m.f. of 0.112 V
when the pH of the buffer used was 4.00. When a buffer of unknown pH was used, the potential was
0.3865 V. What is the pH of unknown buffer?
PROBLEM 519 One hundred mL of 0.01 N KCl are titrated with 0.1 N AgNO 3 . Calculate potential of a
silver electrode in the solution at equivalence point. The solubility product of AgCl is 1.56 ×10 –10 ?
[E ° Ag + / Ag = 0.8 V]
PROBLEM 520 A 2.0 V battery is used in an electrolysis in which an e.m.f. of 1.45 V is developed. If the
resistance of the entire circuit is 10 ohms, determine magnitude of current and quantity of heat produced
by the current flow per faraday of electricity.
PROBLEM 521 A solution is made originally 0.1 M in Ag + ions and 0.25 M in KCN. If the dissociation
constant of the complex Ag(CN) –2 is 3 × 10 –19 , what will be the concentration of Ag + ions in this solution
and what will be the deposition potential of Ag? [ E ° (Ag + / Ag) = 0.8 V]
PROBLEM 522 A solution is 0.1 M in Au + ions and 0.1 M in Ag + ions. The standard reduction
potential of Au + is 1.68 V, while the dissociation constant of Au(CN) –2 is 5 × 10 –39 . What concentration
of NaCN will have to be maintained in the given solution in order to deposit Au and Ag simultaneously?
[E ° Ag + / Ag = 0.8 V].
PROBLEM 523 A Galvanic cell consist of Zn anode dipped into a 1.0 L 0.2 M Zn(NO 3 ) 2 solution and a
silver cathode dipped into a 1.0 L 0.1 M AgNO 3 solution at 25°C and has e.m.f. =1.52 V. Now, KCl(s) is
added to the cathode chamber resulting in precipitation of AgCl and a change in e.m.f. After addition,
e.m.f. was found to be 1.04 V and [ K + ] = 0.3 M. Determine K sp of AgCl.
61
Problems
PROBLEM 524 (a) Calculate E° for:
Ag 2S + 2e
2Ag( s) + S 2– if K sp of Ag 2S = 6 × 10 –50 and E ° Ag + / Ag = 0.8 V
(b) Consider an electrochemical cell having an indicator electrode Ag/Ag 2S coupled with a SCE
(E° = 0.244 V) calculate [S 2– ] if the overall voltage of this cell is 0.766 V.
–
+
PROBLEM 525 An I 2 (s)/ I (0.1 M ) half cell is connected to a H /H 2 (1.0 atm) half cell and e.m.f. found
to be 0.755 V. If E °I 2 /I – is 0.535 V, determine pH of H + / H 2 half cell.
PROBLEM 526 To an aqueous solution of a weak acid (HA), 20 mL of a NaOH is added and connected
to a reference electrode Ag/AgCl/Cl – (0.1 M ) electrode. E.m.f. of the cell was found to be 0.47 V. To the
resulting solution, 30 mL of NaOH of the same strength was further added and new e.m.f. was found to
be 0.5 V. Determine K a of the weak acid. E° (AgCl/Ag, Cl – ) = 0.23 V.
PROBLEM 527 A direct current of 1.25 A was passed through 200 mL of 0.25 M Fe 2 (SO 4 ) 3 solution
for a period of 1.1 hour. The resulting solution in cathode chamber was analyzed by titrating against
acidic KMnO 4 solution. 25 mL permanganate solution was required to reach the end point. Determine
molarity of KMnO 4 solution.
PROBLEM 528 A 80 mL sample solution of KI was electrolyzed for 3.0 minute, using a constant
current. The I 2 produced required 0.25 M, 37.2 mL sodium thiosulphate solution and unreacted KI
required 36.3 mL 0.02 M acidic solution of KMnO 4 . Determine current strength and original molarity of
KI solution.
PROBLEM 529 K sp for AgBr = 8 × 10 –13 . What is the e.m.f. of the cell:
Ag, AgNO 3 (1.0 M) || KBr (1.0 M), AgBr, Ag
PROBLEM 530 E.m.f. of the cell:
Hg, Hg 2Cl 2 , KCl(1.0 M) || R 2 NH 2+ Cl – (0.1 M) | H 2 , Pt is – 0.52 V
Calculate ionization constant of the base R 2 NH ⋅ E ° ( Hg 2Cl 2 | Cl – ) = 0.28 V.
PROBLEM 531
An electrochemical cell is constructed with an open switch as shown below :
V
Sn
Salt Bridge
0.5 M Sn2+
x
0.1 M Xn+
When the switch is closed, mass of tin-electrode increases. If E ° (Sn 2+ /Sn) = −0.14 V and for
E ° (X n + /X) = − 0.78 V and initial emf of the cell is 0.65 V, determine n and indicate the direction of
electron flow in the external circuit.
62
Problems in Chemistry
PROBLEM 532
Consider the following electrochemical cell :
V
Ag
Salt Bridge
Zn
E°Ag+/Ag = 0.8 V
E°Zn2+/Zn = –0.76 V
Zn(NO3)2
0.1 M AgNO3
(a)
(b)
(c)
(d)
Write a balanced net ionic equation for the spontaneous reaction that take place in the cell.
Calculate the standard cell potential E ° for the cell reaction.
If the cell emf is 1.6 V, what is the concentration of Zn 2+ ?
How will the cell potential be affected if KI is added to Ag + half-cell?
PROBLEM 533 You have available a 0.001 M Al(NO 3 ) 3 , 0.1 M Cu(NO 3 ) 2 and 1.0 M Fe(NO 3 ) 2
solutions, Al, Cu and Fe metal strips. Construct a galvanic cell that would have greatest cell potential at
25°C. E ° 3+
= −176
. V, E ° 2+
= − 0.44 V, and E ° 2+ is +0.34 V.
Al
/Al
Fe
/Fe
Cu
/Cu
PROBLEM 534 Magnesium metal is produced commercially by isolation of MgCl 2 from seawater
followed by electrolysis of molten salt.
MgCl 2 → Mg + Cl 2
(a) What mass of Mg can be produced if a current of 430 A is passed for 1.0 hour?
(b) If a current of 500 A is used, how many hours will be required to convert all the 1000 kg MgCl 2
into Mg metal?
PROBLEM 535 Commercial production of Al-metal involves electrolysis of Al 2O 3 obtained from
bauxite ore. The balanced net equation is :
2Al 2O 3 + 3C → 4Al + 3CO 2
(a) How many moles of Al are produced if 12 moles of electrons are passed through this cell?
(b) How many kg of Al metal are produced if a current of 1250 A is passed for 1.25 hours?
PROBLEM 536
Given :
Consider the following redox reaction :
–
+
2IrCl 3–
6 + 3HCOOH → 2Ir + 3CO 2 +12Cl + 6H
CO 2 + 2H 3O + + 2e → HCOOH + 2H 2O
IrCl 3–
6
E ° = – 0.20 V
–
+ 3e → Ir + 6Cl
E ° = 0.77 V
(a) Determine standard state emf of cell.
(b) Is this reaction thermodynamically spontaneous as written? Briefly explain.
PROBLEM 537 An electrical source which might be used on a spacecraft should be light weight. Yet
deliver high voltage and maximum useful energy. A galvanic cell using aluminium, one of the lightest
metal, might be considered.
Al | Al 3+ || Ce 4+ | Ce 3+ | Pt
63
Problems
The cell is constructed with 2.0 L 4 M Al 2 (SO 4 ) 3 and 2.0 L 6 M Ce(SO 4 ) 2 . If
E ° (Al 3+ /Al) = − 1.76 V and E ° (Ce 4+ /Ce 3+ ) = + 1.443 V.
(a) Determine standard state emf and emf of the cell in the given conditions, if the cathode
compartment is initially 1.0 M in Ce 3+ ion.
(b) What is the maximum amount of useful electrical energy which could be derived from this cell?
(c) The instruments on spacecraft require a reasonably steady voltage to operate correctly. What
will be the cell voltage when the concentration of Ce 4+ ion has dropped to half of its original
concentration? Assume this galvanic cell operate under a reversible condition at 25°C.
PROBLEM 538 A Daniell cell was constructed from 2.00 L 4 M aqueous CuSO 4 solution and 2 Lit 4
M aqueous ZnSO 4 solution.
Also, given
e
Cu + ( aq ) → Cu( s)
E ° = + 0.522 V
Cu
2+
+
( aq ) + e → Cu ( aq ) E ° = 0.158 V
Zn
2+
( aq ) + 2e → Zn ( s)
E ° = – 0.76 V
(a) Calculate standard cell potential for this Daniell cell.
(b) Calculate the cell potential when the concentration of Cu 2+ ( aq ) dropped to 0.4 M. Assume
current was drawn slowly, so that this galvanic cell operated under reversible conditions at 25°C.
PROBLEM 539
The half-reactions for a typical rechargable nickel-cadmium (“nicad”) battery are :
Cd(OH) 2 ( s) + 2e → Cd( s) + 2OH – ( aq )
E ° = – 0.81 V
Ni(OH) 3 ( s) + e → Ni(OH) 2 ( s) + OH – ( aq ) E ° = + 0.49 V
(a) Identify cathode and write a balanced equation that shows the reaction that occurs
spontaneously as the cell discharges.
(b) If E° for electrolysis of molten MgCl 2 into Mg and Cl 2 ( g ) is –3.74 V, can this nicad battery be
used to drive decomposition of molten MgCl 2 into Mg metal and Cl 2 gas?
(c) If the nicad battery can deliver 0.1 A for 12 hour, how many grams of Mg can be produced from
MgCl 2 by electrolysis?
(d) The voltage delivered by a nicad battery doesnot change significantly as the battery discharges
and the reaction reaches equilibrium. Using Nernst’s equation, briefly explain, why this is true?
PROBLEM 540 Lead storage battery is used in automobiles. In order to reduce the amount of
hazardous lead and compounds that end up in landfills when these batteries are dumped, it has been
suggested that the lead be replaced with carbon monoxide. The unbalanced cell reaction when the battery
discharge is :
PbO 2 ( s) +CO( g ) + H 2SO 4 ( aq ) → PbSO 4 ( s) + CO 2 ( g ) + H 2O
E ° (PbO 2 /PbSO 4 ) = +1.685 V and
E ° (CO 2 ( g ) /CO( g )) = − 0.12 V
(a) Calculate E° for this lead-storage battery.
(b) If density of sulphuric acid in battery decreases from 1.3 g/ml (40% H 2SO 4 by wt.) to 1.2 g/ml
(20% H 2SO 4 by wt.) during its used, determine the ampere-hour for which battery has been used.
Volume of battery is 3.0 Lit.
(c) Voltage of such battery is maintained constant during its used. Very briefly explain, why this is
so?
64
Problems in Chemistry
PROBLEM 541 Diabetic must monitor their levels of blood glucose to determine the proper insulin
dose. An electronic device that does this uses the reaction below:
OH
OH
HO
HO
O
OH [Fe(CN) ]3–
6
H3O+
O
HO
HO
O + [Fe(CN)6]4–
OH
OH
gluconolactone (gln)
glucose (glu)
E ° (gln/glu) = + 0.29 V and
E ° {[Fe(CN) 6 ]3− / [Fe(CN) 6 ]4– } = + 0.69 V
(a) Balance the cell reaction and determine standard state cell potential.
(b) Is this reaction spontaneous as written?
(c) When the concentration of glucose is increased, what happens to the cell voltage?
PROBLEM 542
The diagram below shows the experimental setup for a typical Zn–Ni galvanic cell :
e
e–
e–
Salt-Bridge
E ° (Zn 2+ /Zn) = – 0.76 V
E ° (Ni 2+ /Ni) = – 0.25 V
MSO 4 = 0.01 M
XSO 4 = 1.00 M
M2+ SO42–
X2+ SO42–
(a) Identify M and X and determine cell potential at 25°C.
(b) If concentration of M 2+ ion changes to 1.0 M during its uses, what would be the new cell
voltage?
(c) Describe, what would happen to cell voltage if salt bridge was removed.
PROBLEM 543
At 25°C, H 2O 2 decomposes according to the following equation :
2H 2O 2 ( aq ) → 2H 2O( l) + O 2 ( q )
E ° = 0.55 V
(a) Determine the value of K eq for decomposition reaction.
4e
(b) If standard reduction potential for O 2 ( g ) + 4H + → 2H 2O is 1.23 V, using this information
in addition to the information provided above, determine the value of E° for
O 2 ( g ) + 2H + + 2e → H 2O 2
PROBLEM 544 In an electrolytic cell, Cu(s) is produced by electrolysis of CuSO 4 (aq ). Calculate
maximum mass of Cu( s) that can be deposited by a direct current of 100 A passed through 2 Lit 2 M
CuSO 4 ( aq ) solution for a period of 1.5 hours. Also determine the final concentration of CuSO 4 in
solution and volume of O 2 ( g ) measured at STP, produced at anode.
65
Problems
(a) Using the following E° values :
PROBLEM 545
°+
° 2+ + = 0.16 V,
= 0.52 V, ECu
ECu
/Cu
/Cu
determine K for : Cu( s) + Cu 2+ ( aq )
2Cu + ( aq ).
(b) If E° for the reaction Cu( s) + Cu 2+ ( aq ) + 2Br – ( aq )
2CuBr( s) is 0.5147 V, determine K sp
of CuBr at 25°C.
PROBLEM 546
Consider the galvanic cell
Zn( s) Zn 2+ (0.04 M)Cl – (5 × 10 −3 M )Cl 2 (0.1 atm ) Pt( s)
(a) Determine emf of cell and equilibrium constant for the net cell reaction.
(b) Now, if NH 3 is added to anode chamber at 1.0 M concentration, what would be the new emf. K f
° 2+ = – 0.76 V and E ° – = 1.36 V.
for [Zn(NH 3 ) 4 ]2+ = 7.8 × 10 8 , E Zn
Cl /Cl
/Zn
2
PROBLEM 547
A galvanic cell was constructed using Ag/Cu electrode as shown below :
CuCu 2+ ( aq ) Ag + (0.01 M) Ag
The cell emf was measured to be 0.382 V. Now some CaCl 2 was added to 250 ml electrolyte present
in cathode chamber such that all Ag + precipitate out as AgCl and finally voltage drop to 0.01 V. If [Cu 2+ ]
was 0.02 M, determine K sp of AgCl.
PROBLEM 548 A solution of M(NO 3 ) 2 is electrolyzed using a current of 2.5 A and 3.06 g metal was
deposited in 35 minutes. Determine molar mass of the metal.
PROBLEM 549 Consider the galvanic cell :
PtSn 4+ ( aq, 0.010 M), Sn 2+ ( aq, 0.10 M)O 2 (g, 1.0 atm); H + (pH = 4)C( gr )
E ° 4+ 2+ = 0.15 V and E ° +
= 1.23 V.
Sn
/ Sn
O 2 , H /H 2 O
Assuming that the cathode chamber is buffered at constant pH = 4, determine (a) Cell-potential,
(b) Concentrations of Sn 4+ and Sn 2+ when cell potential dropped to 0.8 V during its uses at 25°C.
PROBLEM 550 When a pH-meter was standardized with a basic-acid borate-buffer with a pH 9.4, the
cell potential was 0.06 V. When the buffer was replaced by a solution of unknown hydronium ion, the
cell potential was 0.22 V. Determine pH in this solution.
PROBLEM 551 A 200 ml CuSO 4 solution was electrolyzed using a current of strength 4.0 A for 30
min. Determine pH of analyte at the end of electrolysis assuming initial pH = 7.
PROBLEM 552 A solution of Mn(NO 3 ) 2 (0.15 M) and Fe(NO 3 ) 2 (0.1 M) is placed in a beaker and
buffered at pH = 5. Two platinum electrodes are inserted and current is passed through the solution in
order to plate out the metals.
° 2+ = − 0.44 V, E ° 2+
= − 1.18 V, and
E Fe
/Fe
Mn /Mn
EO° , H + /H O = 1.23 V
2
2
Which metal will be deposited first and what minimum voltage would be required for the onset of
electrolysis?
PROBLEM 553 Consider the electroplating of a metal +1 cation from a solution of unknown
concentration according to the half reaction : M + + e → M with a standard potential of E°. When
the half-cell is connected to an appropriate oxidation half-cell and current is passed, M + begins to
66
Problems in Chemistry
plateout at a potential of E1 . To what value ( E 2 ) must the applied potential be adjusted, relative to E1 , if
99.99% of the metal is to be removed from the solution?
PROBLEM 554 A steel pot containing acidic tomato sauce is covered with the aluminium foil; the foil
is in contact with both the pot and sauce. Latter a large excess of aluminium foil has dissolved. It is
believed that the following electrochemical reaction caused aluminium to dissolve :
3Fe 2+ ( aq ) + 2Al
3Fe( s) + 2Al 3+ ( aq )
At some instant, emf was found to be 1.2 V. Determine the area of holes created in the Al-foil at that
° 3+ = − 1.66 V and E ° 2+ = − 0.44 V. Density of Al metal is 2.7 g/cc, thickness of
instant. E Al
Fe /Fe
/Al
aluminium foil and volume of sauce was 100 ml and [Fe 2+ ] = 5 × 10 –4 M.
PROBLEM 555 A balloon is being filled with hydrogen produced by the electrolysis of an aqueous
solution of acid. How long will it take to generate enough hydrogen to lift 1.5 kg by using a current of
8.5 A. Average molar mass of air = 28.8 amu.
PROBLEM 556 A galvanic cell was constructed by dipping a zinc electrode in 0.1 M Zn(NO 3 ) 2 and a
Pt electrode in 0.1 M HIO 3 solution and emf of this cell was found to be 0.72 V at 25°C.
° 2+ = − 0.76 V. Determine acid dissociation constant (K a ) of HIO 3 . Assume partial pressure of
E Zn
/Zn
gaseous species to be 1.0 atm.
PROBLEM 557 A battery was used to supply a constant current of what was believed to be exactly 0.4
A as read on a meter in the external circuit. The cell was based on the electrolysis of a 100 ml 0.1 M
aqueous copper sulphate solution. After a 30 minute duration, concentration of electrolyte dropped to
0.0528 M. Determine the extent to which meter was incorrect?
PROBLEM 558 A galvanic cell with a measured potential of 0.11 V at 25°C, contain a Pt electrode in
a solution which is 0.0135 M in Cr +2 and 2.16 × 10 −4 M in Cr +3 . In second compartment a Ni electrode
dips into a solution of Ni(NO 3 ) 2 . Determine concentration of Ni 2+ second compartment.
E ° (Cr 3+ /Cr 2+ ) = − 0.41 V and E ° ( Ni 2+ /Ni) = − 0.25 V.
PROBLEM 559 The electrolysis of Na 2SO 4 (aq ) is conducted into two separate half-cells joined by a
salt bridge, also containing Na 2SO 4 . The cell diagram for the electrolysis is
Pt Na 2SO 4 ( aq ) Na 2SO 4 ( aq ) Pt
Phenolphthalein indicator is added to each half-cells.
(a) Describe any colour-change occurring in half-cells as electrolysis progresses.
(b) After electrolysis is stopped, the solution of the two half-cells are mixed. Describe and explain
any colour change that occur.
(c) In an experiment 10 ml HCl is added to cathode chamber alongwith phenolphthalein.
Electrolysis is carried out with a 25 mA current and the solution colour becomes pink after 8.00 minutes.
What is the molarity of HCl?
2+
°
PROBLEM 560 Determine formation constant of complex HgI 2–
/Hg) = 0.85 V and
4 . E (Hg
2–
–
.
E ° (HgI 4 /Hg, I ) = − 0.04 V
Problems
67
CHEMICAL KINETICS
PROBLEM 561 A sample of pitch blend has amount of 206 Pb equivalent to one-fifth of the amount of
238
U (by mass). If the disintegration constant of 238 U is 1.54 ×10 –10 Y –1 and all Pb is supposed to be
obtained from uranium, what will be the age of mineral?
PROBLEM 562 The gaseous decomposition of Cl 2O 7 (g ) according to following reaction follow 1st
order kinetics:
7
Cl 2O 7 ( g ) → Cl 2 ( g ) + O 2 ( g )
2
When the decomposition is carried out in a closed container, pressure measured after 15 min and
after a very long time were found to be 240 and 900 mm of Hg respectively. Calculate the pressure that
would be observed in the flask after 140 min under identical conditions of temperature.
PROBLEM 563 The isomerization of cyclobutene to 1,3-butadiene follow first order kinetics and the
rate constant is 2 × 10 –4 s –1 at 150°C in a 530 mL flask. Determine the partial pressure of cyclobutene
and its concentration after 30 min if an isomerization reaction was carried out at 150°C with an initial
pressure of 55 mm of Hg.
PROBLEM 564 The hydrolysis of sugar sucrose to sugars glucose and fructose
C12 H 22O11 + H 2O → C 6 H12O 6 + C 6 H12O 6 follow first order kinetics.
(a) In a neutral solution, k = 2.1 × 10 –11 s –1 at 27°C and 8.5 ×10 –11 s –1 at 37°C. Determine
activation energy, frequency factor and rate constant at 47°C.
(b) When a solution of sucrose with an initial concentration of 0.15 M reaches equilibrium, the
concentration of sucrose is 1.65 ×10 –7 M. How long will it take the solution to reach equilibrium at
27°C?
PROBLEM 565 Fluorine-18 is a radioactive isotope that decays by positron emission to form
oxygen-18 with a half-life of 109.7 min. Physician use 18 F for the study of brain by injecting a quantity
of fluoro substituted glucose into the blood of a patient. The glucose accumulates in the regions where
the brain is active and needs nourishment.
(a) What is the rate constant for the decomposition of fluorine-18?
(b) If a sample of glucose that contains 18 F is injected into the blood, what percentage will remain
after 5.6 hours?
(c) How long does it take for 99% decay of 18 F ?
PROBLEM 566 The nucleidic ratio of 1 H 3 to 1 H1 in a sample of water is 8.0 ×10 –18 :1. Tritium
undergoes decay with a half-life period of 12.3 years. How many tritium atoms would a 10 g sample of
water contains 40 year after the original sample is collected.
PROBLEM 567 The decomposition of N 2O 5 according to the following reaction follow first order
kinetics:
2N 2O 5 → 4NO 2 + O 2
After 30.0 min from start of decomposition in a closed vessel, the total pressure found to be 284.5
68
Problems in Chemistry
mm of Hg and after a very long time, the total pressure found to be 584.5 mm of Hg. Calculate the rate
constant and total pressure after 1.0 hour.
PROBLEM 568 A certain radioactive isotope z X A (half-life =10 days) decays to give z – 2 X
g of X is kept in a sealed vessel, find the volume of He accumulated at S.T.P. in 20 days.
A–4
. If 1.0
PROBLEM 569 A reaction rate increases by a factor of 500 in the presence of a catalyst at 37°C. If the
reaction is carried out in presence of a catalyst at 127°C, by what factor, compared to original rate in
absence of catalyst at 27°C, the rate will increase? Activation energy of original pathway is 106 kJ/mol.
PROBLEM 570 A gaseous mixture containing N 2 and an unknown gas A2 (g ) are present in a sealed
vessel at 700 mm of Hg. Gas A2 decomposes as A2 ( g ) → 2 A ( g ) with progress of time and
decomposition follow first order kinetics. If the total pressure inside the flask measured after 30 min and
60 min are 760 and 800 mm of Hg, calculate t1/ 2 and total pressure after 75 min.
PROBLEM 571 If the rate constant for the decomposition of COCl 2 (g ) according to the following
reaction:
COCl 2 ( g ) → CO + Cl 2
is expressed as log k ( min –1 ) = 15 –
∆H =103 kJ/ mol
11420
. Calculate activation energy for the formation of COCl 2 ( g ) at
T ( K)
the same temperature.
PROBLEM 572 Gaseous decomposition of N 2O 5 follows first order kinetics. Pure N 2O 5 gas is taken in
a flask, sealed and finally heated to 100°C where decomposition occurs as follows:
1
N 2O 5 → 2NO 2 + 2 O 2
After 15 minutes, a leak was developed in the flask. On analysis of the gaseous mixture coming out
initially, mole fraction of NO 2 was found to be 0.5. Calculate half-life for the decomposition reaction.
PROBLEM 573 The half-life of a drug is 3.0 hours. A patient is administered a 5 mg dose of this drug at
an interval of 6.0 hour. Calculate the mass of drug remaining in the patient ’s body just after 6th dose.
PROBLEM 574 0.01 moles of sulphuryl chloride, SO 2Cl 2 (g ) is taken in a sealed tube and heated to 400
K where it decomposes following first order kinetics according to the following reaction:
SO 2Cl 2 ( g ) → SO 2 ( g ) + Cl 2 ( g )
The tube is broken after 4.0 hour and gas passed through a 15 mL of an acidified, 1.0 N iodine
solution where all SO 2 is oxidised to SO 2–
4 . The resulting solution required 7.0 mL 1 M hypo solution.
Calculate the total pressure inside the flask just before it broken down. If the volume of the flask is
200 mL, also calculate its half-life.
PROBLEM 575 A reactant A decomposes to produce two different products simultaneously, but at
different rate as shown below:
k2
k1
E2
E1
C ← A → B
The order of reaction is unity for both reaction and their respective rate constants k1 and k 2 are
1.2 ×10 –2 and 3 × 10 –2 per second respectively. Calculate activation energy for the overall reaction if the
activation energy for individual steps E1 and E 2 are 180 and 200 kJ/mol respectively.
69
Problems
PROBLEM 576 The reaction
cis - Cr(en) 2 (OH) +2
K1
K2
trans - Cr(en) 2 (OH) 2+
is first order in both directions. At 25°C, the equilibrium constant is 0.16 and rate constant k1 is
3.3 ×10 –4 s –1 . In an experiment starting with the pure cis form, how long would it take for half the
equilibrium amount of the trans-isomer to be formed?
PROBLEM 577
40
K decays into
40
Ca and
40
Ar simultaneously as:
t 1/ 2 = 1.5 × 109 y
→
20 K
20Ca
40
+
–1β
0
40
t 1/ 2 = 1.2 × 1010 y
→ 18 Ar 40 + +1β 0
If the atomic concentration of Ar in a sample of rock is 10% that of potassium, calculate the age of
the rock sample considering the only source of Ar is radioactive disintegration of potassium.
PROBLEM 578 2,3-di methyl-3-chloro pentane on heating with methanol undergoes substitution as
well as elimination according to the following reaction:
Cl CH3
CH 3
OMe

 

CH3OH
C 2 H 5 —C— CH —CH 3 → C 2 H 5 —C== C— CH 3 + C 2 H 5 —C—CH—CH 3
∆
 


CH 3CH 3
CH 3
CH 3
A 100 mL 0.1 M solution of dextro-rotatory sample of 2,3-di methyl-3-chloro pentane is heated with
excess of methanol and optical-rotation were measured at different intervals as:
t (min)
0
10
∞
Optical-rotation 40°
30°
5°
Calculate the optical-rotation after 20 minute and volume of 0.15 M aqueous solution of bromine
required after 20 minute for bromination of alkene produced if the molar ratio of alkene to ether is 6 : 4.
PROBLEM 579 The activation energy for the reaction H + CH 4 → H 2 + CH 3 has been found to be
49.8 kJ/mol. Some estimates of enthalpies of formations, ∆H f° are : H = 218, CH 4 = – 74.8 and
CH 3 = 139.5 kJ/ mol. Estimate activation energy for the reverse reaction.
PROBLEM 580 The half-life for the thermal de-naturation of haemoglobin, a first order process, has
been found to be 3460 s at 60°C and 530 s at 65°C. Calculate the enthalpy of activation and entropy of
activation at 60°C assuming the Arrhenius equation to apply.
PROBLEM 581 For a parallel reaction:
A
→ B E a 1 = 150 kJ/ mol
→ C E a 2 = 80 kJ/ mol
→ D E a 3 = 30 kJ/ mol
Mole percentage of B, C and D after one hour was found to be 10, 25 and 15 respectively. Determine
the overall activation energy for this reaction.
70
Problems in Chemistry
PROBLEM 582 You need to convert 1.0 mole of X into product. The reaction follow first order kinetics.
At 1.00 p.m. you start the reaction at 27°C with one mole of X. At 2.0 p.m. you find that 0.8 mole of X
remains. You immediately increase the temperature to 37°C. At 3.0 p.m. you discover that 0.4 mole of X
are still remaining. You want to be able to finish the reaction by 4.0 p.m. but you cannot stop the reaction
until only 0.1 mole of X remains. You decided to increase the temperature again. What is the minimum
temperature required to complete the reaction by 4.0 p.m.?
PROBLEM 583 Gaseous decomposition of A2 (g ) into A(g) follow first order kinetics as:
A2 ( g ) → 2 A ( g )
If the decomposition is carried out in a sealed flask at a constant temperature, the total pressure
measured after 20 minute and 40 minute from the beginning are 2.5 atm and 2.875 atm respectively.
Calculate the total pressure inside the flask after 1.0 hour.
PROBLEM 584 A gaseous reactant A decomposes to produce gaseous products B and C in a parallel
reaction, both by first order, as follows:
K = 2 × 10–3 min –1
1
B (g )
→
A(g )
→
C (g )
–2
–1
K 2 = 3 × 10
min
If the decomposition is carried out in a sealed flask, partial pressure of B(g) at the end of reaction was
found to be 40 mm of Hg. Determine partial pressure of C after 1.0 hour.
PROBLEM 585 Isomerization of cyclobutene into 1,3-butadine follow first order kinetics as:
(g )
(g)
The kinetic study was performed by taking same amounts of cyclobutene in three sealed flasks. First
flask was broken after 20 minute and the reaction mixture was absorbed completely in bromine solution.
16.0 mL 1.0 M bromine solution was required. The second flask was broken after a very long time and
the reaction mixture required 20 mL bromine solution of the same strength. If the third flask was broken
after 30 minute, what volume of bromine solution of same strength would have been required?
PROBLEM 586 Racemization of an optically pure enantiomer in its aqueous solution follow first order
kinetics and optical rotation found after 10 minute and 20 minute from beginning were 40° and 24°
respectively. Determine the optical-rotation after 40 minute.
PROBLEM 587 Decomposition of both A2 (g ) and B 3 (g ) follows 1st order kinetic as :
K1
A2 ( g ) → 2 A ( g )
K2
B 3 ( g ) → 3B ( g )
K1 ( hr
−1
K 2 (hr
−1
2
−
14000
RT
3
−
20000
RT
) = 10 e
) = 10 e
If one mole of each A2 ( g ) and B 3 ( g ) are taken in a 10 L evacuated flask and heated to some
temperature so that they start decomposing at the same rate, determine total pressure in the flask after
1.0 hr.
71
Problems
PROBLEM 588 Decomposition of a non-volatile solute “ A ” into another non-volatile solute B and C,
when dissolved in water follow first order kinetics as:
A → 2B + C
When one mole of A is dissolved in 180 g of water and left for decomposition, vapour pressure after
12 hours was found to be 20.69 mm of Hg. Determine V.P. of the solution after 30 hours. Assume
constant temperature through out to be 27°C and V.P. of pure water at 27°C is 24 mm of Hg.
PROBLEM 589 For the reaction A + B + C → P , following rate informations are available :
Rate (mol L–1 hr –1 )
[A] L mol −1
[B] L mol −1
[C] L mol −1
2.5 × 10 −3
0.1
0.2
0.3
10 −2
0.1
0.2
0.6
1.67 × 10 −3
0.1
0.3
0.2
1.78 × 10 −2
0.2
0.4
0.4
Deduce the rate law.
PROBLEM 590 A sample of 0.42 mg 233 UF6 shows an activity of 9.88 ×10 4 count per second. Find t1/ 2 .
PROBLEM 591 The uranium present in earth today is 99% 238 U and 1% 235 U. The half-lives are
4.5 ×10 9 years and 7 × 10 8 years respectively. How long ago, this mineral was 50% in each isotope?
PROBLEM 592 Mole fraction of A as a function of time in the following reversible first order reaction
A
B are:
Time (hr)
0
1
Infinity
Mole % of A
100
75
30
Determine mole percentage of A after four hour from the beginning.
PROBLEM 593 Isomerization of N-chloroacetanilide to p-chloro acetanilide follow first order kinetics
and isomerization is achieved by adding KI solution to N-chloroacetanilide solution. Liberated iodine is
titrated with standard thiosulphate solution for kinetic study and following results were observed:
Time (hr)
0
1
Vol. of 0.1 M S 2O 32– required
50
35
Determine volume of thiosulphate that would be required after 8.0 hour assuming only
N-chloroacetanilide reacts with KI.
PROBLEM 594 Decomposition of phosgene gas as COCl 2
with:
Kf
Kb
CO + Cl 2 follow first order kinetics
72
Problems in Chemistry
52400 1
4.5 T
26300 1
and
log K b = 8.023 −
.
4.5 T
Determine ∆H ° and ∆S ° for the decomposition reaction.
log K f = 13.48 −
PROBLEM 595 Thermal decomposition of H 2O 2 follows first order kinetics. Three flask, each
containing 25 mL 2.5 V H 2O 2 were taken for kinetic study. Into the first flask, 20 mL of a KI solution
was added after 1.0 hour, heated gently to expel-off liberated iodine and finally titrated against 0.4 M
chromic acid solution. 10.4 mL of chromic acid was required to reach end point. After three hour from
beginning, 20 mL of KI solution of same strength was added to the second flask, heated gently and
finally titrated against chromic acid solution of same strength. After a very long time, similar experiment
was repeated with the third flask and required 16.67 mL chromic acid solution. What was the volume of
chromic acid required for second experiment?
PROBLEM 596 The acid catalyzed hydrolysis of ethyl acetate is first order in ester and overall rate
constant for the hydrolysis reaction is 0.11 L mol –1 s –1 . Determine time required for 30% hydrolysis of
ester at pH 2.5.
PROBLEM 597 A mixture of two different substance A and B undergo simultaneous reaction to produce
the same product as:
K = 2 × 10−2 min −1
1
A 
K = 5 × 10−3 min −1
2
B 
C
After 30 minute from beginning, mole percentage of C was found to be 25. Determine its mole per
cent after 50 minute assuming initially there were no C.
PROBLEM 598 Isotopes of oxygen with mass number less than 16 undergo β + emission. Assuming an
equimolar mixture of O14 and O15 , determine time required for nuclei ratio O14 /O15 to become equal to
0.25. t1/ 2 for O14 and O15 are 71 and 124 second respectively.
PROBLEM 599 A first order chemical reaction has t1/ 2 = 200 hr at 27°C. Adding a catalyst lowers the
half-life to 20 hour. How long the reaction would take to complete 75% at 127°C in presence of this
catalyst. Activation energy of the uncatalyzed pathway is 80 kJ.
PROBLEM 600 The compound Xe(CF3 ) 2 is unstable and decomposes by 1st order reaction as:
Xe(CF3 ) 2 ( g ) → Xe ( g ) + C 2 H 4 ( g ) + F2 ( g ), t1/ 2 = 30 minute at 300 K,
If 1.345 g of Xe(CF3 ) 2 ( g ) is taken in an evacuated 100 mL flask, sealed and left for some time at 300
K, pressure inside the flask was found to be 2.63 bar. How long did the gas stay in flask?
PROBLEM 601 The nuclide Ac 227 decay by β − emission (90%) and α-emission (10%). If 100 g of pure
actinium-227 is taken in a sealed evacuated 1.0 L flask, pressure inside the flask after 5.0 hour was found
to be 0.16 bar at 300 K. Determine half-life of the decay process.
PROBLEM 602 Nitrogen oxides NO x (a mixture of NO and NO 2 ) plays an important role in the
73
Problems
production of photochemical smog. The half-lives for the removal of NO and NO 2 in smoke stack
emission are 1.5 hr. and 2.5 hr. respectively. Starting with 1.5 g of mixture in an experiment, 0.225 g of
NO x was left after 5.0 hour. Determine composition of original mixture assuming both gases are
removed by first order reaction. How many hours of light must elapse to decrease amount of NO x to
2.5 × 10 −6 g.
K
PROBLEM 603 For a chemical reaction A → product, following mechanism has been proposed :
K
E a =190 kJ
K
E a = 80 kJ
1
A →
A*
2
A * →
B*
K3
B * → Product
E a = 40 kJ
K K 
Also, overall rate constant K is related to individual rate constants by the equation: K =  1 2 
 K3 
Determine activation energy for the overall reaction.
2/ 3
.
PROBLEM 604 The first order decomposition of a coloured chemical species X into colourless
products is monitored with a spectrophotometer by measuring changes in absorbance over time. The data
from the experiment are given below:
[ X ] (M)
Absorbance
Time (min)
?
0.600
0.00
0.200
35.0
0.150
44.2
4.00 × 10
−5
3 × 10 −5
Calculate initial concentration of the coloured species and time when absorbance will fall to 0.0075.
The graph below show results from a study of the decomposition of H 2O 2 .
–1.0
ln [H2O2]
PROBLEM 605
–2.0
–3.0
–4.0
0
800
1600
Time (min)
2400
4800
74
Problems in Chemistry
(a) Write rate law for the reaction justifying your answer.
(b) Determine value of rate constant and [H 2O 2 ] after 2000 minute.
PROBLEM 606 The gas phase reaction: 2Cl 2O + 2N 2O 5 → 2NO 3Cl + 2NO 2Cl + O 2 has rate
law: Rate = K [N 2O 5 ]. If the reaction was carried out in a sealed flask with equimolar amounts of
reactant, the total pressure measured after 10 minute and after a very long time were 650 and 750 mm of
Hg respectively. What was the total pressure after 30 minute from beginning? Assume constant
temperature throughout the experiment.
PROBLEM 607 For a chemical reaction A → P, the ratio of amount of reactant left unreacted to the
initial amount ([A] / [A]0 ) were measured after 400 seconds and 964 seconds and were found to be 0.5
and 0.25 respectively. Determine the order of reaction and rate constant if initial moles of reactant was
0.6.
PROBLEM 608 Considering the following concentration of reactant left unreacted Vs time graph for a
first order chemical reaction, determine rate of reaction at 10 sec.
0.4
Molarity
0.3
0.2
0.1
0
20
10
30
40
50
Time in sec
PROBLEM 609 For a first order chemical reaction, variation of rate constants vs temperature is depicted
in the following graph:
–1
log K (s )
3
2
1
x=1
(0, 0)
1
x
2
3
4
5
6
7
8
9
4
–1
1
× 10 (K )
T
Determine the time required for 80% completion of reaction at 500 K.
PROBLEM 610 Rate of a first order chemical reaction measured after 10 min and 30 min were
10.64 × 10 −3 and 8.37 × 10 −3 mol L–1 min –1 . What was the rate after 2 hr. assuming temperature was
constant throughout?
75
Problems
PROBLEM 611
A reversible chemical reaction:
Kf
A
Kb
P
is first order in both A and P and has equilibrium constant of 2 at 300 K. If rate constant for the forward
direction is 2.5 ×10 –2 min –1 at 300 K, determine the time required to form half of equilibrium amount of
P assuming that initially there was no P.
PROBLEM 612 A 1st order chemical reaction was carried out for 1.0 hour in absence of a catalyst and
20% reaction was complete. A catalyst was then added and reaction was allowed to continue for next 30
min when 60% reaction was complete. A second catalyst was then added at this time and reaction was
allowed to continue for further 10 min. when 90% reaction was complete. If activation energy of the
original path was 80 kJ, determine activation energies of catalyzed pathways. Assume constant
temperature throughout to be 300 K.
PROBLEM 613 The half-life of 14 C isotope is 5730 years. The activity of carbon in a living being is
12.5 counts per minute per gram of carbon.
(a) Determine the percentage of carbon in living being that is 14 C.
(b) Determine activity of carbon from remains of an organism that died 50,000 years ago.
(c) Determine age of Egyptian tomb that shows an activity of 7.0 counts per minute per gram of
carbon.
COLLIGATIVE PROPERTIES
PROBLEM 614 A certain aqueous solution of a non-electrolytic solute freezes at – 0.6°C. For this
solution estimate (a) normal boiling point (b) the vapour pressure at 25°C. (c) the osmotic pressure at
20°C. The vapour pressure of pure water at 25°C is 23.75 torr. K f =1.86 and K b = 0.52.
PROBLEM 615 At 25°C, vapour pressure of pure benzene and pure toluene are 93.4 and 26.9 torr
respectively. A solution is prepared by mixing 60 g of benzene and 40 g of toluene. What pressure should
be maintained in the flask containing this solution so that it start boiling at 25°C.
PROBLEM 616 A urea solution in 250 g of water freezes at – 0.744°C. This solution was cooled to
some temperature where some ice is formed. Solution was decanted-off and heated to 100°C where the
vapour pressure was found to be 757.7 mm of Hg. Determine mass of ice formed and temperature to
which solution was cooled. K f of water is 1.86 kg/mol.
PROBLEM 617 At 27°C, 12 L of pure N 2 measured at 1.0 atmosphere are passes through an aqueous
solution of a non-volatile solute, whereby the solution loses 0.25 g in weight. If vapour pressure of pure
water is 23.75 mm of Hg and K b is 0.52 kg/mol, determine boiling temperature of this solution.
PROBLEM 618 An aqueous solution containing 10 g mixture of urea and glucose boils at 100.58°C.
Addition of a further 6.0 g glucose to the above solution causes it to boil at 100.77°C. Determine mass
percentage of urea in the original mixture.
PROBLEM 619 A one litre solution is prepared by dissolving some solid lead-nitrate in water. The
solution was found to boil at 100.15°C. To the resulting solution 0.2 mole NaCl was added. The resulting
76
Problems in Chemistry
solution was found to freeze at – 0.83°C. Determine solubility product of PbCl 2 . Given K b = 0.5 and
K f =1.86. Assume molality to be equal to molarity in all case.
PROBLEM 620 Vapour pressure of a volatile substance A is 400 mm of Hg at 298 K. When some
another volatile solvent B (V.P. = 500 mm at 298 K) is added to A, vapour pressure of the resulting
solution was found to be 470 mm of Hg. To the resulting solution a third volatile solvent C (V.P. = 600
mm of Hg at 298 K) is added so that the vapour pressure of the resulting solution becomes equal to 496
mm of Hg. Determine mole fraction of each A, B and C in the vapour phase in equilibrium with the final
solution.
PROBLEM 621 A solution, prepared by dissolving 38.2 g of Na 2 B4O 7 ⋅ xH 2O in 250 g water has boiling
point 100.582°C. Deduce formula of the salt. K b of the water is 0.52 kg mol −1 . [Atomic mass of B =11,
Na = 23]
PROBLEM 622 A mixture of NaCl and sucrose of combined mass 10.2 g is dissolved in enough water to
make up a 250 mL solution. The osmotic pressure of the solution is 7.32 atm at 23°C. Calculate the mass
percentage of NaCl in the mixture.
PROBLEM 623 A certain organometallic complex contain carbon, hydrogen and a metal from the first
transition series (one metal per molecule). In a non-protic complexing solvent like diethyl ether, this
compound dissociate into ions. In a non-polar solvent like benzene, this compound does not dissociate.
Below are data obtained from freezing point depression and boiling point elevation experiment carried
out on the sample of this compound.
(a) Freezing point of benzene solution containing 1.5 g of complex in 36 g benzene is 4.35°C. (fr. pt
of C 6 H 6 is 5.5°C and K f of benzene is 5.12)
(b) The b.pt of diethyl ether solution containing 0.48 g of the complex in 10 g of the solvent is
36.05°C (b.pt. of pure ether is 34.5°C, K b = 2.02)
(c) Elemental analysis : 64.88% C and 5.45 %H.
In how many ions does the complex dissociated when dissolved in the complexing solvent? Write
the molecular formula of the complex.
PROBLEM 624 A glucose solution in 100 g of water boils at 100.26°C. If this solution is heated to
101°C, determine the mass of water left at equilibrium.
PROBLEM 625 A 100 g 10% by weight urea solution is placed together with a 200 g 10% by weight
glucose solution in a closed jar and allowed to attain the equilibrium. Determine mass percentage of urea
in its solutions at equilibrium.
PROBLEM 626 A 100 g 10% by weight urea solution is placed together with a 200 g glucose solution in
a big jar and left for a very long time. The vapour pressure in the jar at equilibrium was found to be 23.6
mm of Hg. If the vapour pressure of pure water at experimental temperature is 24 mm of Hg, determine
mass percentage of glucose in its solution.
PROBLEM 627 The vapour pressure of a solution of methanol and ethanol at 20°C was found to be 70
mm of Hg. Adding 10 g of urea to 80 g of this solution lowers the vapour pressure to 64.6 mm of Hg.
Determine composition of the original solution.
PROBLEM 628 Vapour pressure of an equimolar mixture of benzene and toluene was found to be 80
torr. If the vapour above the liquid phase is condensed in a beaker, vapour pressure of this condensate
was found to be 100 torr. Determine vapour pressure of pure benzene and pure toluene in the given
condition.
Problems
77
PROBLEM 629 Vapour pressure of methanol and ethanol are 94 and 44 mm Hg at 20°C. To a 50 g
mixture containing equal weight of both methanol and ethanol, 11.85 g of a mixture of NaCl and urea
was dissolved and vapour pressure of the resulting solution was found to be 58 mm of Hg. Determine
mass percentage of urea and NaCl in the mixture.
PROBLEM 630 Determine vapour pressure of water at 50°C if enthalpy of vaporization of water is 40.6
kJ/mol.
PROBLEM 631 An aqueous solution of canesugar ( MW = 342) has osmotic pressure equal to 1.5 atm at
18°C. What will be the V.P. of this solution at 40°C. If 100 g of this solution is cooled to – 2°C, what
mass of ice will be separated out. (V.P. of H 2O = at 40°C = 55.324 mm of Hg, K f =1.86, density
=1.0 g/ mL)
PROBLEM 632 Two elements A and B forms compounds having molecular AB 2 and AB 4 . When
dissolved in 20 g of benzene, 1.0 g of AB 2 lowers freezing point by 2.3 K whereas 1.0 g of AB 4 lowers
the freezing point by 1.3 K. Determine atomic masses of A and B. The molal depression constant for
benzene is 5.1 K kg mol −1 .
PROBLEM 633 1.0 g of a monobasic acid when dissolved in 100 g of water lowers the freezing point by
0.168°C. 0.2 g of the same acid when dissolved and titrated, required 15.1 mL of N/10 alkali. Calculate
degree of dissociation of the acid. K f for water is 1.86.
PROBLEM 634 A complex is represented as CoCl 3 ⋅ xH 2O. Its 0.1 m solution in aqueous medium shows
∆T f = − 0.558 K. K f for water is 1.86 and assume 100% ionization of complex having coordination
number 6, deduce the formula of the complex.
PROBLEM 635 10 g of a weak monobasic acid in 100 g of benzene freezes at 3.35°C. Assume that acid
undergoes complete association in benzene. Find the degree of dissociation of 10 g of acid in 100 g of
water whose freezing point found to be – 1.75°C. K f for water =1.86. K f for the benzene = 5.12 and
freezing point of benzene = 5.51° C.
PROBLEM 636 A saturated solution of a sparingly soluble salt MCl 2 has a vapour pressure of 31.78
mm of Hg at 30°C, while pure water exert a pressure of 31.82 mm of Hg at the same temperature.
Calculate solubility product of the compound at this temperature.
PROBLEM 637 A protein has been isolated as sodium salt with their molecular formula Na x P (this
notation means that xNa + ions are associated with a negatively charged protein P −x ). A solution of this
salt was prepared by dissolving 0.25 g of this sodium salt of protein in 10 g of water and ebulliscopic
analysis revealed that solution boils at temperature 5.93 × 10 −3 ° C higher than the normal boiling point
of pure water. K b of water is 0.52 Kg mol −1 . Also elemental analysis revealed that the salt contain 1%
sodium metal by weight. Deduce molecular formula and determine molecular weight of acidic form of
protein H x P.
PROBLEM 638 A non-volatile organic compound X was used to make-up two solution. Solution A
contains 5.0 g of X in 100 g of water and solution B contains 2.0 g of X in 100 g of benzene. Solution A
has vapour pressure of 754.5 mm of Hg at normal boiling point of water and solution B has the same
vapour pressure at the normal boiling point of benzene. Assuming X undergo partial dimerization in
benzene, determine percentage of X dimerized in benzene solution.
78
Problems in Chemistry
PROBLEM 639 1.32 g of a mixture of cyclohexane and naphthalene is dissolved in 20 g of benzene and
solution freezes at 2.2°C. Calculate composition of the mixture, given K f of benzene is 5.12 kg mol −1
and freezing point of benzene is 5.51°C.
PROBLEM 640 When 1.0 of urea is dissolved in 200 g of an unknown solvent X, the X freezing point is
lowered by 0.25°C. When 1.5 g of an unknown, non-electrolytic solute Y is dissolved in 125 g of same
solvent X, freezing point is lowered by 0.2°C and vapour pressure is lowered by 1%. If freezing point of
X, is 12°C, determine molar enthalpy of fusion of X.
PROBLEM 641 Chloroform boils at 62°C. For a solution containing 0.4 g of naphthalene in 25 g of
chloroform, boiling point is elevated by 0.45 K. If chloroform is to be distilled at 50°C in order to avoid
any decomposition during distillation, what maximum pressure can be maintained in the flask?
PROBLEM 642 Phenol undergoes partial dimerization in bromoform. When 2.5 g of phenol is
dissolved in 100 g bromoform, freezing point is lowered to 6.4°C. Pure bromoform freezes at 8.4°C and
its K f is 14 Kkg mol −1 . Determine K c (equilibrium constant) for the following dimerization reaction if
the density of solution is 0.88 g/mL;
2C 6 H 5OH
(C 6 H 5OH) 2
PROBLEM 643 6.0 g of a mixture of anthracene (C14 H10 ) and naphthalene (C10 H 8 ) when dissolved in
300 g of benzene, freezing point is lowered by 0.7°C. Normal freezing point of benzene is 5.5°C and its
K f is 5.12 Kkg mol −1 . Determine percentage composition of mixture and mass of solid benzene that will
be produced if the above solution is cooled to 4.5°C.
PROBLEM 644 A 1.0 m aqueous solution of HF freezes at – 1.9°C. K f of water is 1.86 Kkg mol −1 . If
the density of solutions is 1.12 g/mL, determine pH of solution and ionization constant of HF.
SOLID STATE
PROBLEM 645 A body centre cubic lattice is made up of two different type of atom A and B. Atom A
occupying body centre and B occupying the corner positions. One of the corner is left unoccupied per
unit cell. Deduce the empirical formula of such solid.
PROBLEM 646 A face centred cubic solid is made up of two different type of atoms A and B. Atom A
occupying the corner and B are at the face centres. One of the face centre is left unoccupied per unit cell.
Deduce the formula of the solid.
PROBLEM 647 A face centred cubic lattice is made up of a single type of atom and one of its corner is
left unoccupied per unit cell. Calculate packing fraction of such solid.
PROBLEM 648 A body centred cubic lattice is made up of hollow spheres whose inner radius is
one-fourth of its outer radius. Calculate packing fraction of such solid.
PROBLEM 649 A hexagonal close packed atomic solid has some defect and its one corner and one face
centre is left unoccupied per unit cell. Calculate packing fraction of such solid.
Problems
79
PROBLEM 650 Al crystallizes in cubic shape unit cell and with edge length 405 pm and density 2.7
g/cc. Predict the type of crystal lattice.
PROBLEM 651 CdO has NaCl structures with density 8.27 g/cc. If the ionic radius of O 2− is 1.24 Å,
determine ionic radius of Cd 2+ .
PROBLEM 652 A compound of mercury chlorine has a crystal structure with two formula weights per
unit cell. The unit cell edges are 4.47 Å, 4.47 Å and 10.89 Å respectively and the unit cell angles are all
90Å. The density of substance is 7.15 g/cc, determine formula of compounds.
PROBLEM 653 Sulphide of cobalt metal has a cubic structure with four formula weights per unit cell. If
density is 4.269 g/cc and edge length of unit cell is 6.93 Å, determine mass of sulphur required to
produce 1.5 kg of this compound.
PROBLEM 654 Graphite has HCP arrangements of carbon atoms and the parallel planes are 3.35 Å
apart. Determine density of graphite.
PROBLEM 655 An atomic solid crystallizes in hexagonal closest packing. Calculate packing fraction.
PROBLEM 656 An ionic solid crystallizes in hexagonal closest packing in which cations occupies the
middle layer of unit cell. Calculate packing fraction of the ionic solid.
PROBLEM 657 An atomic solid crystallizes in a body centre-cubic lattice and the inner surface of the
atoms at the adjacent corner are separated by 60 pm. If the atomic weight of A is 48, calculate density of
the solid.
PROBLEM 658 FeO crystallizes in NaCl type of crystal lattice. The crystal however are always
non-stoichiometric and deficient in iron. Some cation sites are vacant and some contains Fe 3+ ions but
the combination is such that structures is electrically neutral. The formula approximate to Fe 0.95O.
(a) What is the ratio of Fe 2+ to Fe 3+ ion in the solid.
(b) What percentage of cation sites are vacant?
PROBLEM 659 Cadmium oxide crystallizes in NaCl type of crystal lattice. The compound is however
usually non-stoichiometric with approximate formula CdO 0.95 . The defect arises due to some cationic
positions are occupied by neutral Cd-atom instead of Cd 2+ ions and equivalent numbers of anion sites
are vacant:
(a) What percentage of anion sites are vacant?
(b) If the edge length of the unit cell is 470 pm, what would be the density of perfect solid?
(c) What is the density of non-stoichiometric solid.
PROBLEM 660 A hexagonal closed packed solid is made-up of cylindrical atom. Calculate packing
fraction of this solid.
PROBLEM 661 In an atomic BCC, what fraction of edge is not covered by atoms?
PROBLEM 662 CaO crystallizes in face-centred cubic lattice with unit cell edge length of 4.8 Å.
Calculate the closest distance between oppositely charged ions.
PROBLEM 663 In an atomic FCC, all the positions are occupied by A and the body centred octahedral
hole is appropriately occupied by B, without disturbing the FCC of A. Calculate packing fraction of this
solid.
80
Problems in Chemistry
PROBLEM 664 What is the packing fraction of the two dimensional unit cell shown in
figure?
PROBLEM 665 What fraction of the Ca atoms lies on the surface of a cubic crystal that is 1.0 cm in
length. Radius of Ca atoms is 96 pm and it crystallizes in BCC arrangement.
PROBLEM 666 TlAl(SO 4 ) 2 ⋅ xH 2O is BCC with ‘ a’ =1.22 nm. If the density of the solid is 2.32 g/cc,
determine x.
PROBLEM 667 The density of CaF2 is 3.18 g/cc. Calculate the unit cell edge length for CaF2 .
PROBLEM 668 Ar crystallizes in FCC arrangements and density of solid and liquid Ar are 1.59 and
1.42 g/cc respectively. Find percentage of empty space in liquid Ar.
PROBLEM 669 An atomic BCC is made-up of A type of atom and a guest atom B is inserted in the unit
cell of A. If radius of A is R, what could be the maximum radius of B without disturbing the unit cell
dimension of A. Also determine packing fraction of this solid.
PROBLEM 670 Na crystallizes in BCC arrangement with the interfacial separation between the atoms
at the edge 54 pm. Determine density of this solid.
PROBLEM 671 Polonium crystallizes in simple cubic crystal lattice. If the size of largest atom that can
be placed at the body centre of unit cell is 120 pico meter, determine density. M ( Po) = 209.
PROBLEM 672 Density of one form of CaCO 3 solid is 2.93 g/cc and it has orthorhombic unit cell with
unit cell parameter : a = 4.6 Å, b = 5.7 Å and c = 57 Å. Calculate number of Ca 2+ ions per unit cell.
PROBLEM 673 Determine packing fraction of NaCl solid in the following cases:
(a) Ions along an axis connecting opposite face centres are absent.
(b) Ions along an axis connecting opposite edge centres are absent.
(c) Ions along an edge diagonals are absent.
SURFACE CHEMISTRY
PROBLEM 674 In the adsorption of hydrogen gas over a sample of copper powder, 1.36 cm 3 of H 2 (g )
measured over S.T.P. was found to absorb per gram of copper powder. Considering only mono-layer
adsorption, determine specific surface area of copper. Density of liquid H 2 is 0.07 g/cc.
PROBLEM 675 A 0.016 M of an acid solution in benzene is dropped on a water surface, the benzene
evaporates and the acid forms a monomolecular film of solid type. What volume of the above solution
would be required to cover a 500 cm 2 surface area of water with monomolecular layer of acid? Area
covered by single acid molecule is 0.2 nm 2 .
81
Problems
PROBLEM 676 One gram of activated charcoal has a surface area of 1000 m 2 . Considering complete
coverage as well as monomolecular adsorption, how much ammonia in mL at S.T.P. would be adsorbed
on the surface of 25 g charcoal? Diameter of a ammonia molecules is 0.3 nm.
PROBLEM 677 Calculate surface area of a catalyst of which adsorb 100 cm 3 of N 2 (g ) at S.T.P. per
gram in order to form a monomolecular layer of N 2 ( g ). The effective surface area occupied by nitrogen
molecule on the surface of catalyst is 0.16 nm 2 .
PROBLEM 678 Discuss how extent of adsorption varies on
(a) Increasing temperature of system.
(b) Increasing surface area of adsorbent.
(c) Increasing the gas pressure.
PROBLEM 679 Effect produced on extent of adsorption by changing either temperature or pressure are
consistent with Le-Chatelier’s principle. Justify.
PROBLEM 680 Classify adsorption of gases on solids on the basis of force of interaction between
adsorbent and adsorbate. What are their main characteristics?
PROBLEM 681 What are surface active substances? Explain why the surface tension of a liquid is very
much modified in the presence of a surface active substance?
REACTION MECHANISM
PROBLEM 682 Rank the following sets of intermediates in increasing order of their stability giving
appropriate reasons for your choice.
(a) •• CH 2 , •• C(C 6 H 5 ) 2 , CH 3CH •• , C 6 H 5CH •• , O 2 NCH 2CH ••
(b) C 6 H 5+ , p-NO 2 (C 6 H 4 ) + , p-CH 3 —(C 6 H 4 ) + , p-Cl—C 6 H +4
+
+
+
(c)
+
+
(d)
O
O
+
O
O
O
O
O
O





−
−
−
−
(e) CH 2 —C—CH 3 , CH 2 —C—H, CH 2 —C—OCH 3 ,CH 2 —C—NH 2 , CH 2 —C—CH 2 —OCH 3 ,
−
82
Problems in Chemistry
PROBLEM 683 Reorder the following sets of compounds according to increasing pK a .
(a) 1-butyne, 1-butene, 1-butanol, H 2O
(b) 1-butanamine, aniline, butanamide
(c) Acetic acid, oxalic acid, 1,3-propandioic acid.
(d) Protonated forms of pyrrole, pyridine, N-methyl pyrrole.
(e) Phenol, cyclohexanol, cyclohexan carboxylic acid, p-nitrophenol.
PROBLEM 684 2-butenal has sufficient acidic character despite the fact that it lacks enolizable
hydrogen α-to carbonyl group. Explain?
PROBLEM 685 For the following compounds, arrange the labelled proton in increasing order of their
ease of deprotonation :
C≡≡C—H3
O
(b)
(a)
H2
CH2
O
H
(c)
C
H
N
1
H
1
H
H
2
H2C
H
H1
O
3
H
(d)
CH2
H
1
3
2
2
O
SO3H
3
CH2
O
O
PROBLEM 686 In the following pair of compounds, A is insoluble in water but soluble in dilute
sulphuric acid while B is insoluble in both water and dilute acid but soluble in aqueous solution of strong
base, explain :
O
O
O
NH2
A
NH2
B
PROBLEM 687 Draw all proton tautomers of each of the following structures and arrange them in the
increasing order of their enol content.
O
NH2
(a)
N
(b)
N
H
H
O
(c)
O
O
83
Problems
PROBLEM 688 K a 1 of phthalic acid is greater than K a 1 of isophthalic acid while reverse is true for
K a 2, explain.
PROBLEM 689 Benzene sulphonic acid is a stronger acid than benzoic acid, explain.
PROBLEM 690 Which is stronger acid and why-phenol or thiophenol?
PROBLEM 691 Compare the acid strength of benzoic acid with three isomeric hydroxy benzoic acid.
PROBLEM 692 Orthohydroxy benzoic acid is much stronger than ortho methoxy benzoic acid, justify.
PROBLEM 693 Which is a stronger acid, A or B and why?
COOH
COOH
NO2
NO2
A
B
PROBLEM 694 Which is a stronger base, A or B and why?
NH2
NH2
or
NO2
NO2
A
B
PROBLEM 695 Discuss the mechanism of following reactions :
(a)
H
(b) CH 3CH 2CH •2 → CH 3CH==CH 2 + CH 3CH 2CH 3
PROBLEM 696 Discuss the following observations :
(a) C—Cl bond in vinyl chloride is stronger than in chloroethane.
(b) Carbon-carbon bond length in ethene is shorter than in CH 2 == CHOCH 3 .
(c) CH 3SH is stronger acid than CH 3OH.
(d) CH 3CH 2 NH 2 is stronger base than CH 2 == CHNH 2 .
PROBLEM 697 Although trimethyl amine has greater electron density at nitrogen atom than methyl
amine, later is stronger Lewis-base in water than former. Explain.
84
Problems in Chemistry
PROBLEM 698 A cyclobutandicarboxylic acid exist in two stereo-isomeric forms in which one is polar
but non-resolvable while other is non-polar but resolvable into enantiomers. Deduce structures of all
these compounds .
PROBLEM 699 Which of the following rearrangement is thermodynamically favoured and why?
+
+
—CH2
A
+
CH2
+
B
PROBLEM 700 An alkyne is more reactive than alkene in catalytic hydrogenation reaction while
reverse is true for electrophilic addition reaction, justify.
PROBLEM 701 Unlike naphthalene, azulene have significant zwitter ionic character and dissolves in
aqueous acid. Provide a mechanistic explanation.
Azulene
Naphthalene
PROBLEM 702 Label the following 1-4 in order of decreasing acidity. Explain your answer with the
aid of mechanism:
OH
OH
OH
OH
NO2
B
PROBLEM 703
OCH3
CH3O
A
NO2
C
D
Discuss the basic strength of two nitrogens in benzimidazole.
N
N
H
Benzimidazole
85
Problems
PROBLEM 704 Number the following compounds in order of increasing acidity of indicated proton
giving mechanistic reasoning:
CH2—H
O—H
CH2—CH2—H
II
I
O—H
III
IV
PROBLEM 705 For each of the following acid-base reactions, predict the side of the equilibrium that
will be favoured. Briefly explain the choice by describing the features of the species involved that bias
the equilibrium to the chosen side:
O
O
+
H
+
NH + NH4
N
H
O
O
+ CH3—C
+
…(i)
+ NH3
N
+ CH3—C
N
+
…(ii)
NH
N
H
CF3
CF3
CF3
H
+
+
CF3
+
+
N
N
H
H
…(iii)
N
N
H
PROBLEM 706 2,6-di-t-butyl pyridine is more basic the pyridine but later is a stronger nucleophile
than former. Rationalize.
PROBLEM 707 With the help of structure, discuss the positions of Br  C  H and Cl  C  H
planes in 1-bromo-3-chloro-1,2-propadiene.
BrHC==C==CHCl
(1-bromo-3-chloro-1,2-propadiene)
PROBLEM 708 Taking into account hybridization and resonance effects, rank the following C  C
bonds in order of decreasing length:
I
III
II
IV
+
+
+
86
Problems in Chemistry
Which side is favoured at equilibrium, provide quantitative explanation:
PROBLEM 709
O
O
+ H2 O
O
O
O
O
–
+ HO
O
O
–
pKa = 13
pKa = 16
Which of the following reaction would proceed to the right? Explain reason for your
PROBLEM 710
choice?
+
NH3
(a)
+
NO2
NH2
s
NO2
+
H
pKa = 9.33
pKa = 10
+
NH3 +
(b)
pKa = 10.71
PROBLEM 711
acid?
NO2
NO2
NH2 +
s
H
For the following pairs of compounds, predict with clear reasoning, which is stronger
+
N—H or
(a)
—NO2 or HO—
(b) HO—
—NO2
N+
H
O
(c)
O
O
O
(d)
or
MeO
OMe
O
O
COOH
(e)
or
(f)
COOH
or
or
N+
N+
H
H
87
Problems
or
(g)
(h) PhCOOH or PhCO3H
OH
OH
(i)
or
NO2
NO2
PROBLEM 712
From the following pair, select the stronger acid providing clear reasoning:
OH
OH
(a) O N
2
COOH
or
COOH
or
(b)
COOH
NH3Cl
(c)
COOH
NH3Cl
or
PROBLEM 713 From the following pair, select the stronger base:
(a) p-methoxy aniline or p-cyanoaniline
(b) pyridine or pyrrole
(c) CH 3CN or CH 3CH 2 NH 2
CH3
N
O2N
CH3
NH2
O2N
NO2
(d)
NO2
or
NO2
NO2
PROBLEM 714 In each of the following pair of compounds, which is more basic in aqueous solution?
Give an explanation for your choice:
(a) CH 3 NH 2 or CF3 NH 2
NH

(b) CH 3CONH 2 or H 2 N
NH 2
88
Problems in Chemistry
(c) PhNH 2 or CH 3CN
(d) C 6 H 5 N(CH 3 ) 2 or 2,6-dimethyl-N-N-dimethylaniline
(e) m-nitroaniline or p-nitroaniline
PROBLEM 715
(a) Arrange the following in increasing order of acid-strength:
+
(i) H 2SO 4 (ii) CH 3CHCH 3 (iii) CH 3CH 2 O H 2 (iv) CH 3CH 2CH 3

SH
(b) What are the major components at equilibrium if 1.0 mol of each of the four components below
are mixed together?
+
(i) HSO –4 (ii) CH 3CHCH 3 (iii) CH 3CH 2 OH 2 (iv) CH 3CH 2CH 3

S–
PROBLEM 716
Which is the better site of protonation in the following compound and why?
(a)
PROBLEM 717
nitrogen?
l
l
—N
N
l
l
CH3
(b)
CH3
In the following structure, which is better site of protonation and why-oxygen or
O
N—
H+
PROBLEM 718 Rank the following from greatest (most) to smallest (least):
(a) Proton acidity:
O
(i)
H
H
H
;
;
;
C
B
A
H
D
O
O—H
(ii)
N—H
A
C
B
CH3
O—H
89
Problems
(b) Basic strength:
–
CH2
–
NH
O
O–
,
O–
C
,
D
C
B
A
C
,
,
–
–
O
,
F
E
(c) Heat of combustion:
A
B
D
C
H
H
E
(d) Rate of reaction with tertiary butanol:
HCl,
HBr ,
B
A
HI,
HF.
C
D
(e) Rate of reaction with HCl:
OH
OH
CH3OH,
OH
and
A
C
B
D
(f) Relative stability:
+
+
CH2
,
A
B
+
+
,
C
D
,
90
Problems in Chemistry
(g) Rate of reaction with NaCN:
Br
I
,
OMe
,
C
B
A
F
,
,
D
Cl
E
(h) Heat of hydrogenation
(i)
C
B
A
D
,
,
,
(ii)
B
A
C
,
E
D
(i) Rate of reaction with potassium t-butoxide:
A
F,
OTs ,
OMe
B
C
Cl
D
,
Br
E
91
Problems
(j) Rate of reaction in SN-2 reaction:
Br
O
,
Br ,
MeBr ,
B
Br
C
A
,
Br
D
E
PROBLEM 719 Draw the important resonance structures that contribute significantly to the
resonance hybrid of the following molecules. If there is more than one important resonance contributor,
indicate the major one.
CH3
CH3
N
O
(i)
B
(ii)
–
N
l l
l l
(iii)
O
PROBLEM 720
Compare the C  N bond-length in the following species:
H2 N
H2 N
O
(i)
PROBLEM 721
choice.
(ii)
(iii)
Rank the followings in increasing order of acidic strength:
H
+
PROBLEM 722
choice:
H2 N
giving reasons for your
H
H
N
N
I
II
N+
N
+
III
H
H
Rank the followings in increasing order of basic strength, explaining reason for your
H
H
NLi+
N
N
I
II
III
–
92
Problems in Chemistry
Arrange the following in increasing order of electrophilic strength:
PROBLEM 723
+
+
+
II
I
III
PROBLEM 724 In which direction would the following equilibria lie predominantly:
(i) NH 3 + OH –
NH 2 – + H 2O
(ii) CH 3 NH 2 + (CH 3 ) 3 NH +
CH 3 NH 3 + + (CH 3 ) 3 N
(iii) CH 3 NH 2 + H 2O
CH 3 NH 3+ + OH –
Lists the following groups of cations in increasing order of acidity explaining reasons
PROBLEM 725
for your order:
O
+
+
+
NH2
(a)
and why?
N
A
NH2
H
O
+
+
NH2
NH2
(b)
+
NH2
B
A
C
B
C
NH3+
+
(c)
Ph2NH2
PROBLEM 726
C
B
A
+
Ph3NH
Discuss the enhanced basic strength of the following base:
NH
N
N
N
H2 N
N
I
NH2
III
II
PROBLEM 727 For the following compounds, draw all the resonance structures in which every atom
has complete valence shell.
CH3
CH3
N
+
N—CH3
93
Problems
PROBLEM 728 Answer the following questions:
(i) Which of the indicated H is abstracted rapidly by bromine radical and why?
Ph
Hc
Hb
Ha
(ii) One of the indicated proton H a or H b , is approximately 10 30 times more acidic than other,
which is more acidic and why?
H
H
Ha
Hb
(iii) Which proton is more rapidly abstracted by ethoxide ion and why?
O
O
Br
Hb
Ha
(iv) Which compound is more rapidly attacked by a nucleophile and why?
O
O
OCH3
or
(b)
(a)
PRLOBLEM 729
Discuss the relative acid strength of the following substituted phenols:
—OCH3 HO—
HO—
(i)
PROBLEM 730
—Cl
—CHO
HO—
(iii)
(ii)
(iv)
The basic amino acid agrinine has the following structural formula:
NH
NH2
H2N
N
H
HO—
COOH
94
Problems in Chemistry
It has three pKa values. pKa1 = 2.18, pKa 2 = 9.09 and pKa 3 = 13.2. Give the formula of agrinine as it
exist in aqueous solution at pH: (a) below 2.18, (b) between 2.18 − 9.09, (c) between 9.09 − 13.2.
PROBLEM 731
Discuss the relative acidity of the following compounds:
H
NH2
NH2
N
and
(b)
and
(a)
N
O
NO2
H
PROBLEM 732
Amino acid shown below has three pKa values:
O
pKa1 = 1.94
H2 N
OH
NH2
pKa2 = 8.65
pKa3 = 10.76
Give formula of above amino acid in aqueous solution when pH is:
1.9–8.7 (c) between 8.7–10.8 (d) above 10.8.
PROBLEM 733
for your order:
(a) below 1.9 (b) between
Rank the following acids in increasing order of their pKa value, explaining reasons
COOH
COOH
COOH
H2N
(1)
(2)
COOH
NO2
(3)
NO2
(4)
PROBLEM 734
The γ-proton of crotonaldehyde is sufficiently acidic, justify.
PROBLEM 735 The two ammonium ions shown below differ in their pKa value by more than four
units, explain the difference.
+
—CH3
H3N—
A
+
—CH2—NH2—CH3
and
B
95
Problems
STEREOCHEMISTRY
PROBLEM 736 How many stereoisomers are possible for 1,3-dichlorocyclopentane. Draw structural
formulae and discuss the stereochemical relationship among them.
PROBLEM 737 Draw structures of all stereoisomers of 1-bromo-2- chlorocyclopentane and discuss the
stereochemical relationship among them.
PROBLEM 738 For each of the following pair, deduce the stereochemical relationship, i.e., whether
they are enantiomers, diastereomers or identical.
Br
Br
(a)
C
Et
H
CH3
and
C
H
Et
CH3
H
H
CH3
OH
(b) CH3 C—C
and CH3 C—C CH
H
3
OH
OH
OH
H
H
CH
(c) 3
Cl
Cl
Cl
and
C2 H 5
H——C2H5
H——Cl
H
(d)
CH3
H
CH3
H
C==C
C
CH3 and
H Cl
CH3
H
Cl
C==C
C
H
H CH3
PROBLEM 739 (+) 2-butanol has specific rotation of +13.9° when measured in pure form. A sample of
2-butanol was found to have an optical rotation of –3°. What is the stereomeric composition of this
mixture?
PROBLEM 740 (a) Draw structures of all stereoisomers of C 2 HClBrF.
(b) Draw structures of all stereoisomers of 1,3,5-trichloro-1,4-pentadiene and discuss the
stereochemical relationship among them.
(c) How may isomers (including stereoisomers) can be obtained on monochlorination of 2-methyl
butane? Draw structures of all isomers.
PROBLEM 741 A hydrocarbon A(C 6 H 8 ) is resolvable and it decolourise brown colour of bromine
water. A, on treatment with H 2 /Pt yields C 6 H14 ( B ) which can’t be resolved into enantiomers. Deduce
structures of A and B.
PROBLEM 742 An organic compound A has molecular formula C10 H17 Br and it can be resolved into
enantiomers. Also, A decolourises cold, dilute and alkaline solution of KMnO 4 . A on treatment with
alcoholic solution of KOH yields two organic compounds, B and C both have molecular formula C10 H16 .
Only B is resolvable but C is non-resolvable. Reducing either B or C yields the same
4-methyl-1-isopropylcyclohexane. Deduce structures of A to C.
96
Problems in Chemistry
PROBLEM 743 An organic compound A has molecular formula C 4 H10O and it can be resolved into
enantiomers. A on treatment with chromic acid yields B (C 4 H 8O) which is non-resolvable. B on
treatment with CH 3 NH 2 yields C (C 5 H11 N) which exhibit stereoisomerism. C on treatment with H 2 | Pt
yields D(C 5 H13 N) which is a racemate. Deduce structures of A to D.
PROBLEM 744 Bromination of cis-3,4-dimethyl-3-hexene yields a racemic mixture of vicinal
dibromide while catalytic hydrogenation of the same alkene yields a meso compound, explain.
PROBLEM 745 An optically active organic compound A(C 8 H13Cl) does not decolourise bromine water
solution. A on treatment with alcoholic KOH can produce two products in principle but infact only one
product B (C 8 H12 ) is obtained. B on treatment with ozone followed by work-up with H 2O 2 yields
C (C 8 H12O 4 ) which cannot be resolved into enantiomers. C on heating with NaOH/CaO yields
D(C 6 H12 ) which on monochlorination yields C 6 H11Cl as single isomer. Deduce structures of A to D.
PROBLEM 746 An optically active hydrocarbon A has molecular formula C 6 H12 . A on treatment with
H 2 / Pt gives inactive B. B on monochlorination yield four products C, D, E and F, all having molecular
formula C 6 H13Cl. C is enantiomeric while D is diastereomeric. Both E and F are non-resolvable.
Treatment of E with alcoholic solution of KOH yields two possible alkene G and H while similar
treatment with F gives only H. Deduce structures of A to H.
PROBLEM 747 For each of the following molecules, indicate, whether they are chiral, achiral or
meso compound:
CH3
H
HO
A
B
CH3
OH
CH3
H
CH3
H
H
H3C
H
H
Br
E
CH3
H
H
CH3
I
H
G
F
CH3
H
H3C
CH3
Br
H
D
C
CH3
CH3
H
PROBLEM 748 For each of the following pair of structures, indicate, if the compounds are identical,
constitutional isomers, enantiomers, distereomers, different.
OH
(a)
OH
and
HO
OH
H
OH
H
H
(b)
H
CH 3
H 3C
OH
H
and
CH 3
OH
HO
H
CH 3
97
Problems
PROBLEM 749 Discuss the optical activity of the following two compounds and also label them as
polar and non-polar.
Cl Cl
Cl H
H
H
H
Cl
II
I
PROBLEM 750
Discuss the origin of optical activity in 1,3-dichloropropdiene.
PROBLEM 751 From the following four structures select: (a) The optically active isomers,
(b) Optically inactive isomers, (c) Enantiomer pairs, (d) Distereomer pairs.
PROBLEM 752
Cl
Cl
Cl
Cl
I
II
III
IV
Consider the following six structures:
OH
H
N
CH3
CH3
HO
O
I
OH
H
N
CH3
HO
O
CH3
II
OH
H
N
HO
O
CH3
CH3
III
98
Problems in Chemistry
OH
H
N
CH3
HO
O
CH3
IV
OH
H
N
CH3
CH3
HO
O
V
OH
H
N
HO
CH3
O
CH3
VI
Establish the stereochemical relationship between:
and V, (e) IV and VI.
PROBLEM 753
molecule :
(a) I and II, (b) III and IV, (c) II and III, (d) I
Draw structures of all the stereoisomers produced on addition of HCl to the following
CH3
HCl
PROBLEM 754 Draw the energy vs dihedral angle diagram for rotation around C 2  C 3 bond of
2,3-dimethyl butane.
PROBLEM 755 For the following two molecules, draw structures of one set of functional, one set of
positional and one set of stereoisomers.
C 5 H 6O 2
I
and C 5 H11 NO
II
PROBLEM 756 How many stereomers are formed upon hydration of the following compound (in
kinetic condition) :
H3C
CH3
99
Problems
PROBLEM 757 Indicate the stereo centres in the following molecule and total number of stereomers
in the following molecule. Also draw the structures of pair of distereomers.
H
N
O
O
O
I
PROBLEM 758 Draw the structures of stable configuration obtained after acidic hydration of the
following unsaturated compounds: (exclude rearranged products)
CH3
I
CH3
CH3
CH3
III
II
IV
PROBLEM 759 N-methylethenamine as such does not show any stereoisomerism but one of its
resonance form exhibit stereoisomerism. Explain.
PROBLEM 760 Draw the Newman projection formula of important conformers around C 2  C 3
bond of 2,3-dimethylpentane and label them according to the increasing order of stability.
PROBLEM 761
Select the compound which is optically active. Draw the structure of stereoisomers:
CH2CH3
O
(b)
(a)
(c)
(d)
OH
OH
PROBLEM 762 From the following set of compounds, select: (a) enantiomer
(b) distereomer pairs in which both are optically active, (c) label them as D or L sugar.
pairs,
100
Problems in Chemistry
CHO
HO
HOH 2C
H
H
H
OH
OH
HO
CH 2OH
H
H
H
H
HO
HO
HO
H
CH 2OH
H
OH
HO
H
H
H
OH
OH
OH
CHO
CHO
II
I
III
CHO
CHO
OH
H
H
H
HO
HO
CH 2OH
H
H
HO
HO
HO
HO
H
H
H
CH 2OH
OH
IV
V
How many stereoisomers are possible for the molecule shown below–draw their
PROBLEM 763
structures:
O
COOH
OH
Draw resonance structure of the amides shown below and select them which are
PROBLEM 764
stereoisomeric:
O
(a)
(b)
H2 N
H
CH3
(c)
N
H
N
H
PROBLEM 765
Discuss the type of isomerism exhibited by the following pairs:
N
N
N
N
(a)
N
CH3
O
(d)
O
O
and
101
Problems
CH3
(b)
CH3
and
OH
OH
CH3
CH3
(c)
and
O
O
OH
OH
OH
OH
O
CH3
and
(d)
OH
H3C
PROBLEM 766 For the following enol tautomers, write structures of corresponding keto tautomers
and other enol tautomers if possible and label them as chiral/achiral.
OH
OH
I
II
H
(a)
I
F
(b)
F
Cl
F
Cl
CH3
H
II
I
Br
C2H5
H3C
Cl
I
HO
(c)
III
Establish the stereochemical relationship in the given sets of compounds:
PROBLEM 767
H3C
OH
I
H3C
C2H5
Br
II
COOH
OHC
HO
I
H
F
CHO
H
H
Cl
H
HO
HO
II
COOH
102
Problems in Chemistry
H
H 3C
CH3
CH3
OH
H
H3C
OH
H3C
(d)
H
OH
H
CH3
H
OH
I
OH
OH
H
II
III
PROBLEM 768 Both of the quaternary ammonium salts shown below are resolvable into enantiomers
but one when dissolved in water racemizes. Identify that one and explain the reason for its racemization
in aqueous solution.
C6H5
H
H3C
N+
CH3CH2
N+
CH2CH
H3C
CH2
I
PROBLEM 769
II
CH
C2H5
CH2
Discuss the optical isomerism of the following molecule:
O
O
H3CH2C
PROBLEM 770 Both cis and trans isomers of 3-methoxycyclohexanol are chiral. However, one
distereomer reacts with a base and then with methyl iodide to give an optically inactive compound while
the other distereomer gives a racemic mixture of optically active compound. Explain.
PROBLEM 771
conformers:
Indicate, whether each of the following structures is isomers, resonance structures,
Cl
O
OH
and
(a)
(b)
N
N
and
Cl
H
(c)
B
O
H
and
sB
O
+
O
(d)
O
and
H
Problems
103
HYDROCARBONS
PROBLEM 772 A hydrocarbon A has molecular formula C10 H10 and decolourise brown colour of
bromine water. A on treatment with HgSO 4 / H 2SO 4 produces two isomeric compounds B and C, both
having molecular formula C10 H12O. Compound B forms an yellow coloured precipitate on treatment
with alkaline solution of I 2 while C does not. Also A on treatment with O 2 followed by work-up with
H 2O 2 yields D(C 8 H 8O 2 ) as one of the product. Heating D with soda lime (NaOH/CaO) yields E (C 7 H 8 ).
E on treatment with chlorine in presence of sunlight yields F (C 7 H 7Cl) as single isomer while E on
treatment with Cl 2 in dark but in presence of AlCl 3 produces two isomeric compounds with their
molecular formula C 7 H 7Cl. Deduce structures of A to F.
PROBLEM 773 An organic compound A has molecular formula C 8 H12O and it evolve a colourless gas
on treatment with Na-metal. Also A forms a white precipitate with Tollen’s reagent. A on treatment with
HgSO 4 / H 2SO 4 yields B (C 8 H14O 2 ) which of reduction with NaBH 4 yields a resolvable compound
C (C 8 H16O 2 ). A on boiling with aqueous KMnO 4 gives D(C 7 H12O 3 ). D can also be obtained by
treatment of aqueous KCN with cyclohexanone followed by hydrolysing the product with dilute H 2SO 4 .
Deduce structures of A to D.
PROBLEM 774 A hydrocarbon A has molecular formula C10 H10 and decolourises brown colour of
bromine water. A on treatment with ozone followed by work-up with dimethyl sulphide (CH 3 ) 2 S yields
B (C10 H10O 2 ). B gives positive iodoform test as well as positive Tollen’s test. A on boiling with acidic
KMnO 4 yield C (C 8 H 6O 4 ) as one of the product. C on heating with P2O 5 dehydrate to yield D(C 8 H 4O 3 ).
C does not decolourise brown colour of bromine water. Identify A to D.
PROBLEM 775 A hydrocarbon A has molecular formula C11 H18 and it decolourise bromine water
solution. A has a chiral centre and structure of four stereo-isomers can be drawn for A. A on treatment
with H 2 / Pt yields B (C11 H 22 ) which contain two chiral centre. A on treatment with O 3 followed by
work-up with Zn – H 2O yields the following compound and ethanol. Deduce structures of A and B and
O
O


draw structures of all stereo-isomers of A. CH 3CH 2 —CH—CH 2 —C—CH 2 —C—CH 3 .

CHO
PROBLEM 776 A hydrocarbon A has molecular formula C10 H18 and it decolourise purple colour of
cold, dilute and alkaline permanganate solution. A on treatment with HBr yields B (C10 H19 Br) which
can’t be resolved into enantiomers while A on treating with HBr in presence of a peroxide yields, C an
isomer of B, but C can be resolved into enantiomers. Also A on treatment with acidic KMnO 4 produces
D(C 7 H12O 2 ) as one of the product which on heating with NaOH/CaO yields E (C 6 H12 ). E on treating
with Cl 2 ( g ) in presence of sunlight yields C 6 H11Cl as single mono-chloro derivative. Deduce structures
of A to E.
PROBLEM 777 An organic compound A has molecular formula C13 H10O. A decolourises Bayer’s
solution but does not evolve any gas on treatment with Na-metal. A on controlled hydrogenation yields
B (C13 H14O) which does not decolourise brown colour of bromine water solution. A on treatment with O 3
followed by work-up with H 2O 2 yields C (C 8 H 8O 2 ) and D(C 5 H 4O 3 ). Heating C with NaOH/CaO yields
104
Problems in Chemistry
E (C 7 H 8 ) while similar treatment on D yields furan. E on heating with Br 2 yields F (C 7 H 7 Br) F on
treatment with aqueous solution of KCN followed by hydrolysis of product with dilute H 2SO 4 yields C.
Deduce structures of A to F.
PROBLEM 778 An optical active organic compound A has molecular formula C13 H18O and decolourise
brown colour of bromine water solution but does not give any gas on heating with sodium metal. A on
hydroboration-oxidation yields B (C13 H 20O 2 ) which is still resolvable. B on refluxing with dilute
solution of H 2SO 4 yields C (C 7 H 8O) and D(C 6 H14O 2 ) neither of them can be resolved into enantiomers.
C on treating with chromic acid solution gives benzoic acid while D on heating with concentrated
solution of H 2SO 4 yields E (C 6 H12O). E does not evolve any gas on heating with sodium metal. Deduce
structures of A to E.
PROBLEM 779 Propose mechanism of the following reactions :
Br
CH2
(a)
Br
+ HBr
(b)
OH
+
H2SO4
O
CCl3
(c)
+ CHCl3
Peroxide
OH
CH3
H2SO4
(d)
CH3
N
(e)
CH3CH2—C
CH2 + CH2N2
CH3
N
H3C
CH2CH3
PROBLEM 780 Write major product in the following reactions :
(a) F3C—CH==CH 2 + HI →
Peroxide
(b) C 6 H 5 —CH==CHCH 3 + HBr →
105
Problems
CH 3

Peroxide
(c) CH 3CH==C—CH 3 + CHBr 3 →
H 2O 2
(d) C 6 H 5 —CH==CH 2 + B2 H 6 →
NaOH
HCl
50°C
(e) CH 3 —CH==CH—CH==CH 2 →
PROBLEM 781 Select one from the following pair of isomer that has higher heat of combustion,
justifying your choice :
(a)
and
(c)
(b)
or
(d)
and
or
PROBLEM 782 Select one from the following pair of isomers, that has higher heat of hydrogenation,
justifying your choice.
(a)
or
or
(b)
or
(c)
(d)
or
(e)
or
(f)
or
106
Problems in Chemistry
PROBLEM 783 Propose mechanism of the following reactions :
(a)
+ CHCl3
C2H5OK
C2H5OH
Cl
(b)
N
+ CHCl3
Cl
Cl
C2H5OK
C2H5OH
N
H
PROBLEM 784 Write mechanism of chlorination of an alkane using sulphuryl chloride (SO2Cl2) and a
free radical initiator (R2O2).
PROBLEM 785 An organic compound A has molecular formula C8H18 which on monobromination
produced three isomeric products B, C and D. B and C are enantiomeric whereas D is achiral. Treatment
of either C or D with alcoholic KOH solution produced the same product E while B on similar treatment
produced F(C8H16), an isomer of E as major product. Establish structures of A to F.
PROBLEM 786 A hydrocarbon A(C8H16) does not decolourise Br2-water. A on monochlorination
produced four isomeric products B, C, D and E, among which only C and D are enantiomeric. Compound
B is resistant to dehydrohalogenation reaction. Treatment of C with alcoholic solution of KOH produced
F(C8H14) as only product while similar treatment on E produced G, an isomer of F. D on treatment with
alcoholic solution of KOH produces both F and G in comparable amount. Deduce structures of A to G.
PROBLEM 787 A hydrocarbon A(C8H16) does not decolourise Br2 water. A on monochlorination
produced four isomeric products B, C, D and E among which only D is enantiomeric. Compound B is
resistant to dehydrohalogenation reaction. Treatment of either C or D with alcoholic solution of KOH
produced the same alkene F(C8H14) as major product which does not rotate the plane polarized light
while similar treatment on E produced G, an isomer of F which is enantiomeric. Deduce structures of A
to G.
PROBLEM 788 Predict major addition products in the following reactions:
(a) (CH 3 ) 3 CCH==CH 2 + ICl →
(b) C 6 H 5  CH==CHCH 3 + BrCl →
CH3

(c) C 6 H 5  CH==CHCH 3 + IBr →
(d) C 6 H 5  C==C(C 6 H 5 ) 2 + IBr →
PROBLEM 789 Propose mechanism of formation of the indicated products in the following reactions:
Cl
(a)
+ HCl
107
Problems
(b)
+ HCl
Cl
PROBLEM 790 Show how would you convert 1-methylcyclopentanol to 2-methylpentanol.
PROBLEM 791 Predict major products of the following reactions:
(a) Propene + BH 3 /THF →
(b) The product of part (a) + H 2O 2 / NaOH →
H O /NaOH
2 2
(c) 2-methyl-2-pentene + BH 3 /THF →
H2O2 /NaOH
(d) 1-methylcyclohexene + BH 3 /THF →
PROBLEM 792 Show how would you accomplish the following conversion:
(a) 1-butene → 1-butanol
(b) 1-butene → 2-butanol
(c) 2-bromo-2,4-dimethylpentane → 2,4-dimethyl-3-pentanol.
PROBLEM 793 A hydrocarbon A(C6H10) on reduction first gives B(C6H12) and finally C(C6H14). A on
ozonolysis followed by work-up with Zn-H2O gives two molecules of aldehydes C2H4O(D) and one
molecule of aldehyde E(C2H2O). Oxidation of B with acidified KMnO4 gives an acid F(C4H8O2).
Determine structures of A to F with proper reasoning.
PROBLEM 794 A hydrocarbon exist in two stereomeric forms (A) and (B) with their molecular formula
C8H16. A on treatment with cold, dilute and alkaline solution of KMnO4 produces C8H18O2(C) which is a
meso form. B, on the other hand, on similar treatment produces racemic mixture which is isomeric to C.
Either A or B on treatment with O3 followed by work with H2O2 produces D(C4H8O) as the only product.
Identify A to D.
PROBLEM 795 An organic compound A(C13H23Cl) exist as diastereomers and decolourise bromine
water. A on treatment with ethanolic solution of KOH produces isomeric B and C with their molecular
formula C13H22. Treatment of either B or C with Rany Nickel produces 4-isopropyl-1-tertiarybutyl
cyclohexane. A on oxidative ozonolysis gives acetone as one product. Identify A, B and C considering C
to be enantiomeric.
PROBLEM 796 Bring about the following conversions:
(a)
OH
OH
(b)
(c)
COOH
108
Problems in Chemistry
Ph
Ph
(d)
H
(e)
H
D
(f)
Br
(g)
Br
O
O
(h)
BrCH2
O
CH2Br
PROBLEM 797 (a) Propose mechanism of formation of products in the following reaction:
HO
H3O
OH
+
+
+
O
OH
OH
(b) Arrange the following alkenes in increasing order of their heat of hydrogenation explaining
reason for your choice:
109
Problems
(II)
(I)
(IV)
(III)
(V)
PROBLEM 798 Draw the structures of compounds A to C:
(C 2 H 5 ) 3 N •• + •• CCl 2 → A (an unstable adduct)
A → B + C 2 H 4
HO
2
B →
C
PROBLEM 799 Predict mechanism of formation of products in the following reactions:
Ph
(a)
Ph
+ HCl
Cl
(b)
+ HCl
O
O
Cl
Cl

(c) C 6 H 5  CH==CH  CH 3 + HOCl → C 6 H 5  C H  CH  CH 3

OH
PROBLEM 800 An optically active hydrocarbon A has molecular formula C8H18. A on
monochlorination gives five alkyl halide B to F with their molecular formula C8H17Cl. B does not
undergo dehydrohalogenation on treatment with alcoholic solution of KOH. Treatment of either C or D
with alcoholic KOH yields same alkene G(C8H16) which on ozonolysis followed by work-up with
Zn-dimethyl sulphide gives an optically inactive compound C6H12O and ethanal. Also C is enantiomeric
whereas D is diastereomeric. E on dehydrohalogenation yields an alkene, which on reductive ozonolysis
yields H(C7H14O) which is optically inactive. H on treatment with LiAlH4 yields I(C7H16O) which can be
resolved into enantiomers. F on dehydrohalogenation yields an alkene (C8H16) which on reductive
ozonolysis yields J(C7H14O) which is optically active and have same configuration as that of A. Identify
A to J explaining the reactions involved.
PROBLEM 801 A hydrocarbon (A) contains five carbon atoms, reacts with cold, dilute and alkaline
solution of KMnO4 to form B(resolvable). B on oxidation with hot concentrated KMnO4 solution forms
two compounds (C) and a neutral (D). C has molecular formula C2H4O2 and turns blue litmus paper red.
Compound D neither reacts with Fehling solution nor with Tollen’s reagent and its formula is C3H6O. A
on treatment with Cl2 in CCl4 yields another compound E(C5H10Cl2). E on treatment with alcoholic
solution of KOH yields a stable compound F(C5H8). Identify A to F.
110
Problems in Chemistry
PROBLEM 802 An organic compound A(C8H14) on treatment with H2/Pt gives C8H18. Ozonolysis of A
followed by work-up with Zn-dimethyl sulphide yields three products B, C and D. B reduces Tollen’s
reagent and gives a yellow precipitate on treatment with NaOH/I2. C doesn’t reduces Tollen’s reagent
but gives yellow precipitate with NaOH/I2. D has molecular formula C2H2O2 and on treatment with
concentrated solution of NaOH followed by acidification of product yields E(C2H4O3). Deduce
structures of A to E.
PROBLEM 803 An organic compound (A) has molecular formula C5H10, and decolourise
bromine-water solution. When A is treated with cold concentrated solution of sulphuric acid followed by
refluxing of product, B(C5H12O) was formed. B turns the orange colour of chromic acid solution to
blue-green, converting itself into C(C5H10O). B and C both reacts with alkaline solution of iodine
producing yellow precipitate and salt of isobutyric acid. Identify A to C.
PROBLEM 804 A hydrocarbon A(C8H12) is optically active and on treatment with H2/Pt gives (C8H18)
which does not rotate plane polarized light. A does not give any precipitate on treatment with ammonical
silver nitrate solution. A on treatment with Pd/BaSO4 gives C(C8H14) which is also optically inactive but
A on treatment with Na/liq NH3 gives D, an isomer of C, which is able to show enantiomerism. Also A on
treatment with ozone followed by hydrolysis gives C4H6O3(E) which is optically active. Identify A to E
representing their stereochemical structure.
PROBLEM 805 A hydrocarbon A has molecular formula C10H18. A on treatment with H2/Pt gives
B(C10H22). B on monochlorination gives two isomer C and D with molecular formula C10H21Cl. Out of C
and D, only C can undergo dehydrohalogenation with alcoholic solution of KOH as well as only C can be
resolved into enantiomers. A on addition of one equivalent of Cl2 in CCl4 gives a stereomer E(C10H18Cl2).
E on treatment with Bayer’s reagent yields a racemic mixture of molecular formula (C10H20Cl2O2). Also
A on partial hydrogenation with Pd/BaSO4/quinone gives F(C10H20) which on treatment with Bayer’s
reagent gives a meso compound G(C10H22O2). Deduce the structures of A to G.
PROBLEM 806 A hydrocarbon A(C10H12) has no chiral carbon. A gives a white precipitate with
ammonical solution of silver nitrate. A on treatment with H2/Pt gives B(C10H20). A on ozonolysis gives
C(C8H12O4) as one product which on heating with soda lime gives D(C6H12). D on monochlorination
with Cl2/hν gives C6H11Cl as sole isomer. Identify A to D.
PROBLEM 807 An organic compound A(C8H6) gives brick red precipitate with ammonical solution of
Cu2Cl2 . A on treatment with acidic solution of HgSO4 yields B(C8H8O). B gives positive iodoform test. B
can also be prepared by the reaction of benzene with acetic anhydride in presence of AlCl3. Identify A
and B.
PROBLEM 808 Potassium hydroxide is mixed with 2,3-dibromohexane, the mixture is sealed in a
fusion tube and heated to 200oC for one hour. The product mixture (A) is mixed with Cu(I)-ammonia
complex and a precipitate forms. The precipitate (B) and liquid phase (C) are separated. The precipitate
is acidified and the product (D) is distilled at 71oC. Product D is treated with NaNH2, followed by
acetone and then with dilute acid to give alcohol (F). The liquid phase (C) is distilled and the product
alkyne is treated with NaNH2 at 150oC for 1.0 hour and product mixture is distilled to give a pure alkyne
(E) of boiling point 71 oC. Identify A to F.
111
Problems
PROBLEM 809 Propose structures for A to L:
1-bromobutane
KOH/H2O
Mg/Ether
(K)
Pentanal
H3O+
(E)
H2SO4 /heat
Na/heat
CH3Br
( J)
(C)
O3 /(CH3)2S
(F)
Butanal + Pentanal
Br2
(butanone/H+)
(A)
(B)
(D)
(G)
(l)
NaNH2 HBr Peroxide
(i) NaNH2/heat
1-nonyne
(ii) H2O
HBr (two mole)
fused KOH
(H)
mixture
(L)
PROBLEM 810 An optically active compound A has molecular formula C7H11Br. A reacts with HBr, in
absence of peroxide to yield isomeric products, B and C with molecular formula C7H12Br2. Compound B
is optically active; C is not. Treating A with one mole of potassium-t-butoxide yields D(C7H10).
Subjecting one mole of D to ozonolysis followed by treatment with Zn and water yield two moles of
formaldehyde and one mole of 1,3-cyclopentandione. A on addition of HBr in presence of H2O2
produces another isomer of B which is E, and shows diastereomerism. Deduce structure of A to E.
PROBLEM 811 An alkane (A) with formula C6H14 reacts with Cl2 to yield three compounds with
formula C6H13Cl, B, C and D of these only C and D undergo dehydrohalogenation with C2H5ONa/EtOH
to produce alkene (E). Also C can be resolved into enantiomers but D is non-resolvable. E on addition of
HCl produces ‘B’ which on treatment with Zn-acetic acid produces A. Deduce structure of A to E.
PROBLEM 812 Complete the following: (only major product)
(a) Cyclohexene + CHCl 3 (50% NaOH/ H 2O) →
OH
CHBr2
(b)
+ CH2I2
Zn/CuCl
+ 50% NaOH + H O
2
(c)
PROBLEM 813 An organic compound A has molecular formula C10H16 and is known not to contain any
triple bond. On catalytic hydrogenation of A, a new compound B (C10H22) is formed. Ozonolysis of A
followed by treatment with zinc-H2O yields two mole of CH2O, one mole of acetone and a third
compound C(C5H6O3). Deduce the structure of A,B and C considering C being optically active.
PROBLEM 814 Identify A to E (Provide structures)
C10H16Br2(A)
C10H15Br(C)
Br2/CCl4
E(C10H18O2)
H3O+
C10H16O(D)
PhCO3H
C10H16
H2SO4/∆
Br2/H2O
C10H17OBr(B)
O
O3/Zn-H2O
O
112
Problems in Chemistry
PROBLEM 815 An organic compound A (C10H16) on catalytic hydrogenation produces
1-isopropyl-4-methylcyclohexane. Ozonolysis of A followed by work-up with dimethylsulphide
produces formaldehyde and
O
O


CH 3  C  CH  CH 2  CH 2  C  CH 3

CH2CHO
Deduce structure of A.
PROBLEM 816 Three isomeric hydrocarbons A, B and C have molecular formula C6H10. All three
compounds readily decolourise bromine in CCl4, compound A gives precipitate with AgNO3/NH3
solution but compound B and C donot. Compound A and B both yield hexane when treated with excess of
H2/Pt. Under these conditions, C absorbs only one molar equivalent of H2 and give product with formula
C6H12. When A is oxidised with hot basic KMnO4 and resulting solution is acidified, the only organic
product that can be isolated is CH3(CH2)3COOH. Similar oxidation of B gives only CH3CH2COOH and
similar treatment of C gives only HO2C(CH2)4COOH. Identify A, B and C.
PROBLEM 817 Propose mechanism:
OH
(a)
H2SO4
Heat
Br
(b)
Br2, H2O
Br
NaCl
Br
+
OH
Br
+
Cl
PROBLEM 818 An organic compound A has molecular formula C9H14 and on treatment with H2/Pt
gives C9H20. A exist in four stereomeric forms. A on treatment with O3 followed by work-up with Zn-H2O
produced CO2, B (C4H6O3) and C (C4H8O). B rotates plane-polarized light but C doesn’t. C changes the
colour of acidic dichromate solution converting itself into an another optically inactive compound D
(C4H8O2). D on treatment with I2/Red-P produced an optically inactive compound E(C4H7O2I) as the
only isomer. Deduce structures of A to E and draw the structures of all four stereomers of A.
PROBLEM 819 An optically active compound A has molecular formula C9H10. A does not decolourise
aqueous solution of Br2. A on treatment with NBS produced another optically active compound B
(C9H9Br). B on hydrolyzing in aqueous KOH solution gave C (C9H10O) which doesn’t change the colour
of acidic dichromate solution. B on treatment with C2H5ONa/C2H5OH produced D (C9H8) which is
optically inactive. D on oxidative cleavage produced E (C9H8O3) which is optically inactive. E on
heating with soda lime gave C8H8O, which can also be obtained by Friedel Craft’s acylation. Deduce the
structures of A to E describing the reactions involved in each step.
PROBLEM 820 (a) Reagent (two possible isomers) + KMnO 4 / H + → 3,4-diketo hexandioic acid.
Provide structures of the reagent which will lead to the above product.
113
Problems
(b) Propose mechanism:
H
(i)
+
Br
Bu-Li
(ii)
C
Cl
CHCl3/(CH3)3COK
(iii)
N
N
H
PROBLEM 821 (a) An organic compound A(C12H20) on catalytic hydrogenation yields B(C12H24). A on
ozonolysis followed by work-up with H2O2 yields acetone and the following compound. Deduce the
structures of A and B.
O
O
O
OH
HO
(b) Predict product in the following sequence of reaction:
EtOOC
LiAlH4
COOEt
X + 2 EtOH
Conc. H2SO4
NaI
Z
NBS
Y
PROBLEM 822 An organic compound A(C13H21Br) is optically active and decolourise Bayer’s reagent.
A on treatment with EtONa/EtOH yields two and only two isomers B and C both having molecular
formula C13H20 with B as major product. Catalytic hydrogenation of either B or C gave the same product
C13H24 which is optically active. B is stereomeric whereas C is non-stereomeric. C on oxidative cleavage
with acidic KMnO4 produced D(C12H18O5) as one product which on simple heating gave E(C11H18O3). E
gives yellow precipitate with NaOH/I2. E on treatment with NaBH4 gives F(C11H20O3), an optically
active compound. F on heating with catalytic amount of H+ gave the following compound:
O
O
Deduce structure of A to F.
114
Problems in Chemistry
PROBLEM 823 Write structures of single major product of each of the following reactions. If two or
more major products are formed, write structures of both. Show stereochemistry of products where
appropriate.
H
Br
O
(a)
CH3ONa
CH3
CH3OH/E2
CH3
H
(b)
H
H3 C
Br2
H 2O
H
CH 3

(d) CH 3 —C== CH—CH 3 + CH 3CO 3 H →
(i) BH3
(c)
(ii) H2O2/NaOH
50% H SO
2
4
(e) HO—CH 2CH 2 —CH==CH 2 →
Cold, dilute KMnO4
(f)
NaOH
H 2O
CCl4
+ Br2
(g)
Cl2
+ CH3OH
(h)
PROBLEM 824 Hydroboration oxidation of trans-2-( p-anisyl)-2-butene yielded an alcohol A whose
melting point is 60°C while similar treatment on cis isomer yielded an isomeric alcohol which is a liquid
at room temperature. Suggest resonable structures for A and B.
PROBLEM 825 An optically active alkyne A contain 89.25% C and 10.48% H. After hydrogenation
with H 2 /Pt, it is converted into 1-methyl-4-propyl cyclohexane. A gives no gas with CH 3 MgBr.
Treatment of A with acidified KMnO 4 solution yields B which gives positive iodoform test and a salt C.
Acidification of salt C yields an optically active compound D. Identify A to D.
Discuss the stereochemistry of addition of Br 2 to the following alkenes:
PROBLEM 826
H
(a)
Br
H
H
C
(b)
C
C
H3C
H
C
C
H
CH3
Br
C
CH3
H
CH3
Br
H
(c) H C
3
Br
C
C
C
C
H
Br
H
H
H
H
CH3
(d) Br
H
C
C
C
CH3
C
H
CH3
115
Problems
PROBLEM 827
(a)
A
Provide the structures of missing components:
CH3
HO
Br2
H 2O
HgSO4
(b)
H2SO4/H2O
B
Br
OH
Br CH ONa
3
(c)
Excess
H2SO4
C (d)
∆
OH
Br
(e)
H3C
Ph
PROBLEM 828
H
CH3
B2H 6
H2O2/NaOH
E
Propose mechanism of the following reaction:
Br2
OH
O
CCl4
Br
H
PROBLEM 829
Bring about the following transformations:
OH
CH3
(a)
(b)
Br
OMe
OH
Br
(c)
COOH
COOH
D
116
Problems in Chemistry
PROBLEM 830
(a)
Complete the following reactions:
C2H5ONa
NBS
hν
CH3
Bu3SnH
(b)
C2H5OH
AIBN
Br
[AIBN : Azobis isobutyronitrile]
PROBLEM 831
Discuss the relative heat of hydrogenation of following alkenes:
PROBLEM 832
C
B
A
Propose mechanism of the following acid catalyzed reaction:
H2SO4
H2SO4
OH
I
PROBLEM 833
II
Complete the following reactions:
OMe
Br
(a)
tBuOK
tBuOH
Br2
(b)
MeOH
OH
RCO3H
(c)
HO–
H2O
H2SO4
(d)
Heat
H2/Pd
(e)
BaSO4
PROBLEM 834
Convert:
Br
OMe
O
117
Problems
Provide structural formula for the major product in the following reactions:
PROBLEM 835
O
Zn(Hg)
HCl
(a)
DCl
(b)
0°C
O
Ph3P
(c)
BuLi
Br
PROBLEM 836 The reaction of the diene shown below with dry HCl can lead to four products.
Provide structural formula of all the products.
HCl
One-mole
PROBLEM 837 A hydrocarbon A has molecular formula “C14 H 22 ” and it can be resolved into
enantiomers. A on hydrogenation with H 2 /Pt yields B (C14 H 30 ) which can’t be resolved into
enantiomers. Also A on partial hydrogenation with H 2 in presence of Pd/BaSO 4 /PbO yields C (C14 H 26 )
which is still resolvable and decolourise brown colour of bromine water solution. A on treatment with
alkaline permanganate solution yielded D(C 9 H18O 2 ), oxalic acid and propanoic acid. D can also be
resolved into enantiomers. Another organic compound E (C 8 H16O) forms yellow precipitate on
treatment with alkaline iodine solution. E on treatment with NaBH 4 yielded a resolvable organic product
F (C 8 H18O). F on treatment with p-toluenesulphonyl chloride followed by workup with aqueous NaCN
yielded G(C 9 H17 N). Hydrolysing G with dilute sulphuric acid produced D. Identify A to G.
PROBLEM 838 An organic compound A has molecular formula C12 H 22 and can’t be resolved into
enantiomers. A on ozonolysis followed by work-up with Zn/H 2O yields B (C12 H 22O 2 ),. B neither
reduces Tollen’s reagent nor produced any yellow precipitate with alkaline iodine solution. B on
exhaustive oxidation with hot, concentrated, acidic permanganate solution produced C (C 6 H10O 4 ) as
one of the product. C on heating with sodalime yielded isobutane. Identify A, B and C.
PROBLEM 839 An organic compound C 5 H 6O 2 exist in three isomeric forms A, B and C of which
only B can be resolved into enantiomers. All three evolves a gas on heating with sodium metal but only A
and B evolves gas with NaHCO 3 . Also B and C formed white precipitate with ammonical silver nitrate
while A didnot. A on heating with soda-lime yielded D (C 4 H 6 ) which did not yield any precipitate with
ammonical silver nitrate. Hydrolysing C with aqueous alkali followed by heating with sodalime
produced C 3 H 4 . Identify A to D.
PROBLEM 840
Propose synthetic routes to accomplish the indicated transformations:
CH3
OH
CH3
(a)
O
O
CH3
118
Problems in Chemistry
H
Cl
(b)
H
Cl
OH

(c) H—C ≡≡C—H → CH 3 —CH —CH 2 —C ≡≡C—CH 3
CH3
C—CH3
(d) H—C
CH3
Propose mechanism to account for the following transformations:
PROBLEM 841
H2SO4
O
(a)
O
OH
(b)
O
NH3
HO
OH
Br2
(c)
PROBLEM 842
N
OH
O
Br
Propose mechanism:
H2SO4
PROBLEM 843
Propose mechanism of the following reactions:
Cl
(a)
Cl2
CH3OH
H
OCH3
CH3
Br
Br2
(b)
H
KOH, H2O
H
H
O
O
OH
O
119
Problems
H 2O
(c)
H2SO4
OH
O
O
OH
PROBLEM 844 An organic compound A has molecular formula C10 H12O. A forms salt with aqueous
NaOH but does not exhibit stereo-isomerism. A on treatment with HBr yields B (C10 H13OBr) which can
be resolved into enantiomers. On the otherhand, if A is treated with HBr in presence of a peroxide, it
gives C, an isomer of B but C can’t be resolved into enantiomer. C on treating with potassium tertiary
butoxide in tertiary butanol yields D, an isomer of A, which is still non-resolvable. D on hydration in
presence of alkaline H 2O 2 catalyzed by B2 H 6 yields E (C10 H14O 2 ) which can be resolved into
enantiomers. Refluxing E with dilute sulphuric acid yields F (C10 H12O) which does not forms any salt
with aqueous NaOH. Also, A on boiling with alkaline solution of potassium permanganate yields
2-hydroxybenzoic acid. Identify A to F.
PROBLEM 845
Propose mechanism of the following reactions:
H3C
I
I
CH3
CH3
+
+ HI
B
A
PROBLEM 846
Propose mechanism:
Br
+ HBr
Br
+
+
Br
PROBLEM 847 The hydrocarbon shown below is optically active and can be resolved into
enantiomers. A pure enantiomer of this compound on refluxing with conc. H 2SO 4 for sometime
isomerizes into an optically inactive isomer. Explain the observations with the help of mechanism.
CH3
CH
CH3
PROBLEM 848 A hydrocarbon A has molecular formula C 7 H12 . A on treatment with B2 H 6 followed
by work-up with H 2O 2 /NaOH produced B (C 7 H14O) as the only product. Reacting B with torylchloride
followed by workup with KOH yielded C, an isomer of A, in addition to other olefinic products.
Ozonolysis of C followed by work-up with Zn  CH 3COOH produced 2-methyl hexan-1,6-diol.
Identify A to C.
120
PROBLEM 849
Problems in Chemistry
Suggest the most likely site of protonation explaining reason:
ALKYL HALIDES
PROBLEM 850 Considering that following reaction proceed by S N 2 mechanism, select the faster
reaction of each pair and explain your reasoning :
(a) Reaction of cyanide ion with n-iodoheptane or n-chloroheptane.
(b) Reaction of ethanol or sodium ethoxide with n-butyl bromide
(c) Reaction of azide ion with 1-butyltosylate or 2-butyltosylate.
(d) Reaction of tertiary butoxide with ethyl bromide or reaction of ethoxide with tertiary butyl
bromide.
(e) Reaction of cyanide ion with 2-bromopentane or CH 3CH 2 CH—CH(CH 3 ) 2

CH 2 Br
PROBLEM 851 Explain the following observations :
NH2
CH3I
NHCH3
Slower
NH2
and
CH3I
NHCH3
Faster
PROBLEM 852 An organic compound A has molecular formula C 7 H15Cl and it can be resolved into
enantiomers as well as diastereomers. A on treatment with alcoholic solution of KOH yields three
isomeric alkenes B, C and D having molecular formula C 7 H14 , all of them can be resolved into
enantiomers. Treating either B or C with acidic solution of KMnO 4 yield E (C 5 H10O 2 ) and ethanoic acid.
E is still resolvable. Heating E with soda lime (NaOH/CaO) yields C 4 H10 which is non-resolvable.
Identify A to E.
PROBLEM 853 A organic compound A has molecular formula C 6 H13Cl and it can be resolved into
enantiomers. A on treatment with alcoholic solution of KOH yield four alkenes B, C, D and E, none of
them can be resolved into enantiomers. B and C are stereoisomers and treatment of either B or C with Cl 2
yields racemic mixture of C 6 H12Cl 2 . On the other hand D and E are stereoisomers and D on treatment
with Cl 2 yields a meso compound C 6 H12Cl 2 while E on similar treatment yield a racemic mixture of
C 6 H12Cl 2 . Also B has greater heat of hydrogenation than C. Deduce structures of A to E.
121
Problems
PROBLEM 854 Give reason:
(a) Hydrolysis of methylbromide takes place much faster in presence of NaI.
(b) When tertiary butyl chloride undergo hydrolysis in aqueous NaOH, the rate of formation of
tertiary butyl alcohol does not increase appreciably as OH – concentration is increased, however a
marked decrease in concentration of t-butyl chloride is observed.
(c) When methyl bromide reacts with NaCN, major product is CH3CN, whereas when CH3Br reacts
with AgCN, major product is CH3NC.
PROBLEM 855 Explain the mechanism of following reactions:
(a)
OH + HCl
(b)
OH + HCl
Cl
Cl +
Cl
(only product)
PROBLEM 856 Account for the following observations:
(a) t-BuF is solvolysed only in very acidic solution.
(b) t-BuCl is solvolysed more slowly than 2-chloro-2,3,3-trimethylbutane.
(c) t-BuCl is solvolysed much faster than 2-chloro-1,1, 1-trifluoro-2-methy- lpropane.
PROBLEM 857 Explain the following observations in S N 1 reactions:
(a) AgNO3 increases the rate of solvolysis
(b) the more acidic the nucleophilic solvent, the faster is the rate of solvolysis.
PROBLEM 858 Complete the following reaction sequence, showing the major product you expect for
each stage. Also indicate stereo-chemistry where possible.
TsCl
NaN
3
(a) CH 3CHCH 3 → A →
B
Pyridine

OH
CH2CH2CH2OCH3
(b)
H
H3C
CH3SNa
C
C
Br
PROBLEM 859 Explain the relative rate of E-2 reaction of the following compounds:
Cl
Cl
CH3
H 3C
(I)
Cl
CH3
H 3C
(II)
CH3
H 3C
(III)
122
Problems in Chemistry
PROBLEM 860 Explain the followings:
(a) When CH3Br is dissolved in ethanol, no reaction occurs at 25°C. When excess of C2H5ONa is
added, a good yield of ethyl-methyl ether is obtained.
(b) Explain the role of solvent in the reaction of potassium ethanoate with ethyl iodide.
(c) Which nucleophile reacts more rapidly with CH 3I : Et3N or Et3P?
PROBLEM 861 Predict product in the following reaction:
Br
NaN3
(a)
I
(b)
H 3C
BH3-H2O2
SO2Cl CH =CHCH ONa
2
2
NaOH
Br
(CH3)3COK
(c)
(CH3)3COH
Br
Br
NH3
(d) MeO
Br
(e)
Br2
hν
PROBLEM 862 In attempting to synthesize 1-methylcyclohexene, you are trying reaction of
trans-1-iodo-2-methyl cyclohexane with potassium tri-tertiary butoxide. The reaction fails to produce
the desired product and the isolated product is 3-methylcyclohexene. Using a detailed mechanistic
analysis, explain why this alternative product is formed?
PROBLEM 863 The two tosylate shown below react differently under identical conditions. One will
undergo a facile S N 2 reaction while other will easily eliminate via E-2 mechanism. Draw structures of
major product for each reaction and provide mechanistic explanation for them:
OTs
OTs
CH3ONa
OTs
and
CH3ONa
OTs
PROBLEM 864 When piperidine undergoes the indicated series of reactions, 1,4-pintadiene is obtained
as product.
123
Problems
i) excess of CH3I
ii) Ag2O/H2O
iii) Heat
N
i) excess of CH3I
N
ii) Ag2O/H2O
iii) Heat
H
When four different methyl substituted piperidine is subjected to same series of reactions, each
forms a different dienes-1,5-hexadiene, 1,4-pentadiene, 2-methyl-1,4-pentadiene and
3-methyl-1,4-pentadiene. Which methyl substituted piperidine yields which diene?
PROBLEM 865 Give substitution product mentioning their stereochemistry:
H
H
H2O
HO–
(a) H 3C
(b) H 3C
Br →
Br →
C4 H 9
C4 H 9
Cl
H
(c)
CH 3
CH3ONa
H
(d)
H
H 3C
CH3
Cl
H
CH ONa
3
→
C2 H 5
CH 3
(e)
H
Cl
H
CH 3
CH ONa
3
→
C2 H 5
PROBLEM 866 In the following pair of reactions, which will occur at faster rate-explain your choice:
Cl + (CH3)2CHS
or
(b)
Cl
+ HO –
or
(c)
S
Cl + CH3S
(a)
O
S
OH
–
– Cl
O
OH
Cl + H2O
or
–
– Cl
Cl + HO
+ Cl
Cl + H2O
OH
OH
124
Problems in Chemistry
PROBLEM 867 Propose mechanism:
OH
(a)
Br + H2O
∆
OH
+ H2O
(b)
∆
H
Br
CH3
+
CH3
CH3
H
OH
CH3
PROBLEM 868 Explain the following observations:
(a) Azide ion (N –3 ) react with 2-bromopentane thousand times faster than with neopentyl bromide
in a S N 2 reaction though former is a secondary halide while latter is primary.
(b) What will happen to the stereochemistry of product of the following reaction:
Br
H
–
CH3 + N3
S N2
D
(c)
(d)
(e)
(f)
What will happen to the rate if the concentration of alkyl bromide in (b) is doubled?
What will happen to the rate if the concentration of azide ion in (b) is doubled?
How the sign of optical rotation of reactant and product are related in (b)
When allowed to stand in dilute H 2SO4, laevo-rotatory 2-butanol slowly loses optical activity.
PROBLEM 869 Assuming that the following reaction takes place by S N 2 mechanism, select the faster
reaction in each pair and explain reason for your choice:
(a) Reaction of cyanide ion with n-iodoheptane or n-chloroheptane.
(b) Reaction of ethanol or C 2H5ONa with n-butyl bromide.
(c) Reaction of azide ion (N 3–) with n-butyl tosylate or s-butyl tosylate.
(d) Reaction of isopropoxide with ethyl bromide or reaction of ethoxide with 2-bromopropane.
PROBLEM 870 Synthesize the following compounds from indicated starting material:
OMe
(a)
from 1,4-dihydroxy benzene
NH2
(b)
from benzyl alcohol
OMe
S
(c)
from phenol
(d)
from
O
Cl
125
Problems
PROBLEM 871 Explain why allyl bromide (CH2 ==CHCH2Br) undergoes rapid substitution under
either S N 1 or S N 2 reaction condition.
PROBLEM 872 Following reactions are not feasible as indicated. Briefly explain the reason for their
failure:
Br
KCN
(a)
H
+
O H
HO
(c)
H 3C
(b)
CN
–H+
+
CH2
+
Br
HO
OH
H 3C
CH 3
→
CH 3
PROBLEM 873 Provide mechanism:
I
(a)
OH
∆
(b) HO
Br
KOH
O
O
PROBLEM 874 Arrange the following in the increasing order of their ability as a leaving group:
(b) CF3SO 3– , CH 3SO 3– and CH 3COO –
(a) CH 3S – , CH 3O – , CF3– and F –
PROBLEM 875 An organic compound A has molecular formula C6H11Br. A decolorizes brown colour
of bromine water but it can’t be resolved into enantiomers. A on treatment with HBr produces
B(C6H12Br2) which is still non-resolvable. Also A on treatment with HBr in presence of a peroxide
(R2O2) yields C-an isomer of B, but it is resolvable. B on heating with Zn-metal yields D(C6H12) which on
treatment with ozone followed by hydrolysis of product in presence of Zn-powder yield acetone as the
only organic compound. Also A on treatment with ethanolic KOH solution yields E(C6H10)
as only possible product. E on treatment with excess of HBr/R2O2 in cold produces
isomeric F and G with their molecular formula C6H12Br2, in which F is a meso compound
while G is resolvable. Also F on hydrolysis in aqueous NaOH yield a diol C6H14O2 which
O
on refluxing with dilute H 2SO4 yields the adjoining compound.
Deduce structures of A to G.
PROBLEM 876 An organic compound is optically active and has molecular formula C14H23Br. A does
not decolourise cold, dilute and alkaline solution of KMnO4. A on treatment with potassium tertiary
butoxide in tertiary butanol yields B(C14H22) as the major product. B on treatment with acidified
permanganate solution yields C(C14H22O4). C on heating with sodalime yields D(C12H22) which can also
be formed by dissolving cyclohexyl bromide in etheral solution of Mg. Identify A to D.
PROBLEM 877 An optically active organic compound A has molecular formula C7H16O. A neither
changes the colour of bromine water nor evolve any gas on reacting with MeMgBr. A on refluxing with
dilute H2SO4 produces B(C3H8O) and C(C4H10O). B is non-resolvable and gives yellow precipitate with
126
Problems in Chemistry
NaOH/I2 while C is resolvable. C on heating with concentrated solution of H2SO4 yield stereo-isomeric
compounds D and E with their molecular formula C4H8. D on treatment with Br2/CCl4 yields a meso
dibromide while E on similar treatment yields racemic mixture of products. Identify A to E.
PROBLEM 878 An organic compound A(C5H11Cl) is optically active and on treatment with ethanolic
KOH solution yields B(C5H10) as major product which does not show stereo-isomerism. Also A on
treatment with (CH3)2Cu yields C(C6H14) which is optically inactive. Deduce structures of A to C.
PROBLEM 879 An organic compound A has molecular formula C10H17Br and it is non-resolvable. A
does not decolourise brown colour of bromine water solution. A on treatment with
(CH3)3COK/(CH3)3COH yields B(C10H16) as major product. B on treatment with H2/Pt yields C10H18
which on treatment with Cl2/hν yields three monochloro derivative. Also B on boiling with acidic
permanganate solution yields C(C10H16O3). C on heating with soda-lime yields D(C9H16O). D on
reducing with LiAlH4 followed by heating the product with concentrated H2SO4 yields E(C9H16) as
major product. E on treatment with ozone followed by work-up with Zn-H2O yields 6-keto nonanal.
Deduce structures of A to E.
PROBLEM 880 An organic compound A has molecular formula C5H10O and it neither decolourise
bromine water solution nor evolve any gas on heating with Na-metal. A on refluxing with aqueous HI
yields B(C5H10I2) which is non-resolvable. B on treatment with NaCN in acetone yield C(C6H10IN) in
significant amount while B on treatment with aqueous Ag2O yields D(C5H11IO) in significant amount. D
does not change the orange colour of acidic dichromate solution. Deduce structures of A to D.
PROBLEM 881 An optically active organic compound A has molecular formula C11H16O and it gives
off a colourless gas on treatment with CH3MgBr. A on heating with concentrated H2SO4 yields B(C11H14)
which does not show stereo-isomerism. B on treatment with HBr in absence of a peroxide yields
C(C11H15Br) which is resolvable while addition of HBr on B in presence of a peroxide yields D, an
isomer of C but it is non-resolvable. Also B on treatment with ozone followed by work-up with Zn-H2O
yields C8H8O and C3H6O in which former can also be obtained by Friedal Craft’s acylation reaction of
benzene. Both these ozonolysis products gives yellow precipitate on treatment with NaOH/I2. Deduce
structures of A to D.
PROBLEM 882 Bring about the following transformations:
(a) C 6 H 5CH 2OH → C 6 H 5CH 2SCH 2CH 3
(b) n-pentanol → n-pentyl iodide
(c) C 6 H 5SH → C 6 H 5SCH 2CH 2C 6 H 5
(d) n-butyl bromide → 2-heptyne
CH3
CH3
(e)
HO
H5C2
C
H
H5C2
H
O
(f)
C
NH2
O
O
Cl
O
H
N CHCH2OH
CH=CH2
127
Problems
PROBLEM 883 Both mesylates and triflates are good leaving groups. However, triflate
displacements proceed about 5000 times faster than mesylate displacements. Explain why triflate is a
better leaving group than mesylate.
PROBLEM 884 Upon treatment with Ag + ions, on α-halo-ether undergo substitution via SN-1
mechanism. Consider the following two reactions and explain why 1 produces a mixture of two possible
stereoisomers while 2 provides only one.
O
O
O
O
Ph
Br
Br
O
Ph
(1)
(2)
PROBLEM 885 For each of the following of substrate, indicate with reasoning, which will react at
faster rate under the given reaction conditions. In addition, indicate whether you expect each reaction to
follow SN-1 or SN-2 pathways.
Br
(i)
(iii)
—SNa.
Br
Br
(ii)
Br with
or
or
with CH3COOH.
or
Br
Br
with CH3COOH.
PROBLEM 886 Write products in the following reactions explaining mechanism of their formations:
H
I
Et
n-Bu
NaOH
NaCl
(a)
→
(b)
EtOH,
∆
H
Ph
CH3COOH
CH3—CH—CH2—
OTs
i-Pr
(c)
OMs CH3
NaOH
HCOOH, ∆
(d)
OTs
CH3 H
(e)
Me
CH(CH3)2
CH3
H 2O
Br
∆
H
(f)
Et
Et
KOH
EtOH, ∆
COO–
CH3I
128
(g)
Problems in Chemistry
Et
H
H
Cl
PROBLEM 887
(a)
KOH
EtOH
Write products in the following reactions indicating their stereochemistry:
I Et
Me


H 2O
EtONa
E2
(b) CH 3CH 2  C  C  CH 3 → S N 1
∆
EtOH/∆
 
CH 3 H
OMs
CH3
(c)
CH3
(CH3)3COK
(CH3)3COH
OTs
E2 (d)
(CH3)2CHSH
NaH
SN2
CH3
Br
CH 3
Ph
NaCl
Acetone
Et → S N 2
(e) ICH 2
H
(f)
CH 3
OTs
Ph
CH 3
NaOH
∆
→ E–1
CH 3
Cl
CH 3


KOH
(g) CH 2 ==CH  CH 2  CH CH  CH 3 → E–2
EtOH
PROBLEM 888
Write mechanism for the following reaction:
CH 3
Cl
CH 3



heat
CH 2 == C  CH 2  CH  CH 3 + CH 3OH → CH 2 == C  CH  CH 2 CH 3

OCH 3
CH 3

+ CH 3OCH 2  C== CHCH 2CH 3
PROBLEM 889
Bring about the following transformations:
I
CN
129
Problems
PROBLEM 890
Write major products in the following reaction and discuss their stereochemistry:
F
PhNH2
Et2O
H3C
OTs
PROBLEM 891 5-Bromononane on treatment with potassium tertiary butoxide in ethanol produces a
mixture of cis-4-nonene (23%) and trans-4-nonene (77%). Draw Newman’s projections of
5-bromononane, looking down the C5–C6 bond showing the conformational forms that lead to
cis-4-nonene and trans-4-nonene respectively. Based on conformational analysis, suggest reason why
transform of product predominate.
PROBLEM 892
Give stereochemical structures of product formed in the following reactions:
CH2CH3
(a)
H
H3C
+ CH3CH2CH2ONa
SN2
CH3
CH3ONa
(b)
SN2
Br
Cl
CH3
CH3OH
(c)
∆/SN1
Cl
PROBLEM 893 An alkyl bromide A has molecular formula C 8 H17 Br and four different structures can
be drawn for it, all optically active. A on refluxing with ethanolic KOH solution yields only one
elimination product B (C 8 H16 ) which is still enantiomeric. B on treatment with H 2 /Pt yields C (C 8 H18 )
which does not rotate the plane polarized light. B on ozonolysis followed by work-up with H 2O 2 yields
D(C 7 H14O) as one product which is still resolvable. Deduce structures of A to D.
PROLBEM 894 Provide missing products, starting material or reagents/conditions, for each of the
following reactions:
Br
SOCl2
(a)
OH
H
NaCN
(b)
Pyridine
OMe
(c)
(e)
OH
OH
Ph
(d)
(ii) EtBr
TsCl
Pyridine
OH
H
(i) NaH
NaCN
(f)
H2SO4
MeOH/∆
CH3Na
OH
130
Problems in Chemistry
PROBLEM 895
Propose mechanism of the following reaction:
Br
+
+
N
CH3
N
N
H
PROBLEM 896 SN2 reaction of simple carboxylate ions with haloalkanes in aqueous medium
generally do not gives good yields of esters, explain. Reactions of 1-iodobutane with sodium acetate
gives an excellent yield of ester if carried out in acetic acid medium. Why acetic acid is a better solvent
for this process than water?
PROBLEM 897
Provide major substitution product in the following reaction:
heat
heat
(a) 2-butanol + HBr →
(b) 3,3-dimethyl-2-butanol + HBr →
OH
Heat
+ HCl
(c)
OH
(d)
+ HCl
Heat
PROBLEM 898 Provide structure of major product in the following reaction indicating
stereochemistry where appropriate:
D
D
(a)
C
CH3
OH
TsCl
NaCN
(b)
H
HBr
Heat
OH
C
NaCN
H
CH3
PROBLEM 899 How 1-butanol can be converted efficiently into:
(a) 1-iodobutane
(b) N-ethylethanamine
(c) Butylpropanoate
(d) CH 3CH 2CH 2CH 2SH
(e) CH 3CH 2CH 2CH 2OCH 3
PROBLEM 900 Which of the following alkyl halide could be successfully used to synthesize
Grignard reagent and why other fail?
Br
N
Br
HO
II
I
O
Br
OH
III
H2 N
Br
IV
131
Problems
Propose mechanism of the following reactions:
PROBLEM 901
Cl
O
(a)
O
CH3OH
+ CH3O–
O
O
O
(b)
O
CH3O–
Br
O
+ HO–
(c)
O
HO
OH
O
Explain mechanism of formation of products in the following reactions:
PROBLEM 902
(a)
Mg
—Br
—MgBr +
Br
(b)
—OH
Discuss the relative reactivity of reactants in the following pairs:
PROBLEM 903
(a)
HBr
(CH3)3CCl + HO
–
–
and (CH3)3CI + HO
H
H
H
CH3
–
(b)
CH3
H
H3C
I
CH3O
CH3OH
H3C
Br
H
H
H
–
CH3
Br
and
H3C
II
H
CH3O
CH3OH
H3C
CH3
132
Problems in Chemistry
PROBLEM 904 (a) Give structures of products obtained from the reaction of both enantiomers of
cis-1-chloro-2-isopropyl cyclopentane with CH 3ONa.
(b) Are all the products obtained chiral?
(c) How would products differ if the starting material were trans isomer?
(d) Which pair of structure (enantiomers) form substitution product more rapidly-cis or trans and
why?
(e) Which pair of products form elimination product more rapidly cis or trans?
Write elimination products in the following reaction and compare their relative
PROBLEM 905
reactivity:
CH3
CH3
PROBLEM 906
(a)
(CH3)3COK
(CH3)3COK
(CH3)3COH
(CH3)3COH
Cl
Cl
I
II
Compare the following compound for their reactivity in SN–1 reaction:
OCH3
OCH3
Cl
Cl
and
and
(b)
Cl
Cl
Cl
(c)
and
Cl
Cl
(d)
and
N
(e)
N
and
Cl
Cl
Cl
133
Problems
PROBLEM 907
Propose mechanism of the following reaction:
I
NaCl
CH3OH
Cl
+
I
II
+
Cl
OCH3 +
III
IV
OCH3
ALCOHOLS AND ETHERS
PROBLEM 908 Why does ethylene oxide react readily with nucleophiles such as ammonia, whereas
THF is inert to nucleophilic attack by ammonia?
PROBLEM 909 Give the products and mechanism of each of the reaction shown below:
O
CH3ONa/CH3OH
H2SO4/CH3OH
PROBLEM 910 Bring about the following conversion from the indicated starting materials:
O
(a)
CH2OH
1,6-hexan diol
and
(b)
O
O
PROBLEM 911 An optically inactive organic compound A(C7H11Br) is treated with Mg in ether to give
B(C7H11MgBr), which react violently with D2O to give methylcyclohexene with deuterium atom on the
methyl group (C). Reaction of B with acetone followed by hydrolysis gives D(C10H18O). Heating D with
concentrated sulphuric acid gives E(C10H16). Determine structure of A to E.
PROBLEM 912 When ethanol is heated with conc. H2SO4 at 140oC, diethyl ether is obtained, whereas at
180oC ethylene is the major product. Suggest reason.
PROBLEM 913 Compound A is an optically active alcohol. Treatment with chromic acid convert A into
a ketone B. In a separate reaction, A is treated with PBr3, converting it into C. C on reaction with Mg is
added to B to yield D, which after hydrolysis gives 3,4-dimethyl-3-hexanol. Identify A to D.
134
Problems in Chemistry
PROBLEM 914 A compound ‘A’ C10H14O exist in diastereomeric form, liberates a gas on reaction with
CH3MgBr. Treatment of A with aqueous HBr gives B(C10H13Br), which exist in enantiomeric forms. B
on treatment with alcoholic solution of KOH gives a compound C. Ozonolysis of C followed by work-up
with H2O2 gives a compound D and acetophenone. Identify A to D.
PROBLEM 915 Synthesize the followings from the indicated starting materials:
O
from
(a)
O
CH3
from
(b)
O
CN
H
from
+ Br
(c)
O
O
O
O
(i) CH3MgI
A
(ii) H3O+
H2/Pt
H2SO4
B
C
H2SO4
KMnO4
C6H12O (E)
C10H16 (H)
CH3MgBr
H 3O +
H2SO4
C10H18O (G)
(i) Mg/ether
(ii)
(iii) H3O+
C5H9Br (F)
O
BrH
PhCO3H
C5H8O (D)
135
Problems
PROBLEM 916 Determine the structure of compound A through H
OH
(a)
H2SO4
OH
(b)
H2SO4
O
OH
(c)
H2SO4
+
OMe
(d)
+
O
H2SO4
PROBLEM 917 Propose mechanism of the following reactions:
PROBLEM 918 Convert: (a) 2-methyl-2-butene to (CH 3)2C(OH)CHDCH3
(b) Synthesize 2,5-dimethyl-3-hexanone using alcohol containing not more than four carbon.
O
(d)
O
OH
OH
(c) C6H5Br to (C6H5)2C(OH)CH3 using any additional compound containing not more than three
carbon.
OH
Cl
PROBLEM 919 (a) Convert:
(b) Cyclohexyl bromide when treated with potassium ethoxide, the major product is a cyclohexene
whereas when ethyl bromide is treated with potassium salt of hexanol, the major product is
ethyl-cyclohexyl ether. Explain the difference.
OH
(a)
cold, dil, alkaline
KMnO4
A
HIO4
B
(b)
+
H
A
136
Problems in Chemistry
PROBLEM 920 Predict major product:
(c)
OH
OH
HIO4
OCOCH3 (i)
(d)
A
(ii)
LiAlH4
NaBH4
O
PROBLEM 921 An optically active compound A(C10H17Br) when treated with alcoholic solution of
KOH yield two compound B and C of molecular formula C10H16 of which ‘B’ is optically active.
Ozonolysis of A followed by treatment with Zn-H2O gives acetone as one of the product. Hydrogenation
of either B or C yield 4-isopropyl-1-methyl cyclohexane, identify A,B and C.
PROBLEM 922 When cis-2,3-dimethyl oxirane is treated with water containing a trace of HClO4, a
racemic mixture of 2,3-butandiol is formed, trans-2,3-dimethyl oxirane give meso-2,3-butandiol under
identical conditions. Write mechanism for these reactions.
PROBLEM 923 When 3-methyl-2-pentanol is treated with ZnCl2 in concentrated solution of HCl, a
mixture of chloroalkanes forms, including chiefly 2-chloro-3-methylpentane, and 3-chloro-3-methyl
pentane. When 3-methyl-2-pentanol is treated with thionyl chloride in pyridine, only 2-chloro-3-methyl
pentane is formed. Write detailed mechanism that accounts for the observation.
PROBLEM 924 Synthesize the following by Williamson’s method, choosing alkoxide anion and alkyl
halide that will give the best yield:
(a)
(c)
(b)
O
(d)
O
O
O
PROBLEM 925 Treatment of 3-methyl-2-butanol with HCl, only a trace of 2-chloro-3-methyl butane is
formed. An isomeric product was isolated in 97% yield. Suggest a reasonable structure of this product.
PROBLEM 926 Suggest reasonable explanation for the following observations:
(a) The first order rate constant for the solvolysis of (CH3)2C==CHCH2Cl in ethanol is 600 times
greater than that of allylchloride.
(b) After a solution of 3-buten-2-ol in aqueous H2SO4 be allowed to stand for a week, it was found
to contain both 3-bute-2-ol and 2-buten-1-ol.
PROBLEM 927 Propose mechanism of the following reactions:
Ph
O
+
H
Ph
(a)
Ph
OH OH
Ph
OH
(b)
H
+
137
Problems
OH
O
OH
H
(c)
+
H
(d)
+
HO
O
PROBLEM 928 A compound X(C14H14O) on mild oxidation yields C14H12O(Y). If X is treated with a
dehydrating agent, it loses a molecule of water and resulting product on vigorous oxidation yields two
molecules of benzoic acid. Give structure of X and Y.
PROBLEM 929 Convert:
R
R
(a)
(b) Propose mechanism:
H+
OH
O
H 2O
O
(c) Write products:
(i)
OsO4
A
NaIO4
B
(ii)
OMe
(iii)
HI(excess)
D
(iv)
O
PhSNa
H 2O
C
BH3/H2O2/NaOH
E
OMe
PROBLEM 930 Two optically active compounds A and B have same molecular formula C9H12O and do
not decolourise bromine water solution. Both A and B are oxidised by KMnO4 to give benzoic acid and
react with Na to give a colourless, odourless gas. Compound A gives yellow precipitate with I 2 /OH –
whereas B does not. Also A loses optical activity on treatment with PCC while B retain optical activity on
similar treatment. Deduce the structure of A and B.
138
Problems in Chemistry
PROBLEM 931 Compounds A, B and C are isomeric alcohols with formula C5H12O. A and B reacts with
chromic acid solution, B giving an acid D. The three isomeric alcohol reacts with HBr with decreasing
relative rates C > A >> B, all giving same C5H11Br(E). in varying yields. A alone can be oxidised by
I 2 /OH – to F. Write structure of A to F.
PROBLEM 932 C7H14 (A) decolourises Br2 in CCl4 and reacts with Hg(OAc)2 in THF-H2O followed by
reduction with NaBH4 to produce a resolvable compound B. A undergoes reductive ozonolysis to give
the same compound C as obtained by oxidation of 3-hexanol with KMnO 4. Identify A, B and C.
PROBLEM 933 An organic compound A (C7H14) reacts with BH3-THF and then with H2O2/OH – to give
a chiral B. Oxidation of B with KMnO4 affords a chiral carboxylic acid C. Ozonolysis of A followed by
work-up with (CH3)2S produces C6H12O(D) as one product. D on treatment with LiAlH4 produces
another compound E (C 6 H14O) which is optically active. E on heating with conc.H2SO4 produces
F(C6H12) as major product. F on ozonolysis followed by work-up with DMS produces acetone as one
product. Identify A to F.
PROBLEM 934 Treating 3,3-dimethyl-1-butene with dilute sulphuric acid is largely unsuccessful as a
method of preparation of 3,3-dimethyl-2-butanol because an isomeric compound is major product.
What is the isomeric product and how it is formed?
PROBLEM 935 When C2H5ONa reacts with 1-(chloromethyl) oxirane, labelled with 14C as shown by
astric in (*) I, the major product is an epoxide bearing the labelled carbon as shown in II. Provide
mechanism to explain this fact.
Cl
*
*
C2H5ONa
O
I
OC2H5
O
II
PROBLEM 936 An unknown organic compound A(C4H10O2) reacts with sodium metal to liberate one
mole of hydrogen gas per mole of A. Although A is inert towards periodic acid, it does reacts with CrO3 to
form B(C4H6O3). Identify A and B.
PROBLEM 937 Convert:
(a)
OH
D
(b) CH2
CH2
O
O
PROBLEM 938 Synthesize the following compounds starting with isobutane:
(a) ter-butyl bromide
(b) 2-methyl propene
(c) iso -butyl bromide
(d) iso butyl methyl ether
(e) (CH3)2CHCH2OCOCH3
(f) (CH3)2CHCH2CN
PROBLEM 939 A neutral compound A has molecular formula C10H16O2. A does not decolourise
aqueous solution of Br2 and evolve no gas on treatment with Na. A on acidic hydrolysis produces two
compounds B(C4H8O) and C(C6H10O2). B on treatment with SOCl2 gives D(C4H7Cl) which on further
treatment with aqueous solution of KCN followed by hydrolysis of product yielded E(C5H8O2). E on
reducing with LiAlH4 gives F(C5H10O) which on heating with concentrated sulphuric acid solution
139
Problems
yields G(C5H8). G on treatment with HCl yields H(C5H9Cl). C on heating with soda lime gives I(C5H10). I
on monochlorination with Cl2/hν give H as sole product. Identify A to I describing mechanism of their
formation in each step.
PROBLEM 940 An organic compound A has molecular formula C6H10O and known to decolourise
aqueous solution of bromine. A is also resolvable and on catalytic hydrogenation with H2/Pt yields
B(C6H12O) which is optically inactive. Also B does not change the orange colour of an acidic solution of
dichromate ion. B on heating with concentrated sulphuric acid solution yields C(C6H10) which on
reductive ozonolysis gives the following compound:
O
O


H  C  CH 2CH 2CH 2  C  CH 3
Deduce structures of A, B and C.
PROBLEM 941 An organic compound A(C5H8O2) does not decolourise aqueous solution of bromine
and evolve no gas on treatment with CH3MgBr. A on reaction with concentrated solution of HI produced
methylene iodide and an another compound B(C4H6I2). B on heating with sodium metal produced
C(C4H8) which on monochlorination gave a single isomeric product C4H7Cl. Deduce structures of A, B
and C.
PROBLEM 942 Starting with 2-methyl propene and using any other needed reagents, synthesize the
following compounds:
(a) (CH3)2CHCH2OH
(b) (CH3)2CDCH2T
(c) (CH3)2CHCH2T
(d) (CH3)2CHCH2OCH2CH3
PROBLEM 943 Propose mechanism:
OH
+
H
PROBLEM 944 Coniferyl alcohol(X) is not soluble in water or NaHCO3, has molecular formula
C10H12O3. A solution of Br2 in CCl4 is decolourised by X forming C10H12O3Br2(A). Upon reductive
ozonolysis of X, 4-hydroxy-3-methoxy benzaldehyde and B(C2H4O2) are produced. X reacts with
benzoylchloride C6H5COCl in presence of a base to form C(C24H20O5). This product rapidly
decolourises aqueous solution of KMnO4 and is insoluble in NaOH. X reacts with cold HBr to form
D(C10H11O2Br). X reacts with HI to produce E(C9H9O2I) and CH3I. In aqueous base, CH3I and X forms
F(C11H14O3), which is not soluble in strong base but decolourises aqueous solution of Br2. Deduce
structures of X and A to F.
PROBLEM 945 An optically active organic compound A has molecular formula C7H12O2 and does not
decolourise cold, dilute and alkaline solution of potassium permanganate. A on treatment with
concentrated HCl gives CH2Cl2 and B(C6H10Cl2). B on treatment with aqueous solution of KI gives
C(C6H10) which decolourise bromine water. C on treatment with acidic solution of KMnO4 gives
2-methyl pentan-1,5-dioic acid. Deduce structures of A to C.
PROBLEM 946 An optically active organic compound A has molecular formula C9H18O which neither
change colour of Bayer’s reagent nor of acidic dichromate solution, but evolve a colourless gas on
heating with Na-metal. A on dehydration with conc.H2SO4 produces B(C9H16) which exist in two
stereomeric forms. Ozonolysis of B followed by work-up with Zn-H2O produced ethanal and
140
Problems in Chemistry
C(C7H12O). C on treatment with LiAlH4 produced D(C7H14O) which is enantiomeric and change the
colour of acidic dichromate solution from orange to blue-green. D on dehydration produced E(C7H12)
which on ozonolysis followed by hydrolysis of product in presence of Zn produced cyclopentanone.
Deduce structures of A to E showing stereochemical structures of B.
PROBLEM 947 An organic compound A(C11H22) when ozonolyzed yields B(C6H12O) and C(C5H10O).
Mild oxidation of C yields D(C5H10O2). Compound B reacts with NaOCl to yield chloroform and sodium
salt of D. Reduction of B with LiAlH4 yields E(C6H14O). Dehydration of E gives F(C6H12). Ozonolysis of
F yields G(C2H4O) and H(C4H8O). Compound H does not react with Tollen’s reagent or Fehling solution
but does form an oxime and semicarbazide. Write structures of A to H describing the reaction involved.
PROBLEM 948 An optically active organic compound A(C8H16O) does not decolourise the brown
colour of bromine-water but turns orange colour of acidic dichromate solution to blue green converting
itself into another optically active compound B. A on heating with conc.H2SO4 produced C(C8H14) as
major product which does not show stereomerism. C on reductive ozonolysis produced D(C5H8O) as one
of the product which does not change the colour of acidic dichromate solution. D on reduction with
LiAlH4 produced E(C5H10O) which on dehydration followed by reductive ozonolysis of product yields a
non-resolvable F(C5H8O2). F gives positive Fehling’s solution test but negative haloform test. Deduce
structural formula of A to F.
PROLBEM 949
Predict products in the following reactions showing stereochemistry where applicable:
O
1. BH3
(a)
1. NaBH4
(b)
2. H3O+
2. H2O2-NaOH
OCH3
OH
OH
(c)
Ac2O
DMSO
1. TsCl
2. NaI, acetone
(d)
O
(e)
SH
1. NaOH
2. CH3I
1.CH3MgBr (excess)
(f)
OCH2CH3
2. H3O+
OH
O
(g) H
1. Na2Cr2O7, H2SO4
2. CH3CH2MgBr
3. H3O+
(h)
O
CHO
1.LiAlH4
2. H3O+
141
Problems
OH
(i)
1.NaH
2. CH3CH2I
OH
(j)
1. HBr
2. Mg/ether
3. CH3CHO
3. H3O+
PROBLEM 950 Provide reagents that will effect the following transformations (more than one steps
may be required):
CHO
OH
(a)
OH
O
O
O
(b)
OCH3
Enantiomers
(c)
O
O
Br
Br
(d)
OH
OCH3
(e)
(f)
OH
O
OH
HO
O
OH
(g)
CHO
CHO
OCH3
142
Problems in Chemistry
O
OH
(h)
PROBLEM 951 Devise synthesis of the following compound starting from the indicated starting
materials and any other reagents you need.
O
O
Br
from
PROBLEM 952
applicable:
OH
OCH3
CH3CH2Br
NaH
O
(b)
HBr
Excess
S
CH3CH2Br
(c)
S
H2O2 (1 eq)
(d)
CH3COOH
Br
NaOH
(e)
H2O
OH
OH
(f)
O
+
H
Predict all products of the following reactions and show stereochemistry where
(a)
(g)
+
SBr –
BuLi
H2SO4
CH3OH
Cyclohexanone
143
Problems
(h)
(i) CH3CH2SNa
O
(i)
—S—
(j)
O
(2) H3O+
mcpba
excess
(1) CH3MgBr
(2) H3O+
Silicon-oxygen bonds are easily cleaved by fluoride ions as:
PROBLEM 953
H3C
Si
O
CH3
l
l
(CH3)3SiF +
O
H3C
–
F–
Based on the above information, provide a structure of A(C 6 H10O 2 ). Provide mechanism of
formation of A.
CH3
—O—Si—CH3
CF3CO3H
KF
H 3O +
A (C6H10O2)
CH3
Provide reagents that will effect the following transformations:
PROBLEM 954
CH3O
OCH2CH3
B
O
OH
A
C
E
D
S
OCH3
HO
mCPBA (excess)
NaH
CH3I
F
PROBLEM 955 The reaction of a dialkyl ether with excess of HI provides two alkyl iodides. However,
under the similar reaction conditions, aryl-alkyl ethers affords an alkyl iodide and a phenol. Explain the
observations with the help of following reactions.
144
Problems in Chemistry
O
CH3
(1)
+ CH3I
OH
O
CH3
(2)
PROBLEM 956
I
HI
excess
HI
excess
+ CH3I
Bring about the following transformations:
OH
OH
(a)
O
N
OH
(b)
H
OH
(c)
O
OH
(d)
PROBLEM 957 Predict products of the following reactions. Show stereochemistry where important. In
case where more than one stereoisomers are formed, draw both the products and indicate which one will
be the major product.
(a)
PCl3
—OH
Mg
Ether
OH
Ac2O
(b)
NaBH4
(c)
O
D 2O
145
Problems
NaSCH3
TsCl
(d)
OH
H
PCC
(e)
CH3CH2MgBr
H 3O +
OH
CH3
O
O
LiAlH4
(f)
H 3O +
O
O
OH
(g)
TsCl
NaI
(1 eq)
OH
BH3/H2O2
(h)
NaH
CH3CH2I
NaOH
PROBLEM 958
Provide reagents that will bring about the following transformations:
OH
(a)
OH
OH
(b)
(c)
OH
OH
146
Problems in Chemistry
H
H
(d)
OH
OH
CH3
CH3
PROBLEM 959 Reaction A proceeds more quickly than B. Explain why this is so. On the basis of your
explanation of first part, what product would you expect from reaction C?
O
A:
B:
H
CH3
Na2Cr2O7
OH
H2SO4
D
Na2Cr2O7
C:
H
D
O
H2SO4
OH
PCC
CH2Cl2
OH
PROBLEM 960 When benzyl alcohol (A) is treated with TsCl and pyridine at room temperature, the
tosylate (B) is formed. However, when the same reaction is carried out at 60°C, the chloride (C) is
produced. Draw a mechanism of formation of C from B and explain the effect of temperature on reaction.
OH
OTs
TsCl
Cl
60°C
Pyridine
A
B
PROBLEM 961 Predict products in the following reactions :
where appropriate):
mCPBA
excess
(a)
(Provide structures of stereoisomers
NaOH
(b)
SH CH3CH2I
S
(c)
O
H2SO4
t-Butanol
NaH
CH3I
mCPBA
(d)
O
147
Problems
Br
Br
NaHCO3
(e)
excess
(f)
O
(h)
O
(1) NaN3
(2) H3O+
Br
OH
(g)
NaH
(1) PhONa
(2) H3O+
Br
OH
H3
(i)
O+
I
NaH
(j)
HO
OH
CH2N2
(k)
O
PROBLEM 962
HO
Provide the reagents that will effect the following transformations:
Br
CH3O
N3
(a)
HO
Br
CH3O
HO
(b)
PROBLEM 963
OCH2CH3
Bring about the following transformations:
O
HO
OCH2CH3
H
O
I
148
Problems in Chemistry
PROBLEM 964
Hydrolysis of ether proceed via carbocation intermediate as shown below:
H
OCH3
O
O
O—CH3
O
+
H+
+
+ CH3OH
Rate determining step
Based on the above information, rationalize the relative rate of hydrolysis of compounds A to D
below:
O
O
OCH3
I
PROBLEM 965
appropriate:
II
OCH3
O
OCH2CH3
III
OCH3
F
O
IV
OCH3
NH2
Write products in the following reactions. Provide structures of stereo-isomers where
Br
(a)
CH3CH2ONa
m-CPBA
OH
PBr3
(b)
Li
O
CH2
Br
Mg
ether
(c)
D 2O
HBr
excess
OCH3
O
OCH3 CH3MgBr
(d)
excess
Ph
H 2O
H+
H2SO4
Heat
OH
(e) H
CH2CH3
PCC
(i) CH3MgBr
(ii) H3O+
149
Problems
PROBLEM 966
Provide products in the following reactions:
O
OH
(1) NaH
(a)
(2) CH3CH2I
BH3–THF
CH3Li
(b)
H2O2–NaOH
H3O+
OH
O
NaBH4
(c)
K2Cr2O7, H2O
(d)
H2SO4
OCH3
OCH2CH3
(e)
PROBLEM 967
Bring about the following transformations:
OCH3
OH
(a)
OH
(b)
O
(c)
O
O
OCH3
O
(d)
OH
S
O
CH2N2
150
Problems in Chemistry
PROBLEM 968 The conversion A into B by the reaction sequence below does not work well. Explain
why this is the case and provide a better way of doing this.
OH
OTs
LiAlH4
TsCl
PROBLEM 969
Bring about the following transformations:
OH
CN
(a)
O
Br
(b)
OCH3
H
N
OH
(c)
PROBLEM 970
(a)
O
Provide products in the following reactions:
—OH
SOCl2
Pyridine
mCPBA
(c)
CH3
(e)
Hg(OAc)2
(b)
(d)
CH3OH
O
POCl3
OH
Bring about the following transformations:
OH

(a) CH 3 (CH 2 ) 4 CH 2CH 2 Br → CH 3 (CH 2 ) 4 CH  CH 2CN
PROBLEM 971
NaBH4
HI
(CH3)3COK
excess
151
Problems
(b)
—OH
PROBLEM 972
Complete the following reactions:
CH3
SOCl2
(a)
(b)
O
H3 C
CH3MgBr
H 2O
H3 C
OH
Ph
(d)
HI
O
PROBLEM 973
O
OsO4
(d)
NaHSO3
Propose mechanism of the following reaction:
O
PhSNa
H2O
OH
SPh
O
PROBLEM 974
OH
O
O
OH
Explain the following observation:
CH3
CH3
OH
O
CH3MgBr
H2O
H
H
CH3
CH3
CH3
CH3
but
O
CH3MgBr
H2O
OH
H
PROBLEM 975
H
Bring about the following transformations:
(a) PhCH 2CH 2CH 2 Br →
OH

C  CH 2 Ph

CH 3
H 3C
C==C
H
H
152
Problems in Chemistry
O
+
(b)
—C—
—Cl
O
(c)
CH3CH2—C—CH2CN
OH
PROBLEM 976 Starting from propene, propanal methylbromide as the only organic reagents, and any
other inorganic reagent, synthesize the following compound:
OCH 3

CH 2 ==CH  CH 2  CH  CH 2CH 3
PROBLEM 977
Bring about the following transformations starting from indicated materials:
(a)
OH H
+ PhOH
CH3
CH3 Ph
OH
O
O
(b)
C—C2H5
CH3CH2—C—CH2—C
PROBLEM 978
Propose mechanism:
CH2CH3
+ HBr
HOCH2CH2—C—CH2CH3
O
Br
Bring about the following transformations:
OH

(a) CH 3CH 2CH 2CH 2 Br → CH 3CH 2  C  CH 3

CH 2CH 3
PROBLEM 979
(b) PhCH 2CH 3
Br

→ Ph  CH  CH 2OH
CH3
CH3
(c) CH3—C—CH
CH3
CH2
PhI
O
CH3—C—CH2—C—
CH3
SO3H
—I
153
Problems
OH

(d) CH 3CH 2OH + CH 3  → CH 3  CH  C ≡≡C  CH 3
O
CH 3
O

(e) CH 3  CH  CH 2 Br + 
PROBLEM 980
appropriate:
CH 3
OH


→ CH 3  CH  CH 2  CH  CH 2CN
Complete the following reactions and indicate stereochemistry of product where
CH 3

HBr
(a) CH 3  C  CH  CH 2CH 3 →
 
CH 3OEt
+ CH3C
(b)
HO
CMgBr
O
O
KOH
EtSH
(c)
H CH 3
 
PhNH2
TsCl
(d) CH 3  C  C  CH 3 → →
 
OH CH 3
S
H2O2
(e)
H+
OEt
PCl3
(f)
HO
PROBLEM 981
(a)
OMe
NaBr
Et2O
Synthesize the products from indicated reagents and other inorganic reagents required.
OH
Br
O
O
+
+
OH
HO
(b)
Cl
154
PROBLEM 982
Problems in Chemistry
Bring about the following transformations:
Cl
(a)
OH
CH3
Cl
(b)
CH3
CH3
PROBLEM 983 Synthesize the products from indicated starting materials and any other inorganic
reagents needed:
Cl
(a)
+
O
CH2CH2OH
+ CH3OH
CH3
(b)
CH2OCH3
+ PhSNa
SPh
Cl
(c)
CHO +
OH
NC
CH3
(d)
O
+ CH3CH2OH
CH3
OH
PROBLEM 984
OCH2CH3
Show how the following compounds can be synthesized from cyclopentanol?
OH
(a)
OH
C6H5
O
(b)
C6H5
(c)
Ph
OH
PROBLEM 985 When enantiomerically pure (+)-2-phenyl-2-butanol is allowed to stand in methanol
containing a few drops of H 2SO 4 , racemic 2-methoxy-2-phenyl butane is formed. Suggest a plausible
mechanism.
155
Problems
PROBLEM 986
Deduce structures of missing species in the following sequence of reaction:
(i) Mg/Et 2O
Br2
(ii) CH2O
(iii) H3O+
CCl 4
CH 2 ==CH  CH 2 Br → A → B(C 4 H 8 Br 2O)
KOH
25°C
→ C(C 4 H 7 BrO)
Chiral
KOH
heat
D(C 4 H 6O) ←
PROBLEM 987
NaBH4
→ E(C 4 H 8O)
achiral
Bring about the following transformation:
H
OH
OH
PROBLEM 988
H
Bring about the folllowing transformations:
OH
Br
OH
PROBLEM 989
Convert:
CN
(a)
(b)
H
(c) CH3CH2C
CH
H
CH2CH2OH
PROBLEM 990 An organic compound A has molecular formula C 8 H16O 2 and it can be resolved into
enantiomers. A on treatment with acidified dichromate solution yields B (C 8 H14O 4 ) which is achiral. A
on treatment with PCC/CH 2Cl 2 yields another achiral compound C (C 8 H14O 2 ). C on refluxing with
dilute Na 2CO 3 solution yields D(C 8 H14O 2 ) which is distereomeric. D on heating in presence of H 2SO 4
yields E (C 8 H12O) which is still chiral. E on heating strongly in stream of H 2 /Pt yields
1,3-dimethylcyclohexane. Identify A to E.
156
Problems in Chemistry
PROBLEM 991
Provide a detailed, step-by-step mechanism for the following:
OH
TsOH
Heat
OH
O
PROBLEM 992 Provide an ordered sequence of reagents that will convert the single enantiomer
(indicated by *) of a starting material into the desired single enantiomer of product.
*
(a)
H
H
S
OH
H
(b)
CN
OH
PROBLEM 993
H
Bring about the following transformations:
CH3
(a)
OH
+
N
CH3
CH3
(b)
C
PROBLEM 994
(a)
C
Devise synthesis of the following ethers from an alkyl halide and an alcohols:
(b)
O
O
O
(c)
PROBLEM 995
Beginning with alcohols containing not more than three carbons, synthesize:
OH
157
Problems
PROBLEM 996
Starting with alcohol containing not more than six carbon synthesize:
OH
PROBLEM 997
(a)
O
(b)
NH3
O
(g)
O
NaNH2
(c)
(e)
Predict major product in the following reactions:
CH3ONa
(d)
CH3OH
O
(i) LiAlD4
(h)
(ii) H3O+
PROBLEM 998
O
Dil. HCl
(i)
—Li
(ii) H3O+
Predict product in the following reactions:
O
O
(a)
(i) CH3MgBr
(ii) H3O+
(f)
(ii) H3O+
O
O
O
(i) CH3CH2SNa
HBr
excess
OCH3
NaN3
HBr
excess
(b)
O
O
Br
(i) NaBH4
(c)
(d)
(ii) NaOH (aq)
OH
NaH
(e)
O
C6H5CH2Br
(f) O
H2SO4
O
EtOH
O
LiAlD4
NH4Cl
PROBLEM 999 Accounts for the following observations : 2,2-dimethyl oxirane is hydrolyzed at
faster rate than oxirane in acid medium but reverse is true in alkaline medium.
158
Problems in Chemistry
PROBLEM 1000
Propose mechanism:
O
HO
PROBLEM 1001
O
CH2N2
Propose mechanism:
OH
O
O
O
*
OH
HCl
O* +
O
Ph
Ph
Ph
PROBLEM 1002
OH
NaOH/H2O
Identify the missing species:
TsCl
(a)
KCN
A
B
OH
CH3Li
H3
O+
C
PPh3, CH3Br
BuLi
CN
O
H+
(b)
EG
Pd/BaSO4
A
PbO
B
N 2H 4
HO–
C
H 3O +
H 2O
D
H
PROBLEM 1003
Provide reagents necessary for the following transformation:
(a)
OCH3
OH
N3
(b)
OH
(c)
N
N
D
159
Problems
Propose mechanism:
PROBLEM 1004
N 2H 4
H+/Heat
O
O
O
Synthesize the indicated product from the supplied starting materials:
PROBLEM 1005
(a)
OH
H
H
(b)
HOOC
Cl
O
CN
(c)
C
O
and
C
CN
O
O
PROBLEM 1006
Suggest mechanism of the following reaction:
O
OH
CH3COOH
∆
N—NH2
PROBLEM 1007
Propose mechanism of the following reaction:
O
H+
OH
160
Problems in Chemistry
Write product of the following reactions and indicate stereochemistry product where
PROBLEM 1008
appropriate:
O
(a)
H3C
CH3
O
H+
CH3COOH
CH3
CH3
H+
H2O
O
(i) LiAlD4
(c)
O
(b) Ph
Ph
HBr
(d)
(ii) H2O
OH
NaH
(e)
CH3I
CH3
Bring about the following transformations:
PROBLEM 1009
(a)
—OH
O
(b)
from alcohols containing not more than three carbons.
PROBLEM 1010 The following series of reactions was performed during synthesis of a target
molecule. Give structures of each of the indicated compound:
O
OH
Br
PCC
CH2Cl2
HO
A
OH
H 3O +
PROBLEM 1011
reactions:
Mg
B
H+
D
PCC
CH2Cl2
H
C
Et2O
E
NaOMe
MeOH/∆
Predict structures including stereochemistry, of products formed in the following
O
O
(a)
F(C8H12O)
LiAlH4
(b)
Ph
H
(i) PhLi
CH3
(ii) H3O+
161
Problems
CH3
OH
(c)
H
CH3
CH2I2 (1.0 equivalent)
(d)
mCPBA
CH3
PROBLEM 1012
OH
Zn-Cu
Propose mechanism of the following reactions:
OCH3
OH
O
KH
(a)
∆
CH3
H
OMe
H3C
(b)
CH3
CH3
OH
+
CH3
CH3
TsOH
∆
O
ALDEHYDES AND KETONES
PROBLEM 1013 Synthesize 3-pentanone using alkyl bromide containing not more than two carbons.
O
CH3
(a) Br(CH2)5Br
(b)
Br
O
PROBLEM 1014 Convert:
PROBLEM 1015 A hydrocarbon A(C7H14) exist in two stereomeric forms. A on reductive ozonolysis
yields B(C3H6O) and C(C4H8O). B on catalytic hydrogenation yields D(C3H8O) that can’t be resolved
into enantiomers whereas C on catalytic hydrogenation yields E(C4H10O), which can be resolved into
enantiomers. B on treatment with HCN followed by hydrolysis of product yields F(C4H8O3) which can
be resolved into enantiomers. C on treatment with HCN followed by hydrolysis of product yields
G(C5H10O3), which is also resolvable. Identify A to G and show the stereomeric structures of A.
PROBLEM 1016 Which of the following pairs has larger equilibrium constant for addition of water?
O

(a) Cl 3CCHO or Cl 3C  C  CH 3
162
Problems in Chemistry
O
O


(b) Cl 3CCH 2 C  CH 3 or CH 3  CCl 2  C  CH 3
O
O
(c)
H or
C
O
(d)
O
or O N
2
C
NO2
(e)
H
C
MeO
C
O
NO2
NO2
O
or
C
O2N
C
O2N
PROBLEM 1017 An organic compound A(C6H10) on catalytic hydrogenation with H2/Pt yields C6H12. A
on ozonolysis followed by work-up with Zn-H2O yields B(C6H10O2). B on refluxing with dilute NaOH
solution gives C (C6H8O). C on catalytic reduction yields D(C6H12O). D on treatment with chromic acid
yields E (C6H10O2) which on heating with soda lime yields F(C5H10). F on monochlorination yields a
single isomer. Identify A to F.
PROBLEM 1018 Predict products in the following reactions:
(a)
A
+ cold,dil, alk KMnO4
HIO4
–
B
OH
CHO
(b) A + OsO4
(c)
H2O2
B
HIO4
O3
Zn-H2O
–
C
–
A
OH
OH
B
C
163
Problems
PROBLEM 1019 Predict products in the following Cannizzaro reactions:
O

−
(a) CH 3  C  CHO + OH →
PROBLEM 1020 Write products in the following reactions:
(b) H N
2
(c)
CHO + HOC
Zn-H2O
+ O3
–
A
N O2
OH
–
OH
Only cross product
B
PPh
3
(a) Ph  C  CH 2  CH 2  CH 2CH 2Cl →
BuLi

O
PPh3
(b)
BuLi
OHC
(i) O3/Zn-H2O
A
(ii) 50% aq NaOH
B
CH2Cl
PROBLEM 1021 Predict product in the following reactions
O

LiAlH4
H+
H+
(a) CH 3  C  CH 2CH 2CHO + CH 3OH → A →
B → C
Reflux
Excess
O
(b)
HOCH2
CH2OH
+
H
PROBLEM 1022 Predict products in the following reactions:
(a)
O + H2NOH
A
H
+
B
O

–
H+
NH2OH
OH
(b) C 6 H 5  C  CH 3 →
C + D → G → H + I
+
↓H
OH–
E → C 6 H 5 NH 2 + F
164
Problems in Chemistry
NH2OH
(c)
N
+
H
J + K
CHO
OH
N
–
O + P
+
H
L
OH
–
+ M
N
NH2
PROBLEM 1023 An organic compound A has molecular formula C11H14O. A on treatment with
hydroxyl amine yields two stereomers B and C with their molecular formula C11H15NO. B on treatment
with concentrated sulphuric acid solution yields D, an isomer of B where C on treatment with
concentrated sulphuric acid solution yields E, another isomer of C. D on alkaline hydrolysis yields a
white crystalline solid and an optically active amine F(C4H11N). E on alkaline hydrolysis produces
aniline as one product. Identify A to F.
PROBLEM 1024 When 2-chloro-6-methyl cyclohexanone is treated with NaOH, ring contraction
occurs and compound (i) is formed not (ii) explain through mechanism.
COONa
COONa
not
CH4
CH3
(ii)
(i)
PROBLEM 1025 Synthesize (a) 5-nonanone from 1-butanol as only starting organic compound.
(b) 2-butanone from ethyl bromide.
PROBLEM 1026 Outlined below is a synthesis of a compound used in perfumes. Provide all missing
structures.
CH2OH
PCC
A (C11H14O)
Propanal
NaOH
H2/Pd-C
B(C14H18O)
C (C14H20O)
PROBLEM 1027 Show mechanism of the following reaction
O
O
Br
Br +
O
EtONa
O
O
O
PROBLEM 1028 An organic compound A(C12H20) decolourise bromine water, and on reductive
ozonolysis produces two molecules of B(C6H10O). B on aldol condensation followed by dehydration of
165
Problems
product gives C(C12 H18O) which on reduction with N2H4/OH – regenerate A. B on reduction with
Zn(Hg)/C⋅HCl produces C6H12(D) which on monochlorination gives C6H11Cl(E) as a sole product.
Identify A to E.
PROBLEM 1029 Complete the following reactions:
dil. NaOH
(b) C 6 H 5CHO + C 6 H 5COCH 3 →
A+B
O
dil NaOH
(a) C6H5CHO +
A+B
dil. NaOH
(c) C 6 H 5CHO + CH 3 NO 2 →
O
(d)
CHO
dil NaOH
(e) ?
(f) ?
dil NaOH
O
dil NaOH
O
PROBLEM 1030 Carry out the following transformations:
O
(a)
CH=CHCOC(CH3)3
O
(b) C6H5CHO
C6H5CH
NH 2

(c) C 6 H 5CHO → C 6 H 5CH 2 CHCH 3
O
OH
(d)
O
166
Problems in Chemistry
O
(e)
PROBLEM 1031 An alkene (A) on reductive ozonolysis yields acetone and an aldehyde. The aldehyde
is easily oxidised to an acid B. B when treated with Br2/P yields a compound C which on hydrolysis gives
D. D can also be obtained from acetone by reaction with HCN/H 2O. Identify A to D.
PROBLEM 1032 A hydrocarbon A containing C = 88.24% (MW=68) when reacted with Na followed
by n-propylbromide, produces a compound B(C8H14). It gave a ketone C5H16O(C) with Hg2+/H2SO4. B
on refluxing with H+/KMnO4 gave two isomeric acid D and E. Identify A to E.
PROBLEM 1033 A compound having molecular formula C6H12 is treated with Cl2 in CCl4 to form a
compound B(C6H12Cl2). B is treated with alcoholic KOH followed by NaNH2 resulting in the formation
of C (C6H10). C when treated with Pt/H2 gave 2-methyl pentane. Compound C does not react with NaNH2
and with NH4OH/AgNO3. On ozonolysis A gave two aldehydes D and ethanal. Deduce the structure of A
to D.
PROBLEM 1034 Compound A(C7H16O) can be resolved into enantiomers and it reacts with
Na2Cr2O7/H2SO4 to give B(C7H14O). When B is treated with NaOD at 25°C for several hours and then
analyzed by mass spectroscopy, it is found to have MW = 116. Identify A and B.
PROBLEM 1035 An organic compound A(C7H6O) on treatment with hydroxyl amine produces two
isomeric compounds B and C with molecular formula C7H7NO. B and C on acidification rearranges to D
and E respectively. D on alkaline hydrolysis produces a white crystalline solid F(C7H6O2) and ammonia
gas whereas alkaline hydrolysis of E produces G and formic acid. Identify A to G describing mechanism
of their formation.
PROBLEM 1036 The following road-map problem centers on the structure and properties of A, a key
intermediate in these reaction. Give structures for the compounds A through J:
OH
H2CrO4
OH
PCC
A
+
H
H
NaBH4
G
warm H+
I
CH3MgI
H3O +
Tollen's reagent
(ii) H3O +
D
Ethylene glycol
B
(i) PhNHNH2
(ii) H2SO4
OH
(i) J
E
C
Zn(Hg)
HCl
F
167
Problems
PROBLEM 1037 An optically active alcohol A(C5H12O) on oxidation with PCC gives an optically
active aldehyde B(C5H10O). B on treatment with dilute OH – gives C(C10H20O2). C on heating with
H2SO4 gives D(C10H18O) which after reductive ozonolysis gives E(C6H10O2) and F(C4H8O). F on
treatment with I 2 /OH – gives an yellow precipitate. Identify A to F.
PROBLEM 1038 Provide reagents in the following transformations
CHO
?
(a)
?
CHO
CHO
O
?
?
(b)
O
OH
X
(c)
O
Y
OH
CHO
CHO
Z
+
CHO
CHO
O
MeO
(d)
X
CH3
Y
CH3OCH=PPh3
H 3O +
Z
PROBLEM 1039 Propose mechanism of the following reactions:
OH
ROH
(a)
O
OH
O O
+
H
(b)
+
H
O
O
OR
PROBLEM 1040 Synthesize the compounds from indicated starting material:
O
Ph
(a)
COOH
O
(c)
O
(b) PhCHO
Ph
168
Problems in Chemistry
PROBLEM 1041 Predict mechanism of the following reaction:
OH
O
(CH2)10
O
dilute
NaOH
(CH2)9 O
(CH2)9
O
OH
PROBLEM 1042 Give the structures for the indicated compounds in the following reactions:
O
(a) H
CH2CH2OH
CH3CH2MgBr
B
A
+
H
CH3(CH2)5CHO
C
+
H
Pd-CaCO3
H2
D
O
(b)
Br
Mg
Ether
O
(c)
E
H3
O+
F
HOCH2CH2OH
+
H
PCC
G
BH3/H2O2/KOH
H
+
I
H
J
PROBLEM 1043 Complete the following equations.
O
(a)
H
O
+
NaOH
H 3O +
NaOH
H 3O +
O
O
(b)
+
CHO
2 equivalent
1 equivalent
(c) 2
CHO
NaOH
PROBLEM 1044 Plan a synthesis for each of the following compounds through shortest route. You
have been supplied bromobenzene and any organic reagents containing three or fewer carbon atoms and
any inorganic reagents you need.
O
(a)
CH3CH2
C
(b)
H
C
CH2CH2OH
H
169
Problems
O
Br
(c)
(d)
H O
(e)
(f)
H
PROBLEM 1045 A neutral, resolvable organic compound A has molecular formula (C8H16O2). A on
treatment with LiAlH4 gives isomeric B and C (C4H10O) of which only B is optically active. B on
treatment with acidified dichromate solution gives C4H8O(D) which on refluxing with dilute NaOH
followed by acidification of product gave E(C8H14O). E on heating with N2H4 in alkaline medium
affords F(C8H16). F on treatment with B2 H 6 / H 2O 2 /OH – produced a resolvable G(C8H18O). G on
treatment with acidified dichromate solution produced H(C8H16O) which on treatment with
“m-chloroperbenzoic acid” produces A. Deduce structures of A to H.
PROBLEM 1046 An organic compound A has molecular formula (C7H12O) and does not decolourise
Bayer’s reagent and evolves no gas with Na-metal. A also does not react with either ammonical silver
nitrate solution or phenyl hydrazine solution. A on refluxing with methanol in presence of trace amount
of H2SO4 produces a non-resolvable B(C8H16O2). B on treatment with CrO3/pyridine produced
C(C8H14O2) which is non-resolvable. C on refluxing in HI solution produced another non-resolvable
compound D(C7H11OI). D on treatment with alcoholic KOH followed by oxidative ozonolysis of
product produced E(C7H10O4) which does not show optical activity. E on simple heating produced
C6H10O2 which on refluxing with dilute NaOH gave:
O
Deduce structures of A to E.
PROBLEM 1047 Predict the structure of the products in the following reactions and show
stereochemistry when it is known:
O
(a)
C
CH2CH2CH2Cl
(i) HOCH2CH2OH
(ii) TsOH/Benzene
CH2 NH
A
2
H3O
B
+
C
170
Problems in Chemistry
O
NaBH4/H2O
(b)
BH3
(c)
A
NaOH
A
+
Na2Cr2O7
CrO3 H
B
H
(e)
HC
NaNH2
CH
D
KOH/Ethanol
B
+
PhLi
CHO
TsCl
Pyridine
H2O2
THF
(d)
A
C6H5CH2Br
PhNHNH2
CH3CH2MgBr
E
O
(f)
H2NNH2
X
KOH
ethylene glycol
PROBLEM 1048 Bring about the following conversions:
OH
O
(a)
CH3
OMe
CH3
O
(b)
CHO
OMe
Br
O
OMe
(c)
CH3
CH3
PROBLEM 1049 (a) Bring about the following transformation:
CHO
OH
CH2Br
B
O
C
F
C
mCPBA
D
171
Problems
(b) Propose mechanism
Et
O
O
+
OH
H
O
OH
PROBLEM 1050 Propose mechanism:
OH
+
H
(a)
OH
O
(b)
+
H
O
O
(c)
O
+
CH2OH
H
O
PROBLEM 1051 Propose mechanism:
O
(a) MeOOC
COOMe +
H
H
Heat
O
(b) EtOOC
O
O
O
COOEt
+
O
t-BuOK
O
O
(c)
NaOMe
MeNHOH
H
O
N
Me
HO
COOEt
172
Problems in Chemistry
PROBLEM 1052 Bring about the following conversions:
O
(a)
H
O
H
H
H
H
(b)
(c)
OH
OH
Cl
(d)
(e)
+
PROBLEM 1053 An optically active compound A has molecular formula C10H12O. A decolourises
Bayer’s reagent as well as it forms salt with NaOH. A on treatment with HBr forms B(C10H13OBr) as
major product which is still enantiomeric. B on treatment with alcoholic solution of KOH gives a
stereomeric C that is also an isomer of A. A on treatment with acidic solution of KMnO4 gives ortho
hydroxy benzoic acid as one of the product. A on treatment with B2H6/H2O2/NaOH produces
D(C10H14O2) which on refluxing with dilute sulphuric acid produced E(C10H12O), E neither decolourises
Br2-H2O solution nor forms any salt with NaOH. Deduce structures of A to E.
PROBLEM 1054 An organic compound has molecular formula C10H6(A). A on
treatment with HgSO4/H2SO4 gives B(C10H10O2) which forms an insoluble
precipitate with 2,4-DNP but fail to react with ammonical solution of AgNO3. B on
refluxing with dilute NaOH solution produced C(C10H8O). C on further treatment
with alkaline solution of N2H4 gives D(C10H10) which decolourises Br2-H2O
solution. D on treatment with H2/Pt produced the adjoining compound X.
Deduce structures of A to D.
X
173
Problems
PROBLEM 1055 Complete the followings:
O
?
NH
(a)
NH
CH3
H
N
N
?
(b)
O
I
PPh3
(c)
?
BuLi ∆
O
NH2
H
(d)
COOH
?
O
O
O
(e)
Me2CuLi
+
H
(Propose mechanism)
PROBLEM 1056 Provide products in the following reactions:
O
(a)
LDA
O
(CH3)2CHI
OH–
(b)
OH2
Ph
O
PROBLEM 1057 Propose mechanism:
O
CHO
BF3
174
Problems in Chemistry
PROBLEM 1058 Convert:
Ph
OH
O
H3C
H
H
H
Ph
PROBLEM 1059 Propose mechanism:
O
CH2
(a)
OMe
O
(b)
H3O+
X (C4H8O2)
NH2OH
+
H
PROBLEM 1060 Starting from propanone, suggest synthesis of the following compounds:
O
O
(a)
COOEt
Ph
(b)
(c)
CN
O
O
(d)
(e)
(f)
Ph
O
PROBLEM 1061 Offer mechanistic explanation for the formation of products in the following
reaction:
O
O
MgCl
OH
+
OH
+
PROBLEM 1062 Propose mechanism of formation of final products in the following reactions and
identify the labelled product:
COCl
O
CH2N2(excess)
Ag2O/MeOH
(a)
A(C8H12N2O)
O
175
Problems
(b) Me2CO
750oC
MeOH
B(C2H2O
Cool
O
NaN3
O
H2SO4
O
MeOH
C(C4H4O2)
(c)
CH3COOMe
O
D(C6H11NO)
LiAlH4
Me
NH
H2SO4
(d)
O
HO
O
(e)
(f)
O
H2SO4
CH2OH
H2SO4
HO
O
PROBLEM 1063 Supply a suitable technique for the synthesis of following compound from indicated
starting material:
O
OCH3
from
CH3
OCH2CH3
PROBLEM 1064 A sweeting smelling organic compound A has molecular formula “C 9 H12O 3 ” and it
does not change colour of moistened litmus paper. A on acidic hydrolysis yields an optically active
compound B with their molecular formula “C 9 H14O 4 ”. B does not decarboxylate on simple heating. B
on heating with soda-lime (NaOH/CaO) yields an optically inactive compound C “C 8 H14O 2 ”. C does not
reduce Tollen’s reagent but yields a bright orange crystalline substance on treatment with
2,4-dinitrophenylhydrazine. C on oxidation with acidified K 2Cr 2O 7 followed by heating of product with
soda-lime yielded D “C 7 H12O” which is still non-resolvable. D on heating with alkaline solution of
hydrazine yielded methyl cyclohexane. Identify A to D.
PROBLEM 1065 An organic compound A has molecular formula “C 9 H12 ” and decolourises
bromine-water solution. A on treatment with H 2 /Pt gave an optically active hydrocarbon. A on treatment
176
Problems in Chemistry
with acidified solution of KMnO 4 yields B “C 9 H12O 7 ” which can be resolved into enantiomers as well
as gives bright orange crystalline substance with 2,4-dinitrophenyl hydrazine. B on simple heating gives
an optically inactive C(C 7 H12O 3 ) which evolves a gas with aqueous NaHCO 3 and gives a yellow
precipitate with NaOH/I 2 . C on treatment with NaBH 4 yields a resolvable D“C 7 H14O 3 ”. D on heating
with H 2SO 4 yields E “C 7 H12O 2 ” which does not change the colour of moistened litmus paper. Identify
A to E.
PROBLEM 1066 An organic compound A has molecular formula “C 9 H16O” and it gives an yellow
precipitate on treatment with alkaline solution of iodine. A on treatment with hydroxyl amine (H 2 NOH)
yields two stero-isomeric compound B and C with their molecular formula “C 9 H17 NO”. B on treatment
with concentrated sulphuric acid undergo rearrangement to yield a stable product D which is an isomer
of B. C on similar treatment undergo similar type of rearrangement to yield a stable product E which is an
isomer of C. Hydrolysis of E in acid medium yields F(C 7 H15 N) as one of the product. F on treatment with
nitrous acid followed by oxidation with acidified dichromate solution yields G(C 7 H12O 2 ). G on heating
with sodalime yields a product C 6 H12 which on heating with nitric acid affords a single mono-nitro
derivative. Identify A to G.
PROBLEM 1067
important.
Predict all products of the following reactions and show stereochemistry where
CHO
(a)
+ CH3NH2
HCl
O
CH3OH (excess)
(b)
HCl
H
H
HO
(c)
CH2
OH
PPh3
HCl
H 2O
HCl
O
O
(d)
Ph3P/BuLi
Br
(e)
OCH3
HCl
OCH3
H 2O
O
(f)
O
O
Acetone
CH3NH2
N 2H 4
X
Pd/C
H2
KOH
Heat
Y
CH3COCl
HCl
H 2O
Z
177
Problems
Provide reagents for the following transformations :
PROBLEM 1068
OCH3
(a)
OCH3
CH2
(b)
OCH3
NHCH3
OCH3
O
O
OCH3
(c)
NOCH3
OCH3
OCH3
(d)
CH2
OCH3
OH
O
O
PROBLEM 1069 Arrange the following acetals/ketals in increasing ease of their hydrolysis in acidic
medium—(provide mechanistic reasoning).
O
O
O
O
II
I
O
O
III
PROBLEM 1070 The reaction of aldehyde A with HCN gives two cyanohydrin products B and C.
Explain why C in produced in greater amount than B.
OH
OH
CHO
HCN
OH
A
PROBLEM 1071
for your choice.
OH
OH
CN
+
Write the structures of major and minor stereo-isomers of product explaining reasons
PCC
(a)
CH3MgBr
OH
CH3
(b) H
CN
OH
C
OH
B
H
Ph
OH
O
H
CH3MgBr
H3 O +
H3 O+
178
Problems in Chemistry
PROBLEM 1072
Bring about the following transformations:
H
H
I
(a)
H
+
H
H
OH
(b)
O
O
O
O
(c)
OH
H
OH
OH
OH
O
(d)
O
H
O
PROBLEM 1073 Draw the product of the reaction A and B and predict which one proceeds more
efficiently and why?
O
A:
Cl
H
H
B:
CN
HCN
H
HCN
H
CN
O
PROBLEM 1074
OH
Cl
OH
Bring about the following transformations:
O
O
(a)
+
N Cl–
H
CH3
CH3
O
(b)
N—NH2
179
Problems
O
(c)
H
OCH3
O
O
O
Provide a mechanistic explanation for the reaction:
PROBLEM 1075
O
O
O
HCl
+ NH2OH
N
Predict products in the following reaction. Provide stereoisomers where appropriate.
PROBLEM 1076
O
CH3Li
(a)
NaH
CH3I
H 3O +
CH3
O
(CH3)2S
(b)
CH2
NaN3
H 2O
H
OH
(c)
O
(d)
K2Cr2O7
CH3CH2MgBr
H+/H2O
ether
HOCH2CH2OH
Ac2O
HCl
∆
OH
PROBLEM 1077
H 3O +
CH3CH
PPh3
H 2O
HCl
Bring about the following transformations:
O
O
O
N
PROBLEM 1078
Propose mechanism for the reaction:
O
CH3O
OPh
O
H+
CH3OH
O
CH3O
HO
OCH3
OPh
180
Problems in Chemistry
Bring about the following transformations:
PROBLEM 1079
O
HO
(a) F
NH
OH
(b)
O
CHO
O
(c)
COOH
H2N
H
PROBLEM 1080 The equilibrium established when cyclohexanone is treated with methanol and HCl
lies in the favour of ketone. In contrast, the same reaction with cyclopropanone lies heavily in favour of
hemiacetal. Explain.
O
OCH3
CH3OH/H+
H
:
O
H
CH3OH/H+
OCH3
PROBLEM 1081 Within each set, select the compound which is more reactive in nucleophilic addition
at carbonyl carbon:
(b) CH 3CHO or CH 3COCH 3
(a) CH 3COCH 2CH 2 Br or CH 3COCH 2 Br
(d) PhCHO or CH 3CHO
(c) CH 3COCH 3 or CH 3  CO  CF3
PROBLEM 1082
Provide reagents for accomplishing the following transformations:
OCH3
C
Ph
OCH3
CH3
D
OH
CH3
A
C
C
Ph
OH
O
CH2 CH
CH2
Ph
CH3
B
CH3
Ph
N
CH3
CH2
C
Ph
CH3
OCH3
181
Problems
PROBLEM 1083
Bring about the following transformations:
CH3
O
O
O
O
(a)
(a)
CHO
OH
CH3
PROBLEM 1084 Treatment of trans 2-chlorocyclohexanol with NaOH yields 1,2-epoxycyclohexane
while reaction of cis isomer under the same condition yields cyclohexanone. Propose mechanistic
explanation.
PROBLEM 1085 When 4-hydroxybutanal is treated with methanol in the presence of acid catalyst,
2-methoxy tetrahedrofuran is formed. Provide mechanism.
PROBLEM 1086
Give products of the following reactions (major):
O
NH3
(1) LDA
(a)
(b)
(2) CH3Br
HCN/H2O
H
O
O
O
(c)
H
O
OH–
+
Heat
O
(d)
Ph +
H+
N
H
O
(e)
Br2/CH3COOH
KBuO+
PROBLEM 1087 When compound A is treated with CH 3ONa in CH 3OH, isomerization to compound B
occurs. Provide mechanism and explain why equilibrium favour product B?
H
O
H
CH3ONa
CH3OH
H
A
H
B
O
182
Problems in Chemistry
PROBLEM 1088
Bring about the following transformations:
O
OH
(a)
C
CH3
C
H
H
OH
H
(b)
CH3
CH3
(c) H3C
OH
CN
O
O
CH3
O
(d)
PROBLEM 1089
Provide synthesis of the compounds from indicated starting materials:
(a)
OCH3
O
(b)
N
H
PROBLEM 1090 An optically active organic compound A(C 4 H 8O 2 ) gives an orange precipitate with
2,4-dinitrophenylhydrazine and a silver mirror with ammonical silver nitrate solution. A is converted
into B(C 4 H 6O) on treatment with concentrated sulphuric acid. B may exist as cis or trans isomers. B
reacts with HCl to give C(C 4 H 7OCl) as major product and with chlorine to give D(C 4 H 6OCl 2 ).
Reduction of C with Sn/HCl gives E(C 4 H 9OCl) which on hydrolysing with aqueous alkali yields
F(C 4 H10O 2 ). F gives a yellow precipitate with I 2 and NaOH solution but A does not. Identify A to F.
PROBLEM 1091
Synthesize products from indicated reactants and other inorganic reagents needed:
183
Problems
(a) CH 3CH 2 —C ==CH 2 + CH 3  CH  CH 3


OH
Br
CH 3

→ CH 3CH 2  C ≡≡C  C== CH 2
CH 3
CH 3


(b) CH 3  C  OH → CH 3  C  COOH


Br
CH 3
CH 3
CH 3 CH 3



(c) CH 3CH 2  CH  CH==CH 2 + CH 2 == CH 2 → CH 3  CH 2  CH C C ≡≡CH

OH
CH2CH3
(d) CH3CH2—C
O
OH
CH2 +
CH3CH2—CH—C—
CH2CH3
PPROBLEM 1092
Cl
Propose mechanism:
C6 H 5
O
O


CH3OK
C 6 H 5CH 2 —C—CH 2C 6 H 5 + CH 2 ==CH—C —CH 3 →
H3C—
CH3OH
PROBLEM 1093
(a) H3C—
3
C6H5—CHO
NaOH/H2O
PROBLEM 1094
Propose mechanism:
O
O
Br
(a)
O
—C6H5
Predict product in each of the following showing stereochemistry where appropriate:
O
O
CH3
C6H5CH2SH
NaOH
(b)
—CHO + CH3—C—CH3
NaOH/H2O
H2O
O
CH
O
(c)
—OH
(CH3)3COK
184
Problems in Chemistry
O
O O
O
(b) Ph—C—C—Ph + Ph—CH2—C—CH2—Ph
Ph
Ph
KOH
C2H5OH
Ph
Ph
PROBLEM 1095 The acid catalyzed hydrolysis of two acetals shown below occurs at much different
rates. Provide a brief explanations with structures of key intermediate to illustrate why these two acetals
undergo hydrolysis at such different rates.
H
O
H
H 3O +
O
O
H
O
Reaction is complete in 1.0 hour at 20°C
H 3O +
O
H
Reaction is complete in 24 hour at 60°C
O
PROBLEM 1096
Provide a stepwise mechanism for the following acid catalyzed reaction:
CH3
O
H+
2CH3CHO
O
CH3
PROBLEM 1097
Rank the following compounds in increasing order of electrophilicity:
O

F3C
PROBLEM 1098
CH 2

CH 3
H
I
NH

H
CH 3
II
III
CH 3
Provide a stepwise mechanism for the following acid catalyzed reaction:
CHO
H3O+
O
H2O
O
HO
OH
O
O
PROBLEM 1099
H
Br
OH
185
Problems
PROBLEM 1100
Identify A in the following reaction and propose mechanism of its formation:
O
NaOCH3
I
H
C5H10O2 (A)
CH3OH
H 3O +
HO
H 2O
H
O
NaOH — No reaction.
PROBLEM 1101 The enantiomerically pure cyclic ketone, shown below, loses optical activity upon
standing in a solution of EtONa/EtOH. Provide mechanistic explanation.
EtOH
EtONa
O
CH3
H
PROBLEM 1102
Racemic mixture
Bring about the following transformations:
O
CHO
CH3
(a)
CH3
O
(b)
OH
OMe
OH
OH
(c)
OH
Br
OH
PROBLEM 1103 An organic compound A has molecular formula C 6 H12O. A does not decolourize
bromine water solution but evolves a gas on heating with NaH. A can be resolved into distereomers and it
also change colour of acidified dichromate solution from orange to blue green. A on heating with
concentrated sulphuric acid yields B(C 6 H10 ) as a major product which can’t be resolved into
enantiomers. B on oxidation with ozone followed by work-up with Zn  H 2O yields C(C 6 H10O 2 ). C on
refluxing with dilute solution of NaOH gives D–an isomer of C. D also evolves a gas wtih NaH. D on
heating dehydrate to yield E(C 6 H 8O). E on heating with H 2 gas in presence of platinum catalyst at high
temperature and pressure yields F(C 6 H12 ) which gives single monochloro derivative on free radical
chlorination. Identify A to F.
PROBLEM 1104 An organic compound A has molecular formula C 7 H14O 2 and it can be resolved into
enantiomers. A on treatment with H 2CrO 4 yields B(C 7 H12O 2 ) which can’t be resolved into enantiomers.
186
Problems in Chemistry
Refluxing B with dilute alkali solution yields “C” which is an isomer of B. C evolves a gas with NaH but
does not change colour of acidified dichromate solution. C on heating with concentrated sulphuric acid
yields D(C 7 H10O) which on heating with Pt at high pressure yields methyl cyclohexane. Identify A to D.
PROBLEM 1105 An organic compound a has molecular formula C11 H 22 . A on treatment with
concentrated solution of KMnO 4 in acidic medium yields B(C 3 H 6O) and C(C 8 H16O), both form orange
colour precipitate with 2,4-dinitrophenyl hydrazine but neither gives positive Tollen’s test. C on
treatment with LiAlH 4 yields D(C 8 H18O) which on heating with concentrated sulphuric acid solution
gives E(C 8 H16 ). E on further treatment with acidified permanganate solution yielded B and 2,2-dimethyl
propanoic acid. Identify A to E.
PROBLEM 1106
Identify the compounds labeled A to F:
NBS
(CH3 )3 COK
O3
hν/CCl 4
(CH3 )3 COH
Zn  H2O
MeMgBr
A(C 8 H10 ) → B(C 8 H 9 Br) → C → D(C 7 H 6O) →
H 3O +
PCC
CH2Cl 2
N 2H 4
A
→ E → F →
–
H 2O
PROBLEM 1107
OH /∆
Arrange the following hydrates in order of increasing K eq :
O
R
A:
C
H 2O
A
HO
C
R
OH
A
H, (CH 3 ) 2 CH—, ClCH 2 — Ph—.
PROBLEM 1108 Propose a reasonable synthesis of the product in the following reactions starting from
the indicated materials.
CHO
(a)
CHO
O
N
(b) Compound containing 5 or less carbons
PROBLEM 1109
Give structures of all the products in the following reactions:
O
(a)
NaBH4
B2H6
(b)
H2O2/NaOH
NH4Cl
O
187
Problems
O
OH
Ph
(c) H
(i) (CH3)2CHMgBr
Et2O
(ii) H+
H
(d)
(iii) H+/H2O
H 2O
H
CH3 OH
NaBH4
(e)
(f) Ph—CH—CH—CH3
NH4Cl
O
(i) CrO3/H2SO4
(ii) PhCH2Li
(i) PCC
(ii) MeMgBr
(iii) H+/H2O
CH3
O
O
(i) LiAlH4
(g)
(ii) NH4Cl
PROBLEM 1110
Complete the following reactions:
O
O
NH2(excess)
H
(a) H
O
H2/Pt
C—NH2
(b)
CHO
CH3COCl
O
O
Ph
?
(c)
Br2
NaOH
(i) N2H4/OH–
(d)
(ii) O3/Zn–H2O
(iii) EG/H+
NH2
(e)
H+
O
PROBLEM 1111
Bring about the following transformations:
OH
( a)
O
( b)
O
O
NO2
I
188
Problems in Chemistry
(c)
NH2
PROBLEM 1112
N
Propose mechanism:
O
OH
PROBLEM 1113 (a) Draw all the enol forms possible 4-methyl-1,3-cyclopentandione.
(b) Give product of the reaction of above diketo compound with:
(i) Excess of D 2O/H + , (ii) One mole of Br 2 /H + , (iii) LDA/THF then CH 3I.
PROBLEM 1114
Complete the following reactions:
O
KMnO4
(a)
OMe
(b)
H+
OH–
OH
OH
(c) 2PhCH2CHO
O
HO–
Warm
NaBH4
(d)
CH3OH
O
(e)
CH3NH2
CHO
LDA
MeI
CHO
Br2
(f)
H+
O
PROBLEM 1115
Consider the compound shown below:
O
O
H
(a) Give the product of its reaction with excess of D 2O.
(b) Give product of reaction with one mole of Br 2 in dilute acidic medium.
189
Problems
(c) Give product of its reaction with one mole of C 6 H 5CHO in presence of strong base EtONa.
(d) Write structure of its most stable enol form.
PROBLEM 1116 Of the two compounds A and B shown below, A losses its optical activity upon
standing an enantiomerically pure form of it in a solution of NaOEt but no such loss of optical activity is
observed on similar treatment to B. Explain:
A
B
O
CH3
O
Give a complete stepwise mechanism for the reaction:
PROBLEM 1117
O
O
O
NaOH
∆
Complete the following reactions:
PROBLEM 1118
O
(a)
+ LDA
MeI
H3O+
KCN
(b)
H2O
CHO
O
O
O
OEt
(c)
H3O+
O
heat
Br2
(d)
H+/H
2O
O
H2CO
(e)
(f)
Li
HO–/∆
CH2O
HO–/∆
O
O
(g) Cl
Cl
H2O
H+
(h)
H+
+
N
H
190
Problems in Chemistry
PROBLEM 1119
Synthesize the following compound starting from alcohols of three or less carbons:
PROBLEM 1120
Write product in the following reactions:
OH
(a)
O
O
H 3O +
PhMgBr
PCC
CH2CH2
PCC
CH2Cl2
O
(b) H
HO–
D 2O
H
excess
O
(c)
H
HO–
warm
KCN
(d)
H 2O
O
COOH
(e)
CH3Li
H 2O
excess
H+
O
(f)
H2
H+
H + H—N
Ni
H
N
(g)
H+
O
PROBLEM 1121
Give a complete, stepwise mechanism of the following reactions:
O
(a) HO
H+
H
CH3OH
O
OCH3
191
Problems
O
(b)
O
CH3
O
CH3
H2O
HO–
A
H3O+
heat
CH3
O–
CN
NO2
—CHO + H2N—O—
(c)
—NO2
H+
[A]
CH3O–
+
CH3OH
NO2
PROBLEM 1122
O
NO2
Rationalize the difference in enol contents shown below:
O
OH
O
OH
H
I
PROBLEM 1123
H
II
0.001%
O
92%
Propose mechanism of the following reaction:
O
N
NaOH
OH + NC—N
OMe
O
H
N
OMe
H
PROBLEM 1124
Bring about the following transformations:
OH
CHO
(b)
(a)
O
O
OH
(c)
COOMe
O
O
192
Problems in Chemistry
Explain, with the aid of mechanism, the differing outcome in the following pair of
PROBLEM 1125
reactions:
OH
O
HNO2
( a)
H2SO4
NH2
OH
HNO2
but
H2SO4
NH2
CHO
O
O
( b)
CH3
CH3
NaOMe
—P h
O
Ph
OH
O
O
NaOMe
but
O
OH
PROBLEM 1126
Propose mechanism:
TsOH
(a)
Ph
Ph
O
O
O
OH
MeMgBr
(b)
O
HClO4
NaOH
193
Problems
PROBLEM 1127
Explain the following observation:
Br
O
LDA
Et2O
O
O
Br
LDA
Et2O
but
O
O
PROBLEM 1128 An organic compound A has molecular formula C 9 H18O and it can be resolved into
enantiomers. A evolves a colourless gas on heating with Na-metal, but does not change the colour of
acidified dichromate solution or bromine water solution. A on heating with concentrated sulphuric acid
undergo dehydration to yield three isomeric products B, C and D all having molecular formula C 9 H16 . B
shows cis-trans isomerism and on ozonolysis followed by work-up with Zn  H 2O produced
E (C 7 H12O) as one product. E formed an yellow precipitate on treatment with alkaline solution of iodine.
Ozonolysis of C followed by work-up with Zn  H 2O produced F (C 5 H 8O) which did not produce
yellow precipitate with alkaline NaOH but produced a single isomer G(C 5 H 9 NO) on treatment with
H 2 NOH. Also D on ozonolysis followed by work-up with Zn  H 2O produced nonan-2,7-dione.
Identify A to G.
PROBLEM 1129 An organic compound A has molecular formula C 8 H16 . A on treatment with aqueous
(CH 3COO) 2 Hg followed by reduction with NaBH 4 yields yields B (C 8 H18O) which cannot be resolved
into enantiomers. A on treatment with B2 H 6 /H 2O 2 in alkaline medium yields C, an isomer of B, but it can
be resolved into enantiomers. Ozonolysis of A, followed by work-up with dimethylsulphide, yields
D(C 6 H12O) as one product which is non-resolvable. On catalytic reduction with LiAlH 4 , D yields
E (C 6 H14O) which can be resolved into enantiomers. E on heating with concentrated H 2SO 4 yields two
isomeric alkenes F and G, of which F is stereo-isomeric while G is not. Deduce structures of A to G.
CARBOXYLIC ACID AND ITS DERIVATIVES
PROBLEM 1130 How could you use an acid base extraction technique to separate a mixture of
carboxylic acid and phenol into its pure components?
PROBLEM 1131 A carboxylic acid (A) of unknown structure was found to contain only C,H and O. 150
mg of A required 11.9 mL of 0.22 N NaOH to reach the equivalent point. Gentle heating of A evolves
CO2 and a new carboxylic acid (B) was formed whose equivalent weight was found to be 74.Identify A
and B.
PROBLEM 1132 Propose mechanism:
OH
(a)
Br2/CCl4
O
O
O
Br
194
Problems in Chemistry
O
+
(b)
H2SO4
OH
O
C—CH3
O
COCH3
+
OH
(c)
N
H
NHCOCH3
PROBLEM 1133 Predict major product in the following reaction:
(a)
COOH
SOCl2
CH3NH2
COOH
(b)
(CH3CO)2O
Heat
COOH
OH
O
OH
(c)
+ COCl
2
CH3OH
O
(d)
O
OH
ClH
PROBLEM 1134 Bring about the following transformations:
(a)
COOH
(b)
COOH
O
COOH
(c)
PROBLEM 1135 Complete the following reactions:
O
O
(a) MeO
OMe
NaOH(aq)
(b)
O
CH3CH2OH
O
(c) PhCH2COOEt
EtONa
PhCOOEt
(d) EtO2CCH2COOEt
EtONa
CH3CH2I
195
Problems
H3O +
(e) H3C
(i)
LiAlH4
CN
(ii)
PhMgBr
(iii)
+
H
PROBLEM 1136 Convert:
O
COOEt
COOH
(a) HOOC
COOEt
COOEt
EtOOC
(b) CH3COOH
PROBLEM 1137 Bring about the following conversions:
(a) PhCH 2OH → PhCH 2CN
(b) PhCOOH → PhC(CH 3 ) 2 OH
(c) EtOOC
COOEt
COOEt
PROBLEM 1138 Show how would you accomplish the following synthesis in good yield?
O
O
O
(a)
O
(b)
O
O
O
O
(c)
COOH
O
(d)
O
COOH
COOH
O
PROBLEM 1139 Show how would you accomplish the following multisteps synthesis?
(a) 6-hepten-1-ol → caprolactum
CH2Br
NH2
(b)
CHO
(c)
CN
196
Problems in Chemistry
PROBLEM 1140 Synthesize the following compounds from indicated starting materials:
O
(a)
O
from
COOH
O
O
(b)
from EtOOC
COOEt
PROBLEM 1141 Show how would you accomplish the following synthesis:
(a) Isobutyl amine → N-isobutyl formamide
(b) Ethylacetate → 3-methyl-3-pentanol.
(c) Cyclohexyl amine → N-cyclohexylacetamide
(d) Bromocyclohexane → dicyclohexylmethanol
PROBLEM 1142 A neutral organic compound A has molecular formula C5H8O2 and does not
decolourise Bayer’s reagent. A on acidic hydrolysis produced B(C5H10O3) which is resolvable. B on
heating with concentrated H2SO4 undergo dehydration to produce C, an isomer of A. C decolourises
Bayer’s reagent and show stereoisomerism. Also B changes colour of acidic dichromate solution from
orange to blue-green converting itself into D(C5H8O3) which is non-resolvable and forms an yellow
precipitate on treatment with NaOI solution. Deduce structures of A to D.
PROBLEM 1143 A neutral organic compound A(C5H8O2) does not decolourise Bayer’s reagent and on
hydrolysis with dilute H2SO4 produces B(C5H10O3) which is diastereomeric. B on heating with
concentrated H2SO4 undergoes dehydration producing C(C5H8O2) which shows geometrical isomerism.
Also B on treatment with acidic dichromate solution produced D(C5H8O3) which is enantiomeric and
gives an yellow precipitate with NaOI. D on gentle heating produces E(C4H8O) which is nonresolvable.
Deduce structures of A to E.
PROBLEM 1144 An organic acid A(C5H10O2) reacts with Br2 in presence of phosphorus to produce a
resolvable B. B on dehydrobromination yields C. C does not show geometrical isomerism and on
decarboxylation gives an alkene D. D on ozonolysis gives E and F. Compound E gives positive Schiff’s
test but F does not. Give structures of A to F.
PROBLEM 1145 A monobasic acid A has neutralization equivalent 116 and does not decolourise
Bayer’s reagent. Also A is enantiomeric. A on treatment with Br2/red phosphorus produces B which is
still resolvable. B on dehydrobromination produced C which show stereo isomerism. C on
decarboxylation produced D which does not show stereoisomerism. Deduce structures of A to D.
PROBLEM 1146 An organic lactum A on acid hydrolysis produced B, an amino acid. B on treatment
with nitrous acid gives C. C on heating with concentrated H2SO4 produces a lactone D. A can also be
synthesized by the reaction of cyclopentanone with hydroxylamine followed by treatment of product
with concentrated H2SO4. Deduce structures of A to D.
PROBLEM 1147 A neutral organic compound A has formula C9H16O3 and rotates plane polarized light.
A on acid hydrolysis produces B and C(C3H8O). C on partial oxidation with PCC gives D(C3H6O) which
197
Problems
does not gives iodoform test. D on treatment with dilute solution of NaOH gives E(C6H12O2) which is
diastereomeric. E on treatment with acidic solution of KMnO4 produces B. Deduce structures of A to E.
PROBLEM 1148 An organic compound A(C9H6O3) does not reacts with aqueous solution of NaHCO3
and does not change the colour of litmus paper. A on acid hydrolysis gives B(C9H8O4) whereas A on
treatment with methanol in acid medium gives C(C10H14O4) as the major alcoholysis product. B on
heating with excess of soda-lime produces toluene. Deduce structures of A to C.
PROBLEM 1149 Diethyl malonate on treatment with one mole of C2H5ONa followed by reaction with
one mole of 5-bromoethyl pentanoate produces A(C14H24O6). A on heating with dilute H2SO4 yields
B(C7H12O4). B on treatment with excess of C2H5OH in presence of catalytic amount of H2SO4 yields
C(C11H20O4). C on treatment with one mole of EtONa followed by work-up with dilute HCl yields
D(C9H14O3). D on further treatment with one mole EtONa followed by reaction with CH3I yields
E(C10H16O3). E on heating with HCl yields 2-methylcyclohexanone. Deduce structures of A to E.
PROBLEM 1150 An organic compound contain C,H and oxygen and its aqueous solution is neutral.
1.44 g of A on refluxing with dilute H2SO4 produces two organic compound B and C of which only B is
resolvable and gives effervescence with NaHCO3. Also neutralization of entire B require 0.4 g of NaOH.
B on reduction with HI/red phosphorus produces a hydrocarbon D which is non-resolvable. Also D on
monochlorination yields three alkyl halide in which only one is resolvable. Deduce structures of A to D.
PROBLEM 1151 Identify the labelled product and explain their formations:
O
O
Heat
O
A (C8H14O)
NaBH4
B (C8H16O)
PROBLEM 1152 Give structural formula of all possible products in the following reactions:
(a)
MeO
(b)
NH2OH
COOEt
CH2COOH
(c)
(CH3CO)2O
NH2
(CH3CO)2O
heat
COOH
(d)
CHO
+
Ag2O/H2O
X
O
(e)
O
Li
X
H
Y
(CH3CO)2O
Y
198
Problems in Chemistry
PROBLEM 1153 Complete the following reactions:
O
(a)
NaBH4
OMe
SOCl2
A
B
O
O
NaCN
(b)
C
(CH3)2SO4
+
D
CF3
H
heat
E
PROBLEM 1154 Complete the following sequence of reaction:
Li
TsOH
HOCH2
A
BrCH2CH2Cl
B
O
C
NaCN
+
E
H
NaOH
OH2
D
PROBLEM 1155 Complete the following sequence of reaction explaining the formations of products in
the second step:
O
mcpba
A
O
CH3COCl/LiCl H C
3
O
Cl
+ H3C
O
Cl
PROBLEM 1156 Bring about the following transformations:
(a)
H 3C
H
(b)
H 3C
H
H 3C
H
CH2Cl
H
H
CH2CH2Br
H 3C
H
COOC2H5
H
H
CHO
O
(c)
CH2OH
O
HO
199
Problems
O
O
(d)
O
CH2
O
O
NH2
CN
O
(e) HO
COOC2H5
Et
O
O
PROBLEM 1157 Bring about the following transformations:
CH2OCOPh
COOH
O
CD3
O
O
CD3
CD2OCOPh
CD3
PROBLEM 1158 An organic compound exist in three isomeric forms A, B and C with their molecular
formula C10H12O2 and all are resolvable. Decarboxylation of either B or C gives the same D(C9H12)
whereas decarboxylation of A gives E an isomer of D. Both D and E are non-resolvable. Treatment of
either D or E with KMnO4 in acidic medium produces benzoic acid as one product. Monobromination of
E in presence of sunlight yields only two isomeric products whereas D on monobromination yields three
isomers. Also B is more reactive than C towards decarboxylation reaction. Deduce structures of A to E.
PROBLEM 1159 An organic compound A(C6H10) decolourises Bayer’s reagent and it is nonresolvable.
A on treatment with acidic permanganate solution gave B(C6H10O4) which is still nonresolvable. B on
treatment with excess of ethanol in presence of acid catalyst gave C(C10H18O4) which is a neutral sweet
smelling liquid. C on refluxing with ethanolic solution of C2H5ONa gave D(C8H12O3) which is
resolvable. D on treatment with acidic solution of ethylene glycol yields E(C10H16O3). E on treatment
with one equivalent of Grignard reagent followed by reduction of product with Zn(Hg)—HCl and finally
hydrolyzing product yields 2-ethylcyclopentanoate. Deduce structures of A to E.
PROBLEM 1160 Provide products:
OMe
(a)
O
O
(b)
CH3ONa
CH3OH
O
OMe
CH3ONa
Br
+
H OH2
200
Problems in Chemistry
PROBLEM 1161 Predict products and write mechanism of their formations:
O
18
(a)
H2SO4
O
H
N
(b) H2N
heat
KOH
O
O
NH3/H2O
O
(c)
COOH
O
PROBLEM 1162 Provide organic products in the following reactions:
O
COOCH3
MeONa
(a)
CH3CH2CH2Br
O
Y
H3O+
Heat
Z
i) MeONa
ii) CH3(CH2)8CH2I
O
(b) MeO
NaOH
Heat
X
OMe
iii) NaOH/H2O
iv) H+/heat
PROBLEM 1163 Indicate how each of the following compounds can be synthesized from the given
starting material?
O
O
O
(a)
H
O
O
O
(b)
CH3
O
O
C
CH2
O
CH2CH3
(c)
CH2CH2CH2COOH
O
O
O
(d)
OMe
O
O
O
OC2H5
(e) C2H5O
O
201
Problems
PROBLEM 1164 Bring about the following transformations:
Ph
COOH
(a)
H
H
O
(b) Ph
Ph
OMe
(c) H2C(COOCH3)2
COOCH3
Ph
COOCH3
PROBLEM 1165 Propose mechanism:
O
OH
H
(a)
+
H
O
OH
O
O
O
O
(b)
COOEt
O
HCl
PROBLEM 1166 Provide mechanism:
O
O
(i) EtONa
H3O+
O
heat
OEt
EtO
(ii)
PROBLEM 1167 Starting from ethyl propanoate synthesize:
O
O
(a)
(b)
O
O
Ph
(e)
(d)
HOOC
Ph
(c)
N
N
O
O
202
Problems in Chemistry
PROBLEM 1168 Bring about the following conversions:
O
O
(a)
O
OEt
OEt
O
O
O
(b)
OEt
O
O
(c)
O
OEt
O
PROBLEM 1169 Propose mechanism of formation of products in the following reactions:
O
O
(a) MeOOC
COOMe +
H
O
O
O
O
OEt +
(b) EtO
H
H
CH3ONa
heat
H
COOEt
t-BuOK
heat
O
O
HO
O
O
O
(c)
COOEt
PROBLEM 1170
EtONa
COOEt
Propose mechanism of the following transformations:
OH
K2Cr2O7
CH3OH
H2SO4/H2O
H2SO4
COOH
PROBLEM 1171
materials:
Devise synthesis of the compounds shown below from the indicated starting
O
O
(a)
COOCH3
Br
N
203
Problems
NH2
CHO
OH
(b)
NH2
NH2
PROBLEM 1172 Propose an efficient synthesis of the compound shown below from indicated starting
material and methanol.
O
COOH
O
HO
OH
O
PPROBLEM 1173
Propose mechanism of the following reactions:
O
O
OH
(a)
CH3
CH3Li
Excess
SOCl 2
H 2O
(b) R  CN → RCOOH → R  COCl
HCl
O

H 2O
(c) R  C  N 3 → R  NH 2 + CO 2 + N 2
heat
PROBLEM 1174
Provide a mechanism for the following reaction:
H
CH3
N
CH3O
O
POCl3
N
CH3O
CH3
204
Problems in Chemistry
PROBLEM 1175
(a) CH3O
Bring about the following transformations:
O
CH3O
OCH2CH3
O
OCH2CH3
COOH
COOH
(b)
PROBLEM 1176 Arrange the followings in increasing order of reactivity towards nucleophile:
O
O
O
O




(a) H 3C  C  NH 2 (I), CH 3  C  Cl(II), CH 3  C  OAc(III), H 3C  C  OCH 3 (IV)
O
O
O
O




(b) CH 3  C  Cl(I), CH 3CH 2  C  Cl(II), (CH 3 ) 2 CH  C  Cl(III), (CH 3 ) 3 C  C  Cl(IV)
O
O
O



(c) CH 3  C  OCH 3 (I), CH 3  C  OCH 2CH 3 (II), H 3C  C  OCH(CF3 ) 2 (III)
PROBLEM 1177 Convert
(a) CH 3CH 2CH 2COOH → CH 3CH 2CH 2CH 2COOCH 3
O
(b)
O
O
PROBLEM 1178
O
Consider the following acid derivative:
H
O
1
4
O
O
2
O B
3
H
CH3
A
6
5
C
N
O
205
Problems
(a)
(b)
(c)
(d)
Which of the groups in square will hydrolyze first in base catalyzed medium?
How the rate of hydrolysis be affected if a nitrogroup is substituted at C–4?
How the rate of hydrolysis be affected if an amino group is substituted at C–4?
How the rate of hydrolysis be affected if a methoxy group is substituted at C–3?
PROBLEM 1179 Propose mechanism of the following reaction and calculate enthalpy of reaction from
the given bond-energies data:
O
O
O
Ph
OCH3
(1) CH3ONa
+ Ph
OCH3 (2) H3
O
Ph
OCH3 + CH3OH
O+
Ph
BE : C  C = 85 kcal/mol; C  O = 91 kcal/mol; O  H = 102 kcal/mol; C  H = 99 kcal/mol.
PROBLEM 1180
Predict major product in the following giving mechanistic reasoning:
O
O
CH3OH
H3 C
*
O
CF3
PROBLEM 1181 An optically active organic compound A has molecular formula (C 7 H12O 3 ) and
found to produce an yellow precipitate with alkaline solution of Iodine. Also A neither decolourised
brown colour of bromine water nor evolved any gas with aqueous solution of NaHCO 3 but evolved a gas
on heating with sodium metal. On hydrolysing in acidic medium, A produced antoher optically active
compound B(C 7 H14O 4 ) which also produced yellow ppt with NaOH/I 2 . Also B evolved a colourless,
acidic gas on treatment with aqueous NaHCO 3 solution. A on treating with CrO 3 /HCl/Pyridine in
CH 2Cl 2 produced another optically active compound C(C 7 H10O 4 ). C on refluxing with aqueous Ag 2O
produced an optically inactive compound D(C 7 H10O 5 ). D does not decarboxylate (does not evolve CO 2 )
on simple heating. However, heating D with sodalime gave E(C 5 H10O). In a separate analysis, 3-oxo
ethylbutanoate was heated with excess of sodium metal and then with CH 3I to produce X (C 8 H14O 3 ). X
on hydrolysis followed by simple heating of product yielded E. Identify A to E and X.
PROBLEM 1182 An oily liquid A is insoluble in water but on heating with aqueous NaOH for half an
hour it dissolves. From the reaction mixture a liquid B can be distilled, which gave an yellow precipitate
with iodine and NaOH. On careful oxidation, B gives an aldehyde, C, which also gives an yellow
precipitate with iodine and NaOH solution. If sulphuric acid is added to the solution obtained from
heating A with NaOH solution, a white precipitate D is obtained. D liberates a gas on treatment with
aqueous NaHCO 3 . Heating D with sodalime produced benzene. Identify A–D.
PROBLEM 1183 An organic compound A has molecular formula C 9 H 8O 2 and if exist in
stereoisomeric forms. A effervesces with NaHCO 3 . A decolourises brown colour of bromine water
forming B(C 9 H 8O 2 Br 2 ) which is distereomeric. B on hydrolysing with aqueous NaOH followed by
refluxing of product with acid gave C(C 9 H10O 4 ). C on oxidising with acidified dichromate solution gave
D(C 9 H 6O 4 ) which gives orange precipitate with 2,4-dinitrophenylhydrazine but did not react with
ammonical AgNO 3 solution. D on reacting with PCl 5 yielded a steamy fumes of E(C 9 H 5O 3Cl). E on
treatment with aqueous ammonia yielded F(C 9 H 7O 3 N). Both A and F are oxidized by hot aqueous
KMnO 4 solution to yield a white crystalline substance which on heating with sodalime yielded benzene.
Identify A to F.
206
Problems in Chemistry
PROBLEM 1184
O
Propose mechanism of the following reactions:
O
O
(i) NaOH/H2O
(a)
COOH
(ii) H3O
O
NC
(i) TsCl
(b)
COOH
(ii) NaOH
(iii) H3O+
HO
N
H
O
(c)
O
BuLi
N—OTs +
N
O
O
PROBLEM 1185
O
H
Propose mechanism of the following reactions:
O
O
O
K2CO3
(a)
BrCH2CH2Br
OEt
OEt
O
COOEt
Na/excess
(b) EtOOC
COOEt
EtOOC
O
PROBLEM 1186
EtOOC
R
Propose mechanism:
O
N
(i) t-BuOK
CH3
O
OAc
(i) MeMgBr
(iii) Ac2O
N
R
(ii) aq NaOH
(ii) Heat
CH3
O
N
R
207
Problems
(a) Indicate a mechanism of the following reaction:
PROBLEM 1187
COOEt
(i) EtONa/EtOH/heat
COOEt
(ii) CH3COOH/H2O
O
(b) Devise a sequence by which product of the reaction in part “a” converted into the following
comopunds:
O
(i)
(ii)
O
O
O
R
COOH
(iv)
N
(v)
O
PROBLEM 1189
N
O
Bring about the following transformations:
PROBLEM 1188
O
COOCH3
CH3
Bring about the following transformations:
Br
—Br
(a)
COOH
(b) CH 3CH  CH 2CN → CH 3CHCH 2 CH 2OH


C6 H 5
C6 H 5
PROBLEM 1190
Propose mechanism:
O
O
O
H+/H2O
COOH
Ph
(iii)
O
OH
O
208
Problems in Chemistry
PROBLEM 1191
Propose mechanism of the following reactions:
MgBr
+
(a)
H3O+
O
O
OH
HO
MgBr
heat
(b)
SH
S
O
N
O
NH2
PROBLEM 1192
Provide missing reagents/products in the following reactions:
O
Br2/NaOH
NH2
(a)
H2O
NH2
(b)
O
?
O
O
?
(c)
(d)
HO
OCH3
Br
OCH3
?
O
PROBLEM 1193
Bring about the following transformations:
PhBr
PhCH2—N
PROBLEM 1194 A neutral organic compound A(C 4 H 6 ) reacts with Br 2 /CCl 4 solution to form a
compound B(C 4 H 6 Br 2 ). A on treatment with acidified permanganate solution yields C(C 4 H 6O 3 ). C
gives an orange precipitate with 2,4-dinitrophenylhydrazine. C on refluxing with NaHCO 3 evolves a
209
Problems
gas. Treating C with NaBH 4 gave a new compound D(C 4 H 8O 3 ) which also produces a gas with
NaHCO 3 solution. Deduce a structure of A  D.
PROBLEM 1195 2.81 g of an optically active diester A, containing only C, H and oxygen was
saponified using 30 mL 1.0 M NaOH solution. Following saponification, the solution required 6 mL 1.0
M HCl solution to titrate unreacted base. The saponification products were an optically inactive diacid
B, methanol and an optically active alcohol C. Alcohol C reacted with alkaline solution producing an
yellow precipitate and C 6 H 5COONa. Also, the diacid B reacted with Br 2 in CCl 4 to give a single,
optically inactive product D. Ozonolysis of B gave only one product. Also, the diester A reacts with
Br 2 /CCl 4 to give a mixture of E and F, both optically active. Identify A to F.
PROBLEM 1196 The following three reactions occurs by a common mechanism. Write detailed
mechanism for the reaction A. Then indicate, which reaction will be the fastest and which reaction will
be the slowest and explain briefly, why?
Br
O
NH—C—CH3
CH3CN
A:
H2O/∆
H3CO
H3CO
Br
O
NH—C—CH3
CH3CN
B:
H2O/∆
O2N
O2N
Br
O
NH—C—CH3
CH3CN
C:
H2O/∆
NO2
NO2
PROBLEM 1197
Propose mechanism of the reaction:
OH
O
O
LiAlH4
NH4Cl
OEt
TsCl
NaOH
210
Problems in Chemistry
PROBLEM 1198
Complete the following reactions:
COOCH3
COOCH3
∆
(i) NaOH
(a)
(ii)
H3O+
COOCH3
COOCH3
O
O (i) [(CH ) CH] NH
3 2
2
(b)
(ii) H2O
O
O
(i) Br—(CH2)4—Br
(c)
H2O
(ii) Mg/Et2O
I2/H2O
(d)
COOH
NH2
(e)
(i) KMnO4/NaOH
(ii) SOCl2
Cl
(f)
O
PROBLEM 1199
(i) Me2NH
(i) H3O+
(ii) LiAlH4
(ii) NaOH
Propose mechanism of the reaction:
O
OH
NH2
Ph
N
(i) NaOH, PhCOCl
O
O
Ph
(ii)
O
O
O
211
Problems
Provide products of the following reactions:
O
PROBLEM 1200
CH3CH2SNa
(a)
OCH3
H 3O +
H 2O
Et2O
O
(b)
PCl5
—C—OMe
∆
CCl4
(c)
OEt
Peroxide
(d)
H 3O +
COOEt
∆
COOEt
O
O
(e)
OEt
(f) PhCH2Li + HO—
PROBLEM 1201
EtONa
excess
Br(CH2)5Br
—CHO
Provide selective reagents that will bring about the following transformations:
OH
OH
(a)
CHO
COOH
O
O
(b)
O
O
(c)
H
OCH3
Cl
212
Problems in Chemistry
O
(d)
MeO
MeO
OH
H
O
O
PROBLEM 1202
Propose mechanism of the following reactions:
O
OEt
(a)
CH3MgI
CN
O
Et2O
O
(b)
O
H 3O +
O
CH3Li
H 3O +
Et2O
PROBLEM 1203 Synthesize the following compound starting from diethylmalonate and other
necessary reagents:
CONH2
Write product of the following reactions:
PROBLEM 1204
O
O
(a)
ONa
Cl
O
LDA
(b)
Br
OEt
O
(c)
O
N—
Cl
H 3O +
heat
213
Problems
O
(d)
OEt
H3O+
EtO–
EtOH
PROBLEM 1205 An organic compound A reacts with I 2 in basic medium to give a yellow precipitate.
A on treatment with Br 2 in acetic acid medium gives B which readily undergoes reaction with cyanide
ion to give C. C on reduction with NaBH 4 forms D–an optically active substance. Acid hydrolysis of D
followed by treatment with acidic ethanol gives E. Gentle oxidation of E with pyridinium
chlorochromate gives F, which on treatment with potassium ethoxide in ethanol followed by reaction
with ethyl iodide and then heating the final product yields 1-phenyl-1-butanone. Identify A to F.
AMINES
PROBLEM 1206 Provide appropriate reagents for the following transformations:
(a) R —OH → RCH 2 NH 2
(b)
OH
NH2
(c) C 6 H 5 NH(CH 3 ) → C 6 H 5 —N—CH 2CH 3

CH 3
(d) C 6 H 5 NH 2 → C 6 H 5CH 2 NHCH 2 NH 3
PROBLEM 1207 An organic compound A has molecular formula C 5 H13 N and it reacts with benzene
sulphonyl chloride in aqueous potassium hydroxide to give a clear solution, acidification of solution
gives a precipitate. Also A can be resolved into enantiomers. A on refluxing with nitrous acid gives
B (C 5 H12O) which is still resolvable. B on treatment with CrO 3 in pyridine solution of HCl yields another
resolvable compound C (C 5 H10O) which on further treatment with Zn(Hg) in concentrated HCl solution
yields a non-resolvable D(C 5 H12 ). Identify A to D.
PROBLEM 1208 An organic compound A has molecular formula C 7 H 9 N and it forms a clear solution
when dissolved in aqueous KOH solution of benzene sulphonyl chloride. A on treatment with NaNO 2
and HCl at 0°C, and then with 2-naphthol forms an intensity coloured compound. Also A on treatment
with CH 3COCl followed by electrophilic substitution yield a single substitution product. Identify A.
PROBLEM 1209 An organic compound A has molecular formula C 9 H13 NO and it can be resolved into
enantiomers. A does not decolourise bromine water solution. A on refluxing with dilute H 2SO 4 solution
yields another resolvable compound B (C 9 H14O 3 ) which gives effervescence with NaHCO 3 . B on
treatment with NaBH 4 yields C (C 9 H16O 3 ). C on heating with concentrated H 2SO 4 solution yields a
214
Problems in Chemistry
sweet smelling neutral compound D(C 9 H14O 2 ). Also A on reduction with LiAlH 4 yields E (C 9 H19O)
which on treatment with H 2SO 4 yields the following compound :
H
N
Identify A to E.
PROBLEM 1210 Provide structures of A to F in the following sequence of reaction:
Ag2O
Heat
+ CH 3I → A (C 7 H16 NI) → B (C 7 H17 NO) → C (C 7 H15 N)
H 2O
N
CH3
Ag O
CH I
Heat
2
3
C –→
D (C 8 H18 NI) →
E (C 8 H19 NO) → F (C 5 H 8 ) + H 2O + N(CH 3 ) 3
H 2O
PROBLEM 1211 Identify the labelled product in the following sequence of reaction:
Br
CH3NHCH3
(a)
2 equivalent
A
BaO
B
CH3I
2 equivalent
C
Ag2O
H2O
D
Heat
E + 2 (CH3)3N
Br
O
Ph
(b)
Ph
NH2OH
A
LiAlH4
B
CH3I
Excess
C
Ag2O
Heat
D + (CH3)3N
H2O
PROBLEM 1212 In the following sequence of reaction:
Br
+ CH3NHCH3
A
KOH
Heat
Br
Deduce of A and B and explain formation of C.
B
CH3I
KOH
Heat
+ (CH3)3N
C
215
Problems
PROBLEM 1213 Propose mechanism of following reaction:
O
N3
Pd/CaCO3
H2, quinoline
NH
O
PROBLEM 1214 How butylamine and dibutyl amine can be separated?
PROBLEM 1215 An organic compound A has molecular formula C 5 H10 N 2 and it can be resolved into
enantiomers. A on hydrogenation yields optically inactive B (C 5 H12 N 2 ). B on treatment with nitrous acid
forms an oily compound C (C 5 H10 N 2O 2 ) which is yellow coloured. C when warmed with a crystal of
phenol and few drops of concentrated sulphuric acid, turns green which, when made alkaline, turns deep
blue. Also A on treatment with excess of iodomethane forms D(C10 H 22 N 2I 2 ). D on treatment with
aqueous Ag 2O followed by heating yields pyridine. Identify A to D.
PROBLEM 1216 Complete the following reactions:
H
O
(a)
+
N
Heat
A
CH2==CHCOOEt
B
hν
(b) HN 3 →
D
C2H4
Na
C2H5OH
KCN
(c) BrCH 2CH 2 Br →
→ E
PROBLEM 1217 Bring about the following transformation:
(a) R —CN → R —NC
(b) R —NC → RCN
PROBLEM 1218 Complete the following reactions:
H SO
2
4
(i) CH 2 == CH—CN + PhCH 2OH →
?
Et O
2
(ii) PhCH 2COCHN 2 + HCl( g ) →
?
(iii) MeNH 2 + ? → MeNHCOOEt
HBr
Heat
(iv) HOCH 2CH 2CN → ?
PROBLEM 1219 Complete the following reactions:
B H
ClNH
2 6
2
(a) C 2 H 5CH==CH 2 →
A → B
H3O
+
C
216
Problems in Chemistry
H O
– CO
PhNCO
2
2
(b) PhNCO →
C →
D →
E
LiAlH
KOH
4
(c) C 2 H 5 NH 2 + KCN + Br 2 →
F →
G
PROBLEM 1220 MeCOCH 2Cl + CH 2 N 2 → A (C 4 H 7ClO) + B (C 4 H 7ClO)
Ketone
Epoxide
Suggest structures of A and B and mechanism of their formations.
+
−
PROBLEM 1221 An amine A(C 6 H15 N), on treatment with CH 3I and then KOH gives B (C 8 H 20 NOH) B
on heating produces C (C 4 H 8 ) and D(C 4 H11 N). C on treatment with aqueous (CH 3COO) 2 Hg follows by
reduction with NaBH 4 yields E (C 4 H10O) which does not change the colour of chromic acid solution.
Deduce structures of A to E.
PROBLEM 1222 Complete the following reactions:
KHCO3
(a) EtNO 2 + 2CH 2O → ?
(b) RCH 2 NH 2 + 2CH 2O + 2HCOOH → ?
H C==C==O
2
(c) CH 2 ==C==O + HCN → ? →
?
PROBLEM 1223 Provide mechanism of reaction
NaOH
(CH 3 ) 2 CHCONH 2  –→ (CH 3 ) 2 CHNH 2
Br2
PROBLEM 1224 Prepare PhCH 2 NH 2 by Gabriel synthesis and nitrile reduction starting from PhCH 2 Br
in both cases.
PROBLEM 1225 Synthesize following compounds starting from cyclohexanone and any amines
needed.
H
H
N
(a)
N
(b)
N
N
H
H
AROMATIC COMPOUNDS
PROBLEM 1226 A resolvable organic compound A has molecular formula C10H14O. A does not forms
any salt with NaOH but evolve a colourless gas on heating with sodium metal. A gives an yellow
precipitate with NaOH/I2 but does not decolourise Br2-water solution. A on treatment with
CrO3/HCl/pyridine produces B(C10H12O) which is non-resolvable but gives iodoform test. B on
217
Problems
treatment with Cl2/AlCl3 gives C(C10H11OCl) as the only possible isomer. Also A on heating with
alkaline KMnO4 solution yields C8H6O4(D) as one of the product which evolves a gas with NaHCO3. D
on heating with ethan-1,2-diol in presence of an acid catalyst forms a polyester E. Deduce structures of A
to D and write the formula of repeat unit of E.
PROBLEM 1227 Predict products/reagents:
COOH
+
(a)
O
O
Cl
NaOH
CHO
HNO3
(b)
Heat
COOH
H2SO4
COOH
(c)
OH
Br2/CHCl3
Br2/AlBr3
(d)
S
CH3 O
(e)
OMe
CH3
HNO3
(f)
H2SO4
+
H2SO4
CH3
CH3
O 2N
O
(g)
X
C
CH2
CF3
Y
O 2N
CF3
CH2
CF3
PROBLEM 1228 Propose mechanism:
Ph
+
H
HO
PROBLEM 1229 Bring about the following transformation:
Ph
PROBLEM 1230 An organic compound A(C11H14O) exist in two stereo-isomeric forms, none of which
is resolvable. A does not evolve any gas with sodium metal but decolourise bromine water solution. A on
218
Problems in Chemistry
treatment with ozone followed by work-up with (CH3)2S yields B(C9H10O2) as one of the product. B
neither forms any salt with NaOH nor does it evolve any gas with sodium metal. Also B does not react
with Tollen’s reagent. B on treatment with concentrated solution of HI forms C(C8H8O2) as one of the
product which forms salt with NaOH but does not evolve any gas with NaHCO3. B on treatment with
AlCl3/Cl2 gives D(C9H9O2Cl) as the only possible product. C can also be obtained by treatment of an
ester E(C8H8O2) with AlCl3. Deduce structures of A to E and explain the formation of C from E.
PROBLEM 1231 Bring about the following transformation:
O
PhCH2CH2
CHO
(a)
(b)
Cl
COOH
(c)
(d)
SO3H
NO2
PROBLEM 1232 An optically active organic compound A has molecular formula C11H16O2 and it does
not forms salt with NaOH as well as it does not decolourise Br2—H2O solution, but evolve a gas on
treatment with MeMgBr. A on treatment with HI gives another optically active compound B(C9H10O2). B
on treatment with aqueous KOH yields another optically active compound C(C9H12O2). C on treatment
with PCC yields D(C9H8O2) which gives positive iodoform test as well as positive Tollen’s test. D is
optically inactive and on treatment with Cl2/AlCl3 gives one and only one possible monochloro
derivative E(C9H7O2Cl). Deduce structures of A to E.
PROBLEM 1233 Synthesize:
OH
(a) Ph—CH —C
2
C==C
CH3
CH3
from Ph—CH2CH2CH2Br
H H
Cis
Br
(b) O2N
—CH—CH2OH
O
(c)
from
Cl
from benzene, chloroethane
and any inorganic reagent
219
Problems
(d) C6H5CH2COC6H5 from benzene, oxirane and inorganic reagent
O
I from iodo benzene and
(e) (CH3)3CCH2
SO3H
PROBLEM 1234 Complete the following reaction:
O
NO2
(a)
Ph
TiCl4
DMG
X
OH –
TsOH
Y
Z
O
LDA
(b) Ph
O
(c)
X
CuCl2
Y
KCN/AcOH
DIBL-H
X
Ph
H3PO4
Z
Y
TsOH
Z
PROBLEM 1235 Bring about the following transformation:
(a)
Br
O
COOMe
O
O
(b)
Ph
Ph
O
O 2N
(c)
S
S
OH
PROBLEM 1236 (a) Synthesize the following compounds starting from benzene:
F
CH3
COOH
NO2
F
F
F
(b) Write mechanism of the reaction of p-nitrofluoro benzene with (C2H5)2NH. Explain why
m-nitro fluoro benzene does not react with diethyl amine under identical conditions.
220
Problems in Chemistry
(c) Synthesize the following compounds starting from benzene and any other reagents needed:
O
O
O
R
OMe
(i)
(ii)
R'
O
COOCH2CH3
(iv) H2N
Ph
(iii)
PROBLEM 1237 Synthesize the following compounds starting from benzene:
OCOPh
OCOPh
(b)
(a)
O 2N
Br
O
OPh
(c)
NO2
(d)
O2N
O2N
OH
OH
(e)
Br
Br
(f)
NC
PROBLEM 1238 Synthesize the following compounds starting from benzene:
I
I
I
(b)
(a)
(c)
Cl
Cl
Cl
Cl
PROBLEM 1239 Bring about the following transformation:
Br
NO2
Ph
(a)
(b)
NO2
Br
Br
221
Problems
COOH
(c)
O
O
NH2
H2N
O
NH2
PROBLEM 1240 Propose mechanism:
O
H
MeO
N
COOMe
NHNH2
MeO
COOMe
PROBLEM 1241 Bring about the following transformation:
CHO
Br
CH3
CHO
(b)
(a)
OH
OH
CHO
(c)
OH
OH
(d)
CH3
O
OH
(f)
(e)
Ph
O
(g)
H
N
(h)
NHNH2
OH
CHO
O
COOH
Ph
222
Problems in Chemistry
PROBLEM 1242 Bring about the following conversions:
CH3
SO3H
CH3
OMe
(b)
(a)
NH2
CHO
OH
OH
OH
Br
(c)
OH
(d)
Br
Br
CH2OH
OMe
(e)
PROBLEM 1243 Predict product of the following reaction explaining mechanism of their formation:
—CH3 + CBrCl3
hν
PROBLEM 1244 Prepare the following compounds starting with either benzene or toluene:
OH
CH3
(a)
CH3
HO
(b)
HO
(c)
CH3
CN
Cl
OH
(f)
(e)
(d)
CN
Cl
I
CN
(h)
(g)
NH2
CH3
Br
O 2N
(i)
Br
Br
Br
Br
223
Problems
OH
CN
COOH
Br
Br
(j)
(k)
Br
Br
(l)
Br
Br
Br
Br
CH3
Br
OH
NO2
I
(m)
(n)
(o)
Br
NO2
Br
CH3
CH3
CN
(q) H3C
(p)
N N
OH
Br
CH3
(r) H C
3
N N
HO
PROBLEM 1245 Bring about the following transformation:
NO2
NO2
CHO
OHC
(a)
(b)
NO2
Cl
NO2
NO2
Cl
NO2
CN
Cl
(c)
COOH
(d)
NO2
NH2
NO2
CH3
NH2
(e)
CH3
(f)
Br
Br
CH3
CH3
NH2
Br
224
Problems in Chemistry
NO2
CH2NH2
CHO
NO2
CHO
(g)
NO2
(h)
HOOC
CH3
NH2
CH3
Br
Br
(i)
(j)
NO2
NO2
Br
PROBLEM 1246 Bring about the following transformation:
NO2
CONH2
NO2
(a)
Br
Br
(b)
Br
NO2
NO2
NH2
Cl
(c)
(d)
Cl
Cl
Br
PROBLEM 1247 An organic compound A(C8H10) on treatment with fuming sulphuric acid yields two
isomeric products B and C. B on fusion with KCN gives D(C9H9N) which on hydrolysis yields
E(C9H10O2). E on treatment with hot alkaline KMnO4 yields F(C8H6O4). F on heating with P2O5 undergo
intramolecular dehydration to yield G(C8H4O3). Identify A to G.
PROBLEM 1248 Bring about the following transformation:
CH3
OMe
CH3
OMe
Br
(b)
(a)
NH2
Cl
CH2OH
CH3
NH2
NO2
(c)
(d)
Cl
225
Problems
CH3
Br
NO2
CH3
Br
Br
(e)
(f)
NO2
NO2
NO2
NO2
CH3
OH
(g)
OH
PROBLEM 1249 Identify the aromatic product formed in the following reaction:
CN
(a)
C2H5MgBr
HCl/H2O
COOH
COOH
(c)
(d)
HCl/H2O
EtLi
(b)
CHO
(CH3)2Cd
SOCl2
Br
COOEt
Zn/HCl
PROBLEM 1250 Synthesize the following compounds starting from benzene:
(a)
(b)
(c)
(d)
OH
OH
(e)
(f)
226
Problems in Chemistry
PROBLEM 1251 Predict major substitution product in the following reactions:
OMe
N
Br
HNO3
(a)
Na2CO3
NHCOCH3
OMe
CH3COCl
(c)
Cl2
(b)
H2SO4
Br2
(d)
AlCl3
FeBr3
Br
COOCH3
PROBLEM 1252 Starting from benzene synthesize the following compounds as major product:
CH3
NO2
NH2
NO2
(c)
(b)
(a)
Br
COCH3
NH2
NO2
Cl
(e)
(d)
(f)
NO2
PROBLEM 1253 How would you rationalize the following experimental result:
N
N
HNO3
NH3
H2SO4
OH2
NO2
PROBLEM 1254 Propose synthesis of the following compounds starting from benzene:
NO
(a)
O
COOH
(c)
(b)
NO2
COOH
227
Problems
H2N
OH
(d)
(e)
Br
CH3
Cl
NO2
(g)
(f)
Br
O
(h)
(i) Br
SO3H
NO2
N
(j)
CH3
PROBLEM 1255 Identify the starting material in the following sequence of reaction:
O
(a) X
CH3COCl
AlCl3
N
(b) X
(c) X
N2H4
Br2
NaOH
AlBr3
Br2
Br
Sn/HCl
AlBr3
Br
228
Problems in Chemistry
OMe
(d) X
(e) X
Br2
NaOEt
CH3Br
Zn(Hg)
HNO3
Sn/HCl
ClH
H2SO4
Br
H 2N
PROBLEM 1256 Predict the major product in the following reaction:
CH3
(a)
CH3COCl
HNO3
(b)
AlCl3
H2SO4
NO2
O
Cl2
(c)
(d)
AlCl3
CH3
(e)
Cl
OPh
Br2
AlBr3
Br2
AlBr3
PROBLEM 1257 p-nitro phenol is shaken with one equivalent of D2O in the presence of a strong acid
(perchloric acid) at 100°C for a long time. The product has two deuterium atom in it. One of these is lost
instantly when the product is treated with ordinary water. Propose a structure for the original product and
explain the difference in the ease with which the two deuterium atoms can be removed from the
molecule.
PROBLEM 1258 Bring about the following transformation:
CH3
COOH
(a)
(b)
NO2
O2N
COOH
HO
CH3
NO2
(c)
(d)
NO2
229
Problems
O
NH2
(e)
(f)
NO2
PROBLEM 1259 Complete the following reactions:
O
OH
O
O AlCl3
(a)
Cl
+
(b)
excess
Cl
O
O
OMe
O
H 3C
(c)
AlCl3
O
+
AlCl3
H 3C
O
PROBLEM 1260 Synthesize the following compounds starting from benzene:
H
O
H
H3C
C2H5
(a)
(c)
(b)
Ph
OH
PROBLEM 1261 Predict the major substitution product in the following sequence of reaction:
OMe
OH
D
H3PO4
(a)
A
SO3
(b)
H2SO4
Cl
O
(c)
+
AlCl3
O
NH2
CH3
(d)
O
(CH3CO)2O
HNO3
H2SO4
H3O
+
230
Problems in Chemistry
PROBLEM 1262 Predict synthesis of the following products starting from benzene:
Br
CH3
(a)
O
C
H
(b)
CH3
O
NO2
(c) H3C
OH
(d)
COOH
(f)
(e)
(g)
PhCH2
O
CH2Ph
H
H
C 2H 5
PROBLEM 1263 1,3,5-trimethyl benzene undergoes electrophilic aromatic substitution with iodine
monochloride (ICl). Write mechanism for this reaction showing major product.
PROBLEM 1264 When allyl alcohol is treated with HF in presence of benzene, two products are
formed: 3-phenyl-1-propene and 1,2-diphenyl propane. Write equations showing the mechanism for the
formation for these products.
PROBLEM 1265 When 2-hydroxy benzoic acid is heated with isobutyl alcohol in presence of sulphuric
acid, compound A is formed. The same product is formed if tertiary butyl alcohol and sulphuric acid is
used. What is the structure of compound A? Write mechanism of formation of this product.
PROBLEM 1266 Bring about the following synthesis starting from benzene:
(*direct F.C. alkylation of nitro benzene does not succeed)
CH3
CN
*(b)
(a)
Cl
NO2
Br
(d)
(c)
COOH
HO
Cl
F
(e) H3C
NO2
COOH
(f)
NO2
231
Problems
OMe
(g)
(h) HO
I
NH2
PROBLEM 1267 Complete the following reactions:
NO2
MeOH
+
(a)
H
N
+
C6H5SNa
Br
NO2
NO2
(b)
MeOH
F
NO2
NO2
OTs
+
(c)
NH2
NO2
O 2N
H 2N
NO2
N
Cl
+
(d)
O 2N
Heat
CHCl3
NO2
CH3
(e) O2N
+
F + H3N—CH—COO
–
H 2O
NO2
PROBLEM 1268 Give structural formula for all reagents, intermediates and products designated by
letters in the following equations:
CHO OMe
(a)
OMe
NaBH4
A
H3O
+
B
232
Problems in Chemistry
COOH
HOOC
H3PO4
C + D
(b)
CH3
OH
CH3I
NaH
(c)
B2H6
E
CrO3
F
G
H2SO4
H2O2/NaOH
H3PO4
H
LiAlH4/H2O
I
CHO
(d)
OH
NaBH4/H2O
J
NaH
K
CH3I
L
OMe
NO2
Cl
Br
Na
(e) H-CH(COOEt)2
NO2
M
N
COOH
HO
(f)
CH2OH
MeO
COOCH3
MeO
P
O
OH
OMe
OMe
Br
CH2OH
MeO
Q
OMe
PROBLEM 1269 Identify the missing reagents/reactants/products in the following sequence of
reactions:
(a) A
(b) O2N
HNO3
B
Sn/HCl
C
O
NO2
NaNO2
HCl/0oC
D
CuCN
NO2 + H N
OMe
NC
OH2
E+ F
233
Problems
NO2
(c) H2N
(d) CH3CH2O
NH2
G
KI(aq)
H
HCl/0oC
NaNO2
Br2
NH2
(e) HO
NaNO2
ICl
excess
I
NaNO2
J
HCl/0oC
H3PO2
K
KI(aq)
L
HCl/0oC
M
CH3
(f) O2N
CH3
H2/Ni
N
NaNO2
CuCN/heat
O
HCl/0oC
P
H3O +
Q
PROBLEM 1270 Write structural formula for all the products in the following sequence of reaction:
CHO
HNO3
(a)
A
H2SO4
Sn/HCl
NaNO2
B
C
HCl/0oC
CuCl/HCl
NH2
NaNO2
(b)
MeO
H2SO4 , 0oC
OMe
H2O/H2SO4
E
F
CH3
(c)
NO2
O2 N
H2/Ni
NaNO2
G
HCl/0oC
COOH
NH2
(e)
C6H5
N
CH3I
(f)
N
H
C6H5
M
NaNO2
HCl/0oC
KOH
K
J
HCl/0oC
CH3
H2O/H2SO4
N
NaNO2
(d)
H
N
L
CH3I
O
KOH
D
P
+
Q
I
D
234
Problems in Chemistry
PROBLEM 1271 An organic compound A(C8H10) does not decolourise aqueous solution of Br2 and on
treatment with Br2/FeBr3 in dark yields two product in principle, B and C but due to steric reason, B
predominates. A on refluxing with alkaline solution of KMnO4 yields D(C8H6O4) which gives off gas on
treatment with NaHCO3. D on treatment with Br2/FeBr3 reacts very slowly to produce E(C8H5O4Br) as
the only isomer. Also D on treatment with excess of SOCl2 followed by work-up with benzene solution
of AlCl3 produces F(C14H8O2) which does not yields any gas with NaHCO3. F on heating with N2H4
produces G(C14H8N2). Deduce structures of A to G.
PROBLEM 1272 An organic compound A (C9H10O2) does not change colour of Br2-H2O and produces
no gas with NaH. A on treatment with Br2 in presence of FeBr3 yields B(C9H9O2Br) as only isomer. Also
A gives orange precipitate with 2,4-dinitro phenyl hydrazine and produces a resolvable compound
C(C9H12O2) on reduction with NaBH 4. Deduce structures of A, B and C.
PROBLEM 1273 An organic compound A(C9H8O2) does not decolourise bromine water solution and
evolves no gas with CH3MgBr but gives orange precipitate with 2,4-dinitro phenyl hydrazine. A on
refluxing with dilute H2SO4 produces B(C9H10O3) which forms salt with NaOH and on treatment with
excess of CH3COCl yields C13H14O5. B is a non-resolvable compound which on heating with
N2H4/NaOH yields C(C9H12O2). C on dehydrating with concentrated H3PO4 yields D(C9H10O) as major
product. D on ozonolysis followed by work-up with (CH3)2S yields E(C7H6O2) which can also be
obtained by the action of phenol with alkaline solution of chloroform followed by acidification of
product. Identify A to E.
PROBLEM 1274 An oily liquid A is insoluble in water, but on heating with aqueous solution of sodium
hydroxide for one hour, it dissolves. From the reaction mixture, a liquid B can be distilled, which gives a
yellow precipitate with NaOH/I2 as well as it is resolvable. B on treatment with acidic dichromate
solution yields C which also gives positive iodoform test. If sulphuric acid is added to solution obtained
on heating with NaOH, a white precipitate D is obtained. D gives effervescence with NaHCO3 and
heating D with soda lime converts it into toluene. Also A on treatment with Br2/FeCl3 in dark yield single
mono bromo derivative as a substitution product. Compound B on heating with concentrated H2SO4
yields stereomeric alkene, one of which on treatment with cold, dilute and alkaline KMnO4 yields a meso
diol. Deduce structures of A to D.
PROBLEM 1275 An organic compound A has molecular formula C10H14 and it does not decolourize
bromine water solution. A on treatment with Br2/Fe yields three products in principle but in actual
practice, only two of these are produced as mono-bromo derivative. A on heating with Br2 yields B
(C10H13Br) as the mono-bromo derivative which is optically inactive. A on heating with alkaline KMnO4
yields C(C8H6O4) which does not forms anhydride on heating. Deduce structures of A to C.
PROBLEM 1276 An organic compound A(C8H8O3) evolves a gas with NaHCO3 as well as forms salt
with NaOH. A on treatment with HI yields B(C7H6O3) which when treated with excess of acetyl chloride
(CH3COCl) yields C(C11H10O5). B on treatment with Br2/FeBr3 in CCl4 should yield three products D, E
and F in principle but only two products E and F are produced in reality and E is the major product.
Deduce structures of A to F.
PROBLEM 1277 An oily liquid is insoluble in water, but on heating with aqueous solution of sodium
hydroxide for one hour, it dissolves. From the reaction mixture, a liquid B can be distilled, which gives a
yellow precipitate with iodine and NaOH solution. On careful oxidation, B gives an aldehyde C, which
also gives positive iodoform test. If sulphuric acid solution is added to the solution obtained from heating
235
Problems
A with NaOH, a white crystal D is obtained. D liberates a gas with NaHCO3. Heating D with NaOH/CaO,
converts it into C 6H6. Identify A to D.
PROBLEM 1278 Identify the followings as aromatic, antiaromatic or non-aromatic, supporting your
answer with brief explanations.
OH
(a)
(b)
s
+
(c)
(d)
PROBLEM 1279 The pK a of benzylic hydrogen in A is ~ 16, whereas the pK a of benzylic H in B is ~ 25.
Explain the difference.
H
CH3
B
A
PROBLEM 1280
appropriate:
CH3
H
Predict products in the following reactions and show stereochemistry where
OCH3
Br2
(a)
NaNH2
(b)
NH3
FeBr3
Br
O3
Na
(c)
liq. NH3
AlCl3
(d)
CH3COCl
(CH3)2S
NO2
CH3ONa
(e)
(f)
CH3OH
Br
O
HNO3
H2SO4
Cl2
Zn(Hg)
AlCl3
HCl
236
Problems in Chemistry
NO2
O
O
C
HNO3
(g)
H2SO4
AlCl3
(i)
(j)
OH
CH3Li
H3O+
excess
O3
(CH3)2S
O
O
(h)
O
I2/HNO3
t-BuCl
AlCl3
I
PROBLEM 1281 Provide a mechanism for the nitration of benzene using concentrated nitric
acid-sulphuric acid mixture. Draw an energy diagram for the reaction. Will the rate of reaction change if
benzene is replaced by hexadeutrobenzene? Why?
PROBLEM 1282 A by product of sulphonation of benzene is diphenyl sulphone. Propose a mechanism
by which diphenyl sulphone is produced during sulphonation reaction.
PROBLEM 1283
Synthesize the following compounds starting from benzene:
O
O
O
H
H
O
and
H
H
O
PROBLEM 1284
Provide reagents that will accomplish the following transformations:
O
(b)
(a)
Br
OH
O
Br
237
Problems
OH
OCH3
O
OCH3
NO2
H (d)
(c)
OH
O
H
(e)
(f)
O
Br
(g)
O2N
PROBLEM 1285 Reaction of acetophenone with Br 2 /FeBr 3 yield meta-bromo acetophenone. Does
this reaction occur at faster rate or slower rate than the treatment of benzene with bromine under similar
conditions? Why?
PROBLEM 1286 Chloropyridines undergo nucleophilic substitution with sodium methoxide to give
methoxy pyridine as:
CH3ONa
CH3OH
N
Cl
N
OCH3
Order the following derivatives from high to low in term of their reactivity towards nucleophilic
aromatic substitution.
Cl
Cl
N
A
N+ Cl–
N
C
B
CH3
238
Problems in Chemistry
Propose a mechanism for the following reaction:
PROBLEM 1287
+
O
H2SO4
Devise synthesis of indicated compound starting from benzene:
PROBLEM 1288
N
O
H
O
Provide a mechanism for the following reaction:
PROBLEM 1289
CH3O
NH2
CH3COCl
CH3O
HCl
CH3O
NH
CH3O
CH3
Identify the followings as aromatic, antiaromatic or nonaromatic:
PROBLEM 1290
(a)
B—
(b)
(c)
l
(d)
PROBLEM 1291
O
+
—OH
Predict major products in the following reactions:
Cl
(a)
HNO3
H2SO4
(b)
(1) NBS hν
(2) NaOCH3
239
Problems
O
SO3
(c)
AlCl3
NO2
OCH3
O2N
O
O
Na/NH3
(e)
(f)
EtOH
PROBLEM 1292
CH3COCl
(d)
H2SO4
Zn(Hg)
HCl
Predict major products in the following reactions:
KMnO4
(a)
NaOH/∆
H 3O +
(b)
(1) C6H5ONa
(2) Br2/FeBr3
O2N
O
Br2
FeBr3
(c)
—NO2
Br
(d)
—NO2
(f)
—OCH3
(1) Br2/FeBr3
(2) NaNH2/NH3
OCH3
SO3H
(e)
NO2
HNO3
H2SO4
(1) C6H5COCl/AlCl3
(2) Fe(Hg)/HCl
(3) H2/Rh/Heat
PROBLEM 1293 Nitration of toluene gives exclusively the ortho and para substitution products. Using
mechanism, explain the reason for regioselectivity.
PROBLEM 1294 Provide reagents that will effect the following transformations. More than one steps
may be required:
(a)
OCH3
(b)
240
Problems in Chemistry
COOH
(c)
HO3S
(d)
CH3O
PROBLEM 1295
Explain.
Will the following reaction give mostly ortho/para products or mostly meta product.
—B
PROBLEM 1296
OH
HNO3
OH
H2SO4
Devise synthesis of the following compounds from the indicated starting materials:
N3
(a)
Cl
O
SO3H
(b)
Br
PROBLEM 1297
Propose mechanism for the following reaction:
—CH2Br
DMSO
heat
NaOH
—CHO
PROBLEM 1298 When treated with H 2SO 4 , α-methyl styrene gives the indicated dimeric product.
Propose mechanism of its formation:
H2SO4
α-methyl styrene
241
Problems
PROBLEM 1299
Propose a mechanism of the following reaction:
CH3
CH3
D
D2SO4
D
D2O
D
PROBLEM 1300 Would you expect Birch reduction (Na/liq. NH 3 /EtOH) of benzoic acid or anisole to
proceed more quickly, explain.
PROBLEM 1301
Devise a synthesis of the indicated compound starting from benzene:
N—OH
Br
PROBLEM 1302
Write products in the following reactions:
Br
HNO3
(a)
NaNO2
Fe/HCl
CuCN
HCl/0°C
H2SO4
Br
(b)
CH3COCl
N2H4
AlCl3
HCl
Fe/Br2
(c)
H3O+
NaBH4
O
(d)
COOH
Na
NH3
EtOH
SOCl2
242
Problems in Chemistry
PROBLEM 1303
Bring about the following transformation:
O
(a)
Br
OH
O
O
(b)
O2N
PROBLEM 1304 Provide structure of one example of the following species:
(a) An antiaromatic, (b) An aromatic cation, (c) A cyclic non-aromatic compound containing 6 π
electrons, (d) A neutral, aromatic heterocyclic compound.
PROBLEM 1305
(a)
Predict products in the following reactions:
HNO3
Br2
H2SO4
FeBr3
NO2
CH3O
NO2
(b)
Cl2/AlCl3
KMnO4
(c)
H 3O +
NaOH/heat
Na/NH3
EtOH
O
(d)
OH
(1) CH3Li (excess)
(2) H3O+
PPh3
H+
H2SO4
O
O2N
CH2
Zn(Hg)
HCl
243
Problems
(1) NaBH4
CH3COCl
(e)
AlCl3
PROBLEM 1306
(2) H3O+
Predict products of the following reactions:
(a)
H3O+
PhLi
excess
CN
CH3MgBr
HO
D2, Ph
(b)
heat, pressure
(c)
Cl
Br2
NaNH2
FeBr3
NH3
O
F3C
NaOH
(d)
H 3O +
CH3Li
excess
N
CH3O
Na
(e)
NH3/EtOH
O3
HCl
H 2O
(CH3)2S
O
(f)
(1) NBS/hν
(2) NaOCH3
LiAlH4
H 3O +
H3O+
244
Problems in Chemistry
PROBLEM 1307
Provide reagents that can be used to carryout the following transformations:
OCH3
(a)
H3CO
(b)
H
(c)
O
O2N
PROBLEM 1308
Identify A, B and C in the following sequence of reaction:
Ph 3P
C H CHO
HCl
H 2O
6 5
CH 3OCH 2I → A → B → C
BuLi
PROBLEM 1309 Nitration of A can give rise to seven possible products in principle (B H). However,
only three are produced. Indicate, which are formed and explain with the of aid of mechanism.
NO2
NO2
NO2
O2N
HNO3
H2SO4
O2N
A
NO2
NO2
C
B
NO2
NO2
NO2
D
F
E
NO2
NO2
G
NO2
NO2
H
NO2
245
Problems
PROBLEM 1310
Provide reagents that will bring about the following transformations:
O
O
(a)
C
H
O
CH3
H
N
NH2
CH3
N
N
(b)
CH3
CF3
PROBLEM 1311
CF3
Propose mechanism of the following reaction:
OH
OH
CH3
H3 C
+ Br2
CH3
H3 C
FeBr3
Br
SO3H
PROBLEM 1312
In each of the following pair, select the more stable one:
–
+
or
(a)
I
(b)
+
or
–
II
II
I
–
(c)
+
(d)
or
I
II
or
I
II
PROBLEM 1313 Rank the compounds in each group in increasing order of their reactivity towards
electrophilic aromatic substitution reaction:
(a) Chlorobenzene (I), fluorobenzene (II), cyanobenzene (III),
(b) o-dichlorobenzene (I), benzaldehyde (II), methoxybenzene (III),
(c) methoxybenzene (I), o-xylene (II), benzene (III).
PROBLEM 1314 Discuss the orienting effect of nitroso group (  NO) in aromatic electrophilic
substitution reaction.
246
Problems in Chemistry
PROBLEM 1315
Provide synthesis of the following compounds starting from benzene:
(a)
(b)
O2N
COOH
NO2
(c)
Cl
PROBLEM 1316
Write products in the following reaction:
CHO
HCN
KCN
(a)
Br
(b)
Mg
CO2
H3O+
O
OCH3 CH3MgBr
(c)
excess
OH
(e)
PROBLEM 1317
H3O+
OCH3
(d)
Na
NH3/EtOH
+
C6H5N2Cl–
Starting from benzene synthesize the following compounds:
NH2
CH3
OH
(b)
(a)
Br
CH3
(c)
CH3
Br
PROBLEM 1318 Furan undergo electrophilic aromatic substitution. With the help of mechanism,
explain the most probable site for the attack of electrophile on furan ring.
247
Problems
PROBLEM 1319
Bring about the following transformations:
NO2
OH
OCH3
(a)
(b)
CH3
N
(c)
PROBLEM 1320
Which of the following pairs is more stable:
+
(a)
●●
N
I
or
(b)
N
II
or
I
●●
II
PROBLEM 1321 Arrange the following compounds in increasing order of reactivity towards
electrophilic aromatic substitution reaction:
(a) Chlorobenzene (I), benzene (II), nitrobenzene (III),
(b) m-chloroanisole (I), p-chloroanisole (II), anisole (III),
(c) Cyanobenzene (I), p-cyanotoluene (II), p-cyanoanisole (III).
PROBLEM 1322
Bring about the following transformations:
(b)
(a)
PROBLEM 1323
Starting from toluene, synthesize:
CH3
N
CH3
H2N—
—
248
Problems in Chemistry
PROBLEM 1324
Predict the direction of electrophilic attack on the following compound:
CN
PROBLEM 1325
Provide structures of A and B and suggest mechanism of their formation:
Cl
Cl
(a)
O
C10H14(A)
AlCl3
AlCl3
NO2
P(OMe)3
(b)
heat
H2N
C12H9N
B
HCl/NaNO2
0°C, then heat
N
H
PROBLEM 1326
Propose mechanism:
O
OMe
OMe
COOMe
COOMe
NHNH2
then HCl
N
H
PROBLEM 1327
Bring about the following transformations:
Br
CH3
CHO
(a)
CHO
(b)
OH
OH
OH
(c)
OH
(d)
CHO
C
O
CH3
249
Problems
O
OCH3
(e)
OH
CHO
(f)
Ph
COOH
Ph
O
(g)
(h)
N
NHNH2
H
PROBLEM 1328
Provide missing species in the following reactions:
Na
(a)
KMnO4
HO–/heat
(c)
HBr
Peroxide
(b)
liq. NH3
NBS
hν
(d)
OH
Br
NaCN
(e)
(f)
HF
+
Br
O
Cl
(g)
AlCl3
(h)
H2SO4
SO3
PROBLEM 1329
Propose synthesis of the target compound starting with the substrate provided:
OH
(b)
(a)
O2 N
Cl
Br
250
Problems in Chemistry
PROBLEM 1330
Complete the following reaction:
+
NH3
(a)
H2SO4
Br2
(b)
HNO3
FeBr3
Br2
(c)
heat
PROBLEM 1331
Starting from benzene, synthesize:
COOH
—N
O2N—
PROBLEM 1332
N—
—OH
Bring about the following transformations:
(a)
MeO
MeO
Br
(only product)
O
(b)
O
O
NH2
(c)
(d)
(cis/trans)
251
Problems
PROBLEM 1333
Classify each of the followings as aromatic, antiaromatic or nonaromatic:
l
(a)
(b)
(c)
(d)
N
(f)
(e)
PROBLEM 1334 Ordinarily barrier to rotation about carbon-carbon double bond is quite high
(≈ 40 kcal/mol) but the compound shown below was observed to have a rotational barrier of about only
20 kcal/mol, explain.
PROBLEM 1335 Specify whether you expect the benzene ring in the following compounds to be
activated/deactivated for electrophilic aromatic substitution reaction:
O
NO2
(a)
(b)
NO2
O
(d) H2N—
(c)
—CH3
O
SO3H
(f)
(e)
CH3
CD3
PROBLEM 1336
from benzene:
Propose a reasonable synthesis of each of the following multiply substituted arenes
Br
(a) Cl—
—NO2
(b) Br—
252
Problems in Chemistry
(c)
SO3H
Predict products of the following reactions:
PROBLEM 1337
O
Br2
(a)
CH3COCl
(b)
FeBr3
AlCl3
O
O
HNO3
(c)
HNO3
(d)
H2SO4
H2SO4
CH3
O
COOH
HNO3
(e)
PROBLEM 1338
CO, HCl
(f)
H2SO4
AlCl3
Predict product of the following reactions:
Cl
O
(a)
NO2
O2N
NaNH2
NH3
Na/NH3
EtOH
(b)
O
F
OH
(c)
NaOH
Br2
NO2
O2N
(d)
NO2
CH3CH2NH2
253
Problems
Cl
O
LDA
(e)
OPh
(f)
Cl2
FeCl3
[(CH3)2CH]2NH
CF3
Cl
OH
(CH3)2CHCH2Cl
(g)
(h)
H+
AlCl3
OCH3
PROBLEM 1339
Propose mechanism:
NHCH3
O
O2 N
NO2
O2 N
H 2O
NO2
PROBLEM 1340
aromatic.
(b)
⊕
NO2
NO2
Classify each of the following species as either antiaromatic, aromatic or non
(a)
(f)
OH
N
⊕
(g)
(c)
(d)
s
(e)
s
(h)
PROBLEM 1341 Discuss the reasons for low energy barrier to rotation about the double bond
connecting the two rings, as compared to normal double bond.
254
Problems in Chemistry
Suggest a method of synthesis of the following compounds starting from benzene:
NO2
COOH
PROBLEM 1342
(b)
(a)
NO2
COOH
F
OH
NH2
OH
(d)
(c)
CH3
OH
SO3H
(f)
(e)
CH3
NO2
PROBLEM 1343
Predict product of the following reactions:
SO3H
N
Br2
(a)
(b)
FeBr3
O
NO2
HNO3
(c)
O
CH3
COOH
SO3
H2SO4
CH3
FeCl3
H2SO4
Cl
(e)
Cl2
(d)
(f)
HNO3
H2SO4
255
Problems
COCH3
H
CH3
N
(g)
Br2
Na
(h)
NH3/EtOH
FeBr3
O
COCH3
CH3
PROBLEM 1344
Bring about the following transformations:
O
PROBLEM 1345
Synthesize following compounds starting from benzene:
OH
(b) H2N—
(a)
PROBLEM 1346
—COOH
Identify A to F in the following reaction:
O
Cl
(a)
A
AlCl3
Cl
(b)
PROBLEM 1347
E
AlCl3
HNO3
H2SO4
HNO3
H2SO4
B
F
N2H4
NaOH/heat
C
Co(OAc)2
AcOH/HBr/O2
Co(OAc)2
AcOH/HBr/O2
D
No reaction.
Bring about the following transformations:
(b)
(a)
from benzene
O
CH2NH2
256
Problems in Chemistry
F
NO2
Br
(d)
(c)
NH2
On the basis of Huckel rule, discuss the aromaticity of the followings:
PROBLEM 1348
l
(b)
(a)
l
l
l
N
l
l
l
l
l
O
l
l
l
l
l
Give formula of the major organic product in each of the followings:
PROBLEM 1349
OCH3
F
(a)
O
O
O
(b) CH3O—
AlCl3
CH2Cl
NO2
HNO3
(c)
(d)
F
+
H2SO4
AlCl3
Br2
—OH
(e)
H2SO4
—CH3 +
CHCl3
PROBLEM 1350 The following compounds undergo intramolecular Friedel-Craft reaction in presence
of a Lewis acid catalyst. Give structures of most likely product in each case.
O
Cl
(a)
(b)
OMe
(c)
OMe
O
Cl
O
Cl
257
Problems
PROBLEM 1351 Synthesize the following compounds starting from benzene:
(a) 2-bromo-4-nitro toluene, (b) p-bromoanisole, (c) p-nitrobenzene sulphonic acid.
PROBLEM 1352
Bring about the following transformations:
COOH
COOH
(b)
(a)
COOH
SO3H
OCH3
COCH3
OCH3
NO2
(c)
(d)
OCH3
Cl
Cl
CH3O
F
(e)
(f)
NO2
COOCH3
Br
OH
NO2
(g)
(h)
Cl
Br
OMe
(i)
258
Problems in Chemistry
PROBLEM 1353
Starting from benzene, synthesize the following:
Br
Br
(a)
(b)
Br
CN
D
(c)
(d)
NO2
OMe
(e) F—
(f)
—COCH2CH3
CH2OH
CH3
NO2
Br
Cl
(g)
(h)
Br
PROBLEM 1354
Complete the following reactions:
NH2
NaNO2
(a)
HCl/0°C
Br
NO2
CN
(b)
HNO3
H2SO4
CuCl
HCl
Br
259
Problems
O
NaBH4
(c)
NaOH
KMnO4
(d)
CaO/heat
H2O/NaOH
Cl
H+
H 2O
H
N
(e)
AlCl3
H 3O +
O
CHO
(f)
H+
C6H 6
EtOH
1 mol
PROBLEM 1355 A sequence of reaction has been carried out on benzene as described below. Provide
reagents which are needed for each observed transformations and give a structure of product E which
results.
O
OH
H
A
O
O
B
OH
O
H
C
O
OH
CHO
COOH
D
H+
E
PROBLEM 1356 Each of the following reaction is reported to yield single predominant product. Write
structural formula for one of them.
(a)
+
H2SO4
260
Problems in Chemistry
NHOAc
O
CH2CH3
Cl
(b)
AlCl3
CH3
OH
Br2
(c)
CH3
CHCl3
C(CH3)3
HNO3
(d)
CH(CH3)2
Cl2/AlCl3
(e) O2N—
PROBLEM 1357
them?
As written, the following synthesis has certain flaws. What is wrong with each of
CH3
COOH
(i) Br2/Te
(ii) K2Cr2O7/H+
(a)
(iii) Li/Et2O
(iv) CO2/H2O
COOH
CH3
CH3
(i) HNO3/H2SO4
(b)
(ii) NaNO2/HCl/0°C
(iii) CuCN
CN
Cl
(c)
Cl
(i) HNO3/H2SO4/heat
(ii) CH3Cl/AlCl3
CH3
(iii) Sn/HCl
(iv) NaOH/H2O
NH2
261
Problems
Cl
(i) Li/Et2O
(ii) HCHO/H2O
(d)
COOH
(iii) Li/Et2O
(iv) CO2/H2O
PROBLEM 1358
Cl
Bring about the following transformations:
COOH
Cl
COOH
(b)
(a)
COOH
COOH
Cl
F
OCH3
OCH3
(d)
(c)
HOOC
OH
PROBLEM 1359 If we do the electrophilic bromination on the compound below, only one product
(major) will be formed. Predict structure of the predominant product in each case:
O
(a)
Br2
—OCH3
FeBr3
O
(b)
—NO2
Br2
Fe
CH3O
(c)
—OCH3
Br2
FeBr3
262
Problems in Chemistry
PROBLEM 1360
Write product (major) in the following reactions:
O
(a)
NO2
Cl
NH2
Br2
H2SO4
(b)
SO3/100°C
NaOH
MeO
OMe
OMe
NH2
OMe
NaNO2
(c)
HPF6
Heat
(d)
CuCl2
H2SO4/0°C
Br
NO2
O
(e)
I2
Cl
+
AlCl3
PROBLEM 1361 In the following sequence of reaction, every steps has certain flaws. Correct them by
providing appropriate reagents:
Br
Br
NO2
HNO3/H2SO4
NH2
CF3CO3H
(B)
(A)
Br
NHCOCH3
Ethanal
H+ (C)
Br
NH2
NaNO3
H2O
NHCOCH3
LiAlH4
H2SO4/0°C
(F)
CH3CH2CH2Cl
AlCl3(D)
(E)
Br
CuBr
Br
Br
COOH
Br
(i) Li/Et2O
(ii) CO2/H2O
(G)
COOH
Br
Br
F2
Fe
(H)
F
263
Problems
PROBLEM 1362
H
On the basis of Huckel rule, decide whether the following species are aromatic or not:
N
(a)
(c)
(b)
(d)
●●
PROBLEM 1363 The following series of reaction was performed during synthesis of E. Provide
appropriate reagents needed to perform the indicated transformations:
CH3
COOH
A
O
COCl
B
O
C
NEt2
O2N
NO2
NO2
NO2
O
NEt2
O
D
H2N
E
PROBLEM 1364 Two, isomeric, aromatic lactones A and B have molecular formula C 9 H 8O 4 . Both A
and B dissolves in dilute NaOH solution but none gives gas with NaHCO 3 . Both A and B gave violet
colouration with FeCl 3 . Reaction of A with NaOH followed by work-up with CH 3I yields C (C10 H10O 4 ).
Selective demethylation of C with BCl 3 followed by aqueous work-up yields D which is also an isomer
of A. D showed presence of an intramolecularly hydrogen bonded hydroxyl group. Complete hydrolysis
of either A or C with dilute H 2SO 4 produced the same compound E (C 8 H 8O 5 ). Oxidising E with alkaline
permanganate followed by acidification of product yields F (C 8 H 6O 6 ). Heating F with sodalime
followed by acidification of product yields 1,3-dihydroxy benzene. Identify A to F.
PROBLEM 1365
(a) Cl—
Write the major product in the following reactions:
HNO3
H2SO4
(b) O2N—
Cl2
Fe
264
(c)
Problems in Chemistry
CH3Cl
H3CO—
AlCl3
Cl2/Fe
(d) O2N—
NO2
PROBLEM 1366
Bring about the following transformation in good yield.
Cl
NH2
CARBOHYDRATES, AMINO ACIDS AND
POLYMERS
PROBLEM 1367 (a) What do you understand from reducing and non-reducing sugar?
(b) How might you distinguish between a α-glucopyranose and methyl α-D-glucopyranoside?
PROBLEM 1368 Name two disaccharides and write their hydrolysis reaction.
PROBLEM 1369 Name two common polysaccharides and write their hydrolysis products.
PROBLEM 1370 What type of saccharide is lactose? Is it a reducing sugar? Write hydrolysis reaction of
lactose.
PROBLEM 1371 With the help of a chemical reaction show that D-glucose and D-mannose are epimers.
H
HO
H
H
CHO
OH
H
OH
OH
CH2OH
D-glucose
HO
HO
H
H
CHO
H
H
OH
OH
CH2OH
D-mannose
PROBLEM 1372 A, B and C are all aldohexose. Compound A and B yield the same optically active
alditol on reduction with H 2 / Ni but they gives different phenyl hydrazones when treated with
C 6 H 5 NHNH 2 . On the other hand B and C gives same phenyl hydrazones on reaction with C 6 H 5 NHNH 2 ,
but different alditols on reduction. Assuming that all three are D-sugars, deduce structures of A, B and C.
PROBLEM 1373 Why an aminoacid is usually solid at room temperature?
PROBLEM 1374 Write the structures of predominant form of 2-amino-3-methyl butanoic acid (valine)
in its aqueous solution at pH (a) 2 (b) 12.
265
Problems
NH 2

PROBLEM 1375 What form of glutamic acid (HOOC—CH 2 —CH 2 —CH—COOH) predominate in:
(a) Strongly acidic solution?
(b) Strongly basic solution?
(c) At its isoelectric point?
PROBLEM 1376 Glutamic
acid
has
lower
isoelectric
point
O

(H 2 N—C—CH 2CH 2 —CH—COOH) has higher isoelectric pH. Explain?

NH 2
while
glutamine
PROBLEM 1377 An α-amino acid A has molecular formula C 7 H 7 NO 3 . A on treatment with methanol in
presence of HCl yields B (C 4 H10 NO 3Cl). B on further treatment with PCl 5 yields C (C 4 H 9 NO 2Cl 2 )
which on acidic hydrolysis yields D(C 3 H 6 NO 2Cl). D on reduction with Na(Hg) in dilute acidic medium
yields alanine. Deduce structures of A to D.
PROBLEM 1378 Synthesize alanine from acetaldehyde and any other inorganic reagent needed.
PROBLEM 1379 (a) What do you understand from step growth polymerization?
(b) How decaron is synthesized by this method?
PROBLEM 1380 How teflon is synthesized by radical chain polymerization? Device synthesis using
benzoyl peroxide initiator and mention few of its common applications.
PROBLEM 1381 (a) How nylon-6,6 is synthesized?
(b) Why nylon has very high melting point compared to other polymers.
MISCELLANEOUS
Complete the following reactions:
PROBLEM 1382
—NH2
O+
(a)
Major
O
HO
(b) H
(c)
CN
O
O
OH
H+
H3O+
LiAlH4
H3O+
H2O
266
Problems in Chemistry
O
(C2H5)2CuLi
(d)
O
(C2H5)2AlCN
(e)
(CH3COO)2Hg
(f)
H+
Write major products of the following reactions:
PROBLEM 1383
NH2
(1) NaNO2/H+/0°C
(a)
(2) HCl, Hg2Cl2
O
HgO, Br2
(b)
CCl4/∆
OH
(c)
O
O
O
H+/H2O
(d)
Heat
HO
CH3
CH3O
(1) CH3O–
(2) H+/H2O
O
OH SOCl
2
(e)
O
O
N—H
—C—O—CH2—
(f)
H+
H 2O
CH3
Br
(f)
—N
+
N+
—OH
(h)
NaN3
H
LiAlH4
H
PROBLEM 1384 Aminomethylcyclohexane can be prepared from simple starting materials using the
four methods shown below. Provide a suitable starting material for each of the reaction shown:
A
N3– then LiAlH4
LiAlH4
B
O
NH2
N– , H2 O
O
C
D
LiAlH4
267
Problems
PROBLEM 1385
Starting with acetophenone, suggest a synthesis of each of the following compounds.
O
O
(A)
(C)
(B)
N
PROBLEM 1386 Cocaine (A) on alkaline hydrolysis produces benzoic acid, methanol and
B (C 9 H15 NO 3 ). B on treatment with acidified solution of chromic acid produces “C” which on simple
heating yields the following compound:
H3 C
N
O
Deduce structures of A, B and C.
PROBLEM 1387 Starting with benzyl alcohol and any alkylhalides containing five or less carbons,
propose a step-by-step synthesis of the following compound.
OH
Ph
PROBLEM 1388 Propose synthesis of Dimestrol starting from p-methoxypropiophenone. Show
mechanism of each step.
OCH3
H3CO
(Dimestrol)
PROBLEM 1389 Compound A(C 4 H 8O) gives positive iodoform test but does not decolourize bromine
water solution or it does not evolve any gas on treatment with NaH. Reaction of A with excess of
benzaldehyde in alkaline medium gives two products B and C. These products are unstable and readily
dehydrate into stable products D and E. D reacts with I 2 in basic medium to give iodoform; E does not.
Treatment of D with bromine in acetic acid medium yields F. F can be reduced by NaBH 4 to give G.
Warming G in basic solution results in formation of H. Ozonolysis of H followed by reductive work-up,
gives benzaldehyde and a ketone I (C 4 H 6O 2 ). Identify A–I.
268
Problems in Chemistry
PROBLEM 1390
Beginning with benzyl bromide, suggest a synthesis of the following compounds:
O
(a)
and
(b)
OCH3
PROBLEM 1391
appropriate:
Complete the following reaction indicating stereochemistry of product where
O
(i)
(a)
MgBr
CH2CH3
(b)
O
(ii) H+/H2O
O
H+/H2O
(i) CH3CH2O–
(c)
(ii) H+/H2O
PROBLEM 1392 Compound A, on reaction with CH 3Cl/AlCl 3 gives compound B. The mass spectrum
of B is dominated by a peak at m/e = 91, which is identified as tropylium ion. B-undergoes free radical
bromination to give C. C reacts with diethylmalonate in presence of sodium ethoxide to give D. Treating
D with aqueous acid followed by gentle warming gives E and CO 2 . Identify A to E.
PROBLEM 1393
Br
Draw structures of major organic product in each of the reaction below:
O
H
(a)
NaCN, Et2O–H2O
then HCN
(b)
(i) NaBH4
(ii) H2O
O
O
OMe PhMgBr(excess)
(c)
Et2O
CH3
(d)
O
EtOH
H2SO4
H2O
H+
Heat
269
Problems
CH3
Br2–H2O
(e)
NaOH
PROBLEM 1394 Suggest reaction sequences, including reagents and conditions, which could be used
to effect the transformation of 1 to compound 2 and 3. You may use the indicated starting material and
any reagents, which you deem necessary.
OH
HO
OH
OH
2
1
3
Bring about the following transformation:
PROBLEM 1395
O
O
O
OEt
O
PROBLEM 1396 Exposure of 1,3-diketone 1 to aqueous acid catalyzes the formation of enols 2 and 3.
Justifying your answer, indicate, which is thermodynamically more stable enol. Draw a mechanism of
formation of most stable tautomer of 1 in presence of an acid catalyst.
O
OH
O
H3O+
O
O
CH2
+
1
OH
3
2
PROBLEM 1397 For the following carbonyl compounds X, Y and Z, draw the structures of
thermodynamically most stable enolate anion that could be formed in presence of a strong base:
O
O
X
O
Y
Z
270
Problems in Chemistry
PROBLEM 1398
O
Place the following dicarboxylic acids in order of increasing acidity.
O
O
O
O
HO
OH
HO
OH HO
OH HO
A
O
B
O
C
OH
D
O
PROBLEM 1399 Suggest reaction sequence, including reagents and conditions, which could be used to
bring about the following transformations:
O
O
O
A
O
O
OCH3
Br
O
B
O
N
C
PROBLEM 1400
Propose mechanism of the following reaction:
O
O
CH3
O
CH3O–Na+
+
KOH
MeOH
MeOH, Heat
O
PROBLEM 1401
O
Provide structures of missing species:
O
A
Acetone
H+
O
O
Ph3P
O
H
CH2
B
C
O
O
HO
(CH3)2NH
OH
HO
N
CH3
CH3
E
D
271
Problems
PROBLEM 1402 Unknown monosaccharide 1 was converted to tetramethylated aldohexose 4 through
the sequence of reaction shown below:
CHO
CH3OH
HCl
Cu2+
1
Cu2O
precipitated
2
Cu2+
CH3I (excess)
Ag2O, CH3OH
3
H+
H2O
No reaction
CH3O
CH3O
H
H
H
H
OCH3
OH
CH2OCH3
4
Using the information provided, draw the correct structures of 1, 2 and 3.
Bring about the following transformations in good yield:
PROBLEM 1403
O
O
Br
CN
PROBLEM 1404 When the diester (A) was treated with a solution of sodium ethoxide in ethanol and the
reaction mixture then quenched with aqueous acid, only (B) was formed. The alternative product (C) was
not formed under these reaction conditions. Explain the result.
OEt
EtO
O
A
O
O
O
(1) NaOEt/EtOH
O
OEt
OEt
(2) H3O+
O
C
B
PROBLEM 1405 Synthesize the following compounds starting from styrene:
O

(a) Ph  C  CH 3
(b) Ph  CH 2  CHO
(c) PhCH 2CH 2CH 2OH
PROBLEM 1406
Bring about the following synthesis starting from phenyl acetylene:
Br
(a)
HO
—COCH3
(b)
—CH2CH2OH
(c)
QUALITATIVE ANALYSIS
PROBLEM 1407 A is a white compound, on heating gives yellow residue B and colourless gas that turns
lime water milky. If B is heated in air for several hours at 500°C, it is converted into a scarlet powder C. C
on heating with dilute aqueous nitric acid gives a colourless solution D and a brown solid E. If sodium
272
Problems in Chemistry
hydroxide is added to solution D, a white gelatinous precipitate is formed initially, which dissolves in
excess of base. Compound E reacts with concentrated hydrochloric acid to give a white solid F and a
green coloured gas. F is soluble in hot water but insoluble in cold, and forms a soluble complex with
excess of chloride ion. With KI solution, D gives a bright yellow solid G which is insoluble in cold water,
but in hot water a colourless solution, which on cooling gives a shimmering yellow plates like crystal.
Identify A to G.
PROBLEM 1408 A metal reacts with 50% nitric acid solution to give a blue coloured solution ( A ) and a
brown gas ( B ). If the solution A is cautiously treated with dilute NaOH solution, a gelatinous blue
precipitate (C ) is formed, which if warmed, forms a black solid ( D ). Addition of concentrated ammonia
solution to C, gives a deep blue solution that contains the ion ( E ) while addition of concentrated HCl
solution to C gives a green solution of ion ( F ). If brown gas B is passed through water, a mixture of two
monobasic acids are formed. Identify A to F and write reaction of B with water.
PROBLEM 1409 A white powder turns yellow (B ) on heating and evolves a gas which turns lime water
milky, as well as water vapour. The yellow residue turns white on cooling, but will turns yellow again
when heated. B reacts with dilute sulphuric acid to give a colourless solution (C ). If dilute NaOH is added
to C, a white precipitate ( D ) is formed initially, which dissolves on adding excess base. With dilute
ammonia solution, C gives a white precipitate which dissolves in excess ammonia, giving a clear
solution ( E ). Identify A to E.
PROBLEM 1410 A metal ( A ) reacts moderately quickly with dilute sulphuric acid to give a pale green
solution ( B ) and a colourless, neutral gas. If solution B is allowed to crystallize, a pale green solid is
obtained which on strong heating gives a solid (C ) and two acidic, non-metallic gases. A reacts with
steam to give another solid ( D ) and the same gas as obtained in the first experiment with dilute sulphuric
acid. A reacts with dry Cl 2 ( g ) to produce a brown covalent solid ( E ), which sublime on heating. Aqueous
solution of E reacts copper with metal, and for that reason are used to each printed circuit boards in
electronics. Solution B gives a dirty green precipitate ( F ) if sodium hydroxide is added and remains
insoluble in excess NaOH. If F is allowed to stand in air, it forms a foxy-red compound G, which can also
be obtained by adding NaOH( aq ) to aqueous solution of E. Identify A to G.
PROBLEM 1411 A compound of sulphur (one atom per molecule), oxygen and one or more halogen
atoms, was examined. A small amount of the substance reacted with water, it was completely hydrolyzed
without any oxidation or reduction and all reaction products dissolved. 0.1 M solution of a series of test
reagents were added to separate, small portions of a dilute solution of the substance.
(i) Addition of HNO 3 / AgNO 3
(ii) Addition of Ba(NO 3 ) 2
(iii) Adjustment to pH = 7 with NH 3 and addition of Ca(NO 3 ) 2
(iv) Addition of KMnO 4 followed by Ba(NO 3 ) 2 to an acid solution of the substance.
(v) Addition of Cu(NO 3 ) 2 .
The above tests gave the following results:
(i) A yellowish precipitate
(ii) No precipitate
(iii) No visible reaction
(iv) The main features were that purple colour disappeared and a white precipitate was formed on
addition of Ba(NO 3 ) 2 .
(v) No precipitate. Finally 7.2 g of the substance was dissolved in water and volume made to 250
mL. 25 mL of this solution on treatment with excess of AgNO 3 gave 1.425 g precipitate.
Identify the original compound.
Problems
273
PROBLEM 1412 A white salt ( A ) evolves a colourless gas on treatment with dilute HCl solution, which
turned moistened litmus paper red. A small sample of A was moistened with concentrated HCl and
placed on a platinum wire and introduced into a bunsen burner flame, a green colouration was observed.
On strong heating, A decomposes to produce a white solid B, which turned red litmus blue. 1.54 g of B
was dissolved in 250 mL H 2O and its 25 mL required 20.4 mL 0.0985 M HCl solution. Identify A and B.
PROBLEM 1413 A metallic chloride A, when treated with NaOH/H 2O 2 , gives yellow coloured solution
due to formation of B. The colour of this solution changes to orange when dilute H 2SO 4 is added. It is
due to the formation of compound C. When a compound D is heated with C in presence of concentrated
H 2SO 4 , a red volatile liquid E is formed. E, when absorbed in NaOH solution, gives a yellow coloured
solution of B, which when treated with (CH 3COO) 2 Pb solution gives a yellow precipitate. Compound C
when treated with NH 4Cl, forms a compound F, which decomposes on heating, giving a colourless gas,
water and a green residue G. Also D gives golden yellow flame colouration. Identify A to G.
PROBLEM 1414 A white powder A on heating gives a colourless gas B and a solid residue C. The
compound C turns yellow on heating and changes white on cooling. It dissolves in dilute acid and the
resulting solution gives a white precipitate ( X ), with K 4 [Fe(CN) 6 ] solution. Further A dissolves in dilute
HCl with evolution of a gas B, which turns lime water milky. The solution thus obtained gives a white
precipitate D with H 2S in slightly alkaline medium. Another portion of the solution gives initially a white
precipitate E with NaOH, which dissolves on addition of excess base. Identify A to E.
PROBLEM 1415 A white solid A on heating with excess of dilute HCl gave a pungent smelling gas B
and a solution C. Solution C on treatment with aqueous ammonia did not give any precipitate, but on
treatment with NaOH, precipitate D is obtained which dissolves in excess reagent. A on strong heating in
air gave a pungent smelling gas E and a solid residue F. F dissolves completely in dilute HCl solution
and the resulting solution produced a precipitate with BaCl 2 . Identify A to F.
PROBLEM 1416 A white compound A on heating yields a basic oxide B. Compound A on treatment
with dilute HCl evolves a gas which turns lime water milky. A is insoluble in water but dissolves in
mineral acid. When HCl solution of A is made ammonical and a small amount of ammonium oxalate is
added, a white precipitate C is obtained. Solid C decolourises acidic solution of KMnO 4 .Solution of C in
dilute HCl evolves CO 2 with MnO 2 and the resulting solution gives a pink coloured precipitate with
Na 2S solution. Also A gives brick red colouration in bunsen burner flame. Identify A to C.
PROBLEM 1417 A mixture of salt ( A ) is yellow coloured and it did not lose any weight on simple
heating. A dissolved in cold, dilute nitric acid solution giving an orange coloured solution. This solution
on treatment with aqueous NaOH gives a white coloured precipitate ( B ) which dissolves in excess of
reagent leaving some white coloured residue (C ). Filtrate of the above solution forms a brown precipitate
( D ) on treating with H 2O 2 . Compound D reacts with concentrated HCl solution evolving a green
coloured gas and a white solid E is formed which turned black on passing H 2S gas. Also C dissolves in
dilute acid giving, a clear solution, to which if KI is added slowly, a black precipitate ( F ) is formed
initially which dissolves in excess KI forming an orange solution. Identify A to F.
PROBLEM 1418 A green salt mixture ( A ) consists of halides of two metal and both salts are soluble in
water. A on dissolving in dilute HCl and passing H 2S gas yields a black precipitate ( B ). B was removed
by filtration and filtrate was made alkaline by adding NH 3 solution and finally treated with H 2O 2 giving
a yellow coloured solution of ion (C ). On acidifying C with dilute H 2SO 4 followed by treatment with
few drops of amyl alcohol and finally with excess of H 2O 2 , a blue colouration was observed due to
formation of ( D ), which can be extracted into organic phase by gentle shaking. Compound C is insoluble
274
Problems in Chemistry
in boiling, dilute H 2SO 4 , in NaOH as well as in aqueous Na 2S, but dissolves in concentrated solution of
nitric acid, leaving behind a white precipitate. If the above solution is boiled for a long time, white
precipitate dissolved and a clear blue solution was formed. Sodium extract of the original salt mixture
gave a white precipitate with dilute AgNO 3 solution, which dissolved in excess of aqueous ammonia
solution. Identify A to D.
PROBLEM 1419 A mixture consists of a two metal oxide A (green) and B (white). Mixture was
dissolved in 20 mL 2 M NaOH solution containing some H 2O 2 to give a clear yellow solution leaving no
residue. The above solution was separated into two part. One part was acidified with acetic acid and then
treated with (CH 3COO) 2 Pb solution to give an yellow precipitate (C ). C dissolve in dilute nitric acid
forming a clear orange solution. Other part of the solution was acidified with 2 M HCl and the 2 M NH 3
was added till solution became alkaline and finally boiled. A white gelatinous precipitate ( D ) was
obtained. D was then dissolved in dilute HCl and some 6 M ammonium acetate was added. The solution
was finally treated with few drops of “aluminion” reagent and made basic by adding ammonium
carbonate. A red precipitate was obtained. Identify A to D.
PROBLEM 1420 A solid mixture consists of a red metal oxide ( A ) and a white hydrated salt B. The
mixture was dissolved in dilute HNO 3 and a portion of this solution was treated with NH 3 solution, just
to make the solution neutral and then finally treated with NH 3 / NH 4Cl buffer solution when a reddish
brown gelatinous precipitate (C ) was obtained. C was then dissolved in dilute HCl and few drops of
NH 4SCN solution was added when a deep blood red colouration was observed. Filtrate obtained after
removal of C was then treated with excess of (NH 4 ) 2 S solution when a pink coloured precipitate ( D ) was
formed. D was then dissolved in dilute HNO 3 and treated with NaBiO 3 solution. A deep purple coloured
solution was formed.
In a separate experiment a pinch of the original salt mixture was mixed with solid K 2Cr 2O 7 and then
dissolved in concentrated H 2SO 4 and finally boiled. A deep red fumes of ( E ) was obtained which made
aqueous solution of NaOH yellow, when passed through it. Identify A to E.
PROBLEM 1421 A solid substance is a mixture of a scarlet-red oxide ( A ) and a brown substance (B ).
The solid dissolve in boiling solution of dilute nitric acid giving a brown precipitate (C ) and a clear
orange solution. Solution is filtrated off and filtrate as well as precipitates were preserved for further
analysis. Filtrate was further acidified by adding dilute HCl and H 2S gas was passed. A black precipitate
( D ) was obtained. D was then filtered off and filtrate was treated with excess H 2O 2 and finally boiled to
decompose off any unreacted H 2O 2 . The solution was finally cooled and treated with K 4 [Fe(CN) 6 ]
solution when a deep blue coloured precipitate ( E ) was formed. Compound ( D ) was dissolved in dilute
HNO 3 and boiled, resulting in formation of a white precipitate. Brown solid C was analysed by
dissolving in a concentrated HCl solution which resulted in formation of white precipitate ( F ) and a
green coloured gas was evolved. Compound F was dissolved in boiling water and treated with aqueous
sodium chromate solution, resulting in formation of a yellow precipitate. Identify A to F.
PROBLEM 1422 A salt mixture consists of two salts, salt-A is scarlet coloured and insoluble in water,
while salt-B is yellow coloured and soluble in water. The mixture was shaken with water in a test tube
and filtrated off. The precipitate is pure A and filtrate is solution of B. A dissolves in concentrated
solution of KI forming a colourless solution of C, which when treated with aqueous ammonia, forms a
brown precipitate D.
A portion of original filtrate containing ions of salt B was treated with aqueous NaOH solution
resulting in formation of a white gelatinous precipitate ( E ), soluble in excess reagent. E was dissolved in
Problems
275
dilute HCl solution and few drops of this solution was placed on a filter paper, previously moistened with
an alcoholic solution of alizarin. Filter paper, on drying at 100°C, gave a red lake.
In a separate experiment, 1 mL of the original filtrate was mixed with 1 mL starch solution and few
crystals of KNO 2 were added, a deep blue colouration was observed. Identify A to E.
PROBLEM 1423 A mixture consists of a white oxide A and a white salt B in which latter is water
soluble. Mixture was dissolved in water and filtered to separate precipitate A and filtrate containing B. A
was dissolved in dilute hydrochloric acid solution and then treated with aqueous ammonia when a white
gelatinous precipitate C was formed initially, which dissolved in excess reagent, forming a clear solution
of D. A portion of original filtrate was treated with aqueous (NH 4 ) 2 CO 3 solution when a white
precipitate E was formed. E gave a brick-red colouration in Bunsen burner flame. An other portion of
original filtrate was mixed with same concentrated sulphuric acid in a test tube and a freshly prepared
ferrous sulphate solution was add through side wall of the test tube. A brown ring appeared at the
junction of two solution. Identify A to E.
PROBLEM 1424 A white mixture contain two salts A and B, both water soluble. A pinch of this salt
mixture was dissolved in water and treated with excess of ammonium oxalate solution, resulting in
formation of a white precipitate. Precipitate was added to a boiling solution of dilute acetic acid, where a
portion of it went into solution, leaving behind a white solid C. Solution was cooled, filtered and filtrate
was divided into two part. One part of the filtrate was treated with aqueous K 2CrO 4 solution, giving a
yellow precipitate ( D ), insoluble in dilute acetic acid. Other part of filtrate was treated with excess of
(NH 4 ) 2 CO 3 solution giving a white precipitate ( E ). Solid compound C and E was subjected to flame test
separately. C gave brick-red colouration while E gave apple-green colouration in the Bunsen burner
flame.
In a separate analysis, an aqueous solution of original salt mixture was prepared and divided into two
parts. One part of the solution was mixed with concentrated sulphuric acid solution and then a freshly
prepared ferrous sulphate solution was added slowly, giving a brown ring at the junction. Other part of
the solution was treated with aqueous AgNO 3 solution when a white precipitate soluble in aqueous
ammonia, was formed. Also salt B is hydrated one, and on heating lost 49.3% of its weight to become
anhydrous. Identify A to E.
PROBLEM 1425 A red coloured salt ( A ) is insoluble in water and dilute HCl but dissolved in dilute
nitric acid, giving a clear orange solution. This solution when treated with a Na 2CO 3 solution, gave a
white precipitate ( B ), leaving behind a clear orange solution B on strong heating gives off a colourless
acidic gas leaving behind a brown solid (C ), soluble in aqueous ammonia as well as in dilute nitric acid.
After removing precipitate B by filtration, filtrate was mixed with aqueous NaOH and turned into a clear
yellow solution. Yellow solution on acidification with dilute sulphuric acid followed by treatment with
excess of H 2O 2 gave a blue colouration due to formation of a new compound D. D can be extracted into
organic phase by shaking the solution with ether. Identify A to D.
PROBLEM 1426 A white metal ( A ) burns in nitrogen to produce an ionic compound B. B on treatment
with lime water forms a white precipitate (C ), with evolution of a colourless gas ( D ). D when passed
through an alkaline solution of K 2 [HgI 4 ], an yellow precipitate ( E ) was produced. Compound C was
dissolved in dilute hydrochloric acid and some NH 4Cl crystals were added to this solution. Resulting
solution was treated with oxine (8-hydroxy quinoline), when a yellow precipitate of complex salt was
produced. Identify A to E.
PROBLEM 1427 A white hydrated salt ( A ) loses 45% of its weight on heating and rendered anhydrous.
A on strong heating yields a dirty brown solid ( B ) with evolution of two gases (C ) and ( D ). C turned
276
Problems in Chemistry
orange colour of dichromate solution into blue-green. B dissolved in boiling HCl solution to produce an
yellow solution of ( E ). Solution E was separated into two parts and one part was treated with NH 4SCN
solution giving a blood-red solution. To the other part of solution, H 2S was passed after making it
alkaline, where a white turbidity was observed initially and the solution became light green on standing
for some time. An aqueous solution of A gave white precipitate when treated with aqueous barium
chloride solution. Identify A to E.
PROBLEM 1428 A white salt ( A ) does not dissolve in water, but dissolve in dilute nitric acid solution.
Passing H 2S gas through acidic solution of A, a black precipitate was produced which was insoluble in
water, dilute HNO 3 and alkali but dissolved in concentrated solution of Na 2S as well as in aqua-regia.
Adding SnCl 2 -dropwise to the solution of A, gave a white, silky precipitate ( B ), which turned into black
precipitate (C ) on adding excess reagent. Addition of KI to the acidified solution of A gave a red
coloured precipitate ( D ) in the beginning, which dissolved in excess reagent. Also acidified solution of
A, when treated with concentrated sulphuric acid followed by K 2Cr 2O 7 and finally boiled, a dense
brown fume was formed, which turned aqueous solution of NaOH, yellow coloured. Identify A to D.
PROBLEM 1429 A black salt ( A ) is insoluble in dilute acid in cold but dissolve in boiling solution of
concentrated hydrochloric acid, evolving a pungent smelling gas B. B turned a filter paper moistened
with lead acetate, black. The above solution was cooled and separated into two parts. One part was
treated with NaOH solution giving a white precipitate (C ), insoluble in excess of reagent. C, when added
to a concentrated H 2O 2 solution, a clear, yellowish-brown solution of D was formed. Other part of the
original solution was treated with KI solution, giving a black precipitate ( E ), which dissolved in excess
of reagent forming a clear orange solution. Identify A to E.
PROBLEM 1430 A white salt mixture contain two salts (anhydrous). A and B, both are water soluble.
An aqueous solution of the salt mixture was prepared and treated with dilute NaOH when a white
precipitate was obtained. Precipitate was shaken with concentrated NaOH solution and filtered again.
The residue (C ) containing metals of salt B and filtrate were analyzed independently. Passing H 2S gas
through filtrate yielded another white precipitate D. The residue C was dissolved in 5.0 mL diphenyl
carbazide solution, when a violet red precipitate was formed. Also, a small portion of the original
aqueous solution was mixed with some concentrated H 2SO 4 solution and boiled. An irritating gas ( E )
was formed, which gave a white cloud on the surface of a glass road moistened with ammonia solution.
Also, the original aqueous solution did not give test of nitrate and sulphate ion. Identify A to E.
PROBLEM 1431 A solid mixture consists of a red oxide ( A ) and a water soluble, hydrated salt (B )
which loses 51.2% of its weight on heating and becomes anhydrous. Solid mixture was dissolved in
dilute HCl and the solution was used for further analysis. A portion of this solution was mixed with
excess of (NH 4 ) 2 S, when a black precipitate (C ) was obtained. Solution was filtered off and filtrate was
treated with aqueous NaOH when another white precipitate ( D ), insoluble in excess of reagent was
produced. D was dissolved in small amount of dilute HCl and then excess of NaOH was added, followed
by addition of a few drops of magneson-I [4-(4-nitrophenylazo) resorcinol], when a blue coloured
precipitate formed. Solid C, when dissolved in dilute HCl, a white precipitate ( E ) was produced.
Another portions of the original aqueous solution was treated with K 4 [Fe(CN) 6 ] and NH 4SCN
separately, when deep blue, precipitate and deep blood-red colouration were observed respectively.
Soda-extract of the original mixture gave a white precipitate ( F ) with BaCl 2 solution. F when fused
with Na 2CO 3 on charcoal and the residue extracted with water and finally filtered into a freshly prepared
solution of nitropruside {Na 2 [Fe(CN) 5 NO]}, transient purple colouration was observed. Identify A to F.
Problems
277
PROBLEM 1432 A salt mixture contain a black salt ( A ) and a white salt (B ). Salt mixture dissolves in
dilute sulphuric acid solution producing a clear blue solution. Passing H 2S gas through this solution gave
black precipitate (C ). The solution was filtered off and colourless filtrate was treated with aqueous
ammonia giving a white precipitate ( D ), that was readily soluble in aqueous ammonium chloride
solution.
Sodium extract of the salt mixture was prepared and divided into two part. One part of the extract
was treated with BaCl 2 solution, giving a white precipitate ( E ), which dissolved into dilute HCl
evolving a gas ( F ). F turned a filter paper, moistened with acidified K 2Cr 2O 7 solution, green. Other part
of extract was mixed with excess of MnO 2 and concentrated sulphuric acid solution and finally boiled. A
brown coloured vapour was formed which gave orange-red staining when passed through starch paper.
Identify A to F.
PROBLEM 1433 A mixture consists of a black oxide ( A ) and a yellow salt (B ). Mixture was shaken with
dilute sulphuric acid solution resulting in formation of a yellow residue ( B ) and a clear solution on
filtration. Filtrate was treated with excess of H 2O 2 and finally boiled to decompose-off unreacted H 2O 2 ,
cooled and divided into two parts. One part of the solution was treated with aqueous K 4 [Fe(CN) 6 ]
solution giving a blue precipitate (C ), while other part of the solution gave a deep red colouration when
treated with NH 4SCN solution. Residue B was insoluble in excess ammonia but dissolved in KCN
solution as well as in Na 2S 2O 3 solution.
Sodium extract of the original mixture was mixed with chlorine water when a brown colouration
was observed. Brown solution when shaken with chloroform, organic phase turned violet. Identify A to
C.
PROBLEM 1434 A mixture contain a yellow salt ( A ) and a green salt (B ). Heating salt mixture gives-off
a colourless gas which turned lime water milky. The residue thus obtained was shaken with dilute
hydrochloric acid which gave a white precipitate (C ) and a clear blue solution. Solution was filtered off
and H 2S gas was passed through filtrate when a black precipitate ( D ) was obtained. D was insoluble in
boiling dilute H 2SO 4 as well as in NaOH solution but dissolved in a concentrated aqueous potassium
cyanide solution forming a clear and colourless solution of E. Also C dissolved in excess of ammonia as
well as in sodium thiosulphate solution. Sodium extract of the residue, obtained after heating of original
salt mixture, when mixed with concentrated H 2SO 4 and some K 2Cr 2O 7 , a deep brown fume was
produced on boiling, that turned NaOH solution yellow. Identify A to E.
PROBLEM 1435 A mixture consists of an yellow salt ( A ) and a white salt (B ), both anhydrous. Salt
mixture was dissolved in water and few drops of HCl was added to obtain a clear, yellow coloured
solution. The solution was then treated with NH 3 / NH 4Cl solution when a reddish-brown precipitate (C ),
insoluble in NaOH solution, was produced. Solution was filtered off and filtrate was treated with
aqueous NaOH solution when a white precipitate ( D ), soluble in excess reagent, was formed. D was
dissolved in dilute H 2SO 4 solution and few drops of copper (II) sulphate was added to it. To the resulting
solution, 2.0 mL of ammonium tetrathiocyanato mercurate (II) [(NH 4 ) 2 [Hg(SCN) 4 ]] solution was
added when a violet precipitate was formed. Also precipitate (C ), dissolved in dilute HCl and became
blood-red coloured when NH 4SCN solution was added. Sodium extract of the original salt mixture gave
the following results:
(a) Extract solution in concentrated H 2SO 4 , when mixed with K 2Cr 2O 7 , gave a dense brown fumes
on boiling, that turned aqueous NaOH solution yellow.
(b) Extract solution when treated with Hg(NO 3 ) 2 , a yellow precipitate ( E ) was formed. Identify A
to E.
278
Problems in Chemistry
PROBLEM 1436 A white salt mixture contain two salts, salt A is water soluble and salt B is water
insoluble. the salt mixture was leached with water so that soluble portion went into solution, leaving
behind residue of B. Solution was filtered off and filtrate was preserved for further analysis. Precipitate B
was insoluble in dilute HCl as well as in concentrated ammonium acetate solution but dissolved in
boiling solution of concentrated sulphuric acid. The solution was cooled and treated with K 2CrO 4 giving
a yellow precipitate (C ), that was insoluble in dilute acetic acid solution. Also the original filtrate gave a
white gelatinous precipitate ( D ) on treatment with aqueous ammonia. D was dissolved in small volume
of concentrated nitric acid and few drops of cobalt nitrate was added to the above solution. Now a filter
paper was dipped into the solution and burned finally to give a blue fixed ash due to formation of ( E ).
Sodium extract of the original salt mixture gave the following results:
(a) Boiling with concentrated sulphuric acid solution gave off a colourless, pungent smelling gas
which formed a white cloud on the surface of a glass rod moistened with ammonia solution.
(b) Gave a white precipitate with lead acetate solution, that was insoluble in dilute acid but
dissolved in concentrated solution of ammonium acetate. Identify A to E.
CO-ORDINATION COMPOUNDS
PROBLEM 1437 Name the following compounds according to IUPAC convention:
(i) [Co(NH 3 ) 6 ]Cl 3 (ii) [CoCl(NH 3 ) 5 ]2+ (iii) [CoSO 4 (NH 3 ) 4 ]NO 3
(iv) K 2 [OsCl 5 N] (v) Na 3 [Ag(S 2O 3 ) 2 ] (vi) K 2 [Cr(CN) 2 (O) 2 (O 2 )NH 3 ]
(vii) [Ir(NH 3 ) 5 (ONO)]Cl 2 (viii) K 2 [PbCl 4 ] (ix) KCu[Co(CN) 6 ] (x) K 3 [Au(CN) 6 ]
(xi) Li[AlH 4 ] (xii) Na[BH 4 ] (xiii) Na 3 [AlF6 ]
(xiv) [CoCl 2 (H 2O) 4 ]Cl ⋅ 2H 2O
PROBLEM 1438 Name the following complexes according to IUPAC convention:
(i) [Cr(NH 3 ) 6 ][Cr(NCS) 6 ] (ii) [Cu(NH 3 ) 4 ][PtCl 4 ] (iii) [Pt(NH 3 ) 4 ][PtCl 4 ]
(iv) [Co(NH 3 ) 6 ]2 [Ni(CN) 4 ]3 (v) [Pt(NH 3 ) 4 Cl 2 ][PtCl 4 ]
(vi) [Co(NH 3 ) 6 ]4 [Co(NO 2 ) 6 ]3 (vii) [Cr(NH 3 ) 5 NO 2 ]3 [Co(NO 2 ) 6 ]2
(viii) [Pt(Py) 4 ][PtCl 4 ] (ix) [Ni(NH 3 ) 6 ]3 [Co(NO 2 ) 6 ]2
(x) [Co(NH 3 ) 6 ][Cr(NH 3 ) 2 Cl 4 ] (xi) [Co(NH 3 ) 4 Br 2 ]2 [ZnCl 4 ]
(xii) [Co(NH 3 ) 5 CO 3 ]2 [CuCl 4 ] (xiii) [Ag(NH 3 ) 2 ]4 [Fe(CN) 6 ]
(xiv) [Cr(en) 2 Cl 2 ]2 [PdCl 4 ]
PROBLEM 1439 Name the following complexes according to IUPAC convention:
(i) [(NH 3 ) 5 CoNH 2Co(NH 3 ) 5 ](NO 3 ) 5
S
O
O
(ii) [(H 3 N) 4 Co
Co(NH 3 ) 4 ](NO 3 ) 3
NH 2
279
Problems
CO
(iii) [(CO) 3 Fe——CO——Fe(CO) 3 ]
CO
NH 2
(iv) [(H 3 N) 4 Co
Co(en) 2 ]Cl 4
OH
NH 2
(v) [(en) 2 Co
Co(en) 2 ]Br 4
OH
NO 2
(vi) [(en) 2 Co
Co(en) 2 ]Cl 4
NH 2
PROBLEM 1440 Write molecular formula of the following complexes:
(i) pentaaminebromocobalt (III) sulphate
(ii) dichlorobis(ethylendiamine) platinum (IV) sulphate.
(iii) tetrapyridineplatinum (II) tetrachloroplatinate (II).
(iv) potassium carbonylpentacyanoferrate (II).
(v) cesium pentafluorotellurate (IV)
(vi) triammine bromoplatinum (II) nitrate.
(vii) dichlorobis (ethylenediamine) cobalt (III) chloride monohydrate.
(viii) tetraquadibromochromium (III) chloride
(ix) ammonium heptafluorozirconate (IV)
(x) hexaammine nickel (II) hexanitrocobaltate (III).
(xi) dichlorobis (ethylenediamine) chromium (III) tetrachloropalladate (II).
(xii) aluminium tetrachloroacerate (III).
(xiii) Iron (III) hexacyanoberrate (II).
PROBLEM 1441 Two
complexes
of
cobalt
have
same
molecular
formula
Co(NH 3 ) 3 (H 2O) 2 ClBr 2 ⋅ 34.15 g of A loses 1.8 g of weight when heated to 200°C whereas no weight
loss was observed on similar treatment with B. An aqueous solution of A has conductivity equivalent to
that of a compound with two ions per formula unit while conductivity of B in aqueous solution is
equivalent to a compound with three ions per formula unit. Also a 100 mL 0.1 M aqueous solution of A,
on treatment with excess of AgNO 3 solution produced 1.88 g of precipitate while a 100 mL 0.1 M
aqueous solution of B on treatment with excess of AgNO 3 solution produced 3.315 g precipitate. Identify
A and B. (Atomic weight : Co = 59, Br = 80).
PROBLEM 1442 A complex salt has molecular formula Co(NH 3 ) 5 SO 4 Br and exist in two isomeric
forms A and B. An aqueous solution of A produced a white precipitate on treatment with BaCl 2 solution
while B did not produce any precipitate on similar treatment. B on treatment with aqueous AgNO 3
produced a brown precipitate. Identify A and B and predict the type of isomerism being exhibited by
them.
PROBLEM 1443 An octahedral complex of platinum has molecular formula Pt(NH 3 ) 4 Cl 2 ⋅ Br 2 and
exist in two isomeric forms A and B. An aqueous solution of A containing 49.4 g of salt give 37.6 g of
280
Problems in Chemistry
precipitate on treatment with excess AgNO 3 solution whereas an aqueous solution of B containing same
amount of salt gave 28.7 g of precipitate on treatment with excess of AgNO 3 solution. Identify A and B.
[Atomic mass : Pt = 195, Ag =108, Br = 80].
PROBLEM 1444 A, B and C are three complexes of chromium (iii) with their formula H12O 6Cl 3Cr.
Complex A does not react with concentrated H 2SO 4 solution whereas complexes B and C loses 6.75%
and 13.5% of their weight, respectively, on treatment with concentrated H 2SO 4 solution. Deduce
formula of A, B and C.
PROBLEM 1445 An aqueous solution of titanium chloride, when subjected to magnetic measurement,
measured zero magnetic moment. Deduce formula of complex assuming it to be octahedral in aqueous
solution.
PROBLEM 1446 A cobalt (ii) salt when dissolved in excess of concentrated HCl, it forms an octahedral
complex whose magnetic moment was found to be 3.87 Bm. Predict shape of complex.
PROBLEM 1447 How [Co(NH 3 ) 6 ][Cr(NO 2 ) 6 ] and [Co(NO 2 ) 6 ][Cr(NH 3 ) 6 ] can be distinguished by
electrochemical method?
PROBLEM 1448 How [Cr(NH 3 ) 6 ][Cr(NO 2 ) 6 ] and [Cr(NH 3 ) 4 (NO 2 ) 2 ][Cr(NH 3 ) 2 (NO 2 ) 4 ] can be
distinguished from conductivity measurement in their aqueous solution?
PROBLEM 1449 Two complexes of chromium have empirical formulae corresponding to
Cr(NH 3 ) 3 (NO 2 ) 3 . In aqueous solution, one of these is non-conducting while other is electrolytic. What
is the lowest possible formula weight of conducting reagent? What is the highest possible formula
weight of non-conducting reagent?
PROBLEM 1450 Write formula of all possible polymerization isomers of Pt(NH 3 ) 2 (NO 2 ) 2 .
PROBLEM 1451 A metal complex having composition Cr(NH 3 ) 4 Cl 2 Br has been isolated in two forms
A and B. A reacts with AgNO 3 producing a white precipitate which was soluble in dilute ammonia
solution. B reacts with AgNO 3 producing a pale-yellow precipitate soluble in concentrated ammonia
solution. Write formula of A and B and state hybridization of chromium in each. Also calculate their
magnetic moment (spin only).
PROBLEM 1452 A metal complex having composition Cr(NH 3 ) 4 ClBr 2 ⋅ H 2O has been isolated in
three forms A, B and C. Heating 16.7 gram of either A or C loses 0.9 gram of its weight whereas no weight
loss was observed on heating B. Adding 35.35 g of A to a concentrated solution of AgNO 3 resulted in
14.35 gram of precipitate whereas adding same amount of C to a concentrated AgNO 3 solution resulted
in 18.8 g of precipitate. In another experiment, when 6.67 gram of complex B was added to a
concentrated aqueous solution of AgNO 3 , when 7.52 gram precipitate was formed. Deduce formula of
complexes A, B and C.
PROBLEM 1453 An octahedral complex of cobalt has its composition Co(NH 3 ) 3 (H 2O) 3 (SO 4 )(NO 3 )
and it exists in four isomeric forms A, B, C and D. Simple heating of either A or B loses 5.6% of its
original weight whereas similar treatment on C causes 11.2% loss in its original weight D did not lose
any weight on simple heating. Either A or C did not result in any precipitation if treated with aqueous
BaCl 2 solution, whereas B and D resulted a white precipitate when treated with aqueous BaCl 2 solution.
Also C does not conduct electricity in its aqueous solution. Deduce formula of A, B, C and D.
PROBLEM 1454 An octahedral complex of platinum has its composition Pt(NH 3 ) 4 Cl 2 Br 2 and exist in
three isomeric forms A, B and C. A 4.94 gram of compound A on treatment with excess of aqueous
Problems
281
AgNO 3 gave 3.76 g of precipitate. 2.47 gram of B on treatment with excess of AgNO 3 solution gave 1.66
gram of precipitate while 9.88 gram of C gave 5.74 gram of precipitate with aqueous AgNO 3 . Deduce
structural formula of A, B and C. (Atomic weight Pt =195, Br = 80, Ag = 108)
PROBLEM 1455 A complex of cobalt has composition Co(NH 3 ) 3 (H 2O) 2 ClBr 2 and exist in two
isomeric forms A and B. A on simple heating loses 5.3% of its weight whereas B does not lose any weight
on simple heating. A on treatment with excess of AgNO 3 solution gives off precipitate that is 42% by
mass of the complex. Similar treatment of B gives off another precipitate which is 110% by mass of the
complex. Deduce structures of A and B.
PROBLEM 1456 A solution was prepared by dissolving 0.5 g CrCl 3 ⋅ 6H 2O in sufficient water and
volume made upto 100 mL. A 10.0 mL position of this solution when treated with excess of AgNO 3 ,
gave a precipitate, filtered dried and weighed to be 0.538 gram. Deduce structural formula of the
complex.
PROBLEM 1457 A complex [CoA 6 ]3+ is red coloured while [CoB6 ]3+ is green coloured.
(a) Which ligand produces the lager crystal field splitting.
(b) Which complex is expected to be more easily oxidised.
PROBLEM 1458 In which complex [Cu(H 2O) 4 ]2+ or [Cu(NH 3 ) 4 ]2+ , are the bonds from acceptor to the
donor atoms stronger?
PROBLEM 1459 Which complex is more stable [Cr(NH 3 ) 6 ]3+ or [Cr(en) 3 ]3+ ?
PROBLEM 1460 Why are complexes of Ti 4+ and Zn 2+ are typically white?
PROBLEM 1461 Accounts for the followings :
2–
is paramagnetic.
(a) Ni(CN) 2−
4 is diamagnetic but [NiCl 4 ]
(b) [Ni(CO) 4 ] is tetrahedral
(c) [Ni(NH 3 ) 6 ]2+ is an outer orbital complex.
PROBLEM 1462 Why [CuCl 4 ]2– exist but [CuI 4 ]2− does not?
3+
PROBLEM 1463 Both Fe(CN) 3−
6 and Fe(H 2O) 6 are colourless in aqueous medium, explain.
PROBLEM 1464 Calculate the magnetic moment of K 3 [Mn(CN) 6 ] and K 2 [MnBr 4 ] and explain the
geometries of these complexes.
PROBLEM 1465 Select the complexes which will absorb in visible range :
(a) [Ti(H 2O) 6 ]Cl 3 (b) [VF4 ] (c) [Cu(NH 3 ) 4 ]+ (d) [Cu(NH 3 ) 4 ]2+
(e) [Ni(CN) 4 ]2− (f) [Ni(CO) 4 ]
PROBLEM 1466 CoF63− is paramagnetic while Co(CN) 3−
6 is diamagnetic, explain.
PROBLEM 1467 Use an orbital diagram to show the electron distribution in 3d-orbitals of the central
metal ion in each of the following complex ion. If more than one distribution seems possible, indicate
whether you expect low-spin or high-spin state to be favoured and determine spin-only magnetic
moment.
(d) [Mn(CN) 6 ]3–
(c) [CoCl 4 ]2–
(b) [FeCl 6 ]3–
(a) [Cr(H 2O) 6 ]3+
282
Problems in Chemistry
PROBLEM 1468 Of the following two complexes:
[Cr(H 2O) 6 ]Cl 3 and [Cr(NH 3 ) 6 ]Cl 3 , one is yellow coloured while other is violet. Identify them.
PROBLEM 1469 Of the following two complexes :
[Fe(H 2O) 6 ] (NO 3 ) 2 and K 4 [Fe(CN) 6 ] ⋅ 3H 2O, one is green coloured while other is yellow. Identify
them.
PROBLEM 1470 Explain why one of the cis-trans isomers of [CoCl 2 (en) 2 ] exhibit optical isomerism
and other does not.
PROBLEM 1471 Would you expect to find optical isomerism in either of the hypothetical tetrahedral
complexes : [ZnA2 B 2 ]2+ and [ZnABCD ]2+ ? Explain.
PROBLEM 1472 The complex [Pt(NH 3 ) 2 Cl 2 ] displays cis-trans isomerism but [ZnCl 2 (NH 3 ) 2 ] does
not. Why do you suppose these two cases are different?
PROBLEM 1473 The magnetic properties of octahedral complex ion [CrL 6 ]3+ are independent of
identity of the ligand L. How do you account for this fact?
PROBLEM 1474 Four structures are shown in the accompanying sketch. Indicate whether any of these
structures are identical, geometrical isomers or optical isomers:
en
Cl
Co
Cl
3+
Co
Cl
en
Cl
en
3+
Cl
en
3+
Co
en
Cl
3+
en
Co
Cl
en
en
Cl
(a)
(b)
(c)
(d)
PROBLEM 1475 When ethylene diamine is added to a solution of cobalt chloride hexahydrate in
concentrated HCl, a blue crystalline solid separate out. Analysis of this complex shows : N = 14%,
C =12%, H = 5%, Cl = 53.25%. The magnetic moment of the solution was found to be 3.9 Bm. Determine
structural formula and shape of this complex.
PROBLEM 1476 Addition of TiCl 3 to an aqueous solution of urea followed by addition of KI gives a
deep blue crystals of octahedral complex containing Ti, urea and iodine with this magnetic moment
(spin-only) =1.76 Bm. Also 1.0 g of this complex on heating in excess of oxygen gives 0.12 g TiO 2 .
Deduce structures of complex.
PROBLEM 1477 (a) Explain why dissolution of a chromium (III) salt produces an acidic solution?
(b) Explain why the slow addition of hydroxide ions to a solution of Cr(III) ions first produces a
gelatinous precipitate that subsequently dissolves with further addition of hydroxide ion.
PROBLEM 1478 Name the following complex ions:
(c) [Pt(NH 3 )Cl 3 ]–
(a) [Zr(OX ) 4 ]4–
(b) [CuCl 4 (H 2O) 2 ]2–
(d) [Mo(CN) 4 (OH) 4 ]4–
PROBLEM 1479 By considering electronic configuration, suggest a reason why iron (III) compounds
are readily prepared from iron (II), but conversions of Ni(II) and Co(II) to Ni(III) and Co(III) are much
more difficult?
283
Problems
PROBLEM 1480 Suggest a chemical test for distinguishing between:
(a) [Ni(SO 4 )(en) 2 ]Cl 2 and [NiCl 2 (en) 2 ]SO 4
(c) [Ni(en) 2 I 2 ]Cl 2 and [NiCl 2 (en) 2 ]I 2
PROBLEM 1481 (a) Sketch the orbital energy-level diagrams for [MnCl 6 ]4– and [Mn(CN) 6 ]4– .
(b) Which complex is expected to transmits the longer wavelengths of incident electromagnetic
radiations?
PROBLEM 1482 The complex [Co(CN) 6 ]3– is pale yellow.
(a) Is short or long wavelength visible light absorbed?
(b) How many unpaired electrons are available?
(c) If ammonia molecules are substituted for cyanide ions as ligands, will the shift in absorbance of
radiation be towards the blue or red region of electromagnetic radiation?
PROBLEM 1483
Explain why high spin complex Mn 2+ are only faintly coloured.
PROBLEM 1484 The relative thermodynamic stability of two complexes can be predicted from a
comparison of their standard potentials. Determine which complex of the following pair is the more
stable?
(i)
(ii)
[Co(NH 3 ) 6 ]3+ + e → [Co(NH 3 ) 6 ]2+
[Co(H 2O) 6 ]
3+
+ e → [Co(H 2O) 6 ]
2+
E ° = 0.11 V
E ° =1.81 V
PROBLEM 1485 (a) When excess of AgNO 3 is added to a 0.001mol Cr(III) chloride, 0.001 mol AgCl
is formed.
(b) When excess of AgNO 3 is added to a 0.001 mol Cr(III) chloride, 0.003 mol of AgCl is formed.
Discuss the structure of complex based on above information.
PROBLEM 1486 Identify the type of isomers represented by the following pairs :
(a) [Cu(NH 3 ) 4 ][PtCl 4 ] and [Pt(NH 3 ) 4 ][CuCl 4 ]
(b) [Cr(OH) 2 (NH 3 ) 4 ] Br and [CrBr(OH)(NH 3 ) 4 ]OH
(c) [Co(NCS)(NH 3 ) 5 ]Cl 2 and [Co(SCN)(NH 3 ) 5 ]Cl 2
(d) [CrCl(H 2O) 5 ]Cl 2 ⋅ H 2O and [CrCl 2 (H 2O) 4 ]Cl ⋅ 2H 2O
PROBLEM 1487 Following is a chelating ligand called nitrilotriacetic (NTA) acid. It act as a
tetradentate ligand. Sketch the structures of [Co(NTA)(H 2O) 2 ]3+ ion and propose type of isomerism if
present any : N(CH 2COOH) 3 : Nitrilotriacetic acid.
PROBLEM 1488
The complex shown below is called diethylenetriamine (dien) :
••
••
••
H 2 N  CH 2CH 2  NH  CH 2CH 2  NH 2
(a) Sketch the structure of complex [Co(dien) 2 ]3+
(b) Which complex would be expected to be more stable in aqueous solution, [Co(dien) 2 ]3+ or
[Co(NH 3 ) 6 ]3+ ?
PROBLEM 1489 Arrange the following complexes in order of increasing wavelength of visible light
absorbed by them :
(a) [Cr(H 2O) 6 ]3+ (b) [CrCl 6 ]3– (c) [Cr(en) 3 ]3+ (d) [Cr(CN) 6 ]3– (e) [Cr(NH 3 ) 6 ]3+
284
Problems in Chemistry
PROBLEM 1490 Which complex should be expected to absorb light of highest frequency
[Cr(H 2O) 6 ]3+ , [Cr(en) 3 ]3+ or [Cr(CN) 6 ]3– ?
PROBLEM 1491 Which complex should absorb light at longer wavelength?
(b) [Mn(CN) 6 ]3– or [Mn(CN) 6 ]4–
(a) [Fe(OH 2 ) 6 ]2+ or [Fe(CN) 6 ]4–
PROBLEM 1492 Sketch the structures of the following complexes indicating any cis-trans or optical
isomerism:
(b) [Pt(NH 3 )(Py)(Cl)(Br)] square planar, Py = Pyridine
(a) [Pt(H 2O) 2 Br 2 ] square planar
(c) [Zn(NH 3 ) 3 Cl]+
(d) [Ni(H 2O) 4 Cl 2 ]
(e) [Co(C 2O 4 ) 2 Cl 2 ]3–
PROBLEM 1493 Draw diagrams for any cis-trans and optical isomers that could exist for the following
complexes:
(c) [Cr(NH 3 ) 2 (H 2O) 2 Br 2 ]+
(b) [Co(en) 2 Cl 2 ]+
(a) [Co(en) 2 NO 2 (Cl)]+
3+
(d) [Pt(NH 3 ) 2 Cl 4 ]
(e) [Cr(en) 3 ]
PROBLEM 1494 Determine the spin only magnetic moment:
(c) [Mn(CN) 6 ]3–
(b) [Co(en) 3 ]3+
(a) [CoF6 ]3–
(d) [MnCl 6 ]4–
PROBLEM 1495 If trimethylphosphene is added to solution of Ni(II) chloride in acetone, a blue
compound of molar mass 280 is isolated. This compound on analysis gave the following composition :
Ni = 21.5%, Cl = 26%, and (CH 3 ) 3 P = 52.5%
This blue compound does not have any isomeric forms. What are the geometry and molecular formula of
this blue complex? Also determine magnetic moment (spin-only) of the complex.
PROBLEM 1496 The complex Na 2 [Ni(CN) 2 Br 2 ] has zero magnetic moment. Predict geometry and
draw shape of the complex.
PROBLEM 1497 Draw all geometrical isomers of the following complexes:
(a) [Co(NH 3 ) 2 Cl 4 ]–
(b) [Co(NH 3 ) 3 Cl 3 ]
PROBLEM 1498 Draw structures of all the geometrical and optical isomers of each of the following
complex:
(c) [Co(C 2O 4 ) 3 ]3–
(b) [Co(NH 3 ) 5 Cl]2+
(a) [Co(NH 3 ) 6 ]3+
PROBLEM 1499 The [Ni(CN) 4 ]2– ion is diamagnetic while [Ni(Cl) 4 ]2– ion is paramagnetic. Explain
with the help of crystal field splitting diagram.
PROBLEM 1500 Transition metal complexes containing cyanide (CN – ) ligands are often yellow in
colour, whereas those containing H 2O ligands are often green or blue in colour. Explain.
PROBLEM 1501 The absorption maximum for complex ion [Co(NH 3 ) 6 ]3+ occurs at 470 nm.
Determine the crystal field splitting energy.
PROBLEM 1502 For each of the following pair, choose the complex that absorbs light of longer
wavelengths.
(b) [FeF6 ]3– and [Fe(CN) 6 ]3–
(a) [Co(NH 3 ) 6 ]3+ and [Co(H 2O) 6 ]3+
(c) [Cu(NH 3 ) 4 ]2+ and [CuCl 4 ]2–
285
Problems
PROBLEM 1503 A solution made by dissolving 0.875 g of compound Co(NH 3 ) 4⋅Cl 3 in 25 g of water
freezes at −0.56° C. Deduce structural formula of complex if cryoscopic constant of water is
1.86 K kg mol –1 .
PROBLEM 1504 Oxalic acid H 2C 2O 4 is sometimes used to clean rust stains from bathtubs. Explain the
chemistry underlying the cleaning action.
PROBLEM 1505 When aqueous potassium cyanide is added to a solution of Cu(II) sulphate, a white
precipitate is formed in the beginning which dissolves on adding excess of reagent. No precipitate is
formed when H 2S( g ) is bubbled through this solution at the point. Explain.
PROBLEM 1506 A concentrated aqueous copper (II) chloride solution is bright green in colour. When
diluted with water, the solution becomes light blue. Explain.
PROBLEM 1507 In dilute nitric acid solution, Fe 3+ forms a dark-red complex with thiocyanate ion
(SCN – ) according to following reaction :
[Fe(H 2O) 6 ]3+ + SCN –
H 2O + [Fe(H 2O) 5 NCS]2+
In one experiment,1.0 mL of 0.20 M Fe(NO 3 ) 3 was mixed with1.0 mL of a10 −3 M KSCN solution
and 8 mL dilute nitric acid was added. Final concentration of [Fe(H 2O) 5 (SCN)]2+ was found to be
7.3 ×10 –5 M. Calculate formation constant of complex.
PROBLEM 1508 A student has prepared a cobalt complex that has one of the following three structures
: [Co(NH 3 ) 6 ]Cl 3 , [Co(NH 3 ) 5 Cl]Cl 2 or [Co(NH 3 ) 4 Cl 2 ]Cl. Explain how the student would distinguish
between these possibilities by an electrical conductance experiment. At the student’s disposal are three
strong electrolytes—NaCl, MgCl 2 and FeCl 3 .
PROBLEM 1509 Aqueous solution of CoCl 2 are generally either light pink or blue. Low concentrations
favour the pink form while high concentration favours blue form. Adding HCl to a pink solution of
CoCl 2 causes the solution to turn blue; the pink colour is restored by addition of HgCl 2 .Account for these
observations.
PROBLEM 1510 Suggest a method that would allow you to distinguish between cis-Pt(NH 3 ) 2 Cl 2 and
trans-Pt(NH 3 ) 2 Cl 2 .
PROBLEM 1511 You are given two solutions containing FeCl 2 and FeCl 3 at same concentrations. One
solution is light yellow and the other is brown. Identify these solutions based on their colours.
REPRESENTATIVE ELEMENTS
PROBLEM 1512 Write equations for the following process :
(a) B2O 3 + Mg
(b) The result of the addition of aqueous NaOH to a mixture of solid Al 2O 3 and Fe 2O 3 .
(c) CO 2 + Na[Al(OH) 4 ]( aq ) →
286
Problems in Chemistry
PROBLEM 1513 Explain how, during dimerization, each BH 3 molecule acts both a Lewis base and a
Lewis acid.
PROBLEM 1514 Suggest likely products for the following reactions:
(a) BF3 + EtOH →
(b) BCl 3 + PhNH 2 →
(c) BF3 + KF →
PROBLEM 1515 Comment on the following observations:
(a) AlF3 is almost insoluble in anhydrous HF but dissolves if KF is present. Passage of BF3 through
the resulting solution causes AlF3 to reprecipitate.
(b) Borazine is similar to benzene in some ways but dissimilar in other ways.
PROBLEM 1516 (a) Suggest why trimethylamine is pyramidal while trisilylamine is planar.
(b) Suggest reasons why, at 300 K, CO 2 and SiO 2 are not isostructural.
PROBLEM 1517 Write equations for : (a) the reaction of SiCl 4 with aqueous NaOH
(b) the hydrolysis of SiH 3Cl,
(c) the hydrolysis of SiF4 .
PROBLEM 1518 What would you expect to form when:
(a) Sn is heated with concentrated NaOH,
(b) SO 2 is passed over PbO 2 ,
(c) SiH 2Cl 2 is hydrolysed,
(d) 4ClCH 2SiCl 3 + 3Li[AlH 4 ] →
PROBLEM 1519 Discuss the bonding in BeH 2 in its solid state.
PROBLEM 1520 Write balanced equations for the following reactions:
(a) NaH + H 2O →
(b) NaN 3 + Heat →
(c) KO 2 + H 2O →
(d) NaF + BF3 →
PROBLEM 1521 Write balanced equations for the following reactions:
(a) Thermal decomposition of [NH 4 ]2 [BeF4 ]
(b) The reaction between NaCl and BeCl 2
(c) The dissolution of BeF2 in water.
PROBLEM 1522 (a) Suggest a likely structure for the dimer of BeCl 2 , present in the vapour phase.
What hybridization scheme is appropriate for the Be centers?
(b) Suggest structure of BeCl 2 in diethyl ether.
PROBLEM 1523 How do anhydrous CaCl 2 and CaH 2 function as drying agents?
PROBLEM 1524 Among NaF and NaCl, NaF has higher melting point instead of its lower molecular
weight while reverse trend is there between CF4 and CCl 4 , justify.
PROBLEM 1525
4 Be
7
and 6 C11 are positron emitter while 6 C14 is beta emitter, explain.
PROBLEM 1526 Determine nuclear binding energy of He.
Problems
287
PROBLEM 1527 Lithium ion being smallest in its group, is expected to have highest ionic mobility and
hence solution of its salts would be expected to have a higher conductivity than solution of cesium salts.
Explain, why this is not so.
PROBLEM 1528 Ionization energy suggest that Cs should be the most reactive metal but the standard
reduction potential suggest that Li is most reactive. Reconcile these two observations.
PROBLEM 1529 Discuss the colour of alkali metal solution in liquid ammonia.
PROBLEM 1530 0.35 g of a metal ( A ) was dissolved in dilute HNO 3 . This solution gave a red
colouration to a non-luminous Bunsen-burner flame and on evaporation gave 0.75 g of metal oxide ( B ).
A also reacted with hydrogen forming C. On reacting 0.16 g of C with water, a gas D was evolved and a
sparingly soluble compound E was formed, which gave a strongly alkaline reaction and required 200 mL
0.1 M HCl for neutralization. Identify A to E.
PROBLEM 1531 Which is more heavily hydrated LiCl or MgCl 2 ?
PROBLEM 1532 Why do Be salts seldom contain more than four molecules of water of crystallization?
PROBLEM 1533 An element ( A ) reacts quitely with water librating a colourless, odourless gas and a
solution of ( B ). Passing CO 2 gas through B yields a while precipitate (C ) which dissolves on passing
excess CO 2 gas. Precipitate C effervesced with dilute HCl and gave a deep red colouration to a Bunsen
burner flame. Heating C gave a caustic white compound ( D ) which when further heated with carbon at
1000°C, gave a solid ( E ) of immense commercial importance. Identify A to E.
PROBLEM 1534 A white salt ( A ) evolves a colourless gas (B ) when treated with dilute HCl. Bubbling
( B ) through lime water, a precipitate (C ) formed initially which dissolved forming a clear solution on
further passing the gas. A small sample of A was moistened with concentrated HCl, placed on a platinum
wire and introduced into a Bunsen burner flame where it caused a green flame colouration. On strong
heating, A decomposed giving a white solid ( D ) which turned red litmus blue. Identify A to D.
PROBLEM 1535 SnCl 2 is a good reducing agent but PbCl 2 is stable, explain.
PROBLEM 1536 AlCl 3 exist as dimer in benzene. While in aqueous solution it remains in the form of
hydrated ions, explain.
PROBLEM 1537 Discuss the reaction of B2 H 6 with NH 3 under following conditions:
(a) Excess ammonia at low temperature.
(b) Excess ammonia at high temperature.
(c) NH 3 and B2 H 6 in 2 :1 ratio at high temperature.
PROBLEM 1538 Complete the following inorganic reactions:
(a) B3 N 3 H 6 + HCl →
(b) B2 H 6 + H 2O →
(c) B2 N 3 H 6 + H 2O →
(d) B2 H 6 + EtOH →
(e) B2 H 6 + LiH →
(f) B2 H 6 + HCl →
(g) B2 H 6 + Cl 2 →
PROBLEM 1539 An yellowish-white, deliquescent substance ( A ) has its vapour density 133.5. A reacts
quickly with water forming a solution B. A sample of B gave a curdy white precipitate (C ) on addition of
dilute HNO 3 and AgNO 3 solution, but this readily dissolved on addition of dilute NH 4OH, though a
288
Problems in Chemistry
white gelatinous precipitate ( D ) was formed in its place. D dissolves in a concentrated solution of NaOH,
but passage of CO 2 gas through this solution results in reprecipitation of D.
Also compound ( A ) dissolved unchanged in dry diethyl ether and when this solution was reacted
with excess of LiH, a new compound E was formed. Identify A to E.
PROBLEM 1540 (a) How SiCl 4 is prepared from SiO 2 ?
(b) How SiC is prepared from SiO 2 ?
PROBLEM 1541 Complete the following reactions:
(a) Al 4C 3 + H 2O →
(b) CaC 2 + H 2O →
1100°C
(c) CaC 2 + N 2 →
O
2
(d) SiC + NaOH →
(e) Mg 2C 3 + H 2O →
PROBLEM 1542 Synthesize a linear chain silicones starting with SiCl 4 and ethylmagnesium chloride.
PROBLEM 1543 How R 3SiCl acts to control the molecular weight in preparation of polysilicones.
PROBLEM 1544 Draw the structures of:
12–
(a) tri cyclo-dimethyl siloxane (b) Si 3O 6−
9 (c) Si 6O18
(d) Pyroxene (e) Amphiboles.
PROBLEM 1545 Complete the following reactions:
(a) P4 + NaOH →
(b) CaNCN + H 2O →
(c) N 2 H 4 + I 2 →
(d) N 2O + NaNH 2 →
PROBLEM 1546 Discuss, how superphosphate and triple superphosphate fertilizers are manufactured?
PROBLEM 1547 Complete the following reactions:
(a) N 2O 5 + NaCl →
(b) N 2O 5 + Na →
(c) N 2O 4 + H 2O →
(d) NO 2 + Cl 2 →
(e) NO 2 + HCl →
PROBLEM 1548 Explain why nitrogen exist as diatomic molecule while phosphorus exist as tetra
atomic (P4 ) molecule.
PROBLEM 1549 Why NF3 is stable while NCl 3 and NI 3 are explosive?
PROBLEM 1550 Draw the shapes of P4O 6 and P4O10 .
PROBLEM 1551 (a) How nitric acid is prepared by Ostwald’s method.
(b) Give one example of each in which HNO 2 acts as an oxidising agent and reducing agent.
PROBLEM 1552 Explain the following observations:
(a) Liquid oxygen sticks to the pols of magnet but liquid nitrogen does not.
(b) The N—O + ion has a shorter bond than does NO.
PROBLEM 1553 Describe method how H 2O 2 is prepared by electrolysis of H 2SO 4 . Draw structure of
H 2O 2 . Write balanced equations for the reaction of H 2O 2 with:
(a) an acidified solution of KMnO 4 , (b) an acidified solution of K 2Cr 2O 7 ,
(c) aqueous HI,
(d) an acidic solution of K 4 [Fe(CN) 6 ].
289
Problems
PROBLEM 1554 Draw shapes of the following oxyacid of sulphur and indicate oxidation states of
sulphur.
(b) H 2S 2O 5 (c) H 2S 2O 4 (d) H 2S 2O 3
(a) H 2SO 3
(g) H 2S 2O 8 .
(e) H 2S 2O 6 (f) H 2SO 5
PROBLEM 1555 How sulphuric acid is manufactured by contact process?
PROBLEM 1556 Provide two method by which sodium thiosulphate is synthesized. Why thiosulphuric
acid can’t be made by adding acid to aqueous solution of thiosulphate salt?
PROBLEM 1557 Complete the following chemical reactions:
(a) F2 + H 2O →
(b) Cl 2 + CO →
(c) Cl 2 + NH 3 (excess) →
(d) Cl 2 + NaOH →
(e) SiO 2 + HF →
PROBMEL 1558 Complete the following reactions:
(a) Cl 2O + NaOH →
(b) HClO 4 + P4O10 →
(c) ClO 2 + NaOH →
(d) KBrO 3 + F2 + KOH →
(e) Cl 2O 6 + NaOH →
(f) Cl 2O 6 + HF →
PROBLEM 1559 Why it is not possible to produce F2 by electrolysis of aqueous NaF, aqueous HF or
anhydrous HF?
PROBLEM 1560 (a) Draw structures of OF2 , Cl 2O, O 2 F2 and I 2O 5 .
(b) Explain bond angle in OF2 and give a reason why it is different in Cl 2O.
(c) Why are the OF bond in O 2 F2 longer than in OF2 while O—O bond in O 2 F2 is short compared
with that in H 2O 2 .
PROBLEM 1561 Complete the following reactions:
(a) XeF2 + HCl →
(b)
(c) XeF4 + SF4 →
(d)
(e) XeF4 + H 2O →
(f)
(g) XeF6 + XeO 3 →
(h)
XeF4 + KI →
XeF2 + H 2O →
XeF6 + H 2O →
XeO 3 + XeOF4 →
PROBLEM 1562 Ferric bromide (FeBr 3 ) when dissolved in water, converted into Fe 2+ evolving
Br 2 ( g ) while no chlorine gas is evolved when FeCl 3 is dissolved in water. Justify.
PROBLEM 1563
When Cl 2 ( g ) is mixed with F2 ( g ), ClF3 ( g ) is produced not the FCl 3 ( g ). Explain.
PROBLEM 1564 Explain why nitrogen forms extra-ordinary stable N 2 at room temperature but
phosphorus forms P2 only at very high temperature.
PROBLEM 1565 Explain sulphur forms compounds such as SF4 , SF6 , etc while oxygen, of the same
group forms only OF2 ?
PROBLEM 1566 In laboratory, HF and HCl can be prepared by reacting the metal fluorides and
chlorides with concentrated H 2SO 4 but HBr can’t be prepared by the same procedure. Explain.
PROBLEM 1567 Discuss the acidic behaviour shown by boric acid in water and explain why it can’t
be titrated satisfactorily by NaOH. Why does glycerol or other such syn vicinal diols/triols enhances its
acidic properties?
290
Problems in Chemistry
PROBLEM 1568 Explain why CCl 4 is unaffected by water while SiCl 4 is rapidly hydrolyzed? In CCl 4
unreactive towards superheated steam?
PROBLEM 1569
Explain the difference between temporary and permanent hardness in water.
PROBLEM 1570
What is “Boiler-scale” and how it is formed?
PROBLEM 1571
Describe the role of Li 2CO 3 and BaSO 4 in medicinal chemistry.
PROBLEM 1572 Sodium hydroxide has lower melting point (322°C) than sodium chloride (801°C),
electrolysis of NaCl is the primary source of Na-metal not NaOH. Explain.
PROBLEM 1573 Complete the diagram below indicating the reagents and special reaction conditions
needed to produce each substance, using MgCO 3 as the starting material.
[E ]
[F ]
[D ]
[A]
[C ]
[B]
Mg(NO 3 ) 2 ← Mg(OH) 2 ← MgO ← MgCO 3 → MgCl 2 → Mg → Mg 3 N 2
[G]
MgHPO 4
PROBLEM 1574 Consider the following sequence of actions and observations (1) A small chunk of
dry ice is added to an aqueous Ca(OH) 2 solution and a white precipitate forms initially (2). After a short
time, precipitate redissolve. Write chemical equations to explain the observations.
PROBLEM 1575 Although, both boric acid and phosphoric acid have three hydroxy groups,
phosphoric acid is a weak tribasic acid while boric acid is a weak monobasic acid.
PROBLEM 1576
Provide missing reagents below:
[A]
[B]
[C ]
[D ]
Na 2 B4O 7 ⋅ 10H 2O → B(OH) 3 → B2O 3 → BCl 3 → B
[E ]
→ BF3
PROBLEM 1577 A typical baking powder contain baking soda and alum NaAl(SO 4 ) 2 as its active
ingredients. During process, baking powder undergoes a reaction that yields CO 2 . Write a plausible
equation for this reaction.
PROBLEM 1578 What is the principle that underlies the fact that thallium forms both +1and + 3 ions
while aluminium forms only + 3 ion?
PROBLEM 1579 Why can’t aluminium cookware be used in cooking strongly acidic food?
PROBLEM 1580 In the use of aluminium sulphate in water treatment, the water to be treated is usually
kept between pH 5 and pH 8. Why do you suppose this is the case?
PROBLEM 1581 Complete the following reactions:
(a) N 2 H 4 ( aq. ) + HCl →
(b) NO( g ) + O 2 ( g ) →
(c) NO 2 ( g ) + H 2O( l) →
(d) NH 3 ( g ) + O 2 ( g ) →
200°C
(e) NH 4 NO 3 ( l) →
PROBLEM 1582 Write balanced chemical equation for the following reactions:
(a) Action of Al over Fe 2O 3 in presence of electric spark.
(b) Action of water over Na 2O 2 .
(c) Oxidation of Pb 2+ (aq.) to PbO 2 in acidic medium, by ozone.
291
Problems
PROBLEM 1583 Write an equation to represent:
(a) The formation of oxygen by the action of water over KO 2 .
(b) Oxidation of Cl – to ClO –3 by ozone in acidic solution.
PROBLEM 1584
How BF3 is prepared from borax?
PROBLEM 1585 You have available with you elemental sulphur, chlorine gas, sodium metal and
water. Using these materials, prepare (a) Na 2SO 3 (b) Na 2SO 4 (c) Na 2S 2O 3 .
PROBLEM 1586 Why is it that NF3 has no donor properties, but PF3 forms many complexes with
metals? Give one example of such complexes.
PROBLEM 1587 Substance (A) is a gas with vapour density 8.5. On oxidation at high pressure and
temperature with a platinum catalyst, it gave a colourless gas (B), which rapidly turned brown in air
forming a gas (C). B and C were condensed together to give substance (D), which reacted with water
forming a compound (E). E turned blue litmus paper red. On treatment of E with acidified solution of KI,
gas B was evolved, but when E was treated with a solution of NH 4Cl, a stable, colourless gas (F) was
evolved. F did not support combustion, but magnesium continued to burn in it. However, F reacted with
calcium-carbide in an electric furnace, forming a solid (G), which was slowly hydrolyzed by water,
forming a solution of substance A. Identify A to G.
PROBLEM 1588 Explain why P—O bond length in POCl 3 is 1.45 Å whereas the sum of the single
bond covalent radii of phosphorus and oxygen is 1.83 Å?
PROBLEM 1589
Suggest a reason why PF5 is known but PH 5 is not.
PROBLEM 1590 HCl (g) can be prepared from NaCl and H 2SO 4 . HBr and HI can’t be made in a
similar way from NaBr and NaI. Explain why this is so?
PROBLEM 1591 Write Lewis structure for each of the followings:
(a) HCP
(b) PH +4
(c) P2 H 4
(d) PO 3–
(e) PF5
4
PROBLEM 1592 Complete and balance each of the followings:
(a) P4 + Al →
(b) P4 + F2 →
(c) P4 + Na →
(d) P4 + Se →
(e) P4 + Cl 2 →
(f) P4 + O 2 →
PROBLEM 1593
Oxygen forms double bond in O 2 , but sulphur forms single bond in S 8 . Explain.
PROBLEM 1594 Explain why at room temperature fluorine and chlorine are gases while bromine is
liquid and iodine is solid at room temperature.
PROBLEM 1595
(a) XeO 2 F2
Suggest shapes of the following species:
(d) XeO 4–
(c) XeF3+
(b) XeF +
6
(e) XeO 3 F2
PROBLEM 1596 Sodium hydroxide is hygroscopic–that is it absorbs moisture when exposed to the
atmosphere. A student placed a pellet of NaOH on watch glass. A few days later, she noticed that the
pellet was covered with a white solid. What is the identity of this solid?
PROBLEM 1597
PROBLEM 1598
Discuss the reactivity of white P4 .
Complete the following reactions:
Red hot CuO
(a) NH 3 →
Heat
(b) (NH 4 ) 2 Cr 2O 7 →
292
Problems in Chemistry
PROBLEM 1599 Aqueous Cu(II) sulphate is blue coloured. When aqueous potassium fluoride is
added to CuSO 4 solution, a green precipitate is formed. If aqueous KCl is added instead, a bright green
solution is formed. Explain, what happens in each case.
PROBLEM 1600
Write balanced chemical reaction:
Heat
(a) H 3 PO 3 →
(c) HI + HNO 2 →
(b) Li 4C + HCl →
(d) H 2S + Cl 2 →
PROBLEM 1601 Arrange the following species in increasing order of indicated property:
(a) PCl 5 , SiCl 4 , CCl 4 , BCl 3 : Ionic character
(b) F – , C 4– , N 3– , O 2– : Ionic radii
(c) F, Cl, Br, I: Ionisation energies
(d) H 2O, SiO 2 , CO 2 : Acidic character
METALLURGY
PROBLEM 1602 Discuss the principle involved in extraction of Mg from sea water.
PROBLEM 1603 Discuss the method of extraction of Mg from its dolomite ore.
PROBLEM 1604 Why extraction of magnesium is impractical by chemical reduction of its salt?
PROBLEM 1605 Mg 2+ ion being smaller than Na + instead, Mg 2+ has very low ionic mobility and
electrical conductivity compared to Na + in their aqueous solution. Explain.
PROBLEM 1606 (a) How Epsom salt (MgSO 4 ⋅ 7H 2O) is obtained? What is its use?
(b) Write a balanced chemical reaction of Mg metal with dilute nitric acid.
PROBLEM 1607 How Na 2CO 3 is manufactured by Solvay process? Why this method fails in
preparation of K 2CO 3 ?
PROBLEM 1608 Discuss the principle of purification of bauxite Al 2O 3 ⋅ 2H 2O by the following
processes : (a) Baeyer’s process (b) Hall’s process (c) Serpek’s process.
PROBLEM 1609 How Al is extracted from bauxite ore? Describe thermite welding.
PROBLEM 1610 Write the reaction of Al metal with dilute HCl, H 2SO 4 and NaOH. Why concentrated
nitric acid render Al passive?
PROBLEM 1611 Discuss the reactions involved in polymerization of ethene using Zeigler catalyst.
PROBLEM 1612 Discuss the principle involved in extraction of iron from its oxide ore, by carbon
reduction method.
PROBLEM 1613 Discuss the principle of rusting of iron and methods of its protection.
293
Problems
PROBLEM 1614 What happens when iron is treated with:
(a) Steam
(e) Dilute HNO 3
(b) Dilute HCl
(f) Concentrated HNO 3
(g) Boiling sulphur.
(c) Concentrated H 2SO 4
(d) Anhydrous Cl 2 ( g )
PROBLEM 1615 How ferrous ammonium sulphate hexahydrate (Mohr’s salt) is prepared in laboratory.
What are the advantages of using Mohr’s salt as analytical standard of ferrous ion over ferrous sulphate?
PROBLEM 1616 How anhydrous and hydrated ferric chloride is prepared in laboratory. What happens
when anhydrous ferric chloride is (a) heated above 500°C (b) dissolved in water.
PROBLEM 1617 How tin is extracted by carbon reduction method?
PROBLEM 1618 What happens when Sn is reacted with:
(a) Cl 2 ( g ) (b) dilute HNO 3 (c) conc. HNO 3 (d) conc. H 2SO 4
(e) NaOH
PROBLEM 1619 How copper is obtained from copper pyrite, by self-reduction technique?
PROBLEM 1620 When copper is exposed to atmosphere for a longer time, a green mass deposits on its
surface. Discuss the chemical reaction involved.
PROBLEM 1621 Why copper is passive in dilute HCl or dilute H 2SO 4 .
PROBLEM 1622 Discuss the action of pure oxygen gas over copper metal at high temperature.
PROBLEM 1623 What happens when copper is reacted with: (Write complete balanced reaction)
(a) dil HNO 3 ,
(b) warm HCl in presence of oxygen,
(c) warm H 2SO 4 in presence of oxygen,
(d) concentrated HNO 3 ,
(e) with aqueous Fe 2 (SO 4 ) 3 ,
PROBLEM 1624 How Pb is extracted from galena (PbS) by self-reduction method?
PROBLEM 1625 Discuss the reactions involved in action of (i) air and (ii) pure oxygen on lead.
PROBLEM 1626 Lead react with hot concentrated sulphuric acid in the begining but reaction ceases
after some time. Explain.
PROBLEM 1627 What happens when lead is reacted with:
(a) dilute HNO 3
(b) concentrated HNO 3
(c) NaOH
(d) CH 3COOH
PROBLEM 1628 How silver metal is extracted from its sulphide ore?
PROBLEM 1629 Discuss the role of AgBr in photography. How does it react with Na 2S 2O 3 solution?
PROBLEM 1630 Silver ornament gets tarnished when exposed to atmosphere for a long time, why?
PROBLEM 1631 Complete the following chemical reactions:
Heat
(a) Ag + Concentrated HNO 3 →
(b) Ag + Concentrated H 2SO 4 →
(c) AgBr + NaCN( aq ) →
(d) AgCl + NH 4OH →
294
Problems in Chemistry
PROBLEM 1632 How gold is obtained from its ore by cyanide process?
PROBLEM 1633 How does Au react with aqua-regia?
PROBLEM 1634 Both copper and silver forms monovalent ion like alkalimetal but alkali metals are
highly reactive while Cu and Ag are almost inert. What are the reasons for this difference in reactivity?
PROBLEM 1635 How KMnO 4 is manufactured from MnO 2 ?
PROBLEM 1636 How K 2Cr 2O 7 is manufactured industrially from chrome-iron?
PROBLEM 1637 Complete the following reactions:
Heat
(a) K 2Cr 2O 7 →
(b) K 2Cr 2O 7 + KOH →
Heat
(c) K 2Cr 2O 7 + KCl + H 2SO 4 (concentrated) →
PROBLEM 1638
(a) KMnO 4
(b) KMnO 4
(c) KMnO 4
(d) KMnO 4
Write the balanced chemical reaction for the followings:
+ K 2C 2O 4 + H 2SO 4 →
+ O 3 + H 2SO 4 →
+ FeC 2O 4 + H 2SO 4 →
+ HCl →
PROBLEM 1639 Draw shape of FeCl 3 in (a) gaseous state (b) ether (c) water.
PROBLEM 1640 FeCl 3 and FeBr 3 are well known, but FeI 3 has doubtful existence, explain.
PROBLEM 1641 How potash alum is prepared? What is its major application?
297
Solutions
MOLE CONCEPT
1. Empirical formula :
= KAlS 2O 8









Al
K
S
O
Elements
10.5
15.1
24.8
49.6
Mass percentage
0.388
0.387
0.775
3.1
Mole ratio
1
1
2
8
Simple ratio
Empirical formula weight = 258
From weight loss information : 54.4 g anhydrous salt ≡ 45.6 g H 2O
⇒
258 g anhydrous salt ≡ 238.89 g = 13.27 mol H 2O
⇒
Empirical formula of hydrated salt = KAlS 2O 8 ⋅ 13H 2O
1.0mole of KClO 3 ≡ 3.0 mole of Zn
2.
3 × 5.104
3 × 5.104 × 65
5.104
mole KClO 3 ≡
mole of Zn =
= 8.124 g Zn
122.5
122.5
122.5
3. Apply conservation of moles of silver before and after precipitate exchange reaction as :
2.052 − x
1.8
x
=
+
143.5 188
143.5
where, x is mass of AgBr in mixed precipitate.
⇒
x =1.064
x
1
1
Also,
moles of CuBr 2 = moles of AgBr = ×
2
2 188
x
1
(on substituting x)
Mass of CuBr 2 = ×
× 223.5 = 0.6324
⇒
2 188
Mass % of CuBr 2 = 34.18
4. Moles of NaCl in sample = 0.01 = moles of AgCl from NaCl in precipitate.
Total moles of AgCl precipitate =
⇒
⇒
2
= 0.01393
143.5
Moles of AgCl from KCl = 0.00393 = moles of KCl
Mass of KCl in sample = 0.00393 × 74.5 = 0.2928 g
Mass % of KCl in the sample = 29.28
5. Apply conservation of moles of Ag before and after precipitate exchange reaction:
1.94 – x 2.4
x
gives x = 1.483 g AgCl in mixed precipitate.
+
=
188
143.5
188
x
Mass of NaCl =
⇒
× 58.5 = 0.6047 g
143.5
1 1.94 – x
× 200 = 0.243 g
Mass of CaBr 2 = ×
⇒
2
188
Mass % NaCl = 60.47, CaBr 2 = 24.3
⇒
298
Problems in Chemistry
6. Apply conservation of mass : If product contain x g SO 2 , mass of SO 3 = (22 − x ) g.
x 2 (22 − x )
Computing total mass of S : +
= 10
2
5
⇒
x =12 ⇒ SO 2 = 12 g, SO 3 = 10 g
7. Mass of 1.0 L solution = 1025 g
Mass of ethanol in 1.0 L solution = 8 × 46 = 368 g
Mass of solvent in 1.0 L solution = 657 g
8
Molality ( m) =
× 1000 = 12.17
⇒
657
8. 5.0 moles (300 g) of acetic acid is present in 1.0 kg solvent.
⇒ 1300 g solution contain 5.0 moles of solute.
⇒ 1300/1.12 = 1160.7 mL solution contain 5.0 moles of solute.
5
× 1000 = 4.3
⇒ Molarity ( M ) =
1160.7
9. Let the combustion reaction is :
C 8 H18 + nO 2 → xCO + (8 − x ) CO 2 + 9H 2O
1
2650
n = (25 − x ) ; moles of octane =
= 23.246
2
114
Computing mass of product = 23.246 × 28 x + (8 – x ) × 44 × 23.246 + 9 × 23.246 × 18 = 11530
⇒ x =1.125
⇒ n =11.9375
In case of complete combustion :
C 8 H18 +
25
O 2 → 8CO 2 + 9H 2O
2
Efficiency = ( n/12.5) × 100 = 95.5%
10. Moles of BaSO 4 =
⇒
1.846
= moles of BaCl 2 ⋅ xH 2O = 7.923 × 10 –3
233
7.923 × 10 –3 × (208 +18x ) = 1.936
⇒
x≈2
11. Let the mixture contain x g CuSO 4 ⋅5H 2O.
⇒
5−x
x
× 159 +
× 120 = 3
249
246
⇒
x = 3.72
⇒ Mass percentage of CuSO 4 ⋅ 5H 2O = 74.4
12. In the two sample mass ratio of Ca to SiO 2 will remain same since both are non-volatile.
2x
Before heating, let x% CaCO 3 is present ⇒
% Ca is present. ⇒ (85 − x )% SiO 2 is present.
5
After heating, 51.5% SiO 2 and 41.15% CaO are present.
40
× 41.15 = 29.39% Ca is present.
⇒
56
299
Solutions
Now : mass of Ca/SiO 2 before heating = mass of Ca/SiO 2 after heating.
29.39
2x
⇒
⇒ x = 49.97%
=
5 (85 – x ) 51.5
13. In the reaction of ClO 2 with water, products HClO 3 , HCl as well as reactant ClO 2 contain one
chlorine atom each, per formula unit. Hence, total moles of acid produced will be equal to moles of
ClO 2 hydrolysed. Therefore, there is no need of balancing the reaction for computation of moles of
ClO 2 .
Moles of acid = moles of NaOH required = 0.18 = moles of ClO 2
10
Total moles of gas in the flask =
= 0.406
0.082 × 300
Mole fraction of ClO 2 =
0.18
× 100 = 44.33
0.406
14. The balanced chemical reaction is :
C 6 H 5CH 3 + 2KMnO 4 → C 6 H 5COOK + 2MnO 2 + KOH + H 2O
11.5 × 100
For preparation of 11.5 g C 6 H 5COOK, theoretical yield should be
g
71
92 11.5 × 100
Mass of toluene needed =
×
= 9.31 g
160
71
15. Balanced chemical reaction is :
2MnI 2 +13F2 → 2MnF3 + 4IF5
2 × 309 g MnI 2 ≡ 13 × 38 g F2
13 × 38
12 g MnI 2 ≡
× 12 = 9.59 g
2 × 309
Q
Actual requirement of
100
= 12.78 g
75
0.16
100
Ca =
× 40 ×
= 25.6
100
0.25
0.344 32 × 100
S=
×
= 41
233
0.115
0.155 14 × 100
N=
×
= 17.9
17
0.712
F2 = 9.59 ×
16. Mass % of
Mass % of
Mass % of
⇒ Mass % of C = 15.48
Now :
Elements
Mass %
Mol ratio
Simple ratio
Ca
25.6
0.64
1
S
41
1.28
2
N
17.9
1.28
2
C
15.48
1.29
2
300
Problems in Chemistry
Empirical formula = CaC 2 N 2S 2 , empirical formula weight = 156
Hence, molecular formula = CaC 2 N 2S 2
17. Working in backward direction :
In the last step moles of
(AgBr + AgI) = moles of AgI
0.4881 – x
x
0.5868
⇒ x = 0.0933 g
+
=
⇒
188
235
235
100
0.0933
Mass % of NaI =
× 150 ×
= 29.77
0.2
235
Now subtracting mass of AgI from 1st and 2nd precipitate gives :
Mass of (AgCl + AgBr) = 0.3187 g
and
mass of AgBr = 0.3948 g
y
0.3187 – y 0.3948
Again
+
⇒ y = 0.245 g
=
143.5
188
188
100
0.245
Mass % of NaCl =
× 58.5 ×
= 50
⇒
0.2
143.5
Mass % of NaBr = 20.23
18. Let us consider 100 g mixture of NaI and NaCl containing x g NaI.
 x 100 − x  142
Mass of Na 2SO 4 = 
×
=100 ⇒ x = 30.25
+
58.5  2
150
14
3
× 5 = 4.17 g N 2 and
× 5 = 0.83 g H 2
17
17
Therefore, initially 9.17 g N 2 and 5.83 g H 2 were present.
Since N 2 is limiting reagent, maximum mass of NH 3 = 11.135 g
20. Balance the reaction in terms of ‘ n’ as :
19. 5.0 g NH 3 contains
( CH 2CH 2 —
)n
4nXeF2 + —
→ —( CF2CF2 )—
+ 4nHF + 4nXeF4
n
XeF2 + 2F2 → XeF6
Formula weight of one unit of fluorinated polymer =100 ⇒ for 1.0 kg, n = 10
⇒ 40 moles of XeF6 would be required and for this 80.00 moles of F2 would be consumed.
⇒ m (F2 ) = 80 × 38 = 3040 g = 3.04 kg
21. Weight loss is due to conversion of NaHCO 3 into Na 2CO 3 : 31 g weight is lost per mole of
NaHCO 3 .
0.3
0.3
mol of NaHCO 3
moles of Na 2CO 3 .
producing
⇒ 0.3 g wt loss from
31
62
Total moles of carbonate = 15 × 10 –3
3
= 0.01
620
⇒ 42.4% Na 2CO 3 .
⇒ Moles of carbonate in original sample = 0.015 −
Mass of Na 2CO 3 in original sample = 1.06
301
Solutions
3 − x 3.76
x
2.45
× 58.5 = 1.0 g
+
=
⇒ x = 2.45 ⇒ m (NaCl) =
143.5
1435
.
188 188
23. If M is molar mass of (CH 3 ) x AlCl y
22.
m% NaCl = 50%
0.643 x
× 16 = 0.222
M
0.643 y
m (AgCl) =
× 143.5 = 0.996
M
355
. x
M = 15x + 27 + 35.5 y = 15x + 27 +
= 32.75x + 27
2
m(CH 4 ) =
and
dividing :
x
= 2, Also
y
0.643x × 16
= 0.222 ⇒
32.75x + 27
⇒
x = 1.98 ≈ 2 ⇒
24. Element
C
H
N
Cl
O
Mass %
25
4.2
9.8
49.6
11.4
Mol %
2.08
4.2
0.7
1.39
0.7
Simple ratio
2.9
6
1
2
1
y=1
Empirical formula C 3 H 6 NCl 2O
25.
Mass of AgCl = 0.09 × 143.5 = 12.915 gwhich is 95.77% of total ppt.
⇒
⇒
⇒
Total mass of precipitate = 13.485 g and mass of impurity = 0.57 g
Mass of NaCl + KCl = 5.9 g
5.9 – x
x
+
= 0.09 ⇒
58.5
74.5
x = 2.94 g NaCl, 2.96 g KCl
m (Na 2O) = 1.558 g
⇒ m % (Na 2O) = 31.16
m (K 2O) = 1.867 g
⇒ m % (K 2O) = 37.34
26.
Cl 2 + 2KOH → KCl + KClO + H 2O
...(i)
3KClO → 2KCl + KClO 3
...(ii)
4KClO 3 → 3KClO 4 + KCl
...(iii)
Multiply Eq. (i) by 12 and Eq. (ii) by 4 and add Eqs. (i), (ii) and (iii) to obtain
12Cl 2 + 24KOH → 21KCl + 3KClO 4 +12H 2O
852
415.5
1000 g KClO 4 will be produced from
852
× 1000 = 2050.54 g = 2.05 kg
415.5
1
27. 1.598 g TiO 2 loses 0.16 g oxygen therefore 80 g TiO 2 will lose 8 g of oxygen ( 2 mole of oxygen
atom).
⇒
Formula of new oxide = TiO
2–
1
2
= Ti 2O 3
302
Problems in Chemistry
28. Since, the compound contain only one copper atom per formula unit, its formula can be assumed to
be as:
Cu(NH 3 ) x SO 4 ⋅ yH 2O : M = 159.5 + 17x + 18 y
0.195
Moles of compound present in 0.195 g =
M
0.195
Moles of HCl required =
…(i)
⇒
x = 30.63 × 0.1036 × 10 –3
M
Also M g of compound contain 18y g of water
18 y
0.2 g of compound will contain
…(ii)
× 0.2 g of water = 0.015
M
From Eqs (i) and (ii),
18 y
0.015
M
=
⇒ 18 y = 4.6x
× 0.2 ×
M
0.195x 30.63 × 0.1036 × 10 –3
Substituting in Eq. (ii)
18 y × 0.2
= 0.015 ⇒ 61.33x = 159.5 + 21.6x
159.5 + 17x + 18 y
and
x = 4,
y =1
Formula of salt = Cu(NH 3 ) 4⋅SO 4⋅H 2O
Molarity of CuSO 4 solution =1.1 M
Percentage yield :
⇒ moles of Cu(NH 3 ) 4⋅SO 4⋅H 2O precipitate (theoretically) = 25 × 1.10 × 10 –3 = 27.50 × 10 –3
⇒
Theoretical mass of solid = 27.5 × 10 –3 × 245.5 = 6.751g
Percentage yield =
⇒
6.127 × 100
= 90.75%
6.751
29. Mass of water produced in the reaction = 0.09912 g
0.09912
g = 11 mg
⇒ mass of hydrogen =
9
0.3283 × 28.8
1
m mol of CO 2 = 2 m mol of NaOH =
= 4.73 = m mol of carbon
2
⇒ mass of carbon in starting compound = 4.73 × 12 = 56.76 mg
65.12
0.225
mass of nitrogen = 28 ×
×
× 1000 = 22 mg
760 0.082 × 298
and
Now:
mass of oxygen = 115.2 – (56.76 + 11 + 22) = 25.44 mg
Elements
C
H
N
O
Mass mg.
56.76
11
22
25.44
m moles
4.73
11
1.57
1.59
Simple Ratio
3
7
1
1
303
Solutions
Empirical formula = C 3 H 7 NO
Empirical formula wt. = 73
Molecular formula = C 6 H14 N 2O 2
⇒
30. In order to obtain maximum yield from a reaction, the reactants must be supplied in stoichiometric
amount so that no reactant should be left unreacted.
The balanced chemical reaction is,
Pb(NO 3 ) 2 + 2KI → PbI 2 + 2KNO 3
Let x g of KI is taken
x
x
moles of KI =
⇒ moles of Pb( NO 3 ) 2 present =
⇒
2 × 166
166
5– x
x
x
⇒
⇒ x = 2.5 g ⇒ mass of PbI 2 =
× 460 = 3.464 g
=
2 × 166 330
332
31. Let mixture contain x g XH 4 then (5.628 – x) g X 2 H 6 .
x
x
and, moles of X 2 H 6 =
moles of X H 4 =
⇒
M +4
2( M + 4)
x
…(i)
⇒
5.628 – x =
(2 M + 6)
2( M + 4)
Mx
2 Mx
…(ii)
Also
+
=5
M + 4 2( M + 4)
Solving Eqs. (i) and (ii) we get M = 27.8
32. Let one formula unit of Argyrodite contain ‘a’ mol of silver and b mole of Y . Therefore, moles of
a

sulphide per formula unit of mineral must be  + 2b .
2

a
Hence,
Ag a Y b S  a
Ag 2S + bYS + bH 2S
 + bH 2 →
2
+
2
b




2
Also,
⇒
⇒
Solving
moles of H 2 required =
PV
= 8.87 × 10 –3
RT
moles of mineral present in its 10 g =
8.87 × 10 –3
b

a
 
108a + bM +  2 + 2b 32


1127b = 124a + b( M + 64)
108a
11.88
Also
=11.88 ⇒ a =
bM
bM
108
10 =
8.87 × 10 –3
b
Substituting ‘a’ in Eq. (i),
124 × 11.88bM
1127b =
+ b( M + 64) ⇒
108
Empirical formula = Ag 8 YS 6
M = 72.6 g mol –1
…(i)
and
a
=8
b
304
Problems in Chemistry
Mass of uranium in the sample =
33.
1.48
× 238 = 0.894 g
394
Mass % of uranium in the sample = 89.4
Now
UO 2 (NO 3 ) 2 + Na 2C 2O 4 + xH 2O → UO 2 (C 2O 4 ) xH 2O ↓ + 2NaNO 3
mmol 3.756
2.985
Here Na 2C 2O 4 is the limiting reagent, therefore, m mol of UO 2 (C 2O 4 ) ⋅xH 2O formed is 2.985.
⇒
⇒ x=
M(UO 2 (C 2O 4 )) ⋅xH 2O =
1.23
× 1000 = 412 = 238 + 32 + 88 + 18 x
2.985
54
=3
18
35.5
× 0.861 = 0.231g
143.5
mass of iodine in 0.467 g compound = 0.254 g
Iodine
Chlorine
⇒
weight:
0.254 g
0.213 g
mole:
0.002
0.006
simple ratio,
1
3
⇒ Empirical formula = ICl 3
35.
C n H 2n + 2 + O 2 → nCO 2 + H 2O
34. Mass of chlorine in 0.861 g AgCl =
CO 2 + 2NH 2 → H 2 N  CO  NH 2 + H 2O
1000 × 1000
mole =
= 16666.67
60
236 × 1000
moles of CO 2 produced =
× n = 16666.67 ⇒ n =12.5
14n + 2
Formula of hydrocarbon = C12 H 26
36. Weight of NaHCO 3 dissolved at 60° C = 164 ×
250
= 41 g
1000
Weight of NaHCO 3 dissolved at 0°C = 6.9 g
Total weight of NaHCO 3 dissolved = 47.9 g
Weight of NaHCO 3 recovered =150 g
Total weight of NaHCO 3 present originally = 150 + 47.9 = 197.9 g
197.9
% purity =
× 100 = 79.16
250
46
37.
Mole of CuSO4 in solution =
159 + 18x
46x
mass of H 2Oin solution = 54 +
× 18
159 + 18x
⇒
moles of H 2O in solution = 3 +
46x
159 + 18x
305
Solutions
⇒
0.05 =
46
159 + 18x
46x
46
+3+
159 + 18x
159 + 18x
20 × 46
46
(1 + x ) + 3 =
159 + 18x
159 + 18x
⇒
⇒
3=
46
(19 – x ) ⇒ x = 3.97 ≈ 4
159 + 18x
38. The roasting reaction is,
7
Fe 2 O3
+ 2SO2
+ O 2 →
2 × 88 = 176 2
2 × 56 + 48 = 160
2FeS
From the above reaction,
16 g weight is lost by 176 g FeS
4 g weight will be lost by
176
× 4 = 44 g FeS
16
⇒ ore contain 44% FeS by weight.
39. Absorbance of a solution is linear function of concentration.
⇒
⇒
A1 C1
=
A2 C 2
or
40 0.001
=
⇒ C 2 = 00007
.
M
28
C2
mol. wt. of polymer =
40. Volume of one molecule =
Molar mass =
3
= 42857 amu
0.0007 × 0.1
3 2
3
3 3
× 10 –18 cm 3
(10 –6 ) 2 × 3 × 10 –6 cm 3 =
a h=
4
4
4
3 3
× 10 –18 × 6.023 × 10 23 × 1.2 = 939 × 10 3 amu
4
41. Volume of smallest cell = πr 2 l = π(60 × 10 –8 cm ) 2 (6000 × 10 –8 cm ) = 6.785 × 10 –17 cm 3
mass of one smallest cell = 7.6 × 10 –17 g
⇒ Molar mass of mother cell = 7.6 × 10 –17 × 24 × 60 × 6.023 × 10 23 = 6.6 × 10 10 amu
42. Mass of CaSO 4 = 0.5 g
⇒
⇒
⇒
0.5
× 56 = 0.2058 g
136
mass of Mg 2 P2O 7 = 0.5 g
0.5
mass of MgO =
× 2 × 40 = 0.18 g
222
0.2058
mass % of CaO on oven dried basis =
× 100 = 21.8%
0.9437
mass of CaO =
306
Problems in Chemistry
0.18
× 100 = 19.07%
0.9437
On air dried basis : CaO = 20.58% MgO = 18%, H 2O = 5.63%
43. Let the sample contain x g Mohr’s salt [FeSO 4 (NH 4 ) 2 SO 4⋅6H 2O]
mass % of MgO on oven dried basis =
⇒
Solving
Also
0.5 – x 0.75
x
×2+
=
392
132
233
0.23
x = 0.23 g ⇒ Mohr’s salt =
× 100 = 46% , (NH 4 ) 2 SO 4 = 54%
0.50
x
0.2
moles of Fe in 0.2 g sample =
×
= 2.347 × 10 –4
392 0.5
2.347 × 10 –4
× 160 = 18.77 mg
2
44. Smallest volume of AgNO3 would be required when the entire mass is due to highest molecular
weight constituent.
Hence, for smallest volume, the whole mass should be of BaCl 2 ⋅ 2H 2O
0.3
m mol of BaCl 2⋅2H 2O =
× 1000 = 1.229 m mol
244
m mol of AgNO 3 required = 2 × 1.229 = 2.458
2.458
Volume of AgNO 3 required =
= 16.38 mL (smallest)
0.15
Largest volume of AgNO 3 would be required when entire mass is due to lowest molecular weight
constituent, i.e., NaCl.
0.3
m mol of NaCl =
× 1000 = 5.128 = m mol of AgNO 3 required.
58.5
5.128
Volume of AgNO 3 required =
= 34.18 mL (largest).
⇒
0.15
⇒ mass of Fe 2O 3 obtained on ignition of 0.2 g sample =
ACID-BASE TITRATION
Total mmol of AgCl from 20 mL solution =
45.
m moles of AgCl from HCl = 0.8
0.4305 × 1000
=3
143.5
⇒ m moles of AgCl from CaCl 2 = 2.2
⇒ 1.1 m mole of CaCl 2 was consumed for precipitation of oxalate from 20 mL solution.
Hence,
total m mol of oxalic acid in 250 mL solution =
m% of oxalic acid =
1.1
× 250 = 13.75
20
13.75 × 10 −3 × 90
× 100 = 82.5
1.5
46. The balanced disproportionation reaction is:
2P2O 3 + 6H 2O → PH 3 + 3H 3 PO 4
307
Solutions
meq of NaOH added = 6
meq of H 2SO 4 required for back titration = 1.1
⇒ 4.9 meq of NaOH neutralized 4.9 meq of H 3 PO 4
49
m mol H 3 PO 4
⇒ Total meq of H 3 PO 4 produced = 49 =
3
2 49
m mol of P2O 3 present originally = ×
3 3
⇒ Mass % of P2O 3 =
98 × 10 –3
× 110 × 100 = 79.85
9 × 1.5
47. m mol of CO 2–
3 in 10 mL = 10 × 0.05 × 2 = 1
In presence of methyl orange 3.235 m mol H + is consumed in which 2 m mol H + would be
–
+
required for CO 2–
3 , hence 1.235 m mol H for HCO 3 ion.
⇒
m % CaCO 3 =
10 × 10 −3 × 100
× 100 = 40
2.5
m % Ca(HCO 3 ) 2 =
6.175 × 10 –3 × 162
× 100 = 40
2.5
48. m mol of H + required to reach the phenolphthalein end point = 1.12
⇒ 1.12 m mol Na 2CO 3 is present per 10 mL of solution.
Now, V = 17 mL and it contain 1.12 m mol of NaHCO 3 produced in titration in association with
original NaHCO 3 .
Q 5 mL of this solution required 3.53 mL acid to reach the methyl orange end point.
3.53
× 17 = 12 mL acid = 1.92 m mol H +
∴ 17 mL would require =
5
⇒ m mol of H + consumed for neutralization of original bicarbonate = 0.8
m (Na 2CO 3 )/ litre = 112 × 10 −3 × 106 = 11.872 g
m (NaHCO 3 )/ litre = 80 × 10 –3 × 84 = 6.72 g
49. m mol of HCl taken = 7.5
m mol of HCl left unreacted = 5.244
m mol of HCl reacted = 2.256 ≡ 1.128 m mol K 2SO 3 = 0.178 g K 2SO 3
m% K 2SO 3 = 36.6%
50. m mol of NH 3 from glycine = 2.5 – 1.8342 = 0.6658 = m mol of nitrogen.
mass of nitrogen = 9.3212 × 10 –3 g = 18.64%
Theoretical mass% = 18.66
51. Chemical reactions involved in the entire process are :
7
2Ca 3 (PO 4 ) 2 + Mg → Ca 3 P2 + Mg (PO 3 ) 2 + 3CaO + 2 O 2
Ca 3 P2 + 6H 2O → 3Ca(OH) 2 + 2PH 3
2PH 3 + 4O 2 → P2O 5 + 3H 2O
308
Problems in Chemistry
⇒ 2.4 g (0.1 mol) Mg will produce 0.2 mol PH 3 which would require 0.4 mol of oxygen.
⇒ Vol. of O 2 needed = 8.96 L, hence vol. of air needed = 42.66 L. at S. T. P.
52. Let the mixture contain x g of Li 2CO 3 , y g of NaHCO 3 .
30x 106 y
+
+ 9.3 − ( x + y) = 7.37
74
168
44x 62 y
+
=1.93
74 168
From titration information :
2x 2 y 9.3 − ( x + y)
+
+
× 2 = 0.18
74 168
106
32x 62 y
−
= 0.24
⇒
74 168
Solving, Eqs. (i) and (ii) gives x = 2.11g, y = 1.825 g
and
Na 2CO 3 = 5.365 g ⇒ NaHCO 3 = 19.62%, Na 2CO 3 = 57.7%
4
53. M (acid) =
× 1000 = 185.2
8 × 0.27 × 10
⇒ After heating
...(i)
...(ii)
Formula of acid = C n H 2n O 2 ⇒ M = 14n + 32 = 185.2 ⇒ n =11
55
5
Now 5g acid will produce
× 11 =
molCO 2 after complete combustion.
186
186
Total moles of NaOH available = 1.0
2 × 55 76
Moles of NaOH left unreacted = 1 −
in 500 mL
=
186
186
⇒ Molarity of NaOH after precipitation of Na 2CO 3 = 0.817
Therefore,
0.817 × 10 = 0.5 ×V
⇒
V = 16.34 mL
54. Let 10 mL solution contain x m mol NaHCO 3 and y m mol Na 2CO 3 :
⇒
x + 2 y = 1.24
⇒
and
⇒
x = 1.5 – 1.26
m(NaHCO 3 ) = 0.24 × 50 × 10
–3
y = 0.5
× 84 = 1.008g = 20.16%
m(Na 2CO 3 ) = 0.5 × 50 × 10 –3 × 106 = 2.65 g = 53%
m(NaCl) = 1.342g = 26.84%
2NaOH
55. C x H y O 2 + O 2 → xCO 2 → xNa 2CO 3
6.4
M
6.4 x
M
6.4 x
M
Let 10 mL solution contain a m mol NaOH and b m mol Na 2CO 3 :
⇒
a + b =14.5
a + 2b = 18
Also,
25 M
= 32
100
⇒
⇒
b = 3.5
M = 128
6.4x
= 0.35
M
⇒
⇒
x=
0.35 × 128
=7
6.4
...(i)
309
Solutions
Also,
12x + y + 32 = 128 ⇒ y = 12
⇒ Formula = C 7 H12O 2
M (NaOH) = 1.8
56. The balanced chemical reaction is :
Co(NH 3 ) x Cl 3 + xHCl → xNH +4 + Co 3+ + ( x + 3) Cl –
1.58
165.5 + 17x
1.58x
165.5 + 17x
1.58 (x + 3)
165.5 + 17x
1.58 x
23.63 × 1.5
=
165.5 +17x
1000
⇒
⇒
x=6
m(AgCl) = 0.053 × 143.5 = 7.62 g
5x
1
8x
1 77x 1
×
+
×
+
×
= 098
.
⇒ x = 45 g
100 53 100 50 100 40
meq of Na 2CO 3 = 0.6132 × 40 = 24.528 ⇒ m% Na 2CO 3 = 3.25
57.
58.
meq of NaOH = 23.046 × 40 = 922 ⇒ m% NaOH = 92.2
59. Let the sample contain x mol Na 2CO 3 and y mol NaHCO 3 .
2x + y = 18.1396 × 10 −3
⇒
m mol of NaOH combined with NaHCO 3 = 4.1
⇒ Total moles of NaHCO 3 = 16.4 × 10 –3 = y ⇒ x = 0.8698 × 10 −3
16.4 × 10 –3 × 84
× 100 = 91.84
1.5
0.8698 × 10 −3 × 106
m % Na 2CO 3 =
× 100 = 6.15
1.5
Total NaOH = 0.2 mol,NaOH left unreacted = 0.06 mol
⇒
m % NaHCO 3 =
60.
NaOH combined with CO 2 = 0.14 mol producing 0.07 mol of Na 2CO 3 .
Let hydrocarbon is C n H 2n + 2 ⇒ M = 14n + 2
⇒
n
= 0.07
14n + 2
⇒
n = 7, Molecular formula = C 7 H 16
61. Formula = C n H 2n O 2 : M = 14n + 32
Let half-solution contain x m mol Na 2CO 3 and y m mol NaOH.
⇒
x + y = 71.72
2x + y = 123.44
⇒
⇒
⇒
x = 51.72
Total Na 2CO 3 = 103.44 m mol
2n
= 103.44 × 10 –3 ⇒ n = 6
14n + 32
Hence, acid is C 6 H 12O 2
m(NaOH) = 123.44 × 2 × 10 –3 × 40 = 9.87g
62. Moles of NaHCO 3 = 19 × 0.25 × 2 × 10 –3 = 9.5 × 10 −3 = m% NaHCO 3 = 31.92
...(i)
310
Problems in Chemistry
10 –3
= 8.58 × 10 –3
2
⇒
m% Na 2CO 3 = 36.38; m% NaCl = 31.7
63. Let N 1 be normality of HCl and N 2 be the normality of NaOH
...(i)
⇒
20 N 1 = 96 N 2
2
Also,
× 1000 + 8 N 2 × 10 = 50 N 1
50
⇒
4 +8N 2 =5N1
N 2 = 0.25, N 1 = 1.2
64. Let the mixture contain x m mol K 2C 2O 4 , y m mol KHC 2O 4 and z m mol of H 2C 2O 4 ⋅ 2H 2O.
Then,
...(i)
2x + y = 20
...(ii)
y + 2 z = 20
…(iii)
166x + 128 y + 126 z = 2725
166
126
(20 – y) +128 y +
⇒
(20 – y) = 2725
2
2
Solving,
y =10.83
⇒
x = 4.585
Moles of Na 2CO 3 = (26.66 − 9.5) ×
z = 4.585
⇒
m % KHC 2O 4 = 50.87
m % K 2C 2O 4 = 27.93
m % H 2C 2O 4 ⋅ 2H 2O = 21.2
65. Let sample contain x m mol NH 4 NO 3 and y m mol (NH 4 ) 3 PO 4
0077
.
× 2 × 10000 = 2.56
601
⇒
x = 7.32 and m % NH 4 NO 3 = 58.56
66. Let half solution contain a m mol Na 2CO 3 and b m mol NaOH
⇒
x + 3 y = 15
and
y=
⇒
a + b = 0.55 and 2a + b = 0.8
⇒
Total m mol of Na 2CO 3 = 0.5 = m mol of CO 2 produced
Formula of acid = C x H y O 4 ⇒ moles of CO 2 produced =
a = 0.258b = 0.3
10.38 × 10 –3
⋅x
M
Also, 0.168 g acid = 16.18 × 0.125 × 10 –3 equivalent ⇒ M = 166
⇒
Also,
⇒
10.38 × 10 –3 x
= 0.5 × 10 –3 ⇒
166
12x + y + 64 = 166
y=6
⇒
Formula = C 8 H 6O 4
M (NaOH) = 0.016 M
67. Moles of Na 2CO 3 = 0.1 × 10 × 11.32 × 10 –3 = 11.32 × 10 –3
x=8
311
Solutions
⇒
m % Na 2CO 3 =
m moles of AgCl =
11.32 × 10 –3 × 106
× 100 = 40
3
0.306 × 1000
= 2.13 from 10 mL stock solution.
143.5
⇒ Moles of NaCl in original sample = (21.3 − 11.32) × 10 −3 = 9.98 × 10 −3
⇒ Mass% of NaCl = 19.46
Total NaOH consumed for 10 mL stock solution = 0.05 × 42.64 = 2.132
⇒ NaHCO 3 in original sample = (21.32 – 11.32) × 10 –3 = 10 –2 mol.
Mass % of NaHCO 3 = 10 –2 × 84 ×
⇒
100
= 28
3
68. m mol of Na 3 PO 4 = 24.4 × 0.1 × 5 = 12.2
⇒ m % (Na 3 PO 4 ) = 12.2 × 10 –3 × 164 ×
100
= 50
4
Also, if the mixture contain y m mol Na 2 HPO 4 and z m mol NaH 2 PO 4
12.2 × 3 + 2 y + z = 23.572 × 0.5 × 5 = 58.93
Then
⇒
Also,
2 y + z = 22.33
...(i)
142 y + 120 z = 2000
...(ii)
y = 6.9 and z = 8.53
Solving;
m% Na 2 HPO 4 = 6.9 × 142 × 10 −3 ×
⇒
100
= 24.5
4
m% NaH 2 PO 4 = 25.5
69. Weight loss = 5 × 0.265 = 1.325 g
Q 106 g weight is lost from 136 g of LiHCO 3
136
× 1.325 = 1.7 g ⇒
m% LiHCO 3 = 34
⇒ 1.325 g weight will be lost by
106
m moles of H 2SO 4 used for neutralization of Li 2O and Na 2CO 3 = 4 – 0.579 = 3.421
⇒ Total m mol of Li 2O + Na 2CO 3 = 34.21
Also,
moles of Li 2O = 12.5 × 10 –3
⇒
⇒ moles of Na 2CO 3 = 21.71 × 10 –3
m% (Na 2CO 3 ) = 46 ⇒ m% NaCl = 20
70. Let the mixture contain x g LiHCO 3 , y g NaHCO 3
On heating :
2NaHCO 3 → Na 2CO 3 + H 2O + CO 2
168
106
2LiHCO 3 → Li 2O + H 2O + 2CO 2
136
30
CaCO 3 → CaO + CO 2
100
56
312
Problems in Chemistry
Mass of residue =
56
30x 106 y
+
+ (5 − x − y)
= 2.58
100
136 168
30 
 56
 56 106 
−
−

x+
 y = 0.22
 100 136 
 100 168 
⇒
0.34x – 0.07 y = 0.22
Also, m mol of NaOH reacted with bicarbonates = 5 – 1.53 = 3.47
Total m mol of LiHCO 3 + NaHCO 3 = 34.7
⇒
y
x
+
= 34.7 × 10 −3
⇒
68 84
Solving, Eqs. (i) and (ii) gives y =1.686g, x =1
⇒
m% LiHCO 3 = 20, NaHCO 3 = 33.7
...(i)
...(ii)
71. Moles of NaHCO 3 in 1.0 g mixture = 2 × 10 −3 ⇒ m% NaHCO 3 = 16.8
Also moles of HCl consumed by 1 g mixture = 10.53 × 10 –3
Out of this 2 × 10 –3 mol HCl will be used up by NaHCO 3
⇒ 8.53 × 10 –3 mol HCl will be used up by Na 2CO 3 and CaCO 3
⇒ Moles of (Na 2CO 3 + CaCO 3 ) = 4.265 ×10 −3 /g of mixture.
Also 5 g mixture loses 0.75 g wt. and mixture contain 0.84 g NaHCO 3
⇒ 0.31 g weight is lost by NaHCO 3 and remaining 0.44 g by CaCO 3
Therefore, 5.0 g mixture contain 0.01 mol CaCO 3 , m% CaCO 3 = 20
Moles of Na 2CO 3 /g of mixture = (4.265 – 2) × 10 –3 = 2.265 × 10 –3
⇒
m% of Na 2CO 3 = 24 ⇒ m% of NaCl = 39.2
Let
the
original
sample
contained ‘a’ m mol of CaCO 3 and b m mol of NaHCO 3 . Therefore,
72.
CaCO 3 → CaO + CO 2
2NaHCO 3 → Na 2CO 3 + CO 2 + H 2O
b
m mol of CO 2 produced = a +
2
CO 2 + NaOH → NaHCO 3
a+
b
2
a+
b
2
NaHCO 3 + NaOH → Na 2CO 3
a+
x
b
–x
2
Till phenolphthalein end point m mol of HCl = 5 = x
After phenolphthalein end point m mol of NaHCO 3 present in solution = a +
⇒
a+
b
= 15
2
b
2
…(i)
313
Solutions
Also, from precipitation information:
b 0.985 × 1000
=
= 5 ⇒ a = 10, b = 10
2
197
⇒ mass of CaCO 3 = 1g, mass of NaHCO 3 = 0.84 g
m% : CaCO 3 = 50, NaHCO 3 = 42, Impurity = 8%
73. In presence of methyl orange, the whole NaOH and Na 2CO 3 are neutralized.
⇒ meq of HCl = 16 × 0.25 = 4 = meq of (NaOH + Na 2CO 3 ) = meq. of NaOH original
⇒ Total meq of NaOH in original 1.0 g sample = 4 × 5 = 20
20 × 40 × 100
mass % of NaOH (original) =
⇒
= 80
1000
Now, let us assume that in 20 mL, x m mol of NaOH has got converted to Na 2CO 3
m mol of NaOH = 4 – x
⇒ In 20 mL,
x
m mol of Na 2CO 3 =
2
In 2nd titration, HCl used in titration of NaOH + Na 2CO 3 = 5 × 0.1 – 9 × 0.2 = 3.2
x
x
⇒ upto phenolphthalein end point, m mol of HCl required = 4 – x + = 4 – = 3.2
2
2
⇒
⇒
x =1.6
x
5x
Total Na 2CO 3 formed = × 5 =
=4
2
2
m mol of NaOH left unreacted = 20 – 4 × 2 = 12
8 × 40 4 × (106 + 18)
⇒ weight of 1.0 g of exposed sample = 1 –
+
= 1.176 g
1000
1000
4 × 106
× 100 = 36.05%
⇒ weight % of Na 2CO 3 in exposed sample =
1000 × 1.176
74. Molarity (M ) of NH 4SCN solution =
50 × 0.0452
22.98
⇒ m mol of ClCH 2COOH present in beverage = 50 × 0.0452 – 10.43 ×
50 × 0.0452
= 1.234
22.98
⇒ mass of ClCH 2COOH = 1.234 × 94.5 = 116.6 mg
75. Let 20 mL stock solution contain x m mol Na 2C 2O 4 and y m mol H 2C 2O 4
⇒
and
Therefore,
2x + 2 y = 23.34 × 0.04 × 5 = 4.668
2 y = 26.67 × 0.1 = 2.667 ⇒ x =1 and
m% of Na 2C 2O 4 =
m% of H 2C 2O 4⋅2H 2O =
5 × 10
–3
2
× 134
y =1.3335
× 100 = 33.5
1.3335 × 5 × 10 –3 × 126
× 100 = 42
2
314
Problems in Chemistry
76. Let the sample contain x m mol (NH 4 ) 2 SO 4 and y m mol NH 4 NO 3 .
1
In 20 mL, m mol of NH +4 ion = (2x + y)
5
1 70
m mol of NaOH reacted with NH +4 = 5 – 9 × ×
= 3.5
14 30
1
(2x + y) = 3.5
⇒
5
466 100
Also
x=
×
= 6.25 ⇒ From Eq. (i) y = 5
233 32
⇒
mass % of ( NH 4 ) 2 SO 4 =
…(i)
6.25 × 10 –3 × 132
× 100 = 55
1.5
mass % of NH 4 NO 3 =
5 × 10 –3 × 80
× 100 = 26.67
1.5
77. The reaction involved is:
[Zn(NH 3 ) 4 ]Cl 2 + 4NaOH → Na 2 [Zn(OH) 4 ] + 4NH 3 + 2NaCl
10
m mol of NaOH consumed = 15 –
= 13.33
6
13.33
m mol of complex present =
= 3.33
⇒
4
⇒
mass % = 3.33 × 10 –3 × 204 × 100 = 68
Also total m mol of Cl – ions present in final solution = 3.33 × 2 +
⇒ mass of AgCl formed = 8.33 × 10 –3 × 143.5 = 1.195 g
78. From charge balance:
x + y = 2z
Also if M be empirical formula weight then
1.2 1
× × 1000x = 1.1
M 5
1.2 15
×
× 1000 y = 4.95
M 50
Dividing Eq. (iii) by (ii): 3x = y
Now, substituting x =1, y = 3, z = 2
Empirical formula: KH 3 (C 2O 4 ) 2
50 × 0.04
= 0.1
20
In 20 mL, meq. of H 2SO 4 left unreacted = 40 × 0.02 = 0.8
⇒ meq. of Ca 3 (PO 4 ) in 20 mL = 20 × 0.1 – 0.8 = 1.2
⇒ Total meq. of Ca 3 (PO 4 ) 2 = 1.2 × 5 = 6
6 × 10 –3 310
⇒ mass % of Ca 3 (PO 4 ) 2 =
×
× 100 = 15.5%
2
6
79. Normality of original H 2SO 4 solution =
10
= 8.33
6
…(i)
…(ii)
…(iii)
315
Solutions
80. The chemical reaction involved is:
6NH 4ClO 4 +10Al → 5Al 2O 3 + 3N 2 + 6HCl + 9H 2O
3 1
mole of Al reacted =
=
27 9
1 6
1
mole of NH 4ClO 4 present = × =
9 10 15
117.5 100
mass % of NH 4ClO 4 =
⇒
×
= 78.33
15
10
1
Also moles of HCl produced =
15
moles of NaOH taken initially = 0.1
⇒
moles of HCl required to neutralize left over NaOH = 0.1 –
Vol. of HCl required =
1
= 0.033
15
33
= 66 mL
0.5
81. Volume of room = 150m 3 = 150 × 10 3 L
150 × 10 3
⇒ Total mole of CO 2 in the room =
× 0.01 = 60.9756
0.082 × 300
4KO 2 + 2H 2O → 3O 2 + 4KOH
KOH + CO 2 → KHCO 3
⇒ moles of KO 2 in original sample = 60.9756
Reaction of KO 2 with H 2SO 4 is:
2KO 2 + H 2SO 4 → K 2SO 4 + H 2O + 32 O 2
Reaction:
moles of KO 2 required to neutralize H 2SO 4 = 2 × 100 × 0.1 × 10 –3 = 0.02
mass of KO 2 required =
82. Equivalent weight of acid =
5000
× 0.02 = 1.64 g
60.9756
3.25 × 1000
= 63.35 ⇒ Molar mass =190
68.4 × 0.75
Formula of acid = C n H 2n –1 (COOH) 3
⇒ 14n – 1 + 135 = 190
⇒ n=4
Formula = C 7 H 10O 6
REDOX TITRATION
83. (a) 2.525 × 10 –3 mol
(b) meq of VO 2+ = meq of MnO –4 consumed in last step = 0.86 × 0.02236 × 5 = 9.6148 × 10 –2
Total meq of permanganate taken against Fe 2+ and VO 2+ = 1.4087
⇒ meq of KMnO 4 used up by Fe 2+ = 1.4087 – 9.6148 × 10 –2 = 1.3125
⇒ Moles of Fe 2+ titrated with KMnO 4 = 1.3125 × 10 –3
316
Problems in Chemistry
–2
(c) meq of Fe 2+ consumed with Cr 2O 2–
= 2.428
7 = 2.525 – 9.6148 × 10
–3
⇒ Moles of Fe 2+ consumed with Cr 2O 2–
7 = 2.428 × 10
(d) Mass of V = 9.6148 × 10 –2 × 51 × 10 –3
⇒
9.6148 × 51 × 10 –5
× 100 = 0.25
2
2.428 × 10 –3
2.428 × 10 −3
moles of Cr =
⇒ m% of Cr =
× 52 × 100 = 2.1
3
3×2
m% V =
423.3 × 10 –3
= 5.027 × 10 –4
842
in 20 mL = 1.508 × 10 –3
84. In 20.00 mL solution moles of U 3O 8 =
⇒
Moles of U 4+
meq of U 4+ in 20 mL = 3.016
⇒
meq of MnO –4 require for 20 mL = 0.024 × 27.23 × 5 = 3.2676
meq of Fe 2+ = 3.2676 − 3.016 = 0.2516
⇒
⇒
Mass of Fe 2O 3 =
0.2516
× 160 × 10 –3 = 20.128 × 10 –3 g
2
⇒ Mass of Fe 2O 3 associated with 100 g U 3O 8 =
20.128 × 10 –3
423.3 × 10 –3
× 100 = 4.755
4.755
× 100 = 4.54
104.755
85. Let sample contain x m mol Pb 3O 4 and y m mol of PbO 2
⇒
⇒
m% of Fe 2O 3 =
2x + 2 y =
⇒
Also,
⇒
2.7
× 2 × 1000 – 8 × 0.02 × 5 × 25 = 20
134
x + y =10
N (KMnO 4 ) in 2nd titration =
...(i)
5 × 4.48
= 0.4
5.6 × 10
(3x + y) × 2 = 0.4 × 10 × 10 ⇒ 3x + y = 20
… (ii)
Solving, Eqs. (i) and (ii)
x = 5 = y ⇒ m(Pb 3O 4 ) = 3.425 g, m(PbO 2 ) = 1.195 g
⇒
m% Pb 3O 4 = 68.5 m% PbO 2 = 23.9
2x = 2 y + z
86.
Also,
1.7225
1.7225
y+
z = 10 × 2 × 10 –3
M
M
1.7225
1.7225
2y +
z = 15 × 2 × 10 –3
M
M
Solving Eqs. (i) and (ii) gives y = z ⇒ x =1.5 y
Now, substituting y =1, gives Cu 3 (CO 3 ) 2 (OH) 2
...(i)
...(ii)
317
Solutions
87. meq. of KMnO 4 = 0.1 × 22 × 5 = 11 = meq. of CaC 2O 4
⇒ m mol of CaCl 2 = 5.5
Mass of CaCl 2 per gram of mixture = 5.5 × 111 × 10 –3 = 0.61g
Mass of NaCl per gram of mixture = 0.39 g
1000 
1000

+ 2 × 6.67 × 5 × 10 –3 ×
⇒ – ∆T f = K f  3 × 5.5 × 5 × 10 –3 ×
 = 2.92

95 
95
⇒
T f = – 2.92° C
88. Let half solution contain ‘ a’ m mol Na 2CO 3 and ‘b’ m mol NaOH
⇒
a + b =15 and 2a + b = 25
⇒ Total m mol of CaC 2O 4 = 20
Also,
89.
⇒ a = 10 = m mol of Na 2CO 3
20 × 2 + x = 10 × 06
. × 10
⇒ x = 20 meq of Na 2C 2O 4
134 100
m% Na 2C 2O 4 = 20 × 10 –3 ×
×
= 31.53%
2
4.25
50 × 0.25
M(KMnO 4 ) =
= 0.055
45 × 5
⇒ Normality of KMnO 4 in alkaline medium = 0.167 N
25 × 0.1
V=
= 15.00 mL
⇒
0.167
90. I.
meq of KMnO 4 = 3.75 × 0005
.
× 5 = 93.75 × 10 −3
⇒
Total meq. of C 2O 42– = 93.75 × 10 –3 × 5 = 0.46875
⇒
m mol of K 3 [Fe(C 2O 4 ) 3 ] ⋅ 3H 2O = 78.125 × 10 –3
II.
meq of MnO –4 = 17.5 × 0.005 × 5 = 0.4375
⇒ Total meq of Fe 2+ ion = 0.875 = m mol of Fe 2+
⇒ m mol of Fe 2+ from FeCl 3 ⋅ 6H 2O = 0.875 – 78.125 × 10 –3 = 0.7968
⇒ Mass % of FeCl 3 ⋅ 6H 2O =
0.2155
× 100 = 71.84
0.300
91. Let mixture contain x m mol H 2C 2O 4 ⋅ 2H 2O, y m mol Na 2C 2O 4 and z m mol NaHC 2O 4
⇒
2 y + z = 40
...(i)
and
2x + z = 60
…(ii)
Also,
126x + 134 y + 112 z = 6100
Solving, Eqs (i), (ii) and (iii), gives z = 20, x = 20 and y =10
⇒
m% of H 2C 2O 4 ⋅ 2H 2O =
m% of Na 2C 2O 4 =
2.52
× 100 = 41.31
6.1
1.34
× 100 = 21.96 ⇒ m% of NaHCO 3 = 36.73
6.1
…(iii)
318
Problems in Chemistry
92. meq of hypo = 4 = meq of I 2
⇒ 2 m mol of I 2 is produced from unreacted ICl as ICl + KI → KCl + I 2
⇒ m mol of ICl combined with oil = 0.5
Also, 0.127 g oil ≡ 0.5 m mol of ICl.
⇒ 100 g oil ≡
0.5 × 100
0.5
m mol of I 2
× 100 m mol of ICl or
0.127
0.127
⇒ Mass of I 2 required for 100 g oil =
100 × 0.5
× 254 × 10 –3 = 100 g
0.127
93. meq of KMnO 4 reacted with nitrite in 10 mL solution = 2 − 1 = 1
⇒ Total meq of NO –2 = 100 = m mol of NaNO 3
⇒
⇒
100 × 85 × 10 –3
× 100 = 56.67
15
m% of Mg(NO 3 ) 2 = 43.33
Total m moles of O 2 = 100 + 44 = 144
m mol of NO 2 = 44 × 4 = 176
 NO 2 
Molar ratio 
 = 1.22
 O2 
m% of NaNO 3 =
94. meq of As 2O 3 taken = 75 × 0.0125 × 4 = 3.75
meq of MnO –4 used up in back titration = 0.1862
⇒ meq of MnO 2 = 3.5638
⇒
m% of MnO 2 = 3.5638 ×
100
87
× 10 –3 ×
= 68.9
0.225
2
95. Molarity of original K 2Cr 2O 7 solution = 0.85 × 10 –3 M
⇒ m mol of K 2Cr 2O 7 initially = 2.25 × 10 –3
Tinitial = 41.5 ⇒ (Absorbance) initial = 58.5
T final = 43.5 ⇒ (Absorbance) final = 56.6
Q Absorbance ∝ [concentration] or amount.
∴ 56.6% absorbance correspond to
2.25 × 10 –3
× 56.6 = 2.177 × 10 –3 m mol K 2Cr 2O 7
58.5
⇒ m mol of K 2Cr 2O 7 reacted with alcohol = 7.3 × 10 –5
meq of alcohol in air = 7.3 × 10 –5 × 6 = 4.38 × 10 –4
meq of alcohol in 56.5 mL of blood = 4.38 × 10 –4 × 2300 = 1.0074
⇒ meq of alcohol in 100 mL blood = 1.783
46
m(C 2 H 5OH) = 1.783 ×
= 20.5 mg alcohol content is within the legal limit.
4
319
Solutions
96. m moles of BaC 2O 4 lost in washing = 1000 1.5 × 10 –8 = 0.1224
m moles of CaC 2O 4 lost in washing = 1000 2.34 × 10 –9 = 0.0483
meq of MnO –4 used up for oxidising oxalate = 0.05 × 13.94 × 5 = 3.485
3.485
= 1.7425
2
⇒ Total m moles of oxalate produced during precipitation
= 1.7425 + 0.1224 + 0.0483 = 1.9132
Let mixture contain x g Ba(NO 3 ) 2 :
0.3657 – x
x
+
= 1.9132 × 10 −3 ⇒ x = 0.1397 ⇒ m% Ba(NO 3 ) 2 = 38.2
⇒
261
164
97. The balanced redox reaction is: 2KO 2 + 6HI → 2KOH + 2H 2O + 3I 2
meq of hypo = 15 × 0.4 = 6 = meq of I 2 ⇒ Total I 2 liberated = 15 m mol
⇒ m mol of oxalate titrated =
⇒ m mol of KO 2 in original sample = 10
m% = 10 × 10 –3 × 71 × 100 = 71%
⇒
1
2Cu + → Cu 2+ + Cu, Cu 2+ + 2I – → CuI + 2 I 2
98.
8.3
× 1000 = 50
166
meq. of I 2 produced from excess KI = 10
m mol of KI taken initially =
⇒ m mol of KI reacted with Cu 2+ = 40
⇒
m mol of Cu 2+ = 20 = m mol of Cu 2O
m% Cu 2O = 20 × 142 × 10 −3 ×
100
= 94.67
3
99. 3Br 2 + 6OH – → 5Br – + BrO –3 + 3H 2O
⇒
⇒
⇒
10
3
10
meq of BrO −3 =
× 6 = 20 = meq of CaC 2O 4
3
m mol of BrO –3 =
mass of CaC 2O 4 = 20 × 10 –3 × 64 = 1.28 g
⇒
100. Let sample contain x m mol NaCl.
m% = 85.33
⇒ Initially x m mol of AgCl formed.
2AgCl → 2Ag + Cl 2
Now,
0.6 x
and
3Cl 2 → 5Cl
0.3 x
⇒
0.3 x
–
0.5 x
+ ClO –3
→ 5AgCl
0.5 x
Final AgCl = 0.4 x + 0.5x + 0.9x m mol
⇒
0.9 x × 143.5 × 10 –3 = 1.5
⇒
%NaCl = 11.614 × 10 −3 × 58.5 × 100 = 67.94
⇒ x = 11.614
320
Problems in Chemistry
101. m mol of KI taken = 2
meq of hypo 50 × 10 –3 ⇒ meq of I – left unreacted = 50 × 10 –3 × 10 = 0.5
⇒ m mol of I – used up with CuCO 3 = 1.5
Reaction of I – with Cu 2+ is :
2Cu 2+ + 5I – → 2CuI + I –3
⇒ m mol of Cu 2+ =
2
5
× 1.5 = 0.6
⇒ mass % of CuCO 3 = 0.6 × 10 –3 × 123 ×
100
= 18.45
0.4
102. meq. of Fe 2+ required for excess MnO –4 = 19 × 0.1 × 10
⇒ m mol of excess MnO –4 = 3.8
m mol of MnO −4 reacted with CN − = 16.5 − 3.8 = 12.7
meq of MnO –4 = 12.7 × 3 = 38.1 = meq of CN −
⇒
⇒
m mol of NaCN = 19.05
m % NaCN = 19.05 × 10 –3 × 49 × 100 = 93.345
103. meq of oxalate reacted with MnO 2 = 200 – 50.73 × 0.6 = 169.562
169.562
= 84.781
⇒ m mol of MnO 2 produced =
2
⇒ meq of KMnO 4 used for oxidation of toluene = 84.781 × 3
84.781 × 3
⇒ m mol of toluene oxidised =
= 42.39
6
Mass of toluene = 42.39 × 10 –3 × 92 = 3.8998 g
⇒
density (ρ) = 0.78 g / mL
100
meq of KMnO 4 left unreacted = 38 × 0.2 ×
= 72.38
10.5
⇒ m mol of KMnO 4 left unreacted = 14.476
Hence, total m mol of KMnO 4 present in original solution = 99.257
99.257
≈ 1.0 M
⇒ Molarity of original KMnO 4 solution =
100
104. Let mixture contain x m mol CaOCl 2 and y m mol NaOCl
⇒
2x + 2 y = 015
. × 2 × 10 × 10
⇒
x + y =15
0.287 × 10 × 10
= 20
143.5
Solving, Eqs. (i) and (ii) gives x = 5, y = 10
100
m% CaOCl 2 = 5 × 10 –3 × 127 ×
⇒
= 31.75
2
100
m% NaOCl = 10 × 10 –3 × 74.5 ×
= 37.25
2
Also,
2x + y =
…(i)
3
…(ii)
321
Solutions
GASEOUS STATE
8RT
RT
= 4 × 10 2 ⇒
= 2 × 10 4 ⇒ RT = 2πM × 10 4
πM
πM
6 3
9
9
Total K.E. of He = × RT = RT = ⋅ 2π × 4 × 10 −3 × 10 4 = 180 π J
4 2
4
4
12 3
9
9
Total K.E. of Ne =
× RT =
RT =
× 2π × 20 × 10 –3 × 10 4 = 360 π J
20 2
10
10
(360 +180) π
Average K.E. per mol =
= 807.84 J
1.5 + 0.6
105. Average velocity =
106.
− dP
– dP
∝ ( P − 1) ⇒
= K ( P – 1)
dt
dt
−dP
= Kdt ⇒
P −1
⇒
Now,
Now,
∫
P0
t
P −1
− dp
= Kt
= K ∫ dt ⇒ ln 0
P −1
P −1
0
d  − mRt 

 = K ( P − 1)
dt  VM 
Also,
⇒
P
K=
RT (– dm/ dt )
0.0821 × 300 × 1.0
=
= 7 × 10 –7
3
VM ( P –1)
7.28 × 10 × 28 (171.8 – 1)
ln
171.8 − 1
= 7 × 10 −7 × 10 × 24 × 3600
P −1
ln
170.8
= 1.83
P –1
⇒ P = 94.33 atm.
nRT
n2a
0.5 × 0.082 × 300 0.25 × 3.63
−
= 22.06 atmosphere
− 2 =
V − nb V
0.5 − 0.5 × 0.04267
0.25
99b
108. V − b = RT = 100(0.011075V − b) = 1.1075V − 100b
= 921b
⇒ V=
0.1075
107. P =
⇒
⇒
109. n (butane) =
920 b = RT ⇒ b = 24.33 cm 3 mol –1 = 4 ×
4
π r 3 × 6.023 × 10 23
3
r = 13.4 × 10 −9 cm = 134 pm
1 × 40
= 1.626
0.082 × 300
⇒
Mass of butane = 94.308 g
Mass of Ar = 94.308 × 19 = 1791.852 g
n(Ar) = 44.8 Final P =
110.
46.426
= 28.55 bar
1.626
Cl 2O 7 → Cl 2 + 3.5O 2
1−α
α
3.5 α
…(i)
322
Problems in Chemistry
Now, let x be the mole fraction of Cl 2 in the gaseous mixture being analyzed.
rO2
71 0.6
⇒
= 3.5
=
⇒ x = 0.115
rCl 2
32
x
⇒ Mole fraction of Cl 2O 7 = 0.285
rCl 2O7 1 − α
Also,
=
rCl 2
α
N  M 
r 
111.  1 
=  1  2
 r2  n = 1  N 2  0  M 1 
112. (a)
r
rO2
= 1.33 =
Vm =
32
M
n/2
71
0.285
=
⇒ α = 0.2
183 0.115
N 
=  1
 N2 n
⇒
25 10  22 
=
 
75 90  20 
n/2
Solving n = 23
M = 18 amu.
18
= 50 L
0.36
1 × 50
PV
=
= 1.22, repulsive force is dominating.
Z=
RT 0.082 × 500
3
3
KT = × 1.38 × 10 −23 × 1000 = 2.07 × 10 −20 J
2
2
2 × 0.082 × 300
nRT
113. ( P + K ) (V − nb) = nRT ⇒ K =
−P =
− 11 = 1.615 bar
4 − 2 × 0.05
V − nb
Pb
Pb
114.
⇒ 0.5 =
Z =1 +
RT
RT
0.5 RT
⇒
b=
= 0.5 × 0.082 × 273 = 11.193 L mol –1
P
a
Also,
⇒ a = TB ⋅ Rb = 107 × 0.082 × 11.193
TB =
Rb
(b) KET =
a = 98.2 bar L2 mol –2
⇒
115.
Also,
Also,
He
N2
2 mol
6 mol
nHe 1 400 2
=
= ×
nN 2 3 200 3
nHe + nN 2 = 8 ⇒ n He =
P
1
=
300 × 2 200 × 16
24
16
, nN2 =
5
5
⇒ P = 1.066 atm
5
116. Applying concept of uniform pressure :
nN 2
nO2
150
=
=
⇒ nN 2 = 36.16, nO2 = 5.16
(1 − 0.6 3 ) (0.6 3 − 0.3 3 ) (0.3) 3
323
Solutions
⇒
W = 300 + 36.16 × 28 + 5.16 × 32 = 1477.6 g
After adding 50 g H 2 :
175
36.16
5.16
= 3
= 3
3
3
(1 − rB )
( rB − rC )
rC
Solving, rB = 57.6 cm and rC = 28.8 cm
117.
rmix
32
=
= 0.45 ⇒ M = 158
rO2
M
rPCl 3
Also,
⇒
rCl 2
PCl 5
1–α
rPCl 5
rCl 2
n2a
V2
nPCl 3
71
= 0.72 =
nCl 2
137.5
71x + 0.72x × 137.5 + (1 − 1.72x ) × 208.5 = 158
Now,
118.
=
= 002
.
=
1–α
α
⇒
x = 0.268
PCl 3 + Cl 2
α
71
0.539
=
208.5
0.268
α
⇒ α = 0.225
⇒ a = 0.02 × (22.5) 2 = 10.125 atm L2 mol –2
V − nb = nRT ⇒ b = V − RT = 22.5 − 0.082 × 273 = 114 cm 3 mol –1
TB =
a
10.125
=
= 1083.12 K
Rb 0.082 × 0.114
119. 300 × 1 = n × 1.072 × R × 473
600 V = n × 1.375 × R × 273 ⇒ V = 0.37 L
a
8a
120. Tc =
= 282 and Pc =
= 50
27Rb
27b 2
Dividing
282
8a
27b 2
282R
=
×
⇒ b=
= 0.0578 L mol −1
a
50 27Rb
50 × 8
Substituting b in Tc : a =
282 × 27 × 0.082 × 0.0578
= 4.5 atm L2 mol –2
8
121. The second Virial coefficient = b 2 = 2 × 10 –2 ⇒ b = 0.1414 L mol −1
nRT 0.082 × 300
=
= 4.92, V real = Vid + nb = 5.0614 L
5
P
4
122. b = 4.42 × 10 −2 L mol −1 = 44.2 cm 3 mol −1 = 4 × πr 3 × 6.023 × 10 23
3
Videal =
⇒ r = 16.36 × 10 −9 cm
123. Vc =
⇒ Distance of closest approach = 2r = 32.72 × 10 −9 cm = 327.2 pm
V
44
= 97.77 cm 3 mol –1 ⇒ b = c = 32.6 cm 3 mol –1
0.45
3
324
Problems in Chemistry
TC =
8a
= 300 ⇒ a = 2.7 atm L2 mol –2
27Rb
RT
+ B = 2.058 L. Also, for the given equation :
P
10( − 0.1814)
PB
=1 +
= 0.918
Z =1 +
RT
0.082 × 273
a
From van der Waal’s equation, for Z < 1, Z = 1 −
VRT
a
= 0.082 ⇒ a = 3.77 bar L2 mol –2
⇒
VRT
rX 62
40
125.
=
=2
⇒ M = 60 amu.
rAr 38
M
124. V =
2
 100 
nRT
n2a
100 × 0.082 × 313
P=
− 2 =
−
 ⋅ 3.6 = 11.16 bar
100
 44 × 5 
V − nb V
44 (5 −
× 0.044)
44
126.
nRT 100 0.082 × 313
=
×
= 11.666 bar
44
5
V
Pid =
Percentage deviation =
Pid − Preal
× 100 = 4.33%
Pid
∂( PV )
=0
∂P
−B
2.879 × 10 −2
B + 2CP = 0 ⇒ P =
=
= 96 atm
2C 2 × 14.98 × 10 −5
127. At minimum in PV − P curve,
⇒
RT
PV
V
α
α
 β
128. P =
−
⇒
=
−
= 1 − 
2
2
V − β TV
RT V − β RT V  V 
⇒
−1
−
α
RT 2V
β β2
α
α  β2
PV
1
β
= 1 + + 2 + ...−
=
+
−
1
+ ...

+
RT
V V
V 
RT 2V
RT 2  V 2
Comparing with Virial equation :
B =β −
α
RT
2
and TB =
α
Rβ
129. V. P. = 0.6 × 23.76 = 14.256 mm
Moles of gas in 1.0 L dry air =
1
= 0.0409
0.082 × 298
Moles of N 2 = 0.0409 × 0.79 = 0.0323 ⇒ mass of N 2 = 0.9044 g
Moles of O 2 = 0.0409 × 0.21 = 0.0086 ⇒ mass of O 2 = 0.2752 g
325
Solutions
Moles of H 2O( v ) in 1.0 L moist air =
14.256
= 0.76 × 10 –3
760 × 0.082 × 298
⇒ Mass of H 2O( v ) = 0.0138 g
Adding masses of all the gases gives = 1.1934 g / L

n2a 
1 
130.
T=
  P + 2  (V − nb)  = 623.57 K
nR  
V 

P
t
− dP KP
− dP
K
131.
=
⇒ ∫
=
∫ dt
dt
P
M
M 0
P
0
Kt
4K
P 
ln  0  =
⇒ ln 4 =
= 2K
P
M
4
10 K
ln
=
⇒ PHe = 7.07 atm
P 2
Kt
Kt
P 
P 
and ln  0 
ln  0 
=
=
 P  He
 P  CH
2
4
⇒
Also,
4
 P0 
  =
 P  He
⇒
2
 P0 
 
 P  CH
⇒
4
10
100
= 2
7.07 PCH
4
⇒
PCH 4 = 8.4 atm
132. Pideal (V − b) = RT for 1.0 mol
⇒
Pideal =
0.082 × 273
= 438.9
0.075 − 0.024
⇒
Z=
600
= 1.367 (Repulsive force)
438.9
NH 3 ( g ) + HCl( g ) → NH 4Cl( s)
133.
Moles
0.08
0
0.26
0.18
⇒
0
0.08
Q = 0.08 × 4300 = 3440 J
3440
= 955.55
Q = nC v ∆T ⇒ ∆T =
0.18 × 20
Also,
Pfinal
T final = 1255.55 K
0.18 × 0.082 × 1255.55
=
= 10.3 atmosphere
1.8
134. Let P1 be the pressure in methane chamber and P2 be the pressure in He chamber at any instant t.
Then :
− dP1 K1 P1 K 2 P2 K1 P1 K 2 P2
=
−
=
−
dt
4
2
16
4
Also, at equilibrium,
and
− dP
=0
dt
P1 = P2 = 3.7
⇒
K 1 = 2K 2
326
Problems in Chemistry
− dP1 K1
K
=
( P1 − P2 ) = 1 (2P1 − 7)
dt
4
4
⇒
4
∫
⇒
5
135. r.m.s. ( s) =
136. 2NH 3
− dP1
K
= 1
(2P1 − 7)
4
ds
3RT
⇒
=
M
dT
N 2 + 3H 2
1.6 – 2x
∫ dt
Solving,
t=
0
3R
3R 1
3R
= 1558.87 ms –1
⋅
=
=2 ⇒ S =
M 2 T 2 MS
4M
1.6 × 0.082 × 500
=
= 32.8
2
Also,
32.8
1.6
=
⇒ x = 0.383
48.5 1.6 + 2x
n( NH 3 ) = 0.834, n( N 2 ) = 0.383, n( H 2 ) = 1.149
P(dry oxygen) = 729.9 mm ⇒ V (dry oxygen) =
137.
2 ln 3
= 87.88 second
K1
3x = 1.6 + 2x
x
Pinitial
t
Q P1 + P2 = 7
729.9 × 30
= 29.15 mL.
751
PV
= constant, V = 25.82 mL
T
138. Let x mol gas is present in cold flask and y mol in hot flask.
Now,
⇒
applying
2x = 3 y and
x + y =1
Solving:
3
2
x = , y=
5
5
3 0.082 × 200
×
= 9.84 atmosphere.
5
1
139. Boyle’s law : PV = C (constant)
P=
C
P
⇒
V=
⇒
K =−
−C
V
 ∂V 
= 2 =−


 ∂P  n , T
P
P
⇒
1  ∂V 
1
=
= 1.0 atm –1


V  ∂P  n , T
P
140. ∆m = mcold − m hot = [ ncold − n hot ] M =
PVM
R
1
1
T − T 
2
 1
Solving : ∆T = T2 − T1 = 3
∂( PV )
=0
∂P
∂( PV )
P
=
+V
∂P
( ∂P / ∂V )
141. The minimum in curve will occur at
Also,
…(i)
for 1.0 mole, van der Waals’ equation can be written as:
P=
⇒
RT
a
− 2
V −b V
− RT
2a
∂P
=
+
∂V (V − b) 2 V 3
…(ii)
327
Solutions
 ∂P 
Substituting 
 in Eq. (i) and equating to zero gives
 ∂V 
( bRT – a ) V 2 + 2abV – ab 2 = 0
Substituting a, b, R and T gives
− 0.647V 2 + 0.087V − 1.38 × 10 −3 = 0
…(iii)
Solving, Eq. (iii) gives
V = 0.0185 L mol –1
and 0.01158 L mol –1 .
The first solution yields a negative pressure on substitution into Eq. (ii), but usingV = 0.1158 gives
P=
0.082 × 273
1.362
−
= 165.56 atmosphere
0.1158 − 0.032 (0.1158) 2
142. Steam : 24.06 Benzene : 5.464
Toluene : 18.00 Neon
: 0.2107
(Higher the intermolecular force of attraction, higher will be ‘a’)
143. Toluene : 0.1463 Steam : 0.0305
Benzene : 0.1154 Neon : 0.0170
(Larger the van der Waal’s radius, higher will be ‘b’)
THERMOCHEMISTRY
144. Heat consumed by water = 450 × 4.18 × 75 = 141.075 kJ
Heat consumed by copper = 500 × 0.38 × 75 = 14.25 kJ
Total heat required = 155.325 kJ
141.075
% heat used to raise temperature of H 2O =
× 100 = 90.82
155.325
Fe 2O 3 + 2Al → Al 2O 3 + 2Fe;
145.
50 g
∆H = – 851.5 kJ / mol
25 g
Here, Fe 2O 3 is limiting reagent.
Hence,
146.
– 851.5
× 50 = – 266.1 kJ
160
∆H = Σ ∆H comb (reactants) − Σ ∆H comb (products)
Q=
= − 1300 − 2 × 286 + 1560 = − 312 kJ
147. Writing the reactions given as follows :
2NH 4Cl( s) → 2NH 3 ( g ) + 2HCl( g )
2NH 3 → N 2 + 3H 2
N 2 + 4H 2 + Cl 2 → 2NH 4Cl( s)
∆H = 352 kJ
∆H = 92.22 kJ
∆H = − 628.86 kJ
Now add the above equations :
H 2 ( g ) + Cl 2 ( g ) → 2HCl( g )
⇒
∆H °f HCl( g ) = – 92.32 kJ / mol.
∆H = − 184.64 kJ
328
Problems in Chemistry
148. Writing the given reactions as follows :
NO 2 + O 2 → NO + O 3
O 3 →
1
2
3
2
O2
O 2 → O( g )
∆H = + 200 kJ
∆H = −142.7 kJ
∆H = + 249 kJ
Adding above reactions gives :
NO 2 → NO + O( g )
∆H = 306.3 kJ
149. Performing a + b – 2c gives :
4H 2 + 2O 2 → 4H 2O
⇒
H2 +
1
O
2 2
→ H 2O
∆H = −1090
∆H = − 2725
. kJ …(i)
Also a + b + c gives :
3CH 4 + 2O 2 → H 2O + 3CO + 5H 2
5
2
5H 2 + O 2 → 5H 2O
and
∆H = − 349 kJ
∆H = − 1362.5 kJ
Adding the above two equations gives :
9
∆H = −1711.5 kJ
3
∆H = − 570.5 kJ
3CH 4 + 2 O 2 → 6H 2O + 3CO
⇒
CH 4 + 2 O 2 → 2H 2O + CO
150. N 2 ( g ) + 3H 2 ( g )
151.
2NH 3
− 92 = 941.3 + 3 × 436 − 6 B. E. (N – H) ⇒ B. E. (N—H) = 390.25 kJ
306 = 314 + 3 × 121 − 3 B. E. (P – Cl) ⇒ B. E. (P—Cl) = 123.67 kJ
8 = 314 + 3 × 216.5 − 3 B. E. (P – H) ⇒ B. E. (P – H) = 318.5 kJ
152. Multiply the 2nd reaction by 2 and subtract from 1st :
Fe( s) + 2H 2O( l) + 2SO 2 ( g ) → FeS 2 + 2H 2 + 3O 2
987 = ∆H °f ( FeS 2 ) + 2 × 285.83 + 2 × 296.81
⇒
∆H °f FeS 2 = − 178.28 kJ / mol
Also,
−137 = − 178.28 − 2H °f ( H 2S)
⇒
∆H °f (H 2S) = − 20.64 kJ / mol
153.
+ 3H2
;
∆H = 987
∆H = − 119 × 3 = − 357
– 357 = – 156 – ∆H °f (benzene) ⇒ ∆H f ° (benzene) = 201
R. E. = ∆H °f (exp) – ∆H °f (theoretical) = – 152 kJ/ mol
154.
trans-2-butene → cis-2-butene
∆H = + 095
. kcal
cis-2-butene → 1-butene
1-butene + 6 O 2 → 4CO 2 + 4H 2O
∆H = +1771
.
kcal
∆H = − 649.8 kcal
329
Solutions
Adding above three equation gives:
trans-2-butene + 6O 2 → 4CO 2 + 4H 2O
CH 3
HC==CH
+ 6O 2 → 4CO 2 + 4H 2O( l)
H 3C
H 2O( l) → H 2O( v )
CH 3
CH==CH
CH 3
∆H = − 647.079 kcal
∆H °V = +11 kcal
+ 6O 2 → 4CO 2 + 4H 2O( v )
∆H = − 636.079 kcal
– 636.079 = 8 × 98 + 2 × 80 + B. E. (C==C) + 6 × 118 – 8 × 196 – 8 × 110
B. E. (C==C) = 159.921 kcal.
B. E. (C== C) = x, B. E. (C  C) = y ⇒ x − 2 y = − 72
155. Let
6C(gr) + 3H 2 ( g ) →
…(i)
(g )
∆H = 79 = 6 × 715 + 6 × 218 − 3 ( x + y) − 6 × 415 − 152
⇒
x + y = 959
From, Eqs. (i) and (ii) x = 615.33, y = 343.67 kJ.
(g)
156.
…(ii)
(g) + 3H2(g)
∆H = 6 × 415 + 3 × 348 − 3 × 600 − 3 × 436 = 426 kJ
C 6 H 6 ( g ) → C 6 H 6 ( l)
C 6 H12 ( l) → C 6 H12 ( g )
C 6 H 6 ( l)Theo. → C 6 H 6 ( l) Exp.
Adding the above equations yields :
Cyclohexane (l) → C 6 H 6 ( l) + 3H 2
157. Let B. E. of C  C= x, C== C = y and C ≡≡C = z
y − 2x = − 86
z − x − y = −148
y + 436 − x − 2 × 415 = − 132
⇒
y − x = 262
From Eqs. (i) and (iii) x = 348 kJ, y = 610 kJ,
Substituting in Eq. (ii) gives, z = 810 kJ
158. Heat evolved on combustion of one unit of CH 2 = – 158
∆H = − 30.8 kJ
∆H = 33.0 kJ
∆H = −152 kJ
∆H = 276.2 kJ
⇒
…(i)
…(ii)
…(iii)
⇒ ∆H comb (C10 H 22 ) = ∆H comb (CH 4 ) + 9 × ∆H comb (CH 2 ) = – 210 – 9 × 158 = − 1632 kcal
330
Problems in Chemistry
3C 2 H 2 ( g ) → C 6 H 6 ( g )
159.
∆H = 3 × 930 − 3(615 + 348) = − 99 kJ
C 6 H 6 ( g ) → C 6 H 6 ( l)
⇒
∆H = 45 kJ
3C 2 H 2 ( g ) → C 6 H 6 ( l)
⇒
∆H = − 54 kJ = ∆H °f C 6 H 6 ( l) – 3 × 75
∆H °f C 6 H 6 ( l) = 171 kJ
⇒
R.E. = 49 − 171 = − 122 kJ / mol
160. Ca
2+
(g )
+
C 2–
2
(g )
∆H = 95
∆H = 1733
C 2 (g )
U0
∆H = − 614
Ca( g )
2C( g )
∆H = 179
Ca( s)
+
∆H = 1436
∆H = – 60 kJ
2 C(gr) → CaC 2 ( s)
179 + 1733 + 1436 − 614 + 95 + U 0 = − 60 ⇒ U 0 = – 2889 kJ
161. Required combustion reactions are :
C 3 H 8 + 2O 2 → C 2 H 2 + CO + 3H 2O ∆H = − 2220 + 1300 + 285 = − 635 kJ
3
CH 4 + O 2 → CO + 2H 2O
∆H = − 890 + 285 = − 605 kJ
2
10
90
For 100 g L.P.G., Q =
(– 635) + (– 605) = – 1676.98 kJ
16
44
162. Total O 2 consumed in 1.0 hour = 400 × 60 = 24000 mL = 24 L.
n(O 2 ) =
1 × 24
= 0.944; C 6 H12O 6 + 6O 2 → 6CO 2 + 6H 2O
0.082 × 310
Energy produced = 45312
. kJ, available energy = 113.28 kJ,
distance travelled = 1.1328 km.
164. From combustion data : ∆H °f (1, 3-butadiene) = − 4 × 394 − 4 × 285 + 2841 = + 125
⇒
165.
From bond enthalpies : 2 × 348 − 615 = 81 kJ = ∆H °f (1,3-butadiene)–130
∆H °f (1,3-butadiene) = 211
R.E. = 125 – 211= − 86 kJ / mol
2NH 3 ( g ) → N 2 H 4 ( l) + H 2 ( g )
∆H =142 kJ
N 2 H 4 ( l) → N 2 H 4 ( g )
Adding the above two equations yields :
2NH 3 ( g ) → N 2 H 4 ( g ) + H 2 ( g )
⇒
⇒
160 = 2 × 393 − B. E. (N–N) –436
B. E. (N – N) = 190 kJ/ mol
∆H =18 kJ
∆H =160
331
Solutions
166. – 56 = – 285 – ∆H °f (OH – ) ⇒ ∆H °f (OH − ) = − 229
Also for
HF + OH – → F – + H 2O
∆H = − 320 − 285 + 229 + 329 = − 47 kJ
167. Writing the given reactions as follows :
1
2FeO( s) + 2 O 2 → Fe 2O 3 ( s)
∆H = − 284 kJ
2Fe( s) + 4H + → 2Fe 2+ + 2H 2
∆H = – 175.6
3
Fe 2O 3 ( s) → 2Fe( s) + 2 O 2
∆H = 8214
.
2H 2 + O 2 → 2H 2O
∆H = – 570
Adding the above equations yields :
2FeO( s) + 4H + → 2Fe 2+ + 2H 2O
⇒
∆H = − 2082
.
FeO( s) + 2H + → Fe 2+ + 2H 2O
∆H = − 104.1 kJ
168. Q = 1316.7 × 2.36 = 3107.412 J
3107.412
× 1000 J = – 51.79 kJ / mol
60
(C 2 H 5 ) 2 S( g ) + S( g ) → (C 2 H 5 ) 2 S 2 ( g )
∆H = − 202 + 147 − 223 = − 278 kJ
∆H Neut = −
169. For
B. E. (S – S) = + 278 kJ / mol
170. Let B.E. C – C = x, C ≡ N = y and C − H = z
⇒
1
2C(gr) + 2H 2 + 2 N 2 → CH 3CN( g )
1
∆H = 88 = 2 × 717 + 2 × 436 + 2 × 946 − ( x + y + 3 z )
…(i)
x + y + 3 z = 2691 kJ
2C(gr) + 3H 2 → C 2 H 6 ∆H = − 84 = 2 × 717 + 3 × 436 − ( x + 6 z )
or
x + 6 z = 2826 kJ
⇒
x = 2826 − 6 × 410 = 366 kJ / mol
y = 2691 − 366 − 3 × 410 = 1095 kJ/ mol
171. CH 3COOH( g ) + 2O 2 → 2CO 2 + 2H 2O( g )
or
∆H = 3 × 410 + 348 + 728 + 352 + 463 + 2 × 498 − ( 4 × 728 + 4 × 463) = − 647 kJ
⇒ CH 3COOH( l) + 2O 2 → 2CO 2 + 2H 2O( l)
∆H = − 647 + 52 − 82 = − 677 kJ
⇒
∆H reaction = − 677 + 860 = 183 kJ
172. For CH 3OCH 3 ( g ) : – 348 = – 2 × 94 – 3 × 68 – ∆H °f ⇒ ∆H °f = − 44 kcal/ mol
⇒ For
173.
C 2 H 5OH( l) → CH 3OCH 3 ( g )
∆H = − 44 + 66 = + 22 kcal / mol
BH 3 ( g ) → B( g ) + 3H( g )
∆H1°
B2 H 6 ( g ) → 2B( g ) + 6H( g )
∆H 2°
332
Problems in Chemistry
∆H 1° = ∆H f° B( g ) + 3∆H f° H( g ) − ∆H f° BH 3 ( g ) = 563 + 3 × 218 − 100 =1117 kJ
∆H 2° = 2∆H f° B( g ) + 6∆H f° H( g ) – ∆H f° B2 H 6 ( g ) = 2 × 563 + 6 × 218 – 36 = 2398 kJ
⇒ mean B H bond energy in BH 3
=
∆H1°
= 372.33 kJ/mol
3
mean B H bond energy in B2 H 6
∆H 2°
= 399.67 kJ/mol
6
Also, B2 H 6 has four terminal B H sigma covalent bond and two 3-centered 2-electron bond.
=
⇒
2398 = 4 × 372.33 + 2x ⇒ x = 454.34 kJ/mol.
Therefore, average B H bond energy of the bridged B H  B bond
=
Also, BE ∝
454.34
= 227.17 kJ/mol
2
1
; Hence, terminal B H bonds are shorter than bridged B H bond.
Bond length
174. Given:
∆H ° = −120 kJ
+ H2
⇒
∆H ° = − 120 × 3 = − 360 kJ
+ 3H2
Using combustion data:
°
[Benzene + 3H 2 – Cyclohexane] = − 215 kJ
∆H ° = ∆H combustion
Resonance energy = ∆H ° (from bond energy) – ∆H ° (combustion) = −145 kJ
7893
= 171.587
46
7025
moles of octane in 1.0 L =
= 61.62
114
C 2 H 5OH + 72 O 2 → 2CO 2 + 3H 2O
moles of ethanol in 1.0 L =
175.
Also:
∆H = − (2 × 394 + 3 × 286) + 278 = − 1368 kJ mol –1
C 8 H18 + 25
O 2 → 8CO 2 + 9H 2O
2
∆H = − (8 × 394 + 9 × 286) + 208.4 = − 5067.6 kJ mol –1
⇒ Heat produced from 171.587 mol ethanol = − 171.587 × 1368 = − 234.731 × 10 3 kJ
Heat produced from 61.62 mol octane = − 61.62 × 5067.6 = − 312.26 × 10 3 kJ
333
Solutions
⇒
312.26 × 10 3
= 1.33
234.731 × 10 3
i.e., automobile running on gasoline will gas 1.33 times farther than the automobile running in
ethanol, on combustion of 1.0 litre of fuel.
factor =
176. The combustion reaction is:
C 3 H 8 + 5O 2 → 3CO 2 + 4H 2O
10
50
mol of propane,
mole of O 2 would be required.
i.e., for
44
44
50
Therefore,
vol of O 2 required =
× 0.082 × 303 = 28.234 L.
44
100
Vol. of air required =
× 28.234 = 134.45 L.
⇒
21
10
Heat produced in the combustion reaction =
× ( −394 × 3 − 4 × 286 + 104) = − 2222 kJ
44
Q = ms∆T
Since,
⇒
2222 × 10 3 = 8000 × 4.18 ∆T
⇒
∆T = 66.45 ⇒ T2 = 96.45° C.
177. Multiplying both Eqs (i) and (ii) by 2 and finally adding them yields
Adding:
N 2 ( g ) + 2O 2 ( g ) → 2NO 2 ( g )
∆G° =103.56 kJ
2NO 2 ( g ) → N 2O 4 ( g )
∆G° = − 5.77 kJ
N 2 ( g ) + 2O 2 ( g ) → N 2O 4 ( g )
…(iii)
∆G° = 97.79 kJ
178. The combustion reaction is:
1
CO( g ) + O 2 ( g ) → CO 2 ( g )
2
1
∆H ° (from bond enthalpies) = 743 + × 498 − 2 × 743 = − 494 kJ
2
∆H ° (from formation enthalpies) = − 394 + 111 = − 283 kJ
The above values of enthalpy indicates that in carbonmonoxide, the bond between carbon and
oxygen is stronger than the same in CO 2 .
179. The formation reaction is:
N 2 + 3H 2
2NH 3
For reaction of 0.1 mol N 2 , 0.3 mol H 2 is also reacted and 0.2 mol NH 3 formed.
⇒
BE (entered) = 0.1 × 944 + 0.3 × 436 = 130.8 kJ
Energy released on formation of NH 3
= 0.2 × (3 × 388) = 232.8
⇒ Energy transferred to surrounding = 232.8 − 130.8 = 102 kJ.
334
Problems in Chemistry
Heat produced on neutralization = 1316.7 × 2.36 = 3107.412 J
180.
moles of HCl neutralized = 0.15 × 0.4 = 0.06.
3107.412
∆H (neutralization) =
= 51.8 kJ/mol of HCl.
0.06
⇒
CuSO 4 ( s) + ( aq ) → CuSO 4 ( aq )
181. Given:
CuSO 4 ⋅ 5H 2O( s) + ( aq ) → CuSO 4 ( aq )
Subtracting the second equation from first equation yields:
CuSO 4 ( s) + 5H 2O → CuSO 4 ⋅ 5H 2O( s)
182. (a) For the desired equation: (i)–(ii)
CaCl 2 + 2H 2O → CaCl 2 ⋅ 2H 2O
183. Carrying out the following operations and rewriting the equations:
(i) × 2 and reversing
(ii) reversing
(iii) as such
(iv) × 2
2NaCl( s) → 2Na( s) + Cl 2 ( g )
H 2SO 4 ( l) → H 2 ( g ) + S( s) + 2O 2 ( g )
2Na( s) + S( s) + 2O 2 ( g ) → Na 2SO 4 ( s)
H 2 ( g ) + Cl 2 ( g ) → 2HCl( g )
Adding the above equations gives the desired equation:
2NaCl( s) + H 2SO 4 ( l) → Na 2SO 4 ( s) + 2HCl( g )
184. The required reaction is:
Ag( s) + 12 Cl 2 ( g ) → AgCl( s)
∆H ° = − 66.04 kJ
∆H ° = −11.495 kJ
∆H ° = − 54.545 J
∆H ° = 37.628
∆H ° = 821.2 kJ
∆H ° = 810.54 kJ
∆H ° = −1381.5 kJ
∆H ° = −184.42 kJ
∆H ° = 65.82 kJ
∆H ° = ?
(i) + (ii) + 2 × (iii) – (iv) gives
2Ag + Cl 2 → 2AgCl
∆H ° = − 324.4 − 30.56 − 2 × 92.21 + 394 = − 145.38 kJ
1
Ag( s) + 2 Cl 2 ( g ) → AgCl( s)
∆H ° = – 72.69 kJ.
⇒
185. Adding the theory of covalent bonding, the structures of N 6 ( g ) can be predicted as:
N
N
N
N
N
N
N
N
N
N
I
II
N
N
Here I has delocalized π-electrons as well as aromaticity.
N
N
N
3N2
N
N
N
∆H° = 1072 kJ mol–1
335
Solutions
∆H ° (from bond enthalpies) = 3 × 944 − 3 (163 + 409) = 1116 kJ mol –1 .
The observed ∆H °f is 44 kJ mol –1 less than that calculated from bond enthalpies indicating
presence of resonance. Hence, I structure is most probable.
186. If all energy available for muscular activity is consumed in jogging, energy available from 0.5 kg
fat for jogging = 500 × 38 × 0.7 = 13300 kJ
13300
minimum jogging hour =
= 6.65 hr.
2000
3.24
187.
Total energy absorbed = 42.6 ×
= 3 kJ
46
3
1
Energy absorbed in one second =
=
kJ
60 × 10 200
1
m 2 = 50 cm 2 .
200
188. Adding the two given thermochemical equations (i) and (ii) yields
Surface area of beaker =
+ 3H2
∆H° = –208 kJ
…(a)
Had there been no resonance energy in cyclohexadiene, the enthalpy of (ii) reaction would have
been − 170 − 70 = − 240 kJ. This indicates that independent hydrogenation of one double bond
gives off on an average 120 kJ of heat. Therefore, theoretically.
+ H2
∆H°Theoretical = –120 kJ
⇒
° − ∆HTheo
° = 82 = R. E. (benzene) − R. E. (cyclohexadiene)
∆H exp
⇒
R. E. (benzene) = 152 kJ mol –1 .
THERMODYNAMICS
189. − W = nRT ln
V2 − nb
0.5 – 0.5 × 0.02
= 0.5 × 8.314 × 300 ln
= – 1747.8 J
V1 − nb
2 – 0.5 × 0.02
3
190. ∆H = ∆E + P∆V + V∆P : ∆E = nCV ∆T = 2 × 8.314 × 100 = 1247.1 J
P1 = 4.92 atm and P2 = 4.1 atm
⇒ P1 ∆V = 4.92 × 3 = 14.76 L atm and V2 ∆P = 8(4.1 − 4.92) = − 6.56 L atm
⇒ P1 ∆V + V2 ∆P = 8.2 L atm = 831.48 J
⇒
∆H =1247.1 + 831.48 = 2078.58 J
⇒
W =1747.8 J
336
Problems in Chemistry
3
191. ∆H = ∆E + P1 ∆V + V2 ∆P = 2 × 8.314 × 500 + [4.92 × 3 + 8 (8.2 − 4.92)] × 101.4
= 6.236 × 10 3 J
192.
T
V 
100 
20.8 ln 2 + 8.314 ln 2 

T1
28 
V1 
For irreversible adiabatic expansion :
 C + ( P2 / P1 ) 
T2 =  V
 T1 = 217.76 K
CP


∆S =
…(i)
V2 P1T2 30 × 217.76
=
=
= 2.1776
V1 P2T1
10 × 300
Also,
Substituting in Eq. (i) gives ∆S = − 0.692 JK –1
193. ∆S = nCV ln
V
T2
373


+ 8.314 ln 2 = 0.17 JK –1
+ nR ln 2 = 002
. 12.6 ln


300
V1
T1
194. Applying conservation of heat to :
∆S
∆S
1
2
∆S
1
2
3
H 2O( s) →
H 2O( l) →
H 2O( l) ←
H 2O( l)
q
q
q
0° C
0° C
3
TK
363 K
20
10
[6000 + 75.42 (T − 273)] =
× 75.42 (363 − T )
18
18
⇒
T = 306.48 K
Also,
∆S 1 =
10 6000
×
= 12.21 JK −1
18 273
∆S 2 =
306.48
10
× 75.42 ln
= 4.85 JK −1
273
18
∆S 3 =
20
306
× 75.42 ln
= − 14.31 JK −1
18
363
∆S system = ∆S 1 + ∆S 2 + ∆S 3 = 2.75 JK −1
⇒
195. Process involved are :
H 2O( l) → H 2O( l) →
H 2O( s) → H 2O( s)
∆S
– 10°C
∆ S1
∆S 1 = C P ln
∆S 2 =
0° C
2
0° C
∆S 3
– 10°C
T2
273
= 2.814 JK –1
= 75.42 ln
263
T1
− 6000
263
= − 21.98 JK –1 and ∆S 3 = 37.2 ln
= – 1.388 JK –1
273
273
∆S = ∆S 1 + ∆S 2 + ∆S 3 = – 20.554 JK –1
337
Solutions
Zn + H 2SO 4 → ZnSO 4 + H 2
196.
100
× 8.314 × 300 = 3.84 kJ
65
∆V = 0 ⇒ − W = 0
− Wirr = ∆ng RT =
In sealed vessel,
197. P ∝ d ⇒ d (final) = 2.5 m. Also P = kd ⇒ k = 2 atm m –1
− dWrev = P dV = k . d.
⇒ −W =
kπ
2
2.5
∫
d 3 dd =
0.5
π 2
kπ 3
d dd =
d dd
2
2
2 atm m –1 × π
kπ
× 39 m 4
[(2.5) 4 − (0.5) 4 ] =
8
8
= 30.63 atm m 3 = 3.1 × 10 6 J
198. A : P = 10 bar, T = 300 K, B: P = ? T = 300K
3
C : P = 2 bar, T = 250 K, CV = R
2
Connecting reversible adiabatic points B and C :
γ
1−γ
T
PB = PC  C 
 TB 
−W
⇒
AB
− WBC
− WAC
 250 
=2

 300 
A
B
T
–5
2
= 3.15 bar
C
V
10
= 2.88 kJ
3.15
= CV (T1 − T2 ) = 1.5 × 8.314 × 50 = 623.55 J
= 3.50355 kJ
= 8.314 × 300 ln
199. Process is irreversible :
(a)
(b)
4
 2 × R 2R 
− W = Pext ∆V = 2 atm 
−
 × 273 = 2 × × 8.314 × 273 = 3631.55 J
 2
5
10 
∆H = ∆E = 0
∴ ∆T = 0
200. In reversible adiabatic process : TV γ – 1 = constant
⇒
V 
T1 = T2  2 
 V1 
γ −1
= T2 (2) 2/ 5 ⇒ T2 = 225.84 K
5
∆E m = 2 × 8.314 × 72.16 = 1499.84 J
7
∆H m = 2 × 8.314 × 72.16 = 2099.78 J
V −b
RT
dV ⇒ − W = RT ln 2
V −b
V1 − b
0.2 − 0.03
= − 10.155 kJ
− W = 8.314 × 300 ln
10 − 0.03
W = + 10.155 kJ
− dW =
201.
⇒
338
Problems in Chemistry
a 
 RT
− dW = 
− 2  dV
 V
V 
202.
⇒
− W = RT ln
 1
V2
0.384 × 0.15
1
= – 1095 J
+ a  −  = − 1152.56 +
V1
 V2 V1 
10 –3
⇒
W = + 1095 J
203. Work will be done only between A – B
⇒
−WAB = P∆V = 0.8 × 20 L atm = 1622.4 J
0.8 × 20
Initial temperature
TA =
= 390.25 K
0.5 × 0.082
1
TB = 2T A = 780.5 K
TC = 975.625 K
⇒
∆E AB = 0.5 × 1.5 × 8.314 × 390.25 = 2433.4 J
∆E BC = 0.5 × 1.5 × 8.314 × 195.125 = 1216.7 J
⇒
∆E AC = ∆E AB + ∆E BC = 3650.1 J
q = ∆E − WAB = 4461.3 J and ∆H = ∆E + P ∆V = q
204. At constant n and V , P ∝ T ⇒ T2 = 3030 K
⇒
205. − WAB
20 L
q = ∆E = CV ∆T = (75.42 − 8.314) × 2727 = 183 kJ
∆H = C P ∆T = 205.67 kJ
A(1.0 atm, 20 L)
= P∆V = 20 L atm = 2028 J
WBC = 0
− WCA = 2 × 8.314 ln
0.5
= − 11.52
1
Since process in cyclic, ∆U and
∆H = 0 and q = − W = 2016.48 J
206. In adiabatic irreversible expansion :
0 = CV (T2 − T1 ) + Pext (V2 − V1 )
 C + R ( Pext / P1 ) 
Solving for T2 :
T2 =  V
 T1 = 263.74 K
 CV + R ( Pext / P2 ) 
CV (300) = 21.52 + 8.2 × 10 −3 × 300 = 23.98 JK −1
C P (300 K) = CV + R = 32.294 JK −1
V
40 L
B(1.0 atm, 40 L)
C(0.5 atm, 40 L)
V
− W = nCV (T1 − T2 ) = 3.22 × 20.785 = 2293 J
10 × 1
n=
= 0.4
0.082 × 300
∆U = ∆H = 0
− W = 0.4 × 8.314 ln 10 = 7.66 J
⇒
B
P
− WA ↔ A = 2016.48 J
207. Process is isothermal
C
p 0.8 A
339
Solutions
208. dG = VdP − SdT at constant P , dG = − SdT
⇒
∫ dG = − 36.36 ∫ dT − 20.79 ∫ ln TdT
T
∆G = – 36.36 (T2 – T1 ) – 20.79 [T ln T – T ]T 2
1
= − 909 − 20.79 (1543.18 – 1399.73) = − 3891.33 J
CV = 20.266 +1.76 × 10 –2 T
209.
dE = CV dT = (20.266 + 1.76 × 10 −2 T ) dT
⇒
∆E = 20.266(T2 − T1 ) +
1.76
× 10 −2 (T22 − T12 ) = P2 (V1 − V2 ) = R
2

 P2
 T1 − T2 
P

 1
1.76
× 10 −2 T22 + 28.58 T2 – 7869.48 = 0
2
⇒
Solving, T2 = 255.3 K
Also,
⇒
dT
dP
dT
dP
−R
= (28.58 +1.76 × 10 –2 T )
−R
T
P
T
P
T
P
∆S = 28.58 ln 2 +1.76 × 10 –2 (T2 – T1) + R ln 1 = 2.22 JK −1
T1
P2
dS = C p
∆U = ∫ CV dT = ∫ (20.266 +1.76 × 10 –2 T ) dT
= 20.266 (255.3 – 300) +
1.76 × 10 –2
[(255.3) 2 – (300) 2 ] = − 1124.32 J
2
∆H = ∫ C p dT = ∫ (28.58 +1.76 × 10 −2 T ) dT
= 28.58(255.3 – 300) +
1.76
× 10 −2 [(255.3) 2 − (300) 2 ] = − 1496J
2
210. − W = 2 × 30 L atm = 6084 J ⇒ ∆E = (20000 − 6084) J = 13.916 kJ
V
dV
211. −W = ∫ PdV = ∫ 10 ⋅
= 10 ln 2 = 10 × ln 10 L atm = 23 L atm = 2332.2 J
V
V1
q = 420 + 2332.2 = 2752.2 J
212.
∆E = n∫ CV dT = 3 ∫ (30 + 14 × 10 −3 T ) dT
= 3 × 30 (T2 − T1 ) + 42 × 10 −3 (T22 − T12 )
= 9000 + 2940 = 11.94 kJ
− W = (T2 − T1 ) nR = 3 × 8.314 × 100 = 2494.2 J
q = 11.94 × 10 3 + 2494.2 = 14.4342 kJ
213. Applying adiabatic conditions between C and A.
 500 
P =

 300 
–2.5
× 5 = 1.39 atm.
340
Problems in Chemistry
WAB = 0
Q ∆V = 0
A(500 K, 5 atm)
3
WBC = − 8.314 × 300 ln
= − 1918.8 J
1.39
3
WCA = × 8.314 × 200 = 2494.2
2
W = 575.4 J
 500 
PC = 

 300 
214.
P
B
(300 K, 3 atm)
– 2.5
V
× 5 = 1.39 atm
5
× 1.39 = 2.32 atm
3
5
= − 8.314 × 500 ln
= − 3192 J
2.32
=0
= 1.5 × 8.314 × 200 = 2494.2 J
PB =
⇒
WAB
WBC
WCA
A(500 K, 5 atm)
P
B(500 K, PB )
C(300 K, PC )
W = − 697.8 J
215. Applying adiabatic condition between A and D
 300 
P =

 200 
⇒
216.
C(300 K, P )
V
– 2.5
× 5 = 1.82 atm
A(300 K, 5 atm)
W1 = − PV1 = − nRT1 = − 300 R
5
W2 = − nR × 600 ln
= − 606.4 R
1.82
W3 = 1.82 (VC − VD ) = 400 R
W4 = CV × 100 = 150 R
W = − 356.4 R = − 2.963 kJ
P
 RT RT 
. RT
− Wirr = P2 (V2 − V1 ) = P2 
−
 = 08
P1 
 P2
− Wrev = RT ln
2
P
C(600 K, P)
D(200 K, P)
V
P = (2) – 2.5 × 5 = 0.88 atm
5
WAB = − R × 500 ln
= − 868.63 R
0.88
1
WBC = 0.88 (500R − 250R ) ×
= 250R
0.88
WCA = 1.5R × 250 = 375 R
⇒
W = − 243.63 R = – 2.026 kJ
217.
B(600 K, 5 atm)
A(500 K, 5 atm)
P
B(500 K, P)
C(250 K, P)
V
A(500 K, 10 bar)
P
2.0
V
341
Solutions
Adding :
⇒
218. (a)
RT ln
2
+ 0.8RT = 4200
P
P = 1.62 atm
PB = (2) – 2.5 × 10 = 177
. atm
A
WAB = − 1.5 R × 250 = − 375 R
WBC = − 250 R ln
1.77
= − 142.75 R
1
T
(b)
⇒
 250 
P =

 500 
− 2.5
W = − 500R ln
= 5.65 atm
V
10
− 1.5R × 250
5.6
T
A
= − 664.9 R
= − 5.528 kJ
V1 = 0.082 × 300 = 24.6 L,
219.
0.082 × 500
= 1.138 atm
36
∆H = ∆E + P1 ∆V + V2 ∆P = 1.5 R × 200 + [11.4 + 36 × 0.138] × 101.4
= 4.154 kJ
P2 =
dG = C P dT − TdS − SdT = C P dT − T (3 × 10 −3 dT ) − (1.5 + 3 × 10 −3 T ) dT
220.
= (C P − 1.5) dT − 6 × 10 −3 T dT
⇒
∆G = (C P − 1.5) (T2 − T1 ) −
221.
∆S = C P ln
222.
∆V = 0 ⇒ dS = CV
⇒
223.
6
× 10 −3 (T22 − T12 ) = 3377 J
2
T2
P
5
1
+ R ln 1 = 3.5 R ln + R ln = 9.1 J
3
2
T1
P2
dT
dT
dT
= (12 + 28 × 10 −3 T )
= 12
+ 28 × 10 −3 dT
T
T
T
∆S = 12 ln
T2
40 × 1
+ 28 × 10 −3 (T2 − T1 ), T2 =
= 487.8 K
T1
0.082
∆S = 12 ln
487.8
+ 28 × 10 −3 × 187.8 = 11.09 JK –1
300
A → B ∆G ° = − 3 × 1000 − T × 20 = − 9000 J
A → C ∆G ° = − 3600 − 10T = − 6600 J
224.
C
B
W = − 517.75 R = − 4.304 kJ
∆G ° indicates that B is more stable than C.
dG = V dP − S dT = V dP − (10 + 12 × 10 −3 T ) dT
B(500, P)
C(250, 1)
V
342
Problems in Chemistry
∆G = V ∆P − 10 ∆T − 6 × 10 −3 (T22 − T12 )
⇒
⇒
∆G = 24.6 ×
225.
V=
0.082 × 300
= 24.6 L
1
P2 =
400 4
= atm.
300 3
1
× 101.4 − 10 × 100 − 6 × 10 −3 [( 400) 2 − (300 2 )] = − 588.52 J
3
dT
dT
= 2(19.686 + 31 × 10 −3 T )
dS = nCV
T
T
T
∆S = 2 × 19.686 ln 2 + 31 × 10 −3 (T2 − T1 ) = 14.43 JK –1
T1
∆S system = nR ln
226.
V2
= 2 × 8.314 ln 2.5 = 15.236 J
V1
4.17 × 10 3
= − 13.9 J
300
∆S univ = 1.336 J
3
∆S = R ln 3 + 2R ln = 15.87 JK −1
2
T A =122 K and TB = 244 K
∆S surr = −
227.
228.
⇒
WBC = 0,
nR
= − 244 R
1
WCA = 2R × 122 ln 2 = 169.12 R
W = − 74.88 R
= − 622.5 J
A
B
P
WAB = − 1(T2 − T1 )
C
V
b 2
(T2 − T12 ) = 7.4 kJ / mol
2
b
∆H = ∫ C P dT = ∫ (33.314 + bT ) dT = 33.314(T2 − T1 ) + (T22 − T12 ) = 9.063 kJ/ mol
2
3 × 0.082 × 300
V1 =
= 36.9 L
2
3 × 0082
.
× 500
V2 =
= 41 L
3
We can construct a path as:
229. ∆E = ∫ CV dT = ∫ ( a + bT ) dT = a (T2 − T1 ) +
W
1
A (2.0 bar, 36.9 L, 300 K) →
B (2.0 bar, 41 L, T )
W2
→ C (3 bar, 41 L, 500 K)
2 × 41 × 300
T=
= 333.33 K
2 × 36.9
343
Solutions
Then,
⇒
W1 = P ∆V = 2 × 4.1 × 101.4 = 831.48 J
3
and W2 = 0
q = 7.4 × 10 + 831.48 = 8.23 × 10 J
3
dV 
dT
dV 
dT


+R
dS = n CV
 = n  ( a + bT ) T + R V 
V
T





V 
T
∆S = n  a ln 2 + b(T2 − T1 ) + R ln 2  = 59 J
V1 
T1

R
230. WAB = − 5(500 − 400) = − 100 R
5
WBC = 0
Connecting adiabatic points A and D.
P
Applying P 1− γ T γ = K gives TD = 210 K = TC
⇒
A
B
C
PC = 2.1 atm
D
2.1
WCD = − 210 R ln
= − 155.8 R
⇒
1
WDA = 1.5 R × 190 = 285 R
W = WAB + WCD + WDA = 29.2 R
= 242.76 J
231. TD = 388 K, PB = 2 bar,
WAB = − 500 R ln
5
= − 458.14 R
2
V
A
D
P
B
WCD = 1.5R × (388 − 250) = 270 R
W = − 251.14 R
C
= − 2.088 kJ
232. According to second law of thermodynamics:
∆S v =
⇒
(b)
⇒
V
∆H v
Tb
∆H v = ∆S v × Tb = 85 × 353 = 30.005 kJ mol –1 .
Since vaporization process is endothermic, heat lost by surrounding in vaporization of 100 g
benzene is:
30005
Q=
× 100 = 38468 J
78
Q
38468
∆S surrounding = − = −
= − 108.97 JK –1 .
353
T
∆H v = ∆S v ⋅Tb = 85 × 329 = 2796 5 J mol –1
100 × 27965
(b) Heat gained by surrounding =
= 48215.5 J
58
233. (a)
Increasing in entropy of surrounding =
48215.5
= 146.55 J
329
344
Problems in Chemistry
234. For a reaction:
∆GR° = Σ ∆Gf° ( Products) − Σ ∆G °f ( Reactants)
⇒
∆G° (i) = − 2 × 200 + 762 = + 362 kJ
∆G° ( ii) = − 396 + 762 = + 366 kJ
Since, both ∆G° are positive, a reaction with less positive ∆G° will be more likely to occur.
Therefore, reduction of TiO 2 will be more favourable through reaction (i).
235.
∆H r° = − 111 + 242 = 131 kJ
∆S r° = 190 + 131 − (5.7 + 70) = 245.3 JK –1
⇒
∆G ° = ∆H ° − T∆S ° = 131 × 10 3 − 300 × 245.3 = 57.41 kJ
Positive value of ∆G° indicates that reaction is non spontaneous at 300 K. Increasing temperature
will decrease ∆G° and after certain value of T , ∆G ° will become negative, i.e., reaction will turn
from non-spontaneous to spontaneous. That specific temperature at which reaction turns from
non-spontaneous to spontaneous can be determined by equating ∆G° = 0 as:
131000
= 534 K
0 = 131 × 10 3 − T × 245.3 ⇒ T =
245.3
i.e., at temperature above 534 K, reaction will become spontaneous.
236. At 300 K,
∆G ° ( cis-2-butene → trans-2-butene ) = − 3 kJ mol –1
∆H ° ( cis-2-butene → trans-2-butene) = − 4.2 kJ mol –1
⇒
∆S ° ( cis-2-butene → trans-2-butene ) =
=
⇒
Also,
∆H ° − ∆G °
T
−4.2 + 3
× 1000 = − 4 JK –1
300
∆G° (1) at 400 K = − 4.2 × 10 3 + 400 × 4 = − 2600 J
ln K (1) = −
∆G ° (1)
3000
=
8.314 × 300
RT
⇒
K (1) = 3.33 at 300 K
and
K (1) = 2.20 at 400 K
∆G° (2) = − 8 kJ mol –1 at 300 K
⇒
⇒
ln K (2) = −
∆G ° (2)
8000
=
8.314 × 300
RT
K (2) = 24.7 at 300 K.
Now, let at 400 K, mixture contain x% trans-2-butene and y% 2-methyl propene.
x
18
⇒ At
400 K, K (1) = 2.2 =
=
100 − x − y 82 − y
⇒
y = 73.8%
345
Solutions
⇒
K (2) at 400 K =
⇒
⇒
y
73.8
=
=9
100 − x − y 8.2
∆G° (2) at 400 K = − 8.314 × 400 ln q = − 7.3 kJ
∆G° (2) at 300 K = − 8000 = ∆H ° (2) − 300 ∆S ° (2)
∆G° (2) at 400 K = − 7300 = ∆H ° (2) − 400 ∆S ° (2)
⇒
∆S ° (2) = – 7 JK –1
∆H ° (2) = − 8000 + 300 ( −7) = – 10.1 × 10 3 J
K (3) =
Also,
24.7
= 7.41
3.33
∆G° (3) = − 8.314 × 300 ln 7.41 = − 4995 J
9
K (3) =
=4
2.2
∆G° (3) = − 8.314 × 400 ln 4 = − 4610 J
−4995 = ∆H ° (3) − 300 ∆S ° (3)
−4610 = ∆H ° (3) − 400 ∆S ° (3)
⇒ At 300 K:
K (3) =
⇒
At 400 K:
⇒
⇒
⇒
K (2)
K (1)
∆S ° (3) = – 3.85 JK –1
and
∆H ° (3) = – 6.15 kJ.
237. For the decarboxylation reaction:
CH 3COOH
CH 4 + CO 2
Also, above 65 K, reaction is spontaneous
⇒
∆G° = 0 at 65 K
⇒
0 = 15.7 × 10 3 − 65 ∆S °
⇒
∆S ° = 241.54 JK –1 .
∆H ° = −394 − 74.8 + 484.5 = 15.7 kJ
For a spontaneous reaction ∆G° < 0. Since, ∆H ° > 0, spontaneity of reaction is due to ∆S ° > 0.
238. Since,
∆G = ∆G ° + RT ln Q = RT ln Q − RT ln K
Q
∆G = RT ln  
K
⇒
In the present condition:
Q=
⇒
0.66 × 1.2
= 4.06
0.25 × 0.78
2540
 Q  ∆G
=
= 15.275 × 10 –2
ln   =
 K  RT 8.314 × 2000
Q
=1.165 ⇒ K = 3.48
K
Q Q > K , reaction will proceed in backward direction as:
⇒
346
Problems in Chemistry
H2
0.25 + x
+
H 2O( g ) +
CO 2
0.78 + x
K = 3.48 =
0.66 − x
CO
1.2 − x
(1.2 − x ) (0.66 − x )
(0.25 + x ) (0.78 + x )
Solving : x = 0.02.
⇒
PH 2 = 0.27 atm, PCO2 = 0.80 atm, PH 2O = 0.64 atm. PCO = 1.18 atm.
239. For the above reaction:
K p = pCO
2
 K p (2)  ∆H °  T2 − T1 
=
ln 


R  T1T2 
 K p (1) 
Also,
⇒
ln

1830 ∆H ° 
250
=
 = 4.394

22.6
R  973 × 1223 
⇒
∆G ° = − RT ln K p° = − 8.314 × 973 ln
Also,
∆H ° =173.88 kJ mol –1 .
22.6
= 28.43 × 10 3 J
760
∆H ° − ∆G ° 173.88 − 28.43
=
× 1000 = 149.5 JK –1 mol –1 .
973
T
240. The minimum value of ∆S will occur at ∆G = 0 for spontaneous reaction.
∆S ° =
⇒
19000
= 55.07 JK –1 .
345
T − Tc 2200 − 760
Efficiency of engine = h
=
= 0.6545
Th
2200
⇒
0 = 19000 − 345 ∆S °
241.
∆S ° =
⇒
2 × 3.1 × 1000
× 5510 = 299666.67 J
114
Heat converted into work = 299666.67 × 0.6545 = 196132 J
Total heat produced =
⇒
Q = mgh
h=
196132
= 16.67 m
1200 × 9.8
242. Since, expansion occurred at constant temperature,
∆S = nR ln
V2 1
3.0
=
× 8.314 ln
= 0.36 JK –1
V1 32
0.75
Since, this is case of free expansion, Pext = 0.
⇒
− W = Pext ∆V = 0,
q =0
Also, since, ∆T = 0, ∆H = ∆E = 0.
∆S = nC v ln
243. Q
n=
V
T2
+ nR ln 2
V1
T1
3 × 550 × 10 −3
3.5 × 0.73
× 300 = 464.5 K
= 0.067 and T final =
0.082 × 300
3 × 0.55
347
Solutions
⇒
464.5
5
0.73
R ln
+ 0.067 R ln
= 0.767 JK –1
300
2
0.55
5
∆E = nC v ∆T = 0.067 × R × ( 464.5 − 300) = 229 J
2
∆H = ∆E + P∆V + V∆P = 229 + [3(0.73 − 0.55) + 0.73 × 0.5] × 101.4 = 320.767 J.
∆S = 0.067 ×
244. Under isothermal condition,
V2
10
= 0.133 × 8.314 ln
= 1.33 JK –1
3
V1
∆S = nR ln
∆G = nRT ln
P2
3
= 5 × 3 × 101.4 ln
= − 1831.25 J
10
P1
245. (a) In case of adiabatic reversible expansion, dq rev = 0 ⇒ ∆S = 0.
(b) In case of irreversible adiabatic expansion:
0 = Pext (V2 − V1 ) + nC v (T2 − T1 )
T2 =
∆S = C v ln
Pext (V1 − V2 )
−3 × 8
+ T1 =
+ 1000 = 805 K
nC v
0.0821 × 1.5
T2
V
805
+ R ln 2 = 1.5 × 8.314 ln
+ 8.314 ln 2 = 3.06 JK –1 .
1000
T1
V1
(c) In case of free adiabatic expansion Pext = 0
⇒
0 = Pext ∆V + nC v (T2 − T1 )
⇒
T1 = T2
∆S = R ln
⇒
V2
= 5.76 JK –1
V1
246. Since, the two gases have same heat capacities, equal amounts, final temperature will be
(T1 + T2 ) / 2 = 323 K.
Also, final P =1 atm, i.e., partial pressure of each gas in the final mixture will be 0.5 atm.
Now,
⇒
247.
∆S 1 (for cold gas ) = C P ln
1
323
+ R ln
0.5
273
∆S 2 (for hot gas) = C P ln
323
1
+ R ln
373
0.5
 323 × 323 
–1
∆S = ∆S 1 + ∆S 2 = 2R ln 2 + C P ln 
 = 12.03 JK .
273
373
×


dV
dV
dT
dT
+ nR
= n (C p − R )
+ nR
V
V
T
T

dV 
dT

= n  16.6
+ 14.8 × 10 −2 dT − 9 × 10 −5 TdT  + R


V 
T

dS = nC v
348
Problems in Chemistry
⇒

T
V 
9 × 10 −5
∆S = n 166
. ln 2 + 148
. × 10 −2 (T2 − T1 ) −
(T22 − T12 ) + R ln 2 
T1
V1 
2

= n [5.585 + 29.6 − 10.8 + ( −2.965)] = 42.84 JK –1 .
dG = − SdT
248.
at
∆P = 0
= − (25.1 + 29.3 ln T ) dT , on integration
∆G = − (25.1 − 29.3) ∆T − 29.3 (T2 ln T2 − T1 ln T1 )
Substituting, T1 = 298, T2 = 348.
.
∆G = − 9750 J
249. Since, the process is irreversible adiabatic expansion:
0 = ∆E + Pext ∆V = ∆E + Pext (V2 − V1 )
Also,
dE = nC v dT = (18.8 + 0.021 T ) dT
0.021 2
∆E = 18.8 (T2 − T1 ) +
(T2 − T12 )
2
T 

Substituting in Eq. (i), 18.8 (T2 − T1 ) + 0.0105 (T22 − T12 ) + R  T2 − 1  = 0

2
…(i)
…(ii)
Substituting, T1 = 400 K, solving yields T2 = 352.5 K.
dT
dT
dV
dV
Now,
dS = C v
= (18.8 + 0.021T )
+R
+R
T
T
V
V
T
V
Integrating:
∆S = 18.8 ln 2 + 0.021 (T2 − T1 ) + R ln 2
T1
V1
= 18.8 ln
352.5
352.5 × 10
+ 0.021 (352.5 − 400) + R ln
400
400 × 5
= 1.338 JK –1 mol –1
250.
.
0021
[ (352.5) 2 − ( 400) 2 ] = – 1268.3 J
2
dH = ( R + C v ) dT = (27.114 + 0.021 T ) dT
From Eq. (ii),
. (352.5 − 400) +
∆E = 188
⇒
∆H = 27.114 (T2 − T1 ) +
0.021 2
(T2 − T12 ) = – 1663.22 J
2
∆G = nV ∆P = 5 × 0.089 × 99 L-atm = 4467.18 J
251. For isoentropic process:
0 = nC P ln
T2
P
+ nR ln 1
T1
P2
⇒
P  5
T
R ln  2  = R ln 2
T1
 P1  2
Solving, P2 = 1.474 atm.
252. Under isothermal condition:
∆S = nR ln P1 / P2 where P1 = initial pressure, P2 = final pressure
Let, flask A has N 2 at 4 bar and flask B at 2 bar. After mixing final pressure in the combined system
= (8/ 3) bar.
349
Solutions
⇒
∆S A = R ln
12
8
and
∆S B = R ln
6
8
∆S A + ∆S B = ∆S = 0.98 JK –1 .
∆S system =
253. At boiling point,
∆H vap
Tb
=
349 × 10 3
= 111.4 JK –1
3133
Heat change in surrounding = − 349 kJ
∆S surr = −
⇒
349 × 1000
= − 99.71 JK –1
3500
∆S univ = ∆S sys + ∆S surr = 111.4 − 99.71 = + 11.69 JK –1 .
ATOMIC STRUCTURE
V
nh
=
= 6.5 × 10 15 rps.
2π r 4π 2 r 2 m
 4
 1
1 
4
1 1 
255. (a) ν = KZ 2  2 − 2  ⇒ K  − 2  = K  2 − 2  ⇒ in He, required transition in 4 → 2.


1 2
 n1 n2 
 n1 n2 
254. Number of revolutions per second =
 9
9 
1 1 
(b) K  −  = K  2 − 2 
 1 16 
 n1 n2 
⇒ n1 = 3, n2 = 12
 1
256. E ext = 2.18 × 10 −19 1 −  × 6.023 × 10 23 = 116.71 kJ/ mol H
 9
D. E. = 116.71 × 2.67 = 311.62 kJ/ mol H 2
1
PV
=
= 0.04
n=
.
RT 0082
× 300
⇒
257.
T. E. = 0.04 × 311.62 + 0.08 × 116.71 = 21.8 kJ
n( n − 1)
= 10 ⇒ nB = 5 Hence, n A = 3
2
 1
1 
∆E = KZ 2  2 − 2  = 3.868 eV when
 n1 n2 
KZ 2 = 54.4 eV = I. E.
⇒
258. ν =
n1 = 3, n2 = 5
6.626 × 10 −34
h
=
= 7.28 × 10 6 ms –1
mλ 9.1 × 10 −31 × 10 −10
λ=
259. ∆V =
h
2meV
h
= 8.78 × 10 –11 ms –1
4πm∆x
⇒ V=
h2
2λ 2 me
= 150.7 V
350
Problems in Chemistry
1
8
= RH ⋅
λ
9
260.
EThreshold =
⇒ λ =105.14 nm
6.626 × 10 −34 × 3 × 10 8
hc 6.626 × 10 −34 × 3 × 10 8
=
E
=
,
incident
λ
105.14 × 10 −9
230 × 10 −9
K. E. =
6.626 × 10 −34 × 3 × 10 8  1
1 
–18
−
 = 1.026 × 10 J

−9


105.14 230
10
 1
1 
25.7 = KZ 2  2 − 2  = KZ 2
 n1 n2 
261.
1 1 
 − 2
4 n 
…(i)
1 1 
8.7 = KZ 2  − 2 
9 n 
…(ii)
Solving, Eqs. (i) and (ii) gives : n = 5
KZ 2 = 122.38 = I. E.
Also,
1 

13.6 1 − 2  = 10.93 + x
 n 
262.
1 

54.4 1 − 2  = 49.18 + x,
 n 
where x is work-function.
Solving : n = 4, x = 1.82 eV
∆H
E
hc
263. ∆H = 248 × 10 3 J ⇒
⇒ λ = 482.76 nm
=
= 4.12 × 10 −19 J =
λ
molecule N A
264.
∆H reaction (O 3 → O + O 2 ) = 249 − 142 = 107 kJ/ mol
λ=
hc
⋅ N A = 1119 nm
∆H
4 × 10 −17
hc
= 121
= 3.313 × 10 −19 J ⇒ Number of photons =
λ
3.313 × 10 −19
λ 2λ 3
hc hc hc
266.
λ1 =
=
+
⇒
λ2 + λ3
λ1 λ 2 λ 3
h
3RT
267. r. m.s. ( c) =
= 7.38 × 10 −11 m.
= 1351.67 ms −1 ⇒ λ =
M
mc
hc
268. Energy of photon =
= 3.313 × 10 −19 J
λ
E ( required for melting ) = n ∆H = 166.67 kJ
265. E =
Number of photons = 5.03 × 10 23 .
Also, average energy required/molecule =
6000
= 9.96 × 10 −21 J
NA
Average number of H 2O molecules melted/photon = 33.
351
Solutions
269. 3rd I.E. =
KZ 2
⋅ N A = 11817 kJ/ mol ⇒ IInd I.E. = 19612 − (520 + 11817) = 7275 kJ/ mol.
n2
h
270. ∆V =
= 19.85 × 10 −6 ms −1 .
4π m ∆ x
KZ 2 3
271. E = 2 = K b ⋅ T ⇒ T = 20.636 × 10 6 ° C.
2
n
272. Maximum number of emission lines = n( n − 1)/ 2, where n is orbit number ⇒ nB = 5. Possible
values of n A are 2, 3 and 4. Since only one photon of energy greater than 3.066 eV is emitted, n A
must be 2 and only n = 5 to n =1 transition would emit photon of energy greater than 3.066 V.
Ionization energy ( KZ 2 ) :
1 1 
3.066 = KZ 2  − 
 4 25 
⇒
KZ 2 (IE) = 14.6 eV
273. We know that average energy of an oscillator is
hν
E = hν / kt
e
−1
hν 6.627 × 10 −34 × 1014
=
= 4.8
kT 1.38 × 10 −23 × 1000
Q
E =
⇒
⇒
e 4. 8 = 121.5
6.627 × 10 –34 × 1014
= 5.5 × 10 −22 J
120.5
.
1
1 
K.E. = h (ν − ν 0 ) = hc  −

λ λ0 
274.
⇒
1 E
1
where E = 1.5 eV = 1.5 × 1.6 × 10 −19 J = 2.4 × 10 −19 J
=
+
λ hc λ 0
⇒
2.4 × 10 −19
1
1
=
+
34
8
–
λ 6.626 × 10 × 3 × 10
23 × 10 –8
Solving, λ = 1815
.
× 10 −7 m = 1815 Å
λ′ − λ =
275.
⇒
λ
(1 − cos θ ) where λ′ = wavelength of scattered beam = 0.22 Å.
mc
λ = λ′ −
6.626 × 10 −34
h
(1 − cos θ ) = 2.2 × 10 –11 −
(1 − cos 45° ) = 0.2129 Å.
mc
9.1 × 10 −31 × 3 × 10 8
276. de-Broglie equation is:
λ=
or
V=
h2
2meλ 2
=
h
h
=
mv
2meV
⇒
2meV =
h2
λ2
(6.627 × 10 −34 ) 2
2 × 9.1 × 10 −31 × 1.6 × 10 −19 × (3.8 × 10 –11 ) 2
=1044 volt.
352
Problems in Chemistry
mvr =
277. Since,
nh
2π
⇒
v=
nh × Z
6.626 × 10 −34 × 2
hZ
nh
= 1.457 ×10 6 ms –1
=
=
=
2πmr 2πma 0 n 2 2πma 0 n 2π × 9.1 × 10 −31 × 53 × 10 −12 × 3
⇒
h=
h
6.626 × 10 −34
=
= 5 × 10 −10 m = 5 Å.
mv 9.1 × 10 −31 × 1.457 × 10 6
278. The Heisenberg’s uncertainty equation is
h
∆x⋅∆p =
⇒
4π
⇒
∆v ( min ) =
∆x⋅∆v =
h
4πm
h
6.626 × 10 −34
= 0.58 ms –1
=
4πm ∆x ( max ) 4π × 9.1 × 10 −31 × 10 −4
279. The uncertainty in velocity is ∆v = 0.001 km hr –1 = 2.778 × 10 –4 ms –1 .
⇒ The uncertainty in position:
∆x =
6.626 × 10 −34
h
=
= 3.796 ×10 –34 m.
4πm∆v 4π × 500 × 2.778 × 10 –4
The uncertainty in measurement of position ( ∆x = 3.796 × 10 −34 m ) is wholly negligible as
compared to mass and velocity of moving automobile.
hc 6.626 × 10 −34 × 3 × 10 8
280. Energy of incident photon =
=
= 9.94 × 10 –16 J
–12
λ
200 × 10
Energy of emitted electron =
1
1
mv 2 = × 9.1 × 10 −31 (2 × 10 7 ) 2 = 1.82 × 10 −16 J
2
2
⇒ Binding energy of electron = Energy of incident photon – Energy of emitted electron
= 8.12 × 10 –16 J = 5075 eV.
281. Stopping potential is the maximum kinetic energy of photoelectron:
⇒
K.E. (max) = eV = 0.5 × 1.6 × 10 −19 = 8 × 10 −20 J
Energy of photon =
hc 6.626 × 10 −34 × 3 × 10 8
=
= 7.835 × 10 −19 J
−9
λ
253.7 × 10
Work-function = 7.835 × 10 −19 − 8 × 10 −20 = 7.035 × 10 −19 J = 4.4 eV.
282. The maximum uncertainty in position of electron inside the nucleus = 10 −14 m.
⇒ minimum uncertainty in momentum
( ∆ p) =
283. Average
T.E. =
h
= 0.527 × 10 −20 kg ms –1 .
4π ∆ x
3
3
kT = × 1.38 × 10 −23 × 300 = 6.21 × 10 −21 J
2
2
353
Solutions
⇒
p (momentum) = 2mE = 2 ×
λ=
h 6.626 × 10 −34
= 1.46 × 10 −10 m = 1.46 Å.
=
p 4.540 × 10 −24
Potential energy =
284.
10 –3
× 6.21 × 10 –21 = 4.54 × 10 −24 kgms –1
N0
(1.6 × 10 –19 ) 2
e2
= 7.69 × 10 −19 J
=
4πε 0 r 4πε 0 × 3 × 10 −10
285. Probability (P) between 0 and 2a 0 = ∫
2a 0
ψ12s dτ where dτ = 4πr 2 dr.
0
∴
P =∫
2a 0
0
=−
( πa 03 ) −1
e
−2 r/a 0
4
2
4πr dr =
a 03
2a 0
∫0
−
e
2r
a0
r 2 dr
 –4 e − 4 1 
a 03 −4 a 03 
4  3 −4
3 −4
 3e +
−  = 0.7624
a
e
a
e
e
2
=
−
4
+
+
−


0
0
4
4
4
4 

a 03 
286. The radial distribution function for 1s orbital is
R 2 r 2 = 4πr 2 ψ12s
⇒
⇒
For maximum,
d
−2 r/a 0
[ 4πr 2 ( πa 03 ) −1 e
]=0
dr
⇒
r
r
0
0
2r −
Probability = ∫ ψ12s dτ = ∫ ( πa 03 ) −1 e
287.
P=
4
a 03
r
∫0 r
2 −2 r/a 0
e
d[ R 2 r 2 ]
=0
dr
2r 2
= 0 or
a0
−2 r/a 0
r = a0 .
4πr 2 dr
dr
Integrating by parts and substituting r = 10 −12 m and a 0 = 53 × 10 −12 m.
P = 9 × 10 −6 .
⇒
v rms =
288.
3RT
3 × 8.314 × 298
= 1360 ms –1
=
M
4 × 10 −3
de-Broglie equation:
λ=
h
6.626 × 10 −34
=
× 6.023 × 10 23 = 7.34 × 10 –11 m.
mv 4 × 10 −3 × 1360
3
 1
Transition energy = 2.18 × 10 −18 1 −  = 2.18 × × 10 −18 J = 1.63 × 10 −18 J.
 4
4
289.
Translational energy =
⇒
T=
3
kT = 1.63 × 10 −18
2
2 × 1.63 × 10 –18
3 × 1.38 × 10
−23
= 78.744 × 10 3 K .
354
Problems in Chemistry
290. The reduced mass of moving electron
me ⋅ m positron
m
= e
(µ ) =
me + m positron
2
rn = n 2
⇒
4π ε 0 h 2
( me / 2) e 2 Z
= (1.0589 × 10 −10 m ) n 2
r1 = 1.0589 × 10 −10 m = 1.0589 Å.
⇒
CHEMICAL BONDING
291. CO 2 : O==C==O Linear, non-polar molecule.
COS : S==C==O Linear, polar molecule.
Ο

Trigonal planar ion with sp 2 -hybridized carbon. It exists
C
CO 2−
3 :
O
−
in three resonating forms.
O
−
Cl
SiCl4 :
Si
Cl
Cl
Tetrahedral molecule with sp 3 -hybridized silicon.
Cl
Pentagonal bipyramidal with sp 3 d-hybridized phosphorus.
Equatorial bond angle = 120°, Axial-equatorial bond angle = 90° .
Cl
Cl
PCl5 : Cl
P
Cl
H
NH4+
:
Cl
+
N
H
H
H
PH +4 :
Similar to NH +4 .
PCl +4 :
Similar to NH +4 .
Cl
Cl
Cl
PCl6– :
P
Cl
SF6 :
Tetrahedral ion with sp 3 -hybridized nitrogen. All bond angles
are equal and 109° 28′ .
Cl
Same as PCl −6 .
Cl
–
Square pyramidal ion with sp 3 d 2 -hybridized phosphorus.
355
Solutions
F
F
IF7 :
F
72°
F
Pentagonal bipyramidal with sp 3 d 3 -hybridized iodine.
I
F
F
••
292. SO 2 :
Angular (V-shaped) with sp 2 -hybridized sulphur.
S
O
F
O
NH3 :
N
H
Trigonal pyramidal with sp 3 -hybridized nitrogen.
H
H
••
H 2O :
Angular (V-shaped) with sp 3 -oxygen.
O
••
H
H
PCl3 :
P
Cl
Trigonal pyramidal with sp 3 -phosphorus.
Cl
Cl
NCl 3 : Same as PCl 3 .
293. (a) If every things are same, and central atoms are forms same group, bond angle decreases on
descending down the group.
NH 3 > PH 3 > AsH 3 > SbH 3 > BiH 3
(b) H 2O > H 2S > H 2Se > H 2Te
294. AsH 3 < H 2Se < PH 3 < H 2O < NH 3 .
F
295. ClF3 : F
T-shaped with sp 3 d-hybridized chlorine. (In trigonal
bipyramidal geometry, lone pairs are stable at equatorial
positions).
Cl
F
O

C
COCl 2 :
Cl
Cl 2O :
Triangular planar with sp 2 -carbon.
Cl
Two shape can be predicted as :
••
••
O
Cl
••
••
Cl
Cl
V-shaped with oxygen at centre
Cl
••
•
O
•••
V-shaped with chlorine at centre
356
Problems in Chemistry
••
OF2 :
V-shaped with sp 3 -oxygen.
O
••
F
F
Cl
Cl
ICl4– :
–
I
Square planar with sp 3 d 2 -iodine.
Cl
Cl
F
F
F
IF5 :
I
F
Square pyramidal sp 3 d 2 -iodine.
F
O
Cl
IOCl4– :
Cl
Square pyramidal with sp 3 d 2 -iodine.
I
Cl
Cl
Cl
Cl
IOCl4+ :
O
••
+
I
Trigonal bipyramidal with sp 3 d-iodine.
Cl
Cl
296. N −3 :
–
+
•• −
N ≡≡ N → N •• Linear with sp-hybridized nitrogen.
I
I3– :
••
–
Linear with sp 3 d-hybridized central iodine.
I
I
F
F
BrF5 :
F
Square pyramidal with sp 3 d 2 -bromine.
Br
F
F
F
IF4– :
F
–
Square planar with sp 3 d 2 -iodine.
I
F
F
F
BF4– :
B
F
F
F
–
Tetrahedral with sp 3 -boron.
357
Solutions
ICl +2 :




••
I
••
Cl
Cl
+




V-shaped with sp 3 -iodine.
Cl
–
ICl2–
I
:
Linear with sp 3 d-iodine.
Cl
F
F
PF5 :
F
P
Trigonal bipyramidal with sp 3 d-phosphorus.
F
F
297. Relative electron withdrawing power of nitrogen is more in NCl 3 than in NOCl therefore
homolytic cleavage of N—Cl bond would be difficult in NCl 3 .
F
–
+
F
As
AsF2+ :
298. AsF5 : F
As
SnCl3– :
Sn
F
F
Cl
F
Cl
Cl
F
Trigonal pyramidal with sp3-Sn
V-shaped with sp2-As
Trigonal bipyramidal
F
F
F
NOF : O==N
SO32– :
S
O
2
Angular with sp -N
F
TeF5– :
O
O
Te
F
F
Square pyramidal
with sp3d2-Te
Tetrahedral
Cl
–
GeF3–
:
Ge
F
SCl2 :
F
F
3
299. ClF4+ :
F
F
SbCl6– :
F
FClO3 :
F
Cl
V-shaped with sp -S
+
Cl
Sea Saw
S
Cl
Cl
O
O
O
Cl
Cl
3
Trigonal pyramidal with sp -Ge
F
–
Tetrahedral with sp3-Cl
–
Sb
Cl
Cl
Cl
Octahedral with sp3d 2-Sb
358
Problems in Chemistry
O
F2ClO+ :
+
SeO32– :
Cl
F
Se
O
O
F
O
Tetrahedral
Trigonal pyramidal
300. (a) BF3 (µ = 0) = PCl 5 (µ = 0) < POCl 3 < SCl 2 < ICl 3
(b) XeF2 (µ = 0) = SnCl 4 (µ = 0) < PCl 4 F < XeO 3 < SF4
301. Li 3 N < Li 2O < LiF (Fajan’s rule)
302. (a) Al 2 (CO 3 ) 3 < BeCO 3 < MgCO 3 < CaCO 3 < BaCO 3 < Na 2CO 3
(b) CaSO 4 < CaSO 3 < CaCO 3
(c) Al 2 (SO 4 ) 3 < MgSO 4 < SrSO 4 < BaSO 4
(d) FeC 2O 4 < CaC 2O 4 < CaCO 3 < K 2C 2O 4
303. Symmetrical repulsion by the two lone pairs from axial positions cancels the effect of one-another
and bond angles remains intact.
304. AlCl 3 < GaCl 3 < MgCl 2 < CaCl 2 < BaCl 2 < NaCl (Fajan’s rule).
305. (a) NH 3
(b) PF3
(c) AsF3
306. (a) H 2S < PH 3 < H 2O < NH 3
(d) AsCl 3
(b)
GeH –3
(e) H 2O
< SiH –3 < CH –3
F
F
F
Cl
307. (a) F
Cl
(b) Cl
P
(c) Br
P
Cl
Cl
F
Cl
–
I
[Cl
P
Cl
F
F
(d)
Cl
Cl]
+
(e)
I
3
V-shaped with sp -iodine
Cl
Linear
308. PCl 3 F2 (µ = 0) < PF3Cl 2 < PBrF2Cl 2
309. CCl 4 < CHCl 3 < CH 2Cl 2 < CH 3Cl
F
O
310. (a) O
Cl
F
F
Trigonal bipyramidal
with sp3d 2-Cl
O
F
F
(b)
Xe
F
F
Square pyramidal
with sp3d 2-Xe
359
Solutions
O
Cl
Cl
(c)
+
–
I
(d)
I
Cl
Cl
Cl
Cl
V-shaped with sp3-I
Square pyramidal
with sp3d 2-I
311. In solid state, PCl 5 remains as dipolar ion as [PCl 6 ]– [PCl +4 ]
Cl
Cl
Cl
Cl
–
P
P
Cl
Cl
Cl
Cl
Cl
Cl
+
Cl
Cl
In gas phase PCl 5 remains as isolated gas molecule :
Cl
P
Cl
Cl
312. Due to back bonding of a lone pair from p-orbital of oxygen to vacant d-orbital of Si ( pπ − d π
bonding).
313. (a) Linear (b) Square planar (c) Octahedral (d) T-shaped (e) Square pyramidal
(f) Sea-saw (g) Trigonal bipyramidal
CH3
314. (a) F3C
Cl
I
Cl
CH3
CF3
(b) H3C
P
CH3
(h)
CF3
Cl
Cl
I
Cl
Cl
CF3
P
CH3
CF3
Greater electronegativity when bonding through axial positions.
Pπ − Pπ back bonding in BF3 gives some double bond character, which is absent in BF4− .
Pπ − d π back bonding occurs between Si and Cl, which is absent in CCl 4 .
BF3 < BCl 3 < BBr 3 < BI 3
Lone pair of nitrogen in N(SiH 3 ) 3 is involved in pπ − d π back bonding, not available for donation
to a Lewis acid.
H
H
H
Br
320.
C==C
C== C
H
Br
Br
Br
315.
316.
317.
318.
319.
µ≠0
µ=0
321. d (µ = 0) = b(µ = 0) < c < a
322. Due to larger size of 3 p-orbital with Si, there is fewer chance of sidewise overlap giving π-bonds.
360
Problems in Chemistry
F
323.
F
F
N==N
N==N
µ≠0
µ=0
F
324. Due to free rotation about C—C sigma bond, 1,2-dichloro ethane acquire the anti (most stable)
conformation which is non-polar. In dichloroethane, rotation is restricted.
Cl
H H
Cl
Cl
C==C
H H
Cl
H
H
due to restricted
rotation, µ ≠ 0
anti conformation of
1,2-dichloro ethane (µ = 0)
325. In ClHC==C==CHCl, the two C—Cl bonds are in perpendicular plane and hence the molecule is
polar.
In HClC==C==C==CHCl, the two C—Cl bonds are in the same plane and individual C—Cl
dipoles are at 180°, cancelling one another giving zero dipole moment.
Cl
Cl Cl
H H
Cl
326.
C==C
C==C
C==C
H
H H
Cl H
Cl
I
II
III
Dipole moment: II < III < I.
N
327. (a)
N
N
N
(b)
N
N
N
N
N
N
N
N
328. NaF has higher lattice energy than NaCl. Lattice energy is inversely proportional to square of the
interionic distance. Since, fluoride ion is smaller in size than chloride ion, NaF will have higher
lattice energy.
329. The NO 2 exist in two equivalent resonance form as:
●
O
N
●
O
O
N
●
O
O
N
O
both N—O bonds are
identical, therefore equal
bond energies.
• •
• N ==O
It has a one sigma and one pi-bond and bond order is two whereas in NO 2 , the N  O
bond order is less than two. Therefore, N  O bond energy in NO is greater than in NO 2 .
330. Both NO and NO 2 have one odd (unpaired) electron. When the two molecule approach each other,
unpaired electron may be shared to form a covalent bond as:
ON + NO 2 → ON NO 2
both N have complete octet.
361
Solutions
O
O
N—N
Shape of N2O3 :
O
F
O
331. (a)
+
O==S
(b)
S
Cl
F
Triangular planar
F
S
F
One S-is sp3d and other is sp3 hybridized.
332. N has higher electronegativity than C, therefore, energy of orbitals of N is less than that of carbon
and energy level diagram will be as follows:
C(2p)
N(2p)
C(2s)
333. (a)
(b)
(c)
N(2s)
SiF4 is tetrahedral, non-polar molecule while PF3 is trigonal pyramidal, polar molecule.
Therefore, PF3 will be more soluble in polar solvent.
SF6 is square bipyramidal, non-polar molecule while SF4 is a seasaw shaped, polar molecule.
Therefore, SF4 will be more soluble in polar solvent.
IF5 is a square pyramidal, polar molecule while AsF5 is a trigonal bipyramidal, non-polar
molecule. Therefore, IF5 will be more soluble in a polar solvent.
334.
π-bond
σ-bond
CHEMICAL EQUILIBRIUM
335. K =
0.5 × 0.2
2
=
9 0.5(0.2 + x )
⇒ x = 0.7 mol
δ-bond
362
Problems in Chemistry
336. H + B
0.1
0.08
HB
0.1
0.08
337. NH 4Cl( s)
0
0.02
Initial
At equilibrium ⇒ K c = 3.125.
NH 3 ( g ) +
HCl( g )
3.66 × 10–2
3.66 × 10−2
(0.154 – x ) (3.66 × 10–2 – x )
moles at equilibrium K c = 134
. × 10−3
after adding 2.0 g NH3.
⇒ K c = 1.34 × 10 −3 = (0.154 − x )(3.66 × 10 −2 − x ).
Solving, x = 0.163 and
0.026
x cannot be greater than 3.66 × 10 −2 , hence it is 0.026.
Therefore, on adding 0.2 g NH 3 , 0.026 moles of both HCl and NH 3 will combine forming equal
amount of NH 4Cl( s) before equilibrium was re-established.
Thus, amount of NH 4Cl( s) at new equilibrium =1.431g;
% dissociation = 28.45.
2
 0.63 + 2x 
338. K = 6.05 = 
⇒ x = 0.134 ⇒ [H 2 ] = [I 2 ] = 0.366, [HI] = 0.898.

 0.5 − x 
339. (CH 3 ) 3 CCl
(CH 3 ) 2 C==CH 2 + HCl
0.2 – x
x
Molarity
x
K
x2
KC = P = 0.084 =
RT
0.2 − x
340. CO 2 ( g ) + C( s)
⇒
x = 0.094
2CO
79.2 − x
56
x
44
56
x − x = 2.3 ⇒ x = 8.433 g
44
PCO( g ) = 6.286 atm, PCO2 ( g ) = 26.38 atm ⇒ K P (1000 K) = 1.49 atm
56
At 1100 K :
x − x = 5.3 ⇒ x =19.43 g.
44
PCO( g ) = 15.93 atm, PCO2 ( g ) = 24.5 atm; ⇒ K P (1100 K) = 10.35 atm
K (2) ∆H  T2 − T1 
Also
=
ln P
 ⇒ ∆H = 177.25 kJ

K P (1)
R  T1T2 
At 1000 K
341. N 2 + 3H 2
1− x
1 − 3x
2NH 3 Total
2x
2(1 – x )
⇒
KP =
0.85
⇒
(1 − 3x )3 p 2
≈
16x 2
P2
= 64x 2
x = 9.68 × 10 −3 , x NH 3 = 9.78 × 10 −3 .
⇒
342. H 2O( g )
16x 2 (1 − x )
1
H 2 ( g ) + 2 O 2 ( g ) Total moles
0.15
0.075
1.075
K P (T ) = 4.66 × 10 −2
363
Solutions
 4.66 × 10 −2  57.8 × 10 3  T − 1000 
=
ln 


−11 
 1000 T 
2
 8.7 × 10 
⇒
H2 + I 2
343.
x
2
3– x
1−
2HI
x
2
3– x
3−
x
2x
x2
4x 2
=
x  (3 − x ) 2
 x 
1 −   3 − 
 2 
2
⇒
⇒ T = 3284 K
⇒ x = 1.5, K = 4.
344. Iodine in the gaseous state = 0.0386 mole; Iodine in solid state = 0.0114. Minimum H 2 ( g ) would
be required when solid I 2 ( s) has just been exhausted and the following equilibrium is maintained :
H 2 (g ) + I 2 (g )
2HI ( g )
moles at equilibrium
0.0386
x
20 =
⇒
0.0228
2
(0.0228)
x (0.0386)
⇒ x = 0.67 × 10 −3
⇒
moles of H 2 required = 0.012
0.0492
345. I 2 + I –
I −3 K =
= 745.57 L mol −1
−3
1.299 × 10 × 0.0508
Adding water will drive the reaction in backward direction and solubility will decrease.
346.
A(g )
B (g ) + C (g )
0.46 − x
⇒
3=
x
(0.46 + x )(0.082) × 300
5
K P (300 K) =
⇒
x
 0.15 


 0.61 
⇒ x = 0.15
2
 0.31


 0.61
× 3 = 0.357
K P (320 K) = 1.145
K P (350) ∆H 


20
30
1.145 ∆H 
=
=
ln


 and ln

R  300 × 320 
1.145
0.357
R  350 × 320 
Similarly,
Now,
K P (350) = 5.12, P0 (350) = 2.64
Solving,
⇒
Also,
5.12 =
2
P
2.64 − P
⇒ P =1.92
⇒
α(350 K) =
0.15
= 0.326
0.46
0.4012
× 100 = 123.06%
% increase =
0.326
α(300 K) =
1.92
= 0.7272
2.64
364
Problems in Chemistry
347. P(NH 3 ) =175 bar
175
= 43.75 bar
4
= 3 × 43.75 = 131.25 bar
⇒ PN 2 =
and
PH 2
⇒
KP =
(175) 2
(43.75)(131.25)
3
= 3.096 × 10 −4 bar −2
348. Let r be the ratio of P (Cl 2 ) to P (Br 2 ). Thus, P (Cl 2 ) = rP ( Br 2 )
⇒ Pt = P (Cl 2 ) + P ( Br 2 ) + P ( BrCl) = rP ( Br 2 ) + P (Br 2 ) + P (BrCl) = (1 + r ) P ( Br 2 ) + P ( BrCl)
P − P (BrCl)
 P − P (BrCl) 
PBr2 = t
, P (Cl 2 ) = r  t
⇒



r +1
r +1
⇒
KP =
P (BrCl) 2
  Pt − PBrCl    Pt − P (BrCl) 

 
r 

r +1
  r +1  
=
P (BrCl) 2
2
[ P − P (BrCl)]
r t

( r + 1)


Taking logarithm of each side and differentiating w.r.t. r :
ln K p = 2 ln [ P (BrCl)] − ln r − 2 ln [ Pt − P (BrCl)] + 2 ln ( r + 1)

  − ∂P (BrCl) 
1
1
1
 ∂P (BrCl)  1
0=2

 − −2
 +2

 r

r +1
∂r
∂r
P (BrCl) 
 Pt − P (BrCl)  
∂P ( BrCl)
= 0 and solving for ‘ r’ gives r =1
∂r
Hence, stoichiometric amounts of Cl 2 and Br 2 generate maximum yield of BrCl.
349.
F2
2F
Assigning
0.06 − x
2x
2
 2x   0.06 + x 
4x 2
× 2.07 = 9.5 × 10 −3
⋅
KP = 
P
=
 

2
−4
 0.06 + x   0.06 − x 
36 × 10 − x
⇒ x = 2.03 × 10 −3 ⇒ mole fraction of F = 0.065, mole fraction of F2 = 0.935
350.
N 2 (g ) + O 2 (g )
0.8 − x
2NO( g )
0.2 − x
4x 2
K=
= 4 × 10 −4
(0.8 − x )(0.2 − x )
(2x )
⇒ x = 3.95 × 10 −3
⇒ mole fraction of NO( g ) = 0.08, N 2 ( g ) = 0.796, O 2 ( g ) = 0.196
351.
Br 2 + 3F2
0.25
1.497
K=
0.75
1.497
(0.497 × 1.497) 2
0.25(0.75) 3
2BrF3
0.497
1.497
= 5.248
After increasing equilibrium pressure = 2 bar, reaction will shift in forward direction and let x mole
of Br 2 reacted further.
365
Solutions
Then new partial pressure : Br 2 =
0.25 − x
0.75 − 3x
× 2; F2 =
×2
1.497 − 2x
1.497 − 2x
BrF3 =
0.497 + 2x
×2
1.497 – 2x
Substituting these partial pressure in equilibrium constant expression and solving gives x = 0.061.
Therefore, new compositions would be:
Br 2 = 0.189, F2 = 0.567, BrF3 = 0.619 moles.
352. Ag 2CO 3 ( g )
Ag 2O( s) + CO 2 ( g ) K P = 0.0095 bar at 120°C.
P (CO 2 ) at 120° C = K P = 0.0095 which is less than 0.01 bar, the partial pressure of CO 2 in air.
Therefore, Ag 2CO 3 ( s) will not effervess at 120°C.
353. At equilibrium : 2 A + B
3C + 2D
0.4
1.7
0.9
16
.
K c = 6.86
P
354.
PCl 5
PCl 3 + Cl 2 K P =
3
Let x mole of Cl 2 is added out of which y mole reacted with PCl 3 , giving y mol PCl 5 . Also, at new
equilibrium n =12.
⇒
6 + x − y = 12
⇒
x − y=6
2
20
P 2P (2 − y)
and
and
=
⇒
y=
x=
3
3
3
3(2 + y)
355. In the combined system partial pressure of the gases before any reaction occurred are :
250
100
NO = 0.46 ×
= 0.3285; O 2 = 0.86 ×
= 0.2457
350
350
Here, NO is limiting reagent : Partial pressure after 1st step :
NO = 0, O 2 = 0.0814, NO 2 = 0.3285
Now, let p-be the decrease in partial pressure of NO 2 due to equilibrium.
⇒ 0.0814 + 0.3285 − p +
p
= 0.37 ⇒
2
p = 0.0798 ⇒ K P = 0.645 atm –1
356. Let us consider P0 be the initial partial pressure of each Cl 2 ( g ) and F2 ( g ). Then at equilibrium
partial pressure of each gas would be :
PCl 2 = P0 − 0.12, PF 2 = P0 − 0.16
3.2 =
⇒
P (ClF) 2
(0.2) 2
=
P (Cl 2 ) P (F2 ) ( P0 − 0.12)( P0 − 0.16)
Solving, gives P0 = 0.253 and 0.026, where the second value is not acceptable.
⇒ K p for second equilibrium =
357. Cl 2 + F2
0.0665
0.0665
2ClF
0.267
(0.04) 2
(0.133) (0.093) 3
K =16.12
= 14.95
366
Problems in Chemistry
When Br 2 is added :
Cl 2
0.075 − x
+ F2
2ClF
0.075
16.12 =
0.25
2
(0.25)
0.075(0.075 − x )
⇒ x = 0.023
Cl 2
+ Br 2
Kc =
(0.046) 2
= 0.528
0.052 × 0077
.
0.07 − x
2BrCl
0.1 − x
2x
2NOBr. Let initial partial pressure of NO = p0 and decrease in partial pressure
358. 2NO + Br 2
before the 1st equilibrium be p1 . Then :
p1
+ p1 + 1 − p0 = 2.12
2
p0 − p1 + 1.25 −
⇒
p1 = 0.26 and
Kp =
(0.26) 2
( P0 − 0.26) 2 (1.12)
Similarly, if p2 be the total decrease in partial pressure of NO before the second equilibrium was
established then
p0 − p2 + 21.75 −
⇒
p2 = 0.5 and
Equating the two K P ; P0 = 0.69 and
359. Before addition of Br 2 ( v ) :
Cl 2 ( g )
0.492 bar
⇒
−
p2
+ p2 + 1 − p0 = 22.5
2
KP =
(0.5) 2
( p0 − 0.5) 2 (21.5)
K P = 0.326.
+ I 2 (g )
K1
2ICl
2x
0.03
x
0.522 + x = 0.767 ⇒ x = 0.245 and
K 1 = 32.4
After adding Br 2 ( v )
Cl 2 ( g )
0.247 + q − p
I 2 (g ) +
0.03
Also,
⇒
K1
+ I 2 (g )
Br 2 ( g )
K2
0.492 – 0.1476 − p
2IBr ( g )
0.2952
P (Br 2 ) = 0.2 ⇒
K2 =
2ICl( g )
0.49 − 2q
0.03
p = 0.1444
2
(0.2952)
= 14.52
0.03 × 02
367
Solutions
Cl 2 ( g )
+
0.247 + q − p
⇒
2ClBr ( g )
0.492 − 0.1476 − p
K1 = 32.4 =
Also,
K2
Br 2 ( g )
2p
2
(0.49 − 2q )
(0.03) (0.1026 + q )
q = 0.05 ⇒ K 3 =
(0.2888) 2
= 2.73
(0.1526)(0.2)
4x 2
⇒ 2x ( moles of AB ) = 1.87
= 50
(1 − x )(2 − x )
361. m mol of I 2 reacted with hypo = 0.785
360. K =
⇒ At equilibrium : m mol of H 2 = I 2 = 0.785, m mol of HI = 5.93
2
 0.785 
–3
K =
 = 17.52 × 10
 5.93 
362. Let the equilibrium partial pressure of NO 2 = H 2O( v ) = 2 p1
Equilibrium pressure of N 2 = p2 and equilibrium pressure of NH 3 = 2 p3
⇒ Equilibrium pressure of H 2 = 2 p2 − p3 . Then,
N 2 H4
1 − p1 − p2 − p3
+ 3O 2
2NO 2 + 2H 2O
1 − 3 p1
2 p1
2 p1
N 2 + 2H 2
N 2 H4
1 − p1 − p2 − p3
2 p2 − p3
p2
N 2 H4
N2
1 − p1 − p2 − p3
2 p2 − p3
+ 2NH 3
2 p3
K1 = 3
K p (2) = ?
K P (3) = ?
9
1
bar and 2 p3 =
bar
19
38
3
11
and
= − p2
PO2 =
bar
4
38
2 p1 =
Given,
⇒
PN 2H 4
⇒
K1 = 3 =
9
 
 19 
4
3
 11 
  (0.75 − p2 )
 38 
Solving p2 = 0.233
Using values of p1 , p2 and p3 gives p(N 2 H 4 ) = 0.517, p(N 2 ) = 0.233
and
p(H 2 ) = 0.4528
p(NH 3 ) = 0.0263 bar.
⇒
K p (2) = 9.24 × 10 −2 bar 2 and K p (3) = 2.95 × 10 −3 .
2NOCl
363.
P:
mole :
0.64
2NO
0.24
15.2 × 10–3
5.7 × 10−3
+
Cl 2
0.12
2.85 × 10−3
K P = 16.875 × 10 −3 bar .
Total = 23.75 × 10 −3
368
Problems in Chemistry
Let at this stage x mole of Cl 2 is added, out of which y mole react back. Also volume is increased to
1.5 litre but at same temperature and pressure, therefore new total mole
= 23.75 × 1.5 × 10 –3 = 35.625 × 10 –3 .
A new equilibrium : mole Cl 2 = 2.85 × 10 –3 + x − y
NO = 5.7 × 10 −3 − 2 y and
NOCl = 15.2 × 10 −3 + 2 y
Adding moles and equating to total moles gives x − y = 11.875 × 10 −3
Therefore, at new equilibrium, mole-fractions are:
Cl 2 = (2.85 +11.875) × 10 −3 /35.625 × 10 −3 = 0.4133
NO =
5.7 × 10 −3 − 2 y
35.625 × 10 −3
and
NOCl =
15.2 × 10 −3 + 2 y
35.625 × 10 −3
Substituting in expression of K p gives y = 1.108 × 10 −3
⇒ x = 12.983 × 10 –3 mol
N 2O 4
364.
At P =1, ρ = 2.33 : Applying,
2NO 2
Kp =
4α 2 p
1−α2
ρRT
M
where M = molar mass of N 2O 4 .
=
1+α
P
⇒ α = 0.6 and K p = 2.25.
Let β-be the degree of dissociation at new equilibrium. Then,
M 2.33RT
=
1.6
1
and
M
5.08RT
=
1+β
P
⇒ 1 + β = 0.734 P
4β 2 P
4β 2
=
(1 − β)(1 + β) 0.734(1 − β)
⇒ β = 0.468 and
At P =1.5, using K p ; α = 5.22 and ρ =
92 × 1.5
= 3.685 g/ L
1.522RT
⇒ 2.25 =
P = 2 bar.
P2
2(1 − P )
where, P is partial pressure of H 2 ( g ) at new equilibrium.
0.356
= 0.21
⇒
P = 0.356 and α =
1.644
PV
366. (a) n =
= 0.05, i.e., out of 0.03 mole of PCl 5 , 0.02 mol has decomposed into PCl 3 and Cl 2
RT
giving 0.05 mole of gaseous mixture.
2
⇒
α = = 0.67
3
365. H 2S( g )
H 2 ( g ) + S( g )
K p = 0.099 =
369
Solutions
(b) From the information in part ‘ a’; K p =
PCl 5
⇒
367.
I:
0.8 =
PCl 3 + Cl 2
p
p2
2.73 − p
⇒
0.17
0.07
0.03
K1 = 7.563 × 10 .
H 2 ( g ) + CO 2 ( g )
H 2O( g ) + CO( g )
0.17 + y
− y
+ y
n( H 2 ) = 0.01
K1 = 7.563 × 10 −2 =
x = 0.06
⇒
K2 =
0.09
= 9;
0.01
0.09(0.03 – y)
0.01(0.17 + y)
y = 2.83 × 10 −2
K3 =
0.17 + 2.83 × 10 –2
0.03 – 2.83 × 10 –2
2NOBr
n = 0.01 +
x
x/ 2
x PV
=
2 RT
x = 0.0074
If 5 g of NOBr is taken, initial mole =
= 116.64
2NO + Br 2
0.01 – x
Solving,
Co( s) + CO 2 ( g )
0.03 + x = 0.09 ⇒
Solving,
368.
Co( s) + H 2O( g )
+x
K3
CoO( s) + CO( g )
⇒
0.03 − y
0.03 + x
K2
−x
Also,
0.03
–2
CoO( s) + H 2 ( g )
Hence,
p =1.13 atm.
1.13
= 0.42
2.73
H 2 ( g ) + CO 2 ( g )
H 2O( g ) + CO( g )
0.07 − x
Also
p
α=
⇒
II :
= 0.8
1−α2
2.73 − p
P:
⇒
α 2P
⇒ α=
x
= 0.74
0.01
5
= 0.045
110
 α
Moles at equilibrium = 0.045 1 +  = 0.045 × 1.37 = 0.06165

2
370
Problems in Chemistry
nRT
= 0.753 atmosphere.
V
4HCl + O 2
2Cl 2 + 2H 2O
P=
369.
0.405
0.905
0.12
0.905
Equilibrium
Kp =
0.19
0.905
4
(0.19) (0.905) 1
⋅
(0.12) 4 (0.405) P
Also
0.19
0.905
mole fraction
P = P0 (0.905) = 0.905 atm.
4
 0.19   1 
−1
=
 = 15.5 atm .
 
 0.12   0.405 
N 2O 5
370.
N 2O 3 + O 2
x− y
4−x
N 2O 3
Kc
x− y
K = 4.5 =
x+y
N 2O + O 2
( x + y)( x − y)
4−x
x+ y
y
and
x + y = 4.5
5
17
and x =
3
6
4.5 × 5 × 6
Kc =
= 6.428.
3×7
y=
⇒
371. Let initial volume = 1 L.
N 2O 4
0.487
2NO 2
0.0475
K c = 4.633 × 10 −3
After compression moles : N 2O 4 = 0.487 + x; NO 2 = 0.0475 − 2x
⇒
Solving, x = 0.041
Kc =
and
(0.0475 − 2x ) 2 × 2
= 4.633 × 10 −3
(0.487 + x )
0.0068
The 1st value of x is rejectable, therefore [NO 2 ] = 0.0678, [N 2O 4 ] = 0.9876
372. Let the initial partial pressure of NOCl be p0 .
2NOCl
p0 − p1
2NO + Cl 2
p1
p1
= 1.2 ⇒
2
At equilibrium, additional pressure of Cl 2 = 7.8
Total P = 1 +
p1
2
p1 = 0.4
371
Solutions
Therefore, at new equilibrium :
2NO + Cl 2
2NOCl
p0 − 0.4 + p2
0.4 − p2
8−
p2
2
p2
= 8.9 ⇒ p2 = 0.2
2
Now, equating equilibrium constant for the above two stages :
Total P = 9 −
0.2(0.4) 2
=
( p0 − 0.4) 2
2
 p − 0.2 
7.9 × 0.04
⇒  0
 =
−
0.4
0.2
× 0.16
p

 0
7.9(0.2) 2
( p0 − 0.2) 2
⇒
p0 = 0.493
K P = 3.7 bar.
373. The equilibrium reaction is :
PCl 3 ( g ) + Cl 2
PCl 5 ( g )
Kp =
α
2
1−α2
P = 1.78 ⇒ P = 4.8 (Pressure due to PCl 5 , PCl 3 and Cl 2 )
⇒
Now,
PSolvent = 0.7
P (PCl 5 + PCl 3 + Cl 2 ) 0.1(1 + α )
=
P (solvent)
nSolvent
⇒ nSolvent = 0.022,
mass = 3.388 g.
374. At equilibrium P (Cl 2O 7 ) =10, P (O 2 ) = 70, P (Cl 2 ) = 20 ⇒ K P =
100
(20) 2 ( 70) 7
= 3.035 × 10 −14
A when gases are taken in equimolar ratio, mole-fractions are :
Cl 2O 7 = 0.06, Cl 2 = 0.545, O 2 = 0.395
⇒ KP =
(0.06) 2
2
(0.545) (0.395)
7
⋅
1
P7
= 3.035 × 10 −14
⇒ P = 115 bar.
375. Molar ratio of the gases coming out of flask initially will be equal to their rate of effusion.
PCl 5
1−α
PCl 3 + Cl 2
α
α
Here, α = degree of dissociation.
r (Cl 2 )
137.5 0.53
=
=
⇒
r ( PCl 3 )
71
x
where x = mole fraction of PCl 3 in the gaseous mixture coming out initially
Hence, mole fraction of PCl 5 outside of flask = 0.09
r ( PCl 5 ) 1 − α
0.09
71
Now,
=
=
⇒ α = 0.77
r (Cl 2 )
208.5 0.53
α
Kp =
(0.77) 2
= 2.577 bar.
0.23
⇒ x = 0.38
372
Problems in Chemistry
A ( s)
376.
2p
2 p − 0.2 p
Case I
Case II
KP = 4 p3
+ C (g )
2B ( g )
p
p + 1 − 0.1p
K P = (1.8 p) 2 (0.9 p + 1) = 4 p 3
⇒
P = 2.988 bar
P (case II ) = 2.7 p + 1 = 9.06 bar
377. The exchange equilibrium is :
[Ca - EDTA]2– + Pb 2+
[Pb - EDTA]2− + Ca 2+
K=
{[Pb - EDTA]2– }[Ca 2+ ]
2–
{[Ca – EDTA] }[Pb
2+
]
=
1018
10
5 × 10
= 2 × 10 7
The very high K value for the exchange reaction indicate that all Pb 2+ will be converted into
[Pb – EDTA]2− , but since concentration of both Ca 2+ and [Cd – EDTA]2− are very high, will
remain unchanged. Therefore :
Ratio
[Pb – EDTA]2–
[ Pb
2+
]
=
K[Ca – EDTA]2–
[Ca
2+
]
= 2 × 10 7
1
= 8 × 10 6
2.5
378. From the thermodynamic informations : ∆H ° = 90.5 kJ and ∆S ° = 285.5 JK –1
⇒ K = 0.143 = p 2
( p = partial pressure of NH 3 )
⇒
p = 0.378 bar and P (eqm) = 2P = 0.756 bar.
379. ∆G ° (reaction) = 2 × 52 − 98 = 6 kJ = − RT ln K ⇒ K = 0.09
⇒
∆G ° = 4.85 kJ = – RT ln K
Also, for
N 2O 4
K=
4α
2
1−α2
2NO 2
P = 0.09 ⇒ α = 0.1483
Also, since reaction is occurring at constant pressure and temperature :
v ∝ n ⇒ % volume increase would be 14.83.
5
9
Let at new equilibrium partial pressure of AB ( g ) = x and of AB 2 ( g ) = y bar.
380. AB 2 ( g ) + A ( s)
⇒
2 AB ( g )
K=
5 x2
=
y
9
Also
x + y = 0.4
Solving, Eqs. (i) and (ii) gives x = 0.27; y = 0.13
 0.13 
Volume % of AB 2 ( g ) = 
 × 100 = 32.5
 0.4 
Volume % of AB ( g ) = 67.5.
…(i)
…(ii)
373
Solutions
381. Since, volume of container is 1.0 litre, concentrations of each species will be equal to their moles.
Now, setting the equilibrium table
N 2 ( g ) + 3H 2 ( g )
2NH 3 ( g )
1 − x 3 − 3x
2x − y
NH 3 ( g ) + H 2S( g )
K p = 8 × 10 –3 = K c ( RT ) –1
NH 4 HS( g ),
y
⇒ K c = 0.4592
y
also 2x − y = 0.9
K c = 0.4592 =
(1 − y) (2x − y)
2x − y
⇒
1− y
1− y
= 2.42
y
⇒
⇒
y = 0.3
and
(2x − y) 2
=
x = 0.6
(0.9) 2
⇒
K c (for the 1st reaction) =
⇒
K p (for 1st reaction) = K c ( RT ) –2 = 3.55 × 10 –4
27 (1 − x ) 4
27 (0.4) 4
= 1.17
[ H 2S] = 1 − y = 0.7 M
382. Let P0 be the initial pressure of N 2O 4 ( g ), then:
N 2O 4 ( g )
2NO 2 ( g )
P0 − 0.3 P0
0.6 P0
Kp =
18
P0
35
…(i)
On doubling volume, the initial pressure will be halved.
N 2O 4 ( g )
2NO 2 ( g )
P0
–P
2
2P
2
⇒
⇒
⇒
Kp =
 2P 
4P 2
1
=   P02
P0
 P0 
 2P  P0
−P

1 −
2
P0  2

=
α2
18
⋅ 2P0 =
P0
1−α
35
{from Eq. (i)}
35 α 2 + 9 α − 9 = 0
α=
− 9 + 81 + 36 × 35
70
= 0.3945
⇒ 39.45% N 2O 4 dissociated.
383. ∆H ° (Reaction) = ∆H °f ( N 2O 5 ) − ∆H °f (O 3 ) − 2∆H °f ( NO 2 ) = 11 − 143 − 2 × 33 = − 198 kJ
Q Reaction turned spontaneous to non-spontaneous at 1175 K implies that at this temperature
∆G° = 0.
i. e.,
− 198 × 10 3 − 1175 ∆S ° = 0
⇒
∆S ° = −168.5 JK –1
Now at 500 K, ∆G° = − 198 × 10 3 − 500 ( − 168.5) = − 113.75 × 10 3
Also,
∆G ° = − RT ln K = − 113.75 × 10 3
374
Problems in Chemistry
⇒
ln K = 27.36 and
K p = 7.65 × 1011
384. (a) ∆H ° (Reaction) = ∆H °f ( Products) − ∆H °f ( Reactants)
− 55 = ∆H °f ( N 2O 5 ) − 2 × 33.2
⇒
Also,
∆H °f (N 2O 5 ) = + 11.4 kJ
⇒
∆S ° (Reaction) = S ° (Products) − S ° (Reactants)
− 227 = S ° (N 2O 5 ) – 2 × 240 – 12 × 205
⇒
3
S ° (N 2O 5 ) = 355.5
3
∆G° = − 55 × 10 + 298 × 227 = 12.646 × 10 J
(b) Q ∆G° > 0, reaction is non-spontaneous at 298 K.
(c) Q ∆G° > 0, increasing temperature will drive the reaction in backward direction.
kP
CH 3
CH 3
H
CH 3
1
385.
C==C
C==C
∆G °1
H
H
H
CH 3
∆G °2
∆G °3
k P2
k P3
H
H 3C
C==C
H 3C
H
∆G1° = 62.97 − 65.85 = − 2.88 kJ = − RT ln K P1
⇒
ln K P1 =
2880
= 0.693 ⇒ K P1 = 2
8.314 × 500
∆G2° = 58.07 − 65.85 = − 7.78 kJ = − RT ln K P2
⇒
ln K P2 =
7780
8.314 × 500
⇒ K P2 = 6.5
Now let us consider initially there was 1.0 mol of cis-2-butene only and at equilibrium, there are
x-mol trans-2-butene and y mol 2-methyl propene.
y
x
⇒
K P1 =
K P2 =
= 2,
= 6.5
1− x − y
1− x − y
Solving,
x = 0.21 and y = 0.6842
⇒ At equilibrium, mixture contain 21% trans-2-butene, 68.42% 2-methyl propene and rest
10.58% cis-2-butene.
386. Let x mol PCl 5 ( g ) decomposes into PCl 3 ( g ) and Cl 2 ( g ) as:
PCl 5 ( g )
0.02 − x
At equilibrium.:
⇒
PCl 3 ( g ) + Cl 2 ( g )
x
x
Total moles of gaseous species = 0.02 + x + 0.015 = 0.035 + x
Partial pressure : PCl 5 ( g ) =
0.02 – x
× 1.843
0.035 + x
375
Solutions
Cl 2 ( g ) = PCl 3 ( g ) =
⇒
K P = 0.4 =
x
× 1.843
0.035 + x
x 2 × 1.843
(0.035 + x ) (0.02 − x )
Solving, x = 0.01 ⇒ Total moles at equlibrium = 0.035 + 0.01 = 0.045
0.045 × 0.082 × 500
V=
= 1.0 L.
⇒
1.843
387. Let initially, there are 100 moles of gaseous mixture. Then
SO 2 + 12 O 2
SO 3
15
20
65
After changing temperature to T, let x mole of SO 2 is consumed.
Now at new equilibrium : SO 2 = 15 − x
O 2 = 20 − x/ 2
SO 3 = 65 + x
x

From the given condition : 65 + x = 0.8 100 −  ⇒ x =10.7

2
Therefore, partial pressures at new equilibrium are :
75.7
4.3
14.65
atm, SO 2 =
, O2 =
SO 3 =
94.65
94.65
94.65
75.7 94.65
94.65
×
×
= 44.75
94.65
4.3
14.65
22600 21
Also, ∆G ° = − RT ln K P = − 22600 + 21 T
ln K P =
− = 3.8
⇒
RT
R
Solving, T = 429.7 K.
388. Initially : CO( g ) + H 2O( g )
CO 2 ( g ) + H 2 ( g )
1− x
1− x
x
x
⇒
K P (T ) =
 x 
K =9= 

1 − x 
2
⇒ x=
3
4
1
3
and moles of CO 2 = H 2 = . Now, let us assume that
4
4
CO 2 ( g ) flask contained x mole of CO 2 ( g ). Then
CO( g ) + H 2O( g )
CO 2 ( g )
+
H 2 (g )
1
1
1
3







3
Also y =
 + y
 + y
 + x − y
 − y




4
4
4
4
4
0.5 (0.5 − x )
K =9=
⇒
x=4
⇒
(0.5) (0.5)
At this equilibrium, moles of CO = H 2O =
Hence, volume of CO 2 ( g ) flask = 4.00 litre.
376
Problems in Chemistry
H2
I2
2HI
0.5 − x
0.5 − x
2x
0.8 − y
0.8 − y
2y
389.
+
2
 2y 
 2x 
K =

 =
 0.8 − y 
 0.5 − x 
⇒
390. For
2x
= 0.15
V
2y
⇒
=?
V
⇒
2
⇒
y =1.6 x
2 y 3.2x
=
× 0.15 = 0.24 M
V
2x
C( s) + CO 2 ( g )
2CO( g )
KP =
⇒
⇒
2
PCO
= 100 PCO2 = 63
PCO2
PCO2 = 0.63 and PCO = 6.3 atmosphere
P (total) = 6.3 + 0.63 = 6.93 atmosphere.
IONIC EQUILIBRIUM
391. For weak acid HA, the ionization equilibrium is :
HA
C (1 − α )
+
[H ] = 10
Also,
⇒
Ka =
−2
H+ + A −
Cα
Cα
= Cα = 0.1 α
⇒ α = 0.1
Cα 2
= 1.1 × 10 –3
1−α
Now, let us assume that 100 mL 0.1 M HCl is added to V mL of the above weak acid.
m mol of HCl added = 100 × 0.1 = 10
m mol of A − in the solution = x
m mol of HA = 0.1 V
If, in presence of HCl, x m mol of HA is ionized,
m mol of H + in the solution = x + 10
m mol of A − in the solution = x
m mol of unionized HA = 0.1V − x
⇒
and
⇒
Ka =
+
[H ][ A − ]
(x + 10)(x )
=
= 1.1 × 10 −3
[ HA ]
(100 + V )(0.1V − x )
[H + ] = 2 × 10 −2 =
2 × 10 −2 x
= 1.1 × 10 −3
(0.1 V − x )
x + 10
100 + V
⇒ V =192 x
…(i)
377
Solutions
Substituting in [H + ] expression : 2 × 10 −2 =
⇒
392. Let dibasic acid be H 2 A :
V + 1920
192(100 + V )
V = 540 mL
H2 A
HA −
H + + A 2−
[ A 2− ] = Ka 2 = 3 × 10 −6
⇒
[H + ] = [HA − ] + [ A 2− ] =
and
⇒
H + + HA −
[H + ] =
1.4 × 10 –6
+
[H ]
Ka1 [ H 2 A ]
[H + ]
+ 3 × 10 −6
+ [ A 2− ]
⇒ [H + ] = 1.18 × 10 −3
pH = 2.926
393. Enthalpy of ionization of formic acid is 15 kJ/mol
⇒
K ( 47° C) 15 × 10 3
=
K (27° C)
8.314
ln


20
 = 0.376

 300 × 320 
K a ( 47° C) = 2.47 × 10 −4
⇒
[H + ] = K a C = 4.97 × 10 −3
394. [D + ] = 10 −7 +
1.35 × 10 −15
+
⇒ pH = 2.3
⇒ [D + ] = 1.12 × 10 −7
[D ]
395. The hydrolysis equilibrium is :
OCl – + H 2O
Kh =
=
⇒ pD = 6.95
HOCl + OH −
Kw
10 −14
=
= 3.33 × 10 −7
K a 3 × 10 −8
[HO – ][HOCl]
[OCl – ]
=
[HOCl]
[OCl – ]
=
3.33 × 10 −7
[ HO – ]
= 0.528
⇒ % (by mole) HOCl = 34.55, OCl − = 65.45.
396.
PV
= 4.64 × 10 −4
RT
moles of NaOH used up = moles of CH 3COOH produced from decomposition = 7.8 × 10 −4
n(dimer) =
⇒ Moles of dimer dissociated =
397.
7.8 × 10 −4
= 3.9 × 10 −4
2
⇒ α=
3.9 × 10 −4
4.64 × 10 −4
[CO 2 ] = 2.28 × 10 −3 × 3.2 = 7.296 × 10 −3 M
[H + ] = K1 [CO 2 ] = 5.53 × 10 −5
⇒ pH = 4.25
= 0.84
378
Problems in Chemistry
398. [OCl – ] = 0.67 :
OCl – + H 2O
HOCl + OH –
[OH − ] = K h C = 4.72 × 10 −4
⇒
pOH = 3.325 and pH = 10.675
[S ]
pOH = pK b + log
⇒ [ S ] = 0.5
[B ]
399.
m mol of NH 4Cl = 100
m mol of NH 3 = 50
After adding 7.5 mmol KOH; m mol of NH 4Cl = 92.5
m mol of NH 3 = 57.5
92.5
⇒
pOH = pK b + log
= 4.9
57.5
⇒
pH = 9.1 and ∆ pH = 0.1.
400. mol of CH 3COOH = 0.25, mol of CH 3COONa = 0.30
0.3
= 4.78
0.25
306.25
pH = pK a + log
= 4.799
243.75
290
pH = pK a + log
= 4.745
260
pH = pK a + log
(a)
(b)
(c)
401. K a =
Cα 2
; Solving gives α = 0.27
1−α
⇒ van’t Hoff factor ( i) = 1 + α = 1.27
− ∆T f = iK f m = 1.27 × 1.86 ×
ln
402.
⇒
⇒
1.99 × 10 −15
3.16 × 10
ln
−16
=
0.5
× 1000 = 1.26
935.5

∆H 
10


8.314  293 × 303 
K w (50° C)
3.16 × 10 −16
=
⇒ ∆H =136.2 kJ

136.2 × 10 3 
30


8.314  293 × 323 
K w (50° C) = 5.7 × 10 −14
pD = 6.62
403. MOH has lower solubility ( S = 10 −6 ) and will precipitate first of pH = 8
404. K sp = 1.6 × 10 −14 = [Fe 2+ ][OH – ]2
(i) At pH = 8, [OH − ] = 10 −6 , [Fe 2+ ] = Solubility = 1.6 × 10 −2
(ii) At pH = 6, [OH − ] = 10 −8 , [Fe 2+ ] = 160 indicates very high solubility.
379
Solutions
Ca 2+ + 2F −
CaF2
405.
F – + H+
HF
Let ‘α’ be the fraction of total fluoride ion produced in F − form
Ka
[F – ]
[F – ]
α=
= −
=
+
2S
[F ] + [H ] K a + [H + ]
(a) At pH = 7, [H + ] = 10 −7 << K a
⇒ α ≈1 and K sp = 4S 3
S = 2.15 × 10 −4
⇒
(b) At pH = 5, α =
3.45 × 10 −4
3.45 × 10 −4 + 10 −5
= 0.97 and [F − ] = 25α
K sp = 4S 3 ⋅ α 2
⇒
⇒ S = 2.2 × 10 −4
406. Proceeding as in problem no. 405.
(a) α ≈1 and
S = 7.5 × 10 −3
(b) At pH = 4; α =
3.45 × 10 −4
3.45 × 10 −4 + 10 −4
K sp = 4S 3α 2
⇒
407. [Salt] =
and
S = 8.9 × 10 −3
10 −2
= 6.67 × 10 –5 M
150
Kh =
⇒
= 0.775
K
[OH – ]2 (5.12 × 10 −7 ) 2
=
= 3.94 × 10 −9 = w
−
5
C
Ka
6.67 × 10
K a = 2.53 × 10 −6 , α =
5.12 × 10 −7
6.67 × 10 −5
= 7.67 × 10 −3
⇒ % protonation = 0.76
K
[H + ]2
408. K h = w =
⇒ [H + ] = 2.4 × 10 –8 = 1.55 × 10 −4 ⇒ pH = 3.8
Kb
C
409. At first equivalence point, 5 mmol of NaHC 2O 4 will be formed.
Hence,
Kh =
K w [OH – ]2
=
K a1
C
⇒ [OH − ] = 1.3 × 10 −7
⇒
pOH = 6.89 pH = 7.11
At second end point, m mol of Na 2C 2O 4 = 5
K
[OH − ]2
Now,
⇒ [OH − ] = 3.92 × 10 −6
Kh = w =
K a2
C
⇒
pOH = 5.4 pH = 8.6
where S is solubility.
380
Problems in Chemistry
410. For H 2S : K a = K a 1 × K a 2 = 10 −21 =
[H + ]2 [S 2− ]
[H 2S]
At pH = 7, [S 2− ] = 10 −8 and solubility = [ Zn 2+ ] =
1.6 × 10 −24
10 −8
= 1.6 × 10 −16
At pH =10, [S 2− ] = 10 −2 and solubility = 1.6 × 10 −22
K sp ( Ag 2CO 3 )
411.
K sp ( Ag 2CrO 4 )
=
[CO 23− ]
[CrO 24− ]
= 6.75
2–
[Ag + ] = 2[CO 2–
3 ] + 2[CrO 4 ]
Also,
[Ag + ]
⇒
[CO 23−
]
=2+
2
= 2.296
6.75
2– 2
2−
Also, K sp ( Ag 2CO 3 ) = 8.1 × 10 −12 = [Ag + ]2 [CO 2–
3 ] = (2.296 [CO 3 ]) [CO 3 ]
⇒
Similarly,
−4
[CO 2−
= Solubility of Ag 2CO 3
3 ] = 1.15 × 10
−5
[CrO 2−
= Solubility of Ag 2CrO 4
4 ] = 1.7 × 10
412. In both case, solution is an acid-buffer. Let x m mol of acid be present in the given volume and y m
mol of NaOH in 20 mL solution.
y
1.5 y
Then
and 4.18 = pK a + log
3.7 = pK a + log
x− y
x − 1.5 y
Solving,
pK a = 3.7 ⇒ K a = 2 × 10 −4 .
413. The buffer pair in this case is H 3 PO 4 and NaH 2 PO 4 .
⇒
pH = pK a 1 + log
0.15
[H 3 PO 4 ]
⇒ [H 3 PO 4 ] = 0.3 M
Also, m mol of HCl needed = 0.3 × 100 = 30 ⇒ V ( HCl) = 20 mL.
m mol of NaH 2 PO 4 = 0.15 × 100 + 30 = 45
⇒
m(NaH 2 PO 4 ) = 5.4 g
414. [H + ] = 1.48 × 10 −4 M ⇒ [CO 2 ] =
⇒
415. Fe 2+ + H 2O
0.052 = 2.3 × 10 –2 P = 2.26 atmosphere
Fe(OH) + + H +
416. For Ag 2S : 2Ag(CN) –2 + S 2–
K=
⇒
[H + ]2
= 0.052 M
K a1
K w [H + ]2
=
K2
C
−
Ag 2S + 4CN ;
K=
⇒ pH = 4.7
K d2
[CN – ]4
= 1010 =
K sp
[Ag(CN) –2 ][S 2– ]
[S 2− ] = 4 × 10 −10
381
Solutions
Cd(CN) 24− + S 2–
Similarly, for CdS :
K=
CdS + 4CN – ;
Kd
[CN – ]4
=
⇒ [S 2− ] = 7.28 × 10 −12
2–
K sp [Cd(CN) 2–
][S
]
4
Q Lower [S 2− ] is required for CdS, it will precipitate first.
417. [S 2− ] (required for precipitation of CuS) = K sp (CuS)/[Cu 2+ ] = 6 × 10 –35
[S 2− ] (required for precipitation of NiS) = 4 × 10 −18
K(H 2S) = K1 ⋅ K 2 = 10 −21 =
Now,
For NiS; [H + ]2 =
10 −22
4 × 10
−18
= 2.5 × 10 −5
[ H + ]2 [S 2− ]
[ H 2S]
⇒ [H + ] = 5 × 10 −3
Hence, if [H ]is greater than 5 × 10 −3 (pH = 2.3), [S 2− ]will be less than 4 × 10 −18 and NiS will not
precipitate, only CuS will be precipitated.
418. Minimum carbonate ion concentration required for precipitation of carbonates are:
+
For CaCO 3 , [CO 23− ] = 4.5 × 10 −8 and for NiCO 3 , [CO 23− ] = 1.3 × 10 −6
For H 2CO 3 ; K = K1 K 2 = 2 × 10 −16 =
[H + ]2 [CO 23− ]
[H 2CO 3 ]
 2 × 10 −16 × 0.03 
[H + ] for precipitation of CaCO 3 = 

−8

 4.5 × 10
1/ 2
= 1.15 × 10 −5
⇒ pH = 4.93
1/ 2
 2 × 10 −16 × 0.03 
−6
⇒ pH = 5.66
[H ] for precipitation of NiCO 3 = 
 = 2.1 × 10
−6
×
10
1.3


⇒ At pH 4.93, CaCO 3 will start precipitating but NiCO 3 will remain in solution and at pH = 5.66,
NiCO 3 will start precipitating.
419. For the dissolution equilibrium :
+
AgBr + 2NH 3
−x
1 − 2x
Ag(NH 3 ) +2 + Br −
 x 
K = K sp ⋅ K f = 7.5 × 10 −6 = 

 1 − 2x 
x
x
2
⇒
x = 2.7 × 10 −3
420. For MX , solubility = K sp = 1.78 × 10 −5
For
MX 3 ,
 K sp 
421. For M 2 X 3 ; S = 

 108 
⇒
K sp = 27S 4 = 2.76 × 10 −18
1/ 5
= 4.58 × 10 −5 .
Solubility of M 2 X = 2S = 9.16 × 10 −5
⇒ K sp = 3 × 10 −12
382
Problems in Chemistry
422. When CaSO 4 start to precipitate [SO 24− ] =
K sp
[Ca
(a) [Sr 2+ ] when CaSO 4 start to precipitate =
2+
]
= 1.6 × 10 −4 M
K sp (SrSO 4 )
[SO 24− ]
= 2 × 10 −3
(b) % Sr 2+ ion precipitated when CaSO 4 begin to precipitate =
0.15 − 2 × 10 −3
× 100 = 98.67
0.15
423. The dissolution equilibrium is :
Fe(OH) 3 + 6CN −
Fe(CN) 36− + 3OH − : K = K sp ⋅ K f = 1.6 × 10 −8
Let the initial concentration of CN − = x M
1.6 × 10 −8 =
⇒
(0.092)(0.275) 3
( x − 0.55) 6
⇒ x = 7.57 M
⇒ Moles of NaCN dissolved = 7.57 × 1.2 = 9.084
Mass of NaCN = 445 g.
Ag(S 2O 3 ) 3– + Br −
424. AgBr + 2S 2O 2–
3
1.2 − 2x
x
x
x
 x 
K = K sp ⋅ K f = 10 = 

 1.2 − 2x 
2
⇒ x = 0.52 M
⇒ mol of AgBr dissolved = 0.065
mass of AgBr dissolved = 12.22 g
0.278
425. Initial concentration of Ca 2+ =
M = 6.95 × 10 −3 M
40
1.06
Moles of Na 2CO 3 added =
= 0.01
106
Moles of CO 2−
3 left unreacted after precipitation of
CaCO 3 = 0.01 − 6.95 × 10 −3 = 3.05 × 10 −3
⇒ [Ca 2+ ] =
426. 2 × 10
−6
K sp
[CO 23−
]
=
4.5 × 10 −9
3.05 × 10
−3
= 1.475 × 10 −6 M = 0.059 ppm Ca 2+ .
3
= 4S , where S = solubility of PbBr 2
⇒ S = 7.93 × 10 −3 M = [Pb 2+ ]; [Br − ] = 2S = 1.586 × 10 −2 M
8 × 10 −9 = [Pb 2+ ][I − ]2
427. HIn
Y
428. (a)
⇒ [I − ] = 10 −3 M.
R
Y
pH( yellow ) = 8.045 − log 30 = 6.57
pH(red) = 8.045 + log 2 = 8.35 ⇒ pH range = 6.57 ←→ 8.35.
H + + In −
R
pH = pK In + log
Zn(OH) 2 +
2OH −
x − 2 × 10−3
Zn(OH) 24−
2 × 10−3
383
Solutions
K = K sp ⋅ K f = 6 × 10 4 =
( x − 2 × 10 −3 ) 2
x = 2.18 × 10 −3 M.
⇒
3 × 10 −16 = [Zn 2+ ][OH − ]2
(b)
K1 = 2.4 × 10 −3 =
K 2 = 3.3 × 10 −9 =
Ag(NH 3 ) +2 + CN –
AgCN + 2NH 3
⇒
⇒ [OH − ] = 5.47 × 10 −7 .
Ag(NH 3 ) +2 + Cl −
AgCl + 2NH 3
429.
2 × 10 −3
[Ag(NH 3 ) +2 ][Cl – ]
[ NH 3 ]2
[Ag(NH 3 ) +2 ][CN – ]
[ NH 3 ]2
K1
[Cl − ]
Solubility of AgCl
= 7.27 × 10 5 =
=
−
K2
[CN ] Solubility of AgCN
The above ratio indicates that AgCl has very high solubility than AgCN.
⇒
K1 = 2.4 × 10
−3
 S 
1 − 2S 


2
⇒ S = [Cl − ] = 0.045
[CN − ] = 6.13 × 10 −8 .
⇒
x − 50
⇒ x = 450 ⇒ [Acid] = 0.45 M.
x + 50
431. Let x mol of HCl is added. Then moles of C 3 H 6O 3 = x and moles of
430. 4.64 = 4.74 + log
NaC 3 H 5O 3 = 0.1 − x
0.1 − x
3.77 = pK a + log
x
⇒
Solving, x = 0.55 mol
⇒ 55 mL of 1.0 M HCl will give the desired buffer.
432.
Al 3+ + H 2O
Al(OH) 2+ + H +
c−x
⇒
Kh =
y
x
xy
c−x
HSO −4
…(i)
H + + SO 24−
x− y
y
Ka =
y
y2
x− y
⇒ Solving for x and y gives pH = − log y = 3.53
433.
AgCN
Ag + + CN –
CN − + H 2O
HCN + OH −
…(ii)
384
Problems in Chemistry
Applying material balance = [Ag + ] = [CN − ] + [HCN] =
K sp
+
[Ag ]
+
K w K sp
K a [ Ag + ]
Also 1st term on the right hand of the material balance is negligible in comparison to the second
term.
[Ag + ] =
⇒
434. Q α =
Kb
C
K w K sp
⇒ [Ag + ] = 1.66 × 10 −7
K a [ Ag + ]
α2
C1
C1
V
=
=2 ⇒
=4= 2
α1
C2
C2
V1
⇒
⇒ V2 = 4V1 = 200 mL
435. QS 2− hydrolysed completely into HS − : S 2− + H 2O
Kh =
Kw
[ HS − ]2
=1 =
K a2
[S 2− ]
[S 2− ] = [ HS − ]2 = (6.7 × 10 −10 ) 2 = 4.489 × 10 −19
⇒
K sp = [Pb 2+ ][S 2− ] = 3 × 10 −28
⇒
436. (a) K a =
(b) K =
HS − + OH −
Cα 2
1−α
⇒ α = 0.19 and [H + ] = 3 × 10 −3 α
⇒ pH = 3.235
K a ( HL)
1.4 × 10 −4
=
= 311.11
K a 1 ( H 2CO 3 ) 4.5 × 10 −7
(c) (i) HCO −3 + H 2O
(ii) HL +
K w [OH − ]2
=
K a1
C
H 2CO 3 + OH − : K h =
HCO −3
⇒ pH = 11.35
H 2CO 3 + L−
0.024 − 3 × 10−3
3 × 10−3
Q K is very high, all HL will be neutralized.
⇒
K a 1 = 4.5 × 10 −7 =
⇒
[H + ] = 6.428 × 10 −8
K a 1 = 4.5 × 10 −7 = 10 −7
(d)
(e)
Kh =
and
(0.022 − x )
x
pH = 7.2.
⇒ x = 4 × 10 −3
Ca 2+ + CO 23−
CaCO 3 ( s)
H 2O
[H + ][HCO –3 ]
[ H 2CO 3 ]
S
+ CO 23−
S
HCO 3−
S −x
x
2
Kw
x
= 2.12 × 10 −4 =
K a2
S −x
⇒
+ OH −
x
S − x = 3.73 × 10 −5
385
Solutions
x = 8.91 × 10 −5
Also,
⇒
K sp = S ( S − x ) = 1.265 × 10 −4 × 3.73 × 10 −5 = 4.7 × 10 −9
CO 23− + H 2O
(f) For
Kh =
HCO −3 + OH −
K w [OH − ][ HCO 3− ]
=
= 2.12 × 10 −4
2
−
K a2
[CO 3 ]
Substituting [OH − ] and [HCO −3 ], [CO 23− ] = 2.6 × 10 −5
⇒
K sp = 4.7 × 10 −9 = [Ca 2+ ][CO 23− ] ⇒
[Ca 2+ ] = 1.8 × 10 −4 M
437. Let x m mol acid (HA) is present in its 1.7 g.
m mol of salt = x m mol NaA
⇒
m mol of NaOH left unreacted = 10 − x
A − + H 2O
Now,
HA + OH −
m mols x − y
Kh =
Given, [OH − ] = 1.58 × 10 −4 =
y
10 − x + y
Kw
y(10 − x + y)
= 5 × 10 −5 =
Ka
x− y
10 − ( x − y)
100
⇒ x − y = 9.9842
Substituting in Eq. (i) yields y = 3.16 ⇒ x =13.144
1.7
M=
= 129.33
13.144 × 10 −3
438. K sp = 4S 3 = 1.58 × 10 −11 . Also HCl is limiting reagent.
1
[Mg 2+ ] = [HCl] = 0.05
2
⇒
⇒
K sp = [Mg + ][OH − ]2
⇒ [OH − ] = 1.76 × 10 −5
pOH = 4.75 ⇒ pH = 9.25
439. [H 2CO 3 ] = 0.0343 ×
440
10
6
= 1.5 × 10 −5
Ka =
Also,
Solving; α = 0.154 ⇒ [ H + ] = Cα
Cα 2
1−α
⇒ pH = 5.63
440. (a) m mol of HCl consumed = 5 − 2.0265 = 2.9735 = m mol of NH 3
2.9735 × 14 × 10 −3
× 100 = 16.55
0.2515
2.0265
(b) 0.00 mL NaOH : [H + ] =
⇒ pH = 1.39
50
⇒
m% N =
…(i)
386
Problems in Chemistry
9.65 mL NaOH : m mol of HCl left = 2.0265 − 9.65 × 0.1050 = 1.01325
1.01325
[H + ] =
⇒ pH = 1.69
50
19.3 mL NaOH : No HCl left.
NH +4 + H 2O
Now,
Kh =
NH 4OH + H +
K w [H + ]2
=
Kb
C
[H + ] = K h C = 5.8 × 10 −6
⇒
⇒ pH = 5.23
On adding 28.95 mL NaOH : m mol of NH +4 = 2.9735 − 1.01325 = 1.96
m mol of NH 4OH = 1.01325
1.96
= 5.04
pOH = pK b + log
1.01325
pH = 8.96
⇒
10
8
(c)
pH 6
4
2
0
10
20
Vol. of NaOH
30
pH range = 4.23 − 6.23
m mol of N = 32.5 × 0.101 = 3.2825
0.2345
M=
× 1000 = 71.44
3.2825
(d)
(e)
441. At end point, [Ag + ] = K sp ( AgCl) = 1.58 × 10 −5
[CrO 24− ] =
442. pOH = 5 − log 2 + log
1.8 × 10 −12
−5 2
(1.58 × 10 )
0.2
= 4.4
0.4
= 7.3 × 10 −3 M
⇒ [OH − ] = 4 × 10 −5 M
After adding Ca 3+ , total volume =100 mL ⇒ [OH − ] = 2 × 10 −5 M
K sp = 10 −19 = [La 3+ ][OH − ]3
⇒ [La 3+ ] = 1.25 × 10 −5
⇒ m mol of La 3+ in solution = 1.25 × 10 −3
m mol of La 3+ precipitated = 50 × 10 −4 − 1.25 × 10 −3 = 3.75 × 10 −3 % La 3+ precipitated = 75.
387
Solutions
Al 3+ + H 2O
443.
Ka =
x[ H + ]
S −x
Kb =
y [OH − ]
…(ii)
S−y
Al(OH) 2+ + H +
S −x
x
−
HPO 2–
4 + OH
PO 3−
4 + H 2O
S − y
K sp = ( S − x )( S − y) = 10 −20
Also,
…(iii)
Multiplying Eqs. (i) and (ii)
xy([ H + ][OH − ])
= K a K b = 5 × 10 −7
( S − x )( S − y)
⇒
xy = 5 × 10 −13
Now, solving for S : S = 7 × 10 −7 .
444. (a) Mg 2+ + 2OH − → Mg(OH) 2 ( s). At the end of reaction, Mg 2+ = 0.1 mol
0.2
0.2
K sp = 1.6 × 10 −12 = [Mg 2+ ][OH – ]2
⇒ [OH − ] = 4 × 10 −6
(b) Mg 2+ + 2OH − → Mg(OH) 2 ( s) At the end, [Mg 2+ ] = 0.1 −
0.1
⇒ 1.6 × 10
0.04
−12
= 0.08 × [OH − ]2
⇒ [OH − ] = 4.47 × 10 −6
⇒
pH = 8.6
0.04
= 0.08
2
⇒ pH = 8.65
(c) Mg(OH) 2 + 2HCl → Mg 2+ + 2Cl – + 2H 2O
0.02
0.04
0.02
[Mg 2+ ] = 0.1 + 0.02 = 0.12 M
⇒ 1.6 × 10 −12 = 0.12[OH − ]2
[OH − ] = 3.65 × 10 −6
⇒
pH = 8.56
445. The precipitate exchange equilibrium is:
PbSO 4 + S 2–
Also for H 2S : K a = 10 −21 =
Hence,
PbS + SO 2–
4
K=
[SO 24− ]
[S 2− ]
[ H + ]2 [S 2− ]
⇒ [S 2− ] = 10 −22
[ H 2S]
= 6.4 × 1018
⇒ [SO 24− ] = 6.4 × 10 −4
[Pb 2+ ] present = 2.5 × 10 −5 M
Initial [Pb 2+ ] = 1.26 × 10 −4 M ⇒ moles of PbS precipitated = 10 −4
446.
and
m( PbS) = 23.8 mg.
1060 × 2.3
M (urea) =
= 0.4
6000
K C
[H 2S—CO—NH +3 ] = b − = 2.68 × 10 −7
[OH ]
447. Using K sp (AgCl), ∆G° of AgCl( s)
Ag + ( aq ) + Cl − ( aq ) = + 55.33 kJ
Now, 55.33 kJ/ mol = G °f (Ag + ) − 131 + 110
⇒
G °f (Ag + ) = + 76.33 kJ
…(i)
388
Problems in Chemistry
For Ag 2SO 4 :
2Ag + + SO 24−
Ag 2SO 4 ( s)
∆G ° (ionization) = 2 × 76.33 − 742 + 618.5 = 29.16 = − RT ln K sp .
K sp = 7.73 × 10 −6 = [Ag + ]2 [SO 2–
4 ]
⇒
[Ag + ] = Solubility = 8.8 × 10 −2 M
⇒
448.
(177
. × 10 −3 ) 2 3.16 × 10 −6
=
C
C
C + 0.05
After adding 100 mL 0.05 M HOCN, [HOCN] =
2
H + + OCN −
HOCN
⇒
Ka =
Ka =
2 × 10 −5 3.16 × 10 −6
; Solving C = 0.009 M.
=
C + 0.05
C
10 6 × 2.5
= 2.5 × 10 4 g
100
Moles of H 2SO 3 produced = 781.25
449. Mass of sulphur =
Volume of rain-water = 5.2 × 10 7 L. ⇒ [H 2SO 3 ] = 1.5 × 10 −5 M
Such a low concentration of acid indicate α ≈1.
pH = − log (1.0 × 10 −5 ) = 4.8
⇒
[NH 3 ] =
450.
[OH – ]2
3
= 0.0158 ⇒ moles of Mg = × 0.0158 = 0.0237
2
Kb
m = 0.5688 g and % purity = 56.88
451. The balanced equation for reaction between Br 2 and SO 2 is :
Br 2 + SO 2 + 2H 2O → 2HBr + H 2SO 4
0.01
Now,
H 2SO 4 →
HSO –4
0.01 – x
K a 2 = 10 −2 =
0.01
0.02
H
+
+
0.03 + x
H
+
0.03 + x
x (0.03 + x )
0.01 − x
[H + ] = 3.236 × 10 −2
HSO −4
0.01 – x
+ SO 2–
4
x
⇒ x = 2.36 × 10 −3
and
pH = 1.49
Since, Q forms precipitate with both X and Y , it must be K 2CO 3 because Ag 2CO 3 and
BaCO 3 are insoluble salts.
(b) K 2CO 3 on mixing with Pb(NO 3 ) 2 will give precipitate PbCO 3 but no precipitate will be
formed if K 2CO 3 is combined with NaCl solution. Therefore, R is Pb(NO 3 ) 2 and S is NaCl.
(c) If X forms a precipitate with S(NaCl) but not with R = [Pb(NO 3 ) 2 ], it must be AgNO 3 . On
the other hand if Y forms precipitate only with Q but not with R and S , it must be BaCl 2 .
453. Let x mL of acid is taken, then volume of salt solution taken = 20 − x. Now, applying Henderson’s
equation:
452. (a)
389
Solutions
pH = pK a + log
[Salt]
[Acid]
⇒
4.7 = 4.7 + log
0.1(20 − x )
0.15 x
⇒ 2 − 0.1 x = 0.15 x and x = 8 mL, volume of salt solution =12 mL.
454. Let x mL of acetic acid taken, volume of base taken = 50 − x
m mol of acid = 0.15 x, m mol of base = (50 − x ) 0.1
⇒ m mol of salt formed = 0.1(50 − x ), m mol of acid left unreacted = 0.15 x − 0.1(50 − x )
Applying Henderson’s equation:
0.1(50 − x )
5 − 0.1 x
4.7 = 4.7 + log
⇒
=1
⇒
x = 28.6 mL
0.15 x − 0.1(50 − x )
0.25 x − 5
Volume of base required = 21.4 mL.
455. The balanced chemical reaction involved in precipitation reaction is:
3Ba(NO 3 ) 2 ( s) + 2H 3 PO 4 → Ba 3 (PO 4 ) 2 ( s) + 6HNO 3 ( aq )
m mol (Initial) : 6.13
3
0
0
m mol (Final) : 1.63
0
1.5
9
pH of solution will be mainly due to HNO 3 .
9
= 4.5 × 10 –3
⇒
pH = − log ( 4.5 × 10 –3 ) = 2.35
[HNO 3 ] =
⇒
2000
456. The minimum concentrations of sulphide ion required for precipitation of these metals are:
MnS : [S 2– ] = K sp / [ Mn 2+ ] = 2.5 × 10 –8 M
CoS : [S 2– ] = K sp / [Co 2+ ] = 4 × 10 –19 M
Ag 2S : [S 2– ] = K sp / [ Ag + ]2 = 6.3 × 10 –46 M
Now, for H 2S :
H 2S
2H + + S 2–
K a = K a 1 × K a 2 = 10 –21 =
⇒
[H + ]2 [S 2– ]
[H 2S]
[H + ]2 [S 2– ] = 10 –22 .
Since, minimum S 2– ion is required for Ag + , corresponding [H + ]:
[H + ] (to begin precipitation of Ag 2S = 10 –22 /6.3 × 10 –46 = 3.98 ×1011
i.e., to prevent precipitation of Ag 2S, minimum [H + ] = 3.98 × 1011 and at a [H + ] below to this
value Ag 2S will precipitate. Such a large concentration of [H + ] is unachievable, Ag 2S will
precipitate at any practical concentration of H + .
Limiting [H + ] required to begin precipitation of CoS:
[H + ] = 10 –22 / 4 × 10 –19 = 1.58 × 10 −2 M ⇒ pH = 1.8
i.e., at [H + ] above 1.58 ×10 –2 M [S 2– ] will be less than 4 × 10 –19 M and CoS will not precipitate.
Limiting [H + ] required to begin precipitation of MnS:
390
Problems in Chemistry
10 −22
[H + ] =
2.5 × 10 –8
= 6.32 × 10 –8
⇒ pH = 7.2
i.e., at pH below 7.2, MnS will not precipitate.
pH range : At pH below 1.8, only Ag 2S will precipitate.
At pH >1.8 but < 7.2, only CoS will precipitate.
At pH > 7.2, MnS will start precipitating.
457. The cell reactions are:
H 2 → 2H + + 2e
2Ag + + 2e → 2Ag
E ° = 0.8 V
H 2 + 2Ag + → 2H + + 2Ag
E = 0.589 = 0.8 –
+ 2
[H ]
0.059
log
2
[Ag + ]2
S = 1.325 × 10 –4
⇒ [Ag + ] = 2.65 × 10 –4
and 2 S = 2.65 × 10 −4
K sp = [Ag + ]2 [C 2O 42– ] = 4 S 3
⇒
E ° = 0.8 V
K sp = 4 S 3 = 9.3 × 10 –12 .
⇒
458. For the dissolution reaction:
Zn(OH) 2 + 2OH –
0.1 − 2x
x
K = 0.588 =
Zn(OH) 2–
4
x
(0.1 − 2x )
2
K = K sp ⋅K f = 0.588
x
=
x
10
–2
− 4x 2 − 0.4 x
x = 0.0048 M, i.e., 0.0048 × 250 = 1.2 m mol Zn(OH) 2 dissolved.
150
mmol of Zn(OH) 2 taken initially =
= 1.515
99
⇒ mmol Zn(OH) 2 left undissolved = 1.515 − 1.2 = 0.315
⇒ mass of Zn(OH) 2 left undissolved = 31.185 mg
459. Reaction for dissolution of AgCl in ammonia is :
Solving:
AgCl + 2NH 3
x
0.1 − 2x
K = 4 × 10 –3 =
⇒
Ag(NH 3 ) +2 + Cl –
0.1 + x
x
x (0.1 + x )
(0.1 − 2x ) 2
K = K sp ⋅K f = 4 × 10 –3
=
0.1 x + x 2
10 −2 + 4x 2 − 0.4 x
⇒
0.1 x + x 2 = 4 × 10 –5 + 1.6 × 10 –2 x 2 − 1.6 x × 10 –3
or
0.984 x 2 + 0.1016 x − 4 × 10 −5 = 0
⇒
x=
−0.1016 + (0.1016) 2 + 16 × 10 −5 (0.984)
2 × 0.984
460. The order in which halide ions should be added is :
= 3.92 × 10 –4 M
Cl – than Br – than I – .
391
Solutions
When Cl – is added to 0.1 M, [Ag + ] =
When Br – is added to 0.1 M, [Ag + ] =
K sp ( AgCl)
0.10
K sp (AgBr)
= 1.8 × 10 −9 M
0.10
K sp ( AgI)
= 5 × 10 −12 M
= 8.4 × 10 −16
0.10
The [Ag + ] in equilibrium with AgCl in 0.1 M Cl − is1.8 ×10 –9 M, which is more than 5 × 10 −12 M
([Ag + ] required to initiate precipitation of AgBr when [Br – ] = 0.1 M). Hence, the solution
containing [Ag + ] = 1.8 × 10 –9 in equilibrium with AgCl in 0.1 M Cl – , will give precipitate of AgBr
when Br – is added to 0.1 M. Had we added Br – to 0.1 M before Cl – , [Ag + ] would have been
5 × 10 −12 M and on making this solution 0.1 M in Cl – in the subsequent step would not have
yielded AgCl precipitate. Hence the order.
461. As we know, a buffer solution shows its best buffering action near to its ionization constant value.
Therefore, the above buffer can best be prepared by taking H 2 PO 4– and HPO 2–
4 .
When I – is added to 0.1 M, [Ag + ] =
(a) pH = pK 2 + log
[ HPO 24– ]
[H 2 PO 4– ]
= 7 ⇒ log
[ HPO 24– ]
[ H 2 PO 4– ]
= − 0.22 and
[HPO 2–
4 ]
[ H 2 PO 4– ]
= 0.6
(b) In 50 mL buffer, [H 2 PO 4– ] = 0.10 M = 5 m mol ⇒ [HPO 2–
4 ] = 0.06 M = 3 m mol
m mol of NaOH added = 20 × 0.1 = 2
This added NaOH will neutralize 2.0 m mol of H 2 PO –4
⇒
m mol of H 2 PO 4– = 5 – 2 = 3
m mol of HPO 24– = 3 + 2 = 5
pH = pK 2 + log
5
3
= 7.44
462. From the titration curve it is concluded that the 1st end point correspond to completion of the
following reaction :
Na 2CO 3 + HCl → NaHCO 3 + NaCl
⇒ m mol of HCl consumed = 11 × 0.115 = 1.265 = m mol of Na 2CO 3
⇒ mass of Na 2CO 3 = 1.265 × 106 × 10 –3 = 0.134 g
Second end point corresponds to the following reaction :
NaHCO 3 + HCl → NaCl + H 2O + CO 2
Volume of HCl consumed = 35 − 11 = 24 mL
Out of this 24 mL, 11 mL HCl would be consumed by NaHCO 3 produced from Na 2CO 3 .
⇒ m mol of original NaHCO 3 = (24 – 11)0.115 = 1.495
⇒ mass of NaHCO 3 = 1.495 × 84 × 10 –3 = 0.1256 g
0.134
× 100 = 26.8
0.5
0.1256
× 100 = 25.12
⇒ m% of NaHCO 3 =
0.5
m% of KCl = 48.08%.
⇒ m% of Na 2CO 3 =
392
Problems in Chemistry
10 −3
= 8.165 × 10 −2 M
0.15
i.e., to prevent precipitation, [H 2 PO 4– ] should be less than 8.165 × 10 −2 M.
463. K sp = 10 −3 = [Ca 2+ ][H 2 PO –4 ]2
Also
⇒ [H 2 PO –4 ] =
H + + H 2 PO 4–
H 3 PO 4
K a1 =
K a [ H 3 PO 4 ] 7 × 10 –3 × 0.25
[H + ] [H 2 PO 4– ]
⇒ [H + ] = 1
=
[H 3 PO 4 ]
[H 2 PO 4– ]
8.165 × 10 −2
= 2.14 × 10 −2 M
i.e., to prevent precipitation, [H + ]> 2.14 × 10 –2 , pH = 1.66.
i.e., at pH below 1.66, no precipitation of Ca(H 2 PO 4 ) 2 will occur.
464. (a) Addition of HNO 3 will neutralize OH – in solution driving solubility equilibrium in forward
direction and this will increase concentration of Fe 3+ in solution. Also, as Fe 3+ comes in
solution due to neutralization of OH – with HNO 3 , it will form complex with SCN – present
in solution due to high K f value, but still, the overall effect will be increase in concentration
of Fe 3+ in solution.
(b) Addition of NaOH will drive both equilibrium (i) and (iii) in backward direction and it will
decrease concentration of Fe 3+ ion in solution.
(c) Addition of NaSCN will drive equilibrium (ii) in forward direction and concentration of
Fe 3+ ( aq ) will decrease.
(d) Addition of Fe(OH) 3 ( s) will have no effect on Fe 3+ ( aq ) concentration since active mass of
former is unity.
465. The dissociation equilibrium of the complex is:
[Cu(CN) 4 ]2−
Cu 2+ ( aq ) + 4CN – ( aq )
As strong acid is added, following equilibrium will be established parallely :
1
= 1.67 × 10 9
H + + CN –
HCN K =
Ka
The above equilibrium will lie predominantly to right, removing most of the cyanide ion from 1st
equilibrium. This will increase concentration of Cu 2+ in solution.
4 × 1000
466. m mol of NaCl added =
= 68.38
58.5
m mol of Ag + in solution = 25
Now:
AgCl( s) + 2NH 3 ( aq )
x − 0.2
Ag(NH 3 ) +2 ( aq ) + Cl – ( aq )
0.1
K = K sp ⋅K f = 2 × 10 −3 =
⇒
2.735 × 10
x − 0.2
x =13.875 M
468. The solubility equilibrium is:
K = K sp ⋅ K f
0.2735
−2
393
Solutions
Co(OH) 3 + 4SCN –
1 − 0.4
At the given conditions :
Q=
(0.1) (10 –7 ) 3
K = 1.6 × 10 −41
Co(SCN) –4 + 3OH –
10–7
0.1
= 7.716 × 10 −22 >> K (1.6 × 10 –41 )
(0.6) 4
Therefore, precipitation must occur.
469. [OH − ] of buffer = 2 × 10 −5 ⇒ [H + ] = 5 × 10 −10
5 × 10 −10 x
0.1 − x
−3
−10
Solving : x = 8.6 × 10 M
0.1 − x
5 × 10
x
Now, on adding equal volume of 0.1 M Ba 2+ ion, concentration of Ba 2+ ion in final solution will
become 0.05 M and that of carbonate ion will become 4.3 × 10 −3 M.
HCO –3
Now:
K a 2 = 4.7 × 10 −11 =
H + + CO 2–
3
–4
K IP = [Ba 2+ ][CO 2–
>> K sp , precipitation will occur.
3 ] = 2.15 × 10
⇒
470. For the equilibrium :
HCO –3
K a 2 = 4.7 × 10 −11 =
Also
H + + CO 2–
3
[H + ][CO 2–
3 ]
[HCO 3– ]
= 10 −8
[CO 23– ]
[ HCO 3– ]
⇒ [HCO –3 ] = 213 [CO 2–
3 ]
2–
–4
[HCO –3 ] + [CO 2–
3 ] = 0.1 ⇒ [CO 3 ] = 4.67 × 10
[Co 2+ ] =
⇒
[Co 3+ ] =
1.4 × 10 −13
4.67 × 10 −4
1.6 × 10 –44
4.67 × 10
−4
= 3 × 10 −10 M
= 3.4 × 10 −41 M
471. The hydrolysis reaction is :
C 6 H 5 NH +3 + H 2O
C 6 H 5 NH 3OH + H +
C (1 − α )
Cα
Cα
K
10 −14
= 2 × 10 −5
Kh = w =
K b 5 × 10 −10
Also K h =
Cα 2
1−α
⇒ 2 × 10 −5 =
α=
⇒
10 −3 α 2
1−α
⇒ α 2 + 2 × 10 −2 α − 2 × 10 −2 = 0
−2 × 10 −2 + (2 × 10 −2 ) 2 + 8 × 10 −2
2
[H + ] = Cα = 10 −3 × 0.13 = 1.3 × 10 −4
pH = 3.88
472. As, acetic acid is added, it will react quantitatively with OI – as:
CH 3COOH + OI –
CH 3COO – + HOI
m moles: 1
2
0
0
0
1
1
1
= 0.13
394
Problems in Chemistry
K=
K a [ acetic acid ]
= 10 6 > very large value.
K a ( HOI)
Now the conjugate base of weaker acid (HOI) will hydrolyze predominantly as :
K
OH – + H 2O
HOI + OH –
K h = w = 5 × 10 −4
Ka
Concentration of OI – left unreacted after reaction with CH 3COOH = 5 × 10 –3
Concentration of HOI produced by reaction of CH 3COOH with IO – = 5 × 10 –3
Now, if x MIO – is hydrolyzed then :
IO –
+
H 2O
HOI
5 × 10−3 + x
5 × 10–3 − x
K h = 5 × 10 −4 =
(5 × 10
⇒
x=
5 × 10 x + x 2
5 × 10 −3 − x
2
pH = 10.63
ClC 6 H 4 NH 3OH + H +
C (1 − α )
Cα
K h = 2.5 × 10 −3 =
2
Ka =
when 75% ionized
+
Kh =
Kw
= 2.5 × 10 −3
Kb
⇒ α 2 + 0.25 α − 0.25 = 0
−0.25 + 0.0625 + 1
= 0.39
2
[H + ] = Cα = 3.9 × 10 –3
⇒
= 4.22 × 10 −4
Cα
−2
2
Cα
10 α
=
1−α
1−α
α=
and
–
+
and
(not negligible)
pH = 2.4
+
[H ][dnp ] [H ][25] [H ]
⇒ pH1 = 3.48
=
=
75
3
[dnp]
[H + ][ 75]
= 1.1 × 10 −4
25
[H + ] =
Mg(OH) 2
1.1 × 10 −4
3
⇒ pH = 4.53
Mg 2+ ( aq ) + 2OH – ( aq )
K sp = [Mg 2+ ][OH – ]2
⇒
=
x
−3
−5.5 × 10 –3 + 3.025 × 10 −5 + 10 −5
ClC 6 H 4 NH +3 + H 2O
475. For
+ x )x
5 × 10 −3 − x
⇒
pOH = 3.37 and
473. The hydrolysis reaction is :
⇒
−3
x 2 + 5.5 × 10 −3 x − 2.5 × 10 −6 = 0
⇒
474.
OH –
+
⇒ 15
. × 10 −11 = 0.1 [OH – ]2
[OH – ] = 1.224 × 10 –5
This is the minimum concentration of OH – required to begin the precipitation of Mg(OH) 2 .
Therefore, [OH – ] ≤ 1.224 × 10 –5 M.
395
Solutions
Now for
NH +4 + OH –
NH 3 + H 2O
Kb =
[NH +4 ][OH – ]
2 × 10 –5 × 0.1
⇒ [NH +4 ] =
= 0.16 M
[NH 3 ]
1.224 × 10 –5
i.e., to maintain [OH – ] equal to 1.224 × 10 −5 M or less, the concentration of ammonium ion in
solution must be greater than or equal to 0.16 M.
476. The equilibria existing are:
Ba 2+ ( aq ) + SO 2–
4 ( aq )
BaSO 4 ( s)
x+ y
y
SO 2–
4 + H 2O
HSO –4 + OH –
Kb =
Kw
= 8.33 × 10 −13
Ka
The above K b value indicates that tendency of hydrolysis of SO 2–
4 is very less.
Also,
Here x M
HSO –4
HSO –4
H + + SO 2–
4
0.25 − x
x
x+ y
is ionized and y M BaSO 4 dissolved.
K sp = 10 –10 = y( x + y)
⇒
K a = 12
. × 10 −2 =
⇒
83
. × 10 –9 =
y(0.25 − x )
x
…(1)
x ( x + y)
0.25 − x
…(2)
⇒ x >> y and 1.2 × 10 −2 =
x2
0.25 − x
10 −10 10 −10
≈
= 2.24 × 10 −8 M
x+ y
x
10 −5
Solubility in water = K sp = 10 −5 ⇒ factor =
= 446
2.24 × 10 −8
477. Initial mmol of Na 2CO 3 = 500 × 0.1 = 50
Now, let us assume that x mmol of HCl is added to adjust the desired pH. Then
Na 2CO 3 + HCl → NaHCO 3 + NaCl
Solving x = 4.46 × 10 –3
⇒
y=
50 − x
0
x
x
The leftover Na 2CO 3 will hydrolyze to give the desired pH as :
CO 2–
3 + H 2O
Kh =
HCO –3 + OH –
Kw
[OH – ][HCO –3 ] 10 −4 x
= 2 × 10 −4 =
=
K a2
50 − x
[CO 23– ]
100 1
= ×V
3
3
⇒
V =100 mL
i.e., to prepare the desired solution, 100 mL of the given HCl will have to be added and finally
solution will be diluted to 1000 mL.
⇒
x=
396
Problems in Chemistry
478.
Initial (M) :
Final (M) :
⇒
Ag + + 2CN –
Ag(CN) –2
0.03
0.1
≈0
0.1 − 0.06
0.03
K f = 25
. × 1018 =
[ Ag(CN) 2 ]−
+
– 2
[Ag ][CN ]
=
0.03
+
[Ag ] [0.04]2
[Ag + ] = 7.5 × 10 –18 M
⇒
479. (a) For CaF2 : 3.4 × 10 –11 = [Ca 2+ ][F – ]2 = [Ca 2+ ] ( 0.05) 2
⇒ [Ca 2+ ] required to begin precipitation of CaF2 = 13.6 × 10 –9 M
2+
For CaCO 3 : 9.5 × 10 –9 = [Ca 2+ ][CO 2–
][0.02]
3 ] = [Ca
2+
⇒ [Ca ] required to begin precipitation of CaCO 3 = 4.75 × 10 –7 N
i.e., higher concentration of Ca 2+ is required to start precipitation of CaCO 3 , CaF2 will precipitate
first.
(b) When CaCO 3 start precipitating, the required [Ca 2+ ] is 4.75 ×10 –7 . To commence
simultaneous precipitation of CaF2 and CaCO 3 at this stage, the minimum [F – ] can be
evaluated as:
3.4 × 10 −11 = 4.75 × 10 −7 [F – ]2
[F – ] = 8.46 ×10 –3 M.
⇒
480. (a) For
PbI 2 ( s)
Pb 2+ ( aq ) + 2I – ( aq )
∆G ° (at 17° C) = − RT ln K sp = − 8.314 × 290 ln ( 7.47 × 10 –9 ) = 45.1166 kJ
and
∆G ° (at 27° C) = −RT ln K sp = − 8.314 × 300 ln (1.4 × 10 −8 ) = 45.1056 kJ
Now
∆G ° (at 17° C) = 45116.6 = ∆H ° – 290 ∆S °
∆G ° (at 27° C) = 45105.6 = ∆H ° – 300 ∆S °
From Eqs. (i) and (ii),
11 = 10 ∆S ° ⇒ ∆S ° = 1.1 JK –1
Substituting, ∆S ° = 1.1 in Eq. (i):
∆H ° = 45116.6 + 290 × 1.1 = 45.4356 kJ mol –1
(b) ∆G ° (at 77° C) = ∆H ° – 350 × 1.1 = 45.8206 kJ = − RT ln K sp
⇒
K sp = 1.45 × 10 −7 = 4 S 3
481. Normality of solution =
Q
2.54 × 2
= 0.0318 N
159.5
K ×1000
N
Λ eq =
=
Also
⇒ S = 3.3 × 10 –3 M
⇒ k (specific conductance)
91 × 0.0318
= 2.89 × 10 –3 Ω –1cm –1
1000
1
 A
=k  
 l
R
⇒ R=
1
1 l
 1
 =
  = 346 Ω.
k  A  2.89 × 10 –3  1
…(i)
…(ii)
397
Solutions
482. For a weak electrolyte:
Λc
10
=α =
= 0.042
238
Λ∞
For NH 3 :
NH +4 + OH –
NH 3 + H 2O
C (1 − α )
Kb =
Cα
−2
2
Cα
2
10 (0.042)
Cα
=
= 1.8 × 10 –5
1−α
1 − 0.042
483. The ionization reaction of acetic acid is:
CH 3COO – + H +
CH 3COOH
C (1 − α )
Cα
K a = 2 × 10 –5
α 2 + 2 × 10 –3 α − 2 × 10 –3 = 0
⇒
α=
⇒
Now
⇒
Also
⇒
484. G =
Cα
Cα 2 10 –2 α 2
=
=
1−α
1−α
−2 × 10 −3 + 4 × 10 −6 + 8 × 10 −3
2
= 0.0437
Λc
= 0.0437
Λ∞
Λ c =17.043 cm 2 Ω –1eq –1
k ×1000
Λc =
C
Λ C 17.043 × 0.01
k= c =
= 1.704 × 10 –4 Ω –1cm –1 .
1000
1000
1
1
 A
= k   = k ⋅ , where K is cell constant ( l/A ).
 l
R
K
K = Rk = 25 × 0.0116 = 0.29 cm –1
⇒
For acetic, acid solution, K = 0.29 cm –1 = kR
⇒
⇒
k (CH 3COOH) =
Λm =
0.29 cm –1
= 1.463 × 10 –4 Ω –1cm –1
1982 Ω
k × 1000 1.463 × 10 −4
=
× 1000 = 1.463 cm 2 mol –1 Ω –1 .
0.1
C
485. Initially, only NaOH was present at 0.1 M.
⇒
Λ m (NaOH) =
k × 1000 0.022 × 1000
=
= 220 Ω –1cm 2 mol –1 .
C
0.1
After adding equal volume of HCl, NaOH will be neutralized completely giving NaCl of
concentration 0.05 (since volume is doubled, concentration of salt will be halved).
398
Problems in Chemistry
0.0056 × 1000
=112 Ω –1cm 2 mol –1 .
0.05
On further adding HCl, no neutralization will occur and conductivity will be due to NaCl and HCl
2
2
together. Now the concentrations of NaCl = × 0.05 and concentration of HCl = × 0.05 M.
3
3
C
Now :
k = k ( NaCl) + k ( HCl) = ( Λ NaCl + Λ HCl ) ×
1000
1000 × 0.017
⇒
× 3 = 112 + Λ HCl
2 × 0.05
⇒
Λ NaCl =
Λ HCl = 510 − 112 = 398 Ω –1cm 2 mol –1
⇒
486. For NaCl solution :
k × 1000
…(i)
= 126.5
C
Here the 1st objective is to determine k (specific conductance) of NaCl and then C (NaCl).
1
 A k
Also; conductance
…(ii)
G= =k   =
 l K
R
Λm =
Λm =
For KCl solution :
⇒
k (KCl) =
k ×1000
C
138.3 × 0.02
= 2.766 ×10 –3 Ω –1cm –1
1000
Now from Eq. (ii),
K (cell constant) = kR = 2.766 × 10 −3 × 85 = 0.235 cm –1
Now; conductance of water
⇒
GH 2O =
Conductance of
NaCl( aq ) = GNaCl( aq ) =
⇒ Conductance due to NaCl alone =
⇒
1
9200
1
7600
1
1
−
= 22.88 × 10 –6 Ω –1
7600 9200
k (NaCl) = GK = 22.88 × 10 –6 Ω –1 × 0.235 cm –1 = 5.38 × 10 −6 Ω –1cm –1
Substituting, in Eq. (i) :
126.5 =
⇒
⇒
5.38 × 10 −6 × 1000
C
C (NaCl) = 4.25 ×10 –5 M
Volume of swimming pool =
Number of moles of NaCl
500
=
Molarity
58.5 × 4.25 × 10 −5
= 2.011 × 10 5 L
399
Solutions
ΛC 195 × 0.05
=
= 9.7 × 10 −3 Ω –1cm –1
1000
1000
1
1.5
 A
= 0.0291 Ω –1
(G ) = = k   = 9.7 × 10 −3 ×
 l
0.5
R
487. k (specific conductance) =
Conductance
R = 34.36 Ω =
⇒
⇒
488. Since, Λ eq
⇒
V
I
5
V
=
= 0.1455 A.
R 34.36
k × 1000 K × 1000
where K is cell constant.
=
=
N
RN
K × 1000
Λ eq ( KCl) =
= 138.3
85 × 0.02
K × 1000
Λ eq ( PbCl 2 ) =
= 106
70 N
⇒
I=
70 N
138.3
=
106 85 × 0.02
N = 3.168 × 10 −2
⇒
⇒ Molarity of PbCl 2 in its saturated solution =
N
= 1.58 × 10 −2
2
K sp = 4 M 3 = 1.6 × 10 −6 .
⇒
ELECTROCHEMISTRY
0.15 × 8 × 3600
faraday
96500
Volume of F2 ( g ) produced =11.2 F = 0.5 L
(b)
Volume of O 2 ( g ) produced = 5.6 F = 0.25 L
490. Partial pressure of O 2 = 722 − 19.83 = 702.17 torr.
Q = It =
489. (a)
⇒ Moles of O 2 ( g ) produced =
25 × 10 −3
702.17
×
= 9.548 × 10 −4
760
0.082 × 295
Gram equivalents of O 2 = 3.82 × 10 −3 =
491. Total surface area =11.2 cm 2
Mass of Ag = 11.76 × 10 –3 g
⇒ I = 0.2 ampere
⇒ V = 11.2 × 10 −4 cm 3
⇒ g eq. of Ag = 1.088 × 10 –4 =
⇒
492. The cell reaction is :
I × 30 × 60
96500
t =105 seconds.
2Sn
2+
→ Sn 4+ + Sn
Initial moles of SnCl 2 =
19
= 0.1
190
It
96500
400
Problems in Chemistry
0.119
= 0.001 = moles of SnCl 4 produced.
119
Moles of SnCl 2 left unelectrolysed = 0.1 − 0.002 = 0.098
m(SnCl 2 ) 0.098 × 190
=
= 71.34.
m(SnCl 4 ) 0.001 × 261
Moles of Sn deposited =
⇒
Moles of Br 2 required =
493.
5 × 10 5
94
10 6 It 20 × 10 3
= =
t
94 F
96500
t = 14.258 hr.
pH
(from
Henderson
equation)
= 0.28,
= 4.188
Gram equivalent of Br 2 required =
494. E ° cell
E = E ° − 0.059 log [H + ][Cl − ] = E ° + 0.059 pH = 0.527 V
495. The spontaneous cell reaction is Cu + Br 2
Cu 2+ + 2Br –
E ° = 0.73 V
0.059
log (0.05)(0.1) 2 = 0.827 V
2
In order to reverse the cell reaction, minimum voltage requirement = 0.827 V
After deposition of 2.827 g Cu :
m mol of CuBr 2 left = 50 − 44.87 = 5.13 ⇒ [CuBr 2 ] = 0.01026 M
0.059
E cell = 0.73 −
log (0.01026)(0.02052) 2 = 0.888 V
⇒
2
Thus, at this stage, minimum voltage required to reverse the reaction is 0.888 V.
nE ° F
496. For
I2 + I−
I 3− E ° = − 0.001 V ⇒ ln K =
⇒ K c = 0.925
RT
At the beginning : E cell = 0.73 −
⇒
0.925 =
[I –3 ]
−
0.25[I ]
⇒ 0.925 =
x
0.25(0.25 − x )
Solving, x = 0.158 M.
Tl → Tl + + e
497.
Cd
Cell reaction :
2+
0.34
+ 2e → Cd − 0.40
2Tl + Cd 2+ → 2Tl + + Cd
E = − 0.06 −
E ° = − 0.06
+ 2
[Tl ]
0.059
log
= − 0.036 V
2
[Cd 2+ ]
498. The cell reaction before adding KCl is :
M + H + → M + +
1
H2
2
E = − 0.4 = E ° − 0.059 log
[M + ]
[H + ]
401
Solutions
⇒
E ° (M + / M) = + 0.4177 V
[Cl − ] = 0.1 M
After adding KCl,
Cell reaction :
M + Cl − → MCl + e
H + + e →
1
2
H2
M + Cl − + H + → MCl + 2 H 2
1
⇒
⇒
⇒ E ° for MCl
⇒
− 0.05 = E ° − 0.059 log
1
[Cl ][H + ]
E ° = 0.05 ⇒ E ° ( MCl / M ,Cl – ) = − 0.05
M + + Cl – = − 0.05 − 0.4177 = − 0.4677 V = 0.059 log K sp
K sp = 1.18 × 10 −8 .
0.987 = 0.8 − 0.059 log
499.
⇒
–
[H + ] = 5.4 × 10 −4 M
Ka =
[H + ]
0.8
[H + ]2
= 2.09 × 10 −4
[HOCN]
500. [Ag + ] in a saturated solution of AgCl = 1.4 × 10 −9
[Ag + ] in a saturated solution of AgBr = 3.3 × 10 −10
The spontaneous direction of reaction will increase concentration of Ag + in AgBr/KBr solution
and decrease Ag + in AgCl/KCl solution.
Therefore, spontaneous cell reaction is
Ag → Ag + ( R )
Ag + ( L) → Ag
Ag + ( L) → Ag + ( R )
E = 0 − 0.059 log
[Ag + ]R
[ Ag + ]L
0.76 = − E ° Zn 2+ / Zn −
501.
⇒
[Zn 2+ ]
0.059
log
2
[H + ]2
E ° Zn 2+ / Zn = − 0.76 V
0.24 = − E ° Ni 2+ / Ni −
⇒
= 0.037 V
E ° Ni 2+ / Ni = − 0.24 V
[Ni 2+ ]
0.059
log + 2
2
[H ]
402
Problems in Chemistry
Zn + Ni 2+
⇒ For
Zn 2+ + Ni
1.5 − x
E ° = 0.52 V =
x
[Zn 2+ ]
x
2.303RT
= 1.99 ×1017
log
⇒
2+
1.5 − x
2F
[Ni ]
Such a large value of equilibrium constant indicate that all Ni 2+ has been quantitatively reduced to
Ni.
Amount of Zn consumed = 1.5 × 0.75 = 1.125 moles
E = E° −
502. (a)
[Mg 2+ ]
0.059
0.059
log
log = 3.14 V
= 3.17 −
2
+
2
2
[ Ag ]
(b) As the reaction proceed, [Ag + ] will decrease and [Mg 2+ ] will increase.
m mol of Ag + initially present = 34.6
m mol of Ag + present finally = 34.6 − 11.11 = 23.48
[Ag + ] = 0.0678 M
⇒
m mol of Mg + produced = 5.55
[Mg + ] = 0.119 M
0.059
0.119
log
= 3.128 V
E = 3.17 −
2
(0.0678) 2
⇒
E = E° −
503. (a)
[Zn 2+ ]
0.059
0.059
0.2
= 1.1 V
log
log
= 1.1 −
2+
0.2
2
2
[Cu ]
(i) On adding NH 3 ( aq ) to Cu 2+ solution, complex Cu(NH 3 ) 24+ will be formed reducing [Cu 2+ ],
E cell will decrease (ii) On adding NH 3 ( aq ) to Zn 2+ solution, complex Zn(NH 3 ) 24+ will be formed
reducing [Zn 2+ ], E cell will increase.
(b) After adding NH 3 , following equilibrium will establish:
Cu 2+ + 4NH 3
0.68 = 1.1 −
Now,
Cu(NH 3 ) 2+
4
[Zn 2+ ]
0.059
log
2
[Cu 2+ ]
[Cu 2+ ] = 1.16 × 10 −15 M
Solving,
Such low concentration of Cu 2+ indicate that almost all Cu 2+ has been consumed in complex
formation.
[Cu(NH 3 ) 2+
4 ] = 0.1 M, [NH 3 ] = 1.1 M
⇒
Kf =
504. E ° =
∆S °
∆H °
T−
nF
nF
0.1
4
(1.1) × 1.16 × 10 –15
= 5.88 × 10 13 .
Solving for E ° at 25°C and 20° gives
∆S ° = − 30.88 JK −1
and
∆H ° = − 77.23 kJ and
∆G° = − 68.03 kJ
403
Solutions
505. For the given cell, Nernst equation can be written as :
E = C + 0.059 pH, where C is a constant
⇒
0.129 = 0.059(6.86 − pH1 ) ⇒ pH1 = 4.68
0.059
506.
0.348 = 0.13 −
log [Pb 2+ ] ⇒ [Pb 2+ ] = 4.075 × 10 −8
2
K sp = [Pb 2+ ][ F − ]2 = 4.075 × 10 −8
507. The spontaneous cell reaction will be :
Zn 2+ (T f = − 2)
E =0−
Zn 2+ (T f = − 0.9)
0.059
0.9
= 0.01 V
log
2
2
508. The discharging reaction is :
Pb + PbO 2 + 2H 2SO 4 → 2PbSO 4 + 2H 2O
23.92
⇒
m(PbSO 4 ) produced =
× 2 × 302 = 60.7 g
238
23.92
g eq. of PbO 2 reduced =
× 2 = 0.201
238
⇒
0.201 × 96500 = 3 × t ⇒ t = 6465.5 second = 1.8 hour
2ClO –3
509. E ° for
ClO –4 + ClO –2 = – 0.03 V
0.1 − 2x
x
x
K = 9.66 × 10 −2
⇒
2
Therefore,
. × 10
966
−2
 x 
=
 ; Solving x = 0.019 M
 0.1 − 2x 
0.035 = 0.77 − 0.059 log
510.
[Fe 2+ ]
⇒
[Fe 3+ ]
NiO 2 + 4H + + 2e
2e + NiO 2 + 2H 2O
–
Adding
Ni 2+ + 4H 2O
Ni(OH) 2 + 2OH
512. Q
E ° = −1.678 V
E ° = − 0.49 V
Ni(OH) 2 + 4H + + 2OH –
2H 2O
Subtracting :
[Fe 3+ ]
= 2.87 × 10 12
Ni 2+ + 2H 2O
511.
[Fe 2+ ]
Ni 2+ + 2H 2O
dE ° ∆S °
=
dT
nF
2H + + 2OH –
Ni(OH) 2 + 2H +
E ° = − 2.168 V K1 = 4.5 × 10 −74
K 2 = 10 −28
K = K1 / K 2 = 4.5 × 10 −46
⇒ ∆S ° = − 0125
.
× 2 × 96500 = − 24.125 kJ K −1
∆G ° = − nE ° F = − 2 × 0.0372 × 96500 = − 7179.6 J ⇒ ∆H ° = − 7196.43 kJ
404
Problems in Chemistry
513. Let x be the concentration of Ca 2+ ion in diluted sea water.
1
0.059
log
− 0.053 = E ° −
2
x
0.059
50
and
log
− 0.0422 = E ° −
2
50x + 0.05
0.0108 =
⇒
50x + 0.05
0.059
log
2
50x
⇒ x = 7.55 × 10 −4
[Ca 2+ ] in sea water = 10x = 7.55 × 10 −3 M
⇒
514. Moles of Cl 2 ( g ) produced =
9.6 × 10 3
= 82.05
58.5 × 2
P=
515. E = E ° − 0.059 log
516. E = 0 − 0.059 log
[H + ]
+
[Ag ]
[Ag + ]L
[ Ag + ]R
nRT
n2a
− 2 = 21.66 atm.
V − nb V
= 0.8 − 0.059 log
0.22
= 0.26 V
1.6 × 10 −10
[ Ag + ]L
⇒ 0.4312 = − 0.059 log
0.1
[Ag + ]L = 4.9 × 10 −9
⇒ K sp = [Ag + ][Cl − ] = 4.9 × 10 −9 × 0.05
= 2.4 × 10 −10
E = 0.28 −
517.
[H + ]2
RT
= 0.6885 ⇒ [H + ] = 2.57 × 10 −7
ln
pH 2
2F
or
pH = 6.6
518. For the given cell :
E = E ° + 0.059 pH ⇒ 0.112 = E ° + 0.059 × 4
…(i)
0.3865 = E ° + 0.059 pH
…(ii)
Solving, Eqs. (i) and (ii) gives pH = 8.65
519. E = E ° − 0.059 log
1
[Ag + ]
= 0.8 + 0.059 log K sp = 0.51 V
520. Voltage drop due to resistance by the circuit = 0.55 V
I=
⇒
∆V
= 0.055 A
R
Q = 055
. × 96500 = 53.075 kJ
Ag(CN) −2
521.
0.1 − x
3 × 10 −19
Ag + + 2CN −
x
x (0.05 + 2x ) 0.05x
=
≈
0.1 − x
0.1
0.05 + 2x
⇒ x = 6 × 10 −19
405
Solutions
E = 0.8 − 0.059 log
522. E ° Au + / Au − 0.059 log
1
+
[Au ]
= E ° Ag + / Ag − 0.059 log
1
[Ag + ]
= − 0.27 V
1
[Ag + ]
Substituting values of E ° and [Ag + ] = 0.1 M gives [Au + ] = 1.21 × 10 −16
2x
0.1 − x
K = 5 × 10 −39 =
1.21 × 10 −16 [CN − ]2
0.1
0.48 =
⇒ [CN − ] = 2.02 × 10 −12 M.
Zn + 2Ag +
Zn 2+ + 2Ag.
E1 = 1.52 = E ° −
[Zn 2+ ]
0.059
log
2
[Ag + ]12
E 2 = 1.04 = E ° −
[Zn 2+ ]
0.059
log
2
[Ag + ]22
523. The cell reaction is
⇒
Au + + 2CN −
Au(CN) –2
Now,
[Ag + ]12
0.059
log
2
[Ag + ]22
⇒ [Ag + ]2 = 7.31 × 10 −10
[Cl − ] = 0.2 M ⇒ K sp = 1.46 × 10 −10
524. (a) For
2Ag + + S 2–
Ag 2S
E° =
0.059
log K sp = − 1.425 V
2
2Ag + + 2e
2Ag
E ° = 0.8 V
…(ii)
Adding, Eqs. (i) and (ii) gives :
Ag 2S + 2e
2Ag + S 2−
0.766 = 0.896 −
⇒
[S 2− ] = 3.92 × 10 −5 M
⇒
E ° = − 0.652 V
0.059
1
log 2−
2
[S ]
(b)
525. Cell reaction is : H 2 ( g ) + I 2 ( s)
2H + + 2I −
E ° = 0.535
0.755 = 0.535 − 0.059 log [H + ][I – ] ⇒ pH = 2.72
526. For the cell reaction :
1
2
H 2 + AgCl
…(i)
Ag + H + + Cl –
0.47 = 0.23 − 0.059 log [H + ][Cl – ] ;
pH = 3.07
0.5 = 0.23 − 0.059 log [H + ][Cl – ];
pH = 3.58
406
Problems in Chemistry
If there is x mol of acid and y mol of NaOH in its 20 mL, then
y
2.5 y
and 3.58 = pK a + log
3.07 = pK a + log
x− y
x − 2.5 y
0.51 = log
⇒
pK a = 3.07 − log
and
527. Faraday of electricity passed =
⇒
528. 2I –
2.5( x − y)
x − 2.5 y
y
= 3.89 ⇒ K a = 1.3 × 10 −4
6.6 y
1.25 × 1.1 × 3600
= 0.0513
96500
meq of KMnO 4 required = 51.3 ⇒
I 2 + 2e F =
⇒ x = 7.6 y
M (KMnO 4 ) =
I × 3 × 60
It
=
= 9.3 × 10 −3
96500
96500
⇒ I = 4.98 A
Total meq of I − = 9.3 + 36.3 × 0.1 = 19.93 ⇒
529. [Ag + ]R = 8 × 10 −13
⇒ E = 0 − 0.059 log
530. For the cell reaction
2Hg + 2Cl − + 2H +
− 0.52 V = − 0.28 − 0.059 log
For salt R 2 NH 2Cl : K h =
51.3 1
× = 0.41 M
25 5
M (KI) = 0.16
[ Ag + ]L
= − 0.71 V
[ Ag + ]R
Hg 2Cl 2 + H 2
1
+
[H ][Cl – ]
⇒ [H + ] = 8.5 × 10 −5 M
Kw
[H + ]2
=
⇒ K b = 1.36 × 10 −7
K b [ R 2 NH 2Cl]
531. Since mass of Sn increasing, Sn-electrode is working as cathode and X-metal electrode as anode
and electrons are flowing from X-electrode to Sn-electrode in the external circuit. The half cell
reactions are :
Sn 2+ + 2e
X
Cell reaction :
n+
+ ne
nS 2n+ + 2 X
Sn
E ° = − 0.14 V
X
E ° = − 0.78 V
nS n + 2 X
n+
Now applying the Nernst equation:
E=
⇒
0.65 =
⇒
0.01 =
⇒
E ° – 0.0591
[ X n + ]2
log
2n
[Sn 2+ ]n
1
0.64 − 0.0591 
−2
 log 10 − n log 

2n
2
0.0591
(2 − 0.3 n)
2n
n=3
Solving :
° = 0.64 V
E cell
407
Solutions
532. (a)
The spontaneous cell reaction is :
Zn + 2Ag + ( aq )
(b)
(c)
° = E cathode
°
° = 0.8 − ( −0.76) = 1.56 V
E cell
− E anode
Applying Nernst’s equation:
1.6 = 1.56 −
⇒
log
⇒
(d)
Zn 2+ ( aq ) + 2Ag( s)
[ Zn 2+ ]
+ 2
[Ag ]
[Zn
2+
[Zn 2+ ]
0.059
log
2
[Ag + ]2
= − 1.356 ⇒
] = 4 × 10
−4
[Zn 2+ ]
10
−2
…(i)
= 4 × 10 −2
M
As we add KI to cathode chamber, some Ag + will precipitate out as:
Ag + + I – → AgI
The above reaction reducing [Ag + ] from cathode chamber. This will reduce E cell according
to Nernst’s equation (i).
533. In the above conditions, two Galvanic cell can be constructed as:
(i) Al/Cu and (ii) Fe/Cu.
° /Cu = + 0.34 + 1.76 = 2.1 V and
E Al
° /Cu = + 0.34 + 0.44 = + 0.78 V
E Fe
° , cell (i) is more likely to produce higher voltage.
Q E cell depends predominantly on E cell
E cell (i) = 2.1 −
[Al 3+ ]2
0.059
0.059
10 –6
= 2.1295 V
log
=
2.1
−
log
6
6
10 –3
[Cu 2+ ]3
E cell (ii) = 0.78 −
⇒ Cell :
[Fe 2+ ]2
0.059
1
0.059
log
= 0.78 −
log
= 0.75 V
2+
2
2
0.1
[Cu ]
Al Al 3+ ( aq ) Cu 2+ ( aq )Cu will produce greater voltage.
534. Q = It = 430 × 60 × 60 = 1.548 × 10 6 C
Q
= 16.04 = number of equivalents of Mg deposited.
96500
24
mass of Mg deposited = 16.04 ×
= 192.5 g
2
⇒ number of faraday passed =
⇒
(d) Number of gram equivalents of MgCl 2 required to be electrolyzed =
⇒
⇒
Q=
10 6
.
47.5
10 6
× 96500 = It = 500 t
47.5
t = 4.063 × 10 6 sec = 1128.6 hours.
535. (a) 2Al +6
2 +12e → 4Al ⇒ 12 moles of electron will produce 4 moles of Al.
1250 × 1.25 × 3600
(b)
Number of faradays passed =
= 58.29
96500
408
Problems in Chemistry
⇒ Mass of aluminium metal deposited = 58.29 ×
27
= 524.6 g = 0.5246 kg.
3
°
536. (a) E cell
= E ° (IrCl 63– /Ir, Cl – ) – E ° (CO 2 , H 3O + /HCOOH) = + 0.97 V
(b) For thermodynamic spontaneity of a cell reaction, ∆G° < 0.
Also
∆G ° = − nE ° F < 0 since, E ° > 0
⇒ Reaction is thermodynamically spontaneous.
537. (a) The cell reaction is :
Al + 3Ce 4+ → Al 3+ + 3Ce 3+
…(i)
° = E ° (cathode) − E ° ( anode) = + 1.443 + 1.76 = 3.203 V
E cell
°
°
E cell
= E cell
−
[Ce 3+ ]3 [Al 3+ ]
0.059
8
0.059
log
= 3.23 V
log
= 3.203 −
4
3
+
3
3
(6) 3
[Ce ]
(b) Since, Ce +4 is in limited quantity.
∆G ° = − nE ° F = − (1 mol e per mol Ce 4+ ) (3.203) × 96500 = − 309 kJ/mol Ce 4+
⇒
∆G° ( total) = − 309 × 12 = − 3708 kJ
(c) Initially [Al
Also
3+
] = 8 M, [Ce 4+ ] = 6 M, [Ce 3+ ] = 1 M
total mole of Ce 4+ present initially = 6 × 2 = 12
total mole of Al 3+ present initially = 8 × 2 = 16
total mol of Ce 3+ present initially = 2
As the concentration of Ce 4+ drop to half, 6 mol of Ce 4+ will be converted to Ce 3+ .
8
⇒
[Ce 4+ ] = 3 M, [Ce 3+ ] = 2 M = 4 M
Also, for 3 mol Ce 4+ reduction one mol Al 3+ comes in solution.
⇒ for 6 mol Ce 4+ reduction, two mol Al 3+ comes in solution.
[Al 3+ ] final = 9 M
⇒
⇒
E cell = 3.203 −
( 4) 3 (9)
0.059
log
= 3.177 V
3
(3) 3
538. (a) First of all, we need to determine E ° for Cu 2+ /Cu half-cell.
E°
Given:
Add:
⇒
Note :
+
e + Cu ( aq ) → Cu( s)
∆G°
+ 0.522 V
− 0.522 F
Cu 2+ ( aq ) + e → Cu + ( aq ) + 0.158 V
− 0.158 V
Cu 2+ ( aq ) + 2e → Cu( s)
° = − 0.68 F
∆G ° = − 2 E cell
° = + 0.34 V
E cell
Here E° is determined using ∆G ° = − nE ° F relationship since, E ° is an intensive
property and can’t be manipulated algebraically.
409
Solutions
Now, for the Daniell cell:
Zn + Cu 2+ ( aq )
Zn 2+ ( aq ) + Cu( s)
° = E cathode
°
° = 0.34 + 0.76 = 1.1 V
E cell
− E anode
Cu 2+ = 2 × 4 = 8
(b) Initial moles of
⇒
moles of Cu 2+ finally = 2 × 0.4 = 0.8
⇒
moles of Cu 2+ deposited = 8 − 0.8 = 7.2
⇒
⇒
moles of Zn 2+ produced during deposition of Cu 2+ = 7.2
final moles of Zn 2+ in anode chamber = 2 × 4 + 7.2 = 15.2
finally [Zn 2+ ] =
⇒
⇒
° −
E cell = E cell
15.2
= 7.6 M
2
[Zn 2+ ]
0.059
7.6
0.059
= 1.062 V.
log
= 1.1 −
log
+
2
0.4
2
2
[Cu ]
539. (a) Ni(OH) 3 ( s) is cathode:
Ni(OH) 3 ( s) + e → Ni(OH) 2 + OH –
Cd(OH) 2 + 2e → Cd + 2OH –
…(i)
…(ii)
2 (i) – (ii)
2Ni(OH) 3 ( s) + Cd → 2Ni(OH) 2 + Cd(OH) 2
(b) E ° ( Nicad ) = 0.49 + 0.81 = 1.3 V
E ° for MgCl 2 → Mg + Cl 2 is − 3.74 V, –ve because the reaction is being driven in a
thermodynamically non-spontaneous direction. To make the reaction proceed as written, the nicad
would have to supply atleast 3.74 V. The voltage calculated for nicad battery is 1.3 V, so that nicad
battery can’t drive this electrolysis. It would take 3 nicad batteries in series to have enough voltage
to do this:
3 × 1.3 = 3.9 V > 3.74 V.
(c) Q = 0.1 × 12 × 3600 C = 4320 C
number of faradays = Q/96500 = 0.0447 F
⇒
mass of Mg deposited = 0.0447 × 12 = 0.54 gram.
⇒
0.059
(d) Nernst equation is:
E = E° −
log Q.
2
All the reactant in nicad battery is solid and the products are also solid and not part of Q. Therefore,
as the cell approach equilibrium, there is no change in Q and no change in voltage. This is a
fundamental idea on the design of most modern batteries.
540. (a) The balanced cell reaction is:
PbO 2 + H 2SO 4 + CO → PbSO 4 + H 2O + CO 2
E ° = E ° (PbO 2 /PbSO 4 ) – E ° (CO 2 /CO) = 1.685 + 0.12 = 1.905 V
(b) Initial moles of H 2SO 4
= 1.3 × 3000 ×
40
1
×
= 15.918
100 98
410
Problems in Chemistry
Initial moles of H 2SO 4 = 1.2 × 3000 ×
20
1
×
= 7.347
100 98
moles of H 2SO 4 consumed = 8.571
⇒
Q It is evident from the given cell reaction that per mole of H 2SO 4 consumed involves transfer of
two moles of electrons, total electron transferred =17.142.
⇒
Q(withdrawn) = 17.142 × 96500 C
Ampere-hour = Q/3600 = 459.5
(c) The Nernst equation tells us that voltage depends on Q(reaction quotient).
PCO2
In the given cell Q =
.
[H 2SO 4 ] PCO
541. (a) Glu → Gln + 2e + 2H + (
CH(OH) →
C==O + 2H + + 2e) E ° = − 0.29 V…(i)
4–
Fe(CN) 3–
6 + e → Fe(CN) 6
Cell-reaction:
(i) + 2 (ii) Glu + 2[Fe(CN) 6 ]
3–
E ° = + 0.69 V …(ii)
+
→ Gln + 2H + 2[Fe(CN) 6 ]4– E ° = + 0.40 V
(b) Q E ° = 0.40 > 0, ∆G° < 0, cell is spontaneous.
(c) Nernst equation is:
E = E° −
[Gln][H + ]2 {[Fe(CN) 6 ]4– }2
.
0059
log
2
[Glu] {[Fe(CN) 6 ]3– }2
Increasing concentration of glucose will increase cell-potential.
° > 0 ⇒ M → Zn and X → Ni and net cell reaction is:
542. (a) For spontaneous cell reaction, E cell
Zn + Ni 2+
° −
E cell = E cell
(b)
Zn 2+ + Ni
° = + 0.51 V
E cell
[Zn 2+ ]
0.059
0.059
log
= 0.51 −
log 0.01 = + 0.569 V
2+
2
2
[ Ni ]
During use, [Zn 2+ ] increases and [Ni 2+ ] decreases by the same factor. Therefore, when
[Zn 2+ ] = 1.0 M, [Ni 2+ ] = 0.01 M
0.059
1
log
= + 0.451 V
2
0.01
Note : Equal volume of electrolytes have been considered in the two half-cells.
(c) If salt bridge is removed, electrolysis will stop immediately due to accumulation of Zn 2+ ion
around Zn-electrode making solution positively charged. This will prevent further oxidation
of Zn into Zn 2+ ion. Hence, cell voltage will become zero.
nE ° F 2 × 0.55 × 96500
543. (a)
∆G ° = − nE ° F = − RT ln K ⇒ ln K =
=
= 42.84
RT
8.314 × 298
E cell = 0.51 −
⇒
K = 4.046 × 1018 .
411
Solutions
E°
1
O (g )
2 2
(b) Given:
+
+ 2H + 2e → H 2O
+1.23 V
+ H 2O → H 2O 2
− 0.55 V
1
O (g )
2 2
O 2 ( g ) + 2H + + 2e → H 2O 2
Adding:
⇒
∆G°
− 2 × 1.23 F
+ 0.55 F
∆G ° = − 2E ° F = − 1.91 F
E ° = 1.91/ 2 = 0.955 V.
Q = It = 100 × 1.5 × 3600 = 5.4 × 10 5 C
544.
⇒
Cu 2+
Also,
Q
= 5.596 = number of gram equivalent of Cu 2+ deposited mass of
96500
63.5
deposited as Cu = 5.596 ×
=177.673 g
2
vol. of O 2 ( g ) produced at anode at STP = 5.596 × 5.6 = 31.34 L.
number of faradays =
initial moles of Cu 2+ = 4.0
Also;
final moles of Cu 2+ = 4.0 −
⇒
molarity of CuSO 4 in final solution =
5.596
= 1.2
2
1.2
= 0.6 M
2
Cu( s) → Cu + ( aq ) + e
545. (a)
Cu
adding
2+
E ° = − 0.52 V
+
( aq ) + e → Cu ( aq )
Cu( s) + Cu
2+
E ° = 0.16 V
+
( aq )
E ° = − 0.36 V
2Cu ( aq )
Negative E ° value indicates that the above reaction is non-spontaneous.
Also
∆G ° = − nE ° F = − RT ln K
nE ° F − 0.36 × 96500
= − 14.02 ⇒ K = 8.14 × 10 –7
=
ln K =
⇒
8.314 × 298
RT
(b) The reaction is combination of:
Cu( s) + Cu 2+ ( aq )
2Cu + ( aq ) + 2Br – ( aq )
Adding : Cu( s) + Cu 2+ ( aq ) + 2Br – ( aq )
2Cu + ( aq )
K = 8.14 × 10 −7
2CuBr( s)
K=
2CuBr( s)
K=
1
K
sp 2
8.14 × 10 –7
k sp 2
Now, E ° for the given cell is 0.5147 V.
⇒
⇒
E ° F = RT ln K
⇒ ln K =
K = 5.086 × 10 8 =
0.5147 × 96500
= 20.04
8.314 × 298
8.14 × 10 −7
K sp 2
⇒ K sp =
8.14 × 10 –7
5.086 × 10
8
= 4 × 10 –8 .
412
Problems in Chemistry
° = 1.36 + 0.76 = 2.12 V
E cell
546. (a)
⇒
. −
E cell = 212
(0.04)
[Zn 2+ ]
.
0059
0.059
log
log
= 2.12 −
= 2.025 V
–
2
2
2
[Cl ]
(5 × 10 –3 ) 2
(b) On adding ammonia, following reaction will go to right quantitatively:
Zn 2+ ( aq ) + 4NH 3
Initial
Eqn.
⇒
Now :
K f = 7.8 × 10 8 =
E = E° −
Zn(NH 3 ) 2+
4
0.04
1M
0
x
1 − 0.16
0.04
[Zn(NH 3 ) 2+
4 ]
2+
[Zn ][NH 3 ]4
=
(0.04)
x (0.84)
4
K f = 7.8 × 10 8
⇒ x = [ Zn 2+ ] = 10 −10 M
[Zn ]
0.059
10 −10
0.059
2.12
–
log
=
= 2.28 V
log
2
2
(5 × 10 −3 ) 2
[Cl – ]2
2+
547. The galvanic cell reaction is :
2Ag + + Cu
and
E = E° −
[Cu 2+ ]
0.059
0.059
log
log [Cu 2+ ] + 0.059 log [Ag + ]
= E° −
+ 2
2
2
[Ag ]
E = K + 0.059 log [Ag + ]
⇒
where
⇒
⇒
2Ag( s) + Cu 2+
K = E° −
0059
.
log [Cu + ] = constant in the given problem.
2
0.382 = K + 0.059 log [Ag + ]1
0.372 = 0.059 log
[Ag + ]1
+
[Ag ]2
and 0.01 = K + 0.059 log [Ag + ]2
⇒
0.01
[Ag + ]2
= 2.0187 ×10 6
[Ag + ]2 (final) = 4.95 × 10 –9 M
⇒
Also, initial mmol of Cl – (added) = 2 × mmol of Ca 2+ = 2 × 0.02 × 250 = 10
Initial mmol of Ag + = 250 × 0.01 = 2.5
⇒ mmol of Cl – left after precipitation = 7.5 ⇒ [Cl – ] final = 0.03 M
⇒
K sp ( AgCl) = [Ag + ]2 (0.03) = 4.95 × 10 –9 × 0.03 = 1.485 × 10 –10 .
Q = It = 2.5 × 35 × 60 = 5250 C
548.
Q
= 0.0544 = number of equivalent of metal deposited.
96500
3.06
Equivalent mass of metal =
= 56.25 ⇒ Molar mass =112.5 amu.
⇒
0.0544
549. (a) The spontaneous cell reaction is:
⇒ Number of faradays =
2Sn 2+ ( aq ) + O 2 + 4H +
2Sn 4+ ( aq ) + 2H 2O
° = 1.08 V
E cell
413
Solutions
Applying Nernst equation :
2
. −
E = 108
4
.
0059
 0.01   1 
log 
 
 = 0.8735 V
 0.10   10 –4 
4
At any instant of uses, if x M Sn 2+ concentration is decreased, [Sn 4+ ] will increase by the
same amount.
  0.01 + x  2

0.059
16
log  
0.8 = 1.08 −
⇒
 × 10 
4
  0.10 − x 

4+
Solving
x = 0.0966 M ⇒ [Sn ] = 0.1066 M; [Sn 2+ ] = 0.0034 M
550.
E1 = E ° + 0.059 pH1 and E 2 = E ° + 0.059 pH 2
⇒
E 2 − E1 = 0.16 = 0.059 (pH 2 – pH1 ) ⇒ pH 2 = 12.11
551. The electrolysis reaction is :
2Cu 2+ + 2H 2O → 2Cu + 4H + + O 2
4 × 30 × 60
Number of faradays of electricity passed =
96500
= 0.0746 = number of gram equivalent of H + produced.
0.0746
molarity of H + =
⇒
= 0.373 ⇒ pH = 0.43
0.20
552. Iron will be deposited first. The cell reaction would be :
(b)
Fe 2+ + H 2O → Fe + 2H + + 12 O 2
E ° = − 0.44 – 1.23 = − 1.67 V
⇒
E = − 1.67 –
[H + ]2
0.059
log
= − 1.4 V
2
[Fe 2+ ]
⇒ A minimum of 1.4 V would be required for the onset of electrolysis.
553. Reduction half-reaction is :
M (+aq ) + e → M( s)
E1 = E ° − 0.059 log
⇒
1
[M + ]1
= E ° + 0059
.
log [M + ]1
E 2 = E ° + 0.059 log [M + ]2 = E ° + 0.059 log 10 −2 [M]1+
= E1 + 0.059 log 10 –2 = E1 − 0.118 V
i.e., lowering applied potential by 0.118 V.
° = 1.66 – 0.44 = 1.22 V
E cell
554.
E cell = 1.2 = 1.22 −
⇒
 [Al 3+ ]2 
log  2+ 3  = 2.033
[Fe ] 
[Al 3+ ]2
0.059
log
6
[Fe 2+ ]3
414
Problems in Chemistry
[Al 3+ ]2
⇒
1.25 × 10
⇒ moles of Al
3+
= 108 ⇒ [Al 3+ ] = 1.16 × 10 –4 M
–10
in sauce = 1.16 × 10 –4 × 0.1 = 1.16 × 10 –5
⇒ mass of Al dissolved = 1.16 × 10 –5 × 27 = 3.132 × 10 –4 g
⇒ vol. of of Al-foil dissolved =
3.132 × 10 –4
= 1.16 × 10 −4 cm 3
2.7
V
1.16 × 10 –4
=
= 1.16 × 10 −3 cm 2 .
thickness
0.1
1500
555.
mol H 2 required
1500 = n(28.8 − 2) = 26.8 n ⇒ n =
26.8
⇒ 2n faraday of electric current would be required
1500
⇒
Q =2×
× 96500 = 8.5 t. ⇒ t = 353 hours.
26.8
556. The electrodes reaction are :
⇒ area of holes =
Zn → Zn 2+ + 2e
0.76 V
+
2H + 2e → H 2
Cell reaction:
0.00 V
Zn + 2H + → Zn 2+ + H 2
E ° = + 0.76 V
Now, applying Nernst equation :
E = 0.72 = 0.76 –
For HIO 3 :
0.059
0.1
log + 2
2
[H ]
⇒ [H + ] = 0.0664 M
HIO 3
H + + IO –3
0.1 − x
x
Ka =
x
(0.0664) 2
x
=
= 0.13
01
. − x 0.1 – 0.0664
2
557. mmol of Cu 2+ electrolyzed in 30 min = (0.1 – 0.0528)100 = 4.72
meq of Cu 2+ deposited = 9.44 = number of mF. Q Q = It
⇒
9.44 × 10 –3 × 96500 = I × 30 × 60 ⇒ I = 0.5 A
i.e., meter is showing only 80% of the actual current.
558. The spontaneous cell reaction is :
2Cr 2+ + Ni 2+
E = 0.11 V = 0.16 −
⇒
2Cr 3+ + Ni
+3 2
E ° = + 0.16 V
[Cr ]
0.059
2.56 × 10 –4
0.059
log
=
0.16
−
log
2
2
[Ni +2 ]
[Cr +2 ]2 [Ni +2 ]
[Ni 2+ ] = 5.16 × 10 –6 M.
415
Solutions
559. (a) The electrode reactions occurring are:
2H 2O + 2e → H 2 + 2OH –
: at cathode
2H 2O → 4H + + O 2 + 4e : at anode
(b)
(c)
Since, phenolphthalein is colourless below a pH value of 8 and pink above a pH values of 10.
At the start of electrolysis, both the solutions will be colourless. As electrolysis proceeds,
cathode compartment will become basic and will acquire pink colouration while anode
compartment will become more and more acidic and remains colourless.
If electrolysis is stopped at any stage, the two compartments, cathode and anode
compartment, will have same number of gram equivalent of hydroxide and hydronium ions.
Therefore, if the content of two compartments are mixed, the net result will be formation of a
neutral solution and it will be colourless.
As electrolysis proceed, HO – will be produced and will be neutralized with HCl added from
out side. At end point when the whole HCl is neutralized, solution will acquire alkaline
nature if electrolysis is carried on further and will become pink coloured.
⇒
Q=
25 × 10 −3 × 8 × 60
F = 1.24 × 10 –4 = equivalents of HCl in 10 mL.
96500
⇒
Molarity of HCl =
1.24 × 10 –4
= 0.0124 M
0.01
HgI 2–
4 + 2e
560. Given :
Hg + 4I –
Hg 2+ + 2e
Hg
Adding :
E ° = − 0.04 V
HgI 2–
4
Hg
2+
+ 4I
E ° = − 0.85 V
–
E ° = − 0.89 V
Writing Nernst equation :
E = − 0.89 −
E = 0 = − 0.89 −
At equilibrium
⇒
[Hg 2+ ][I – ]4
0.059
log
2
[HgI 2–
4 ]
K d = 6.768 × 10 –3 =
1
Kf
0.059
log K d
2
⇒ K f = 1.47 × 10 30 .
CHEMICAL KINETICS
WU
=5
WPb
561.
U 238 →
N0 − N
Also,
⇒
Pb 206
NU
5 × 206
:
=
N Pb
238
⇒
N
λt = ln
N0
N0 − N
N 0 − N 1030
=
238
N
…(i)
416
Problems in Chemistry
Q
N
235
=
N 0 − N 1030
⇒
N0
238 + 1030 1268
=
=
N0 − N
1030
1030
t=
⇒
1
1.54 × 10
−10
ln
1268
= 1.35 × 10 9 year.
1030
7
80
p = 240 mm ⇒ p =
mm
2
7
200
200
and 140 K = ln
15 K = ln
⇒
80
200 − p′
200 −
7
Solving p′ = 84.5 mm
7
⇒
PTotal ( after 140 min ) = 200 + p′ = 495.75 mm.
2
563. Let after 30 min, partial pressure of 1,3-butadiene be p mm.
55
30 × 60 × 2 × 10 −4 = ln
⇒ p =16.63
⇒
55 − p
562. P0 = 200 nm
After 15 min : P0 +
⇒
p(cyclobutene) = 38.37 mm
[cyclobutene] = 1.45 × 10 −3 M
564. (a)
ln
K 2 E a  T2 − T1 
=


K1
R  T1T2 
Also, ln K1 = ln A −
⇒
(b)
Ea
RT1
8.5 × 10 −11
2.1 × 10
−11
=
Ea 

10


8.314  300 × 310 
⇒ E a = 108.1 kJ
108.1 × 10 3
8.314 × 300
A = 13.97 × 10 7
⇒ ln 2.1 × 10 −11 = ln A −
ln K ( 47° C) = ln A −
Kt = ln
Ea
R × 320
0.15
1.65 × 10 −7
λ=
565. (a)
(b)
⇒ ln
⇒
⇒ K ( 47° C) = 3.155 × 10 −10 S −1
⇒ t = 6.53 × 10 11 second
ln 2
= 6.32 × 10 −3 min −1
t1/ 2
5.6 × 60 × 6.32 × 10 −3 = ln
100
X
⇒
X = 12%
6.32 × 10 −3 t = ln 100 ⇒ t = 12.14 hours
(c)
566. 1 H 3 in fresh 10 g H 2O :
x = 5.35 × 10 6
x
= 8 × 10 −18
10
× 2 × 6.023 × 10 23 − x
18
⇒
5.35 × 10 6
ln 2
× 40 = ln
12.3
N
⇒
N = 5.6 × 10 5 1 H 3
417
Solutions
2N 2O 5 → 4NO 2 + O 2
567.
p0 − p
2p
PT = p0 +
p
2
3
p
2
5
p0 = 584.5 mm ⇒ p0 = 233.8 mm
2
3
After 30 min : PT = 284.5 = 233.8 + p ⇒ p = 33.8
2
233.8
1
ln
K=
= 5.2 × 10 −3 min −1
⇒
30
200
233.8
After 1.0 hour; 60 × 5.2 × 10 −3 = ln
⇒ p′ = 62.71
233.8 − p′
After infinite time PT =
3
p′ = 327.87 mm.
2
568. Mass of X decayed in 20 days = 0.75 g = mass of He ( g ) produced.
PT = p0 +
⇒
Volume of He (S.T.P.) =
ln
569.
Kc
E − Ec
= uc
K uc
RT
ln K c = ln A −
⇒
⇒ ln 500 =
90 × 10 3
8.314 × 400
and
0.75
× 22.4 = 4.2 L
4
(106 − E c ) × 10 3
8.314 × 310
ln K uc = ln A −
 K (127° C) 
106 × 10 3
90 × 10 3
=
−
ln  c

 K uc (27° C)  8.314 × 300 8.314 × 400
⇒
⇒ E c = 90 kJ
106 × 10 3
8.314 × 300
K c (127° C)
= 5.05 × 10 6
K uc (27° C)
570. Let p0 be the initial partial pressure of A2 ( g ).
A2 ( g )
2 A(g )
p0 − p1
p0 − p1
I. at 30 min
II. at 60 min
2 p1
2 p2
I.
PTotal = p0 − p1 + 2 p1 + 700 − p0 = p1 + 700 = 760 ⇒
II.
PTotal = p2 + 700 = 800 ⇒ p2 = 100
p0
p0
and 60 K = ln
30 K = ln
p0 − 60
p0 − 100
Now,
Solving :
p0 = 180 mm
After 75 min : 75 K = ln
⇒ K = 13.5 × 10 −3 min −1
180
180 − p3
⇒
⇒ t1/ 2 = 51.28 min.
p3 =114.68 ⇒ PT = 700 + p3 = 814.68 mm
E af = 11420 × 2.303R = 218.66 kJ
571.
⇒
p1 = 60
E ab = E af − ∆H = 115.66 kJ mol −1
418
Problems in Chemistry
1
N 2O 5 → 2NO 2 + O 2
2
572.
1−α
2α
α
2
Let x be the mole fraction of O 2
r (NO 2 ) 4 32 0.5
=
=
46
r (O 2 )
x
r (N 2O 5 ) 1 − α 46 0.5 − x
=
=
2α 108
r (NO 2 )
x
Kt = ln
Now,
 1
573. m = m0  
 2
⇒
1
1−α
⇒
ln 2 1
1
= ln
t1/ 2 15 0.69
⇒ x = 0.15
⇒ α = 0.31
⇒ t1/ 2 = 27.17 min.
n
where, n = number of half-lives.
 1  1 2  1 3  1 4  1 5 
m = 5 1 + +   +   +   +    = 6.66 mg
 4  
 4
 4
 4  4 
574. Meq I 2 taken = 15, meq of I 2 reacted with hypo = 7
⇒ meq of I 2 consumed by SO 2 = 8 = Meq of SO 2
⇒ m mol of SO 2 produced = 4
⇒ After 4.0 hours, total m mol of gases =14
14 × 10 −3 × 0.082 × 400
= 2.296 atm
0.2
ln 2 1 10
= ln
⇒ t1/ 2 = 5.43 hour
t1/ 2 4
6
P=
ΣE i K i 180 × 10 3 × 1.2 × 10 −2 + 200 × 10 3 × 3 × 10 −2
= 194.28 kJ
=
ΣK i
1.2 × 10 −2 + 3 × 10 −2
576. Let at equilibrium degree of conversion be x.
x
= 0.16 ⇒ x = 0.138
⇒
1− x
575. E =
Also,
K1
= 0.16 ⇒ K 2 = 6.25 K1
K2
Now,
dx
= K1 (1 − x ) − K 2 x = K1 (1 − 7.25x )
dt
⇒
⇒
dx
= K1 dt
1 − 7.25x
1
1
ln
= K1 t
7.25 1 − 7.25x
419
Solutions
For half equilibrium reaction : t =
1
ln
7.25K1
1
1 − 7.25 ×
x
2
= 290 second
K (Ca) t1/ 2 ( Ar )
n(Ca)
=
=8=
= n( K ) = 52.64
K (Ar) t1/ 2 (Ca )
n(Ar)
577.
K ( K ) = K (Ca ) + K (Ar ) = 5.198 × 10 −10 y −1
⇒
t=
1
100
ln
= 1.23 × 10 9 years
K 52.64
α 0 = α ( t = ∞ ) − α ( t = 0) = 35° ⇒ 10 K = ln
578.
20 K = ln
and
35
35
= 2 ln
α
25
35
25
⇒ α =17.85
0 ⋅ R after 20 min = 17.85 + 5 = 22.85° .
m mol of reactant consumed after 20 min = 10 ×
17.15
= 4.9 m mol.
25
m mol of alkene = 2.94 = m mol of Br 2 = 0.15 V ⇒ V = 19.6 mL
∆H R = − 3.7 kJ and
579.
E af − E ab = − 3.7 ⇒ E ab = 53.5 kJ
 (T )  E 

5
ln  1/ 2 1  = a 

 (T1/ 2 ) 2  R  333 × 338 
580.
E=
581.
E=
⇒
E a = 351.13 kJ
ΣE i K i E1 K1 + E 2 K 2 + E 3 K 3
=
ΣK i
K1 + K 2 + K 3
K1 10
=
K 2 25
Also,
and
and
K1 10
=
K 3 15
E1 K1 + 2.5K1 E 2 + 1.5K1 E 3
= 79 kJ
K1 + 2.5K1 + 1.5 K1
1
= 0.223 hr. −1
0.8
0.8
K (37° C) = ln
= 0.693 hr. −1
0.4
K (27° C) = ln
582.
⇒
ln

10
0.693 E a 
=


0.223 R  300 × 310 
K (T ) = ln
⇒
ln =
⇒ E a = 87.63 kJ
0.4
= 1.386 hr. −1
0.1
1.386 87.63 × 10 3  T − 310 
=


 310T 
0.693
8.314
⇒ T = 316.45 K
420
Problems in Chemistry
20K = ln
583.
⇒
ln
p0
p0 − p1
and
p0
p0 − p2
40K = ln
p0
p0
= 2 ln
p0 − (2.875 − p0 )
p0 − (2.5 − p0 )
p0
p02
=
2 p0 − 2.875 (2 p0 − 2.5) 2
⇒
Solving, p0 =1.5625, 2
If
p0 = 2 then 60K = ln
pB
1
=
pC 15
584.
⇒
60( K1 + K 2 ) = ln
2
3
= 3 ln
2− p
1.5
⇒
p =1.156,
⇒
⇒
p0 = 15 × 40 + 40 = 640
pC =15 pB
640
640 − P
PT = 3.156 atm.
⇒ P = 546.17 = pB + pC = 16 pB
⇒
pB = 34.135
pC = 512.025 mm of Hg.
585. Initial m mol of cyclobutene =10
Let after 20 min, x m mol cyclobutene isomerized.
m mol of cyclobutene left = 10 − x and m mol of diene formed = x
m mol of Br 2 required after 20 min = 10 − x + 2x = 10 + x = 16
10
20K = ln
⇒
x =6
⇒
4
If y m mol of cyclobutene isomerized after 30 min.
10
30K = ln
10 − y
From Eqs. (i) and (ii)
y = 7.47
⇒ m mol of Br 2 required = 10 + y = 17.47
⇒ Vol. of bromine solution required = 17.47 mL
586.
A
→ A ′
α
α 0 − α1
− α1
⇒ α 0 − 2α1 = 40 ⇒ α1 = 0 − 20
α0 −α2
2
α0
− 12
⇒ α 0 − 2α 2 = 24 ⇒ α 2 =
2
−α2
10K = ln
⇒
α0
0.5α 0 + 20
and 20K = ln
α0
0.5α 0 + 12
Solving, α 0 = 59°
⇒
10K = ln
59
49.5
and
40K = ln
59
59 − α 3
⇒ α 3 = 29.77
⇒ Optical-rotation after 40 min = α 0 − 2α 3 = 59 − 29.77 = − 0.54° .
…(i)
…(ii)
421
Solutions
2
10 e
587.
−
14000
RT
3
= 10 e
⇒ Rate constant K1 = K 2 = 0.464 hr.
⇒
−
20000
RT
Solving, T = 313.42 K
−1
K1 t = ln
n0 ( A2 )
n0 ( A2 ) − n1
⇒ n1 = 0.37
K 2 t = ln
n0 ( B 3 )
n0 ( B 3 ) − n2
⇒ n2 = 0.37
⇒ Total moles of gases after 1.0 hr. =1.37 +1.74 = 2.11
P = 5.42 atm.
588. Let α be the degree of dissociation after 12 hr.
⇒ P = P0 x1 From Raoult’s law.
20.69
10
10
 10 
=
=
⇒ α = 0.3 ⇒ 12K = ln  
⇒
 7
24
10 + 1 + 2α 11 + 2α
30K = ln
1
1−β
where β is degree of dissociation after 30 hr.
⇒ β = 0.6 and moles of solute after 30 hr. = 1 + 2β = 2.2
 10 
P = 24 
⇒
 = 19.67 mm of Hg.
 12.2 
589.
R2
= 4 = (2) z
R1
y
R3
 3  3
= 0.668 =    
 2  2
R1
⇒ Z(order w.r.t. C ) = 2
2
R4
 4
= 10.65 = (2) x ×   × (2) 2
 3
R3
⇒
y (order w.r.t. B) =1
⇒ x(order w.r.t. A) =1
Rate Law : R = K [ A][ B][C ] 2
590.
0.42 × 10 −3
dN
= KN = 9.88 × 10 4 = K ×
× 6.023 × 10 23
dt
347
ln 2
⇒ K=
= 1.35 × 10 −13 sec −1
t1/ 2
⇒ t1/ 2 = 5.13 × 1012 sec = 1.628 × 10 5 year
591. If x mol of U 238 and y mol of U 235 has decayed, then presently
U = 100 − ( x + y)
100 − ( x + y)
50 − x
Also,
= 50 − y ⇒
= 99
100
50 − y
422
Problems in Chemistry
Also,
t ( K 2 − K1 ) = ln
⇒
t=
ln 99
= 5.5 × 10 9 year.
K 2 − K1
K1
A
592.
50 − x
= ln 99
50 − y
1−x
K2
K1 7
=
K2 3
B,
x
dx
 10 
= K1 (1 − x ) − K 2 x = K1 1 − x 

7 
dt
7
1
= K1 t
ln
10
 10 
1 − x 

7 
⇒
…(i)
Solving at t =1 gives K1 = 0.309 hr. −1
After 4.0 hr, ln
594. log
Kf
Kb
10K1 t
1
=
⇒ x = 0.58 ⇒ mole % of A after four hour = 42%
10
7
1− x
7
= 5.457 −
5800
∆S °
∆H °
=
−
T
2.303R 2.303RT
⇒ ∆S ° = 104.48 JK −1 , ∆H ° = 111 kJ
595. Normality of chromic acid =1.2 N
Normality of KI = 1 N
meq of KI left = 10.4 × 1.2 = 12.48
Ist flask :
meq of KI ≡ H 2O 2 = 20 − 12.48 = 7.52
Initial meq of H 2O 2 = 25 ×
2.5
11.16
= 11.16 ⇒ K = ln
5.6
7.52
IInd flask : Let x meq of H 2O 2 left undecomposed.
3K = ln
11.16
x
⇒ x = 3.41 = meq of KI consumed
meq of KI reduced with chromic acid =16.59
Vol. of chromic acid required = 13.825 mL
596. R = K [H + ][ester] = K ′ [ester ] where K ′ = K [ H + ] = 3.478 × 10 −4 s −1
⇒
t=
1
100
ln
= 17 minutes
70
K′
597. Let the initial mole percentage of A is x and that of B is (100 − x ). Let after ‘ t’ time, p mole of A and
q mole of B are converted into C.
⇒
2 × 10 −2 × 30 = ln
x
x− p
⇒
x = 2.22 p
423
Solutions
5 × 10 −3 × 30 = ln
and
100 − x
100 − x
Q p + q = 25
= ln
100 − x − q
75 − x + p
= ln
100 − x
75 – 0.55x
2 × 10 −2 × 50 = ln
After 50 min:
36
36 − p
5 × 10 −3 × 50 = ln
⇒
⇒
64
64 − q
p = 22.74%
⇒ q =14.2%
mole percentage of C after 50 min = p + q = 36.94
= 71 sec.
t
1/ 2
Required nuclei ratio
O14 →
598.
solving x = 36
1− x
1− x
= 0.25
1− y
t1/ 2 = 124 sec.
O15 →
1− y
ln 2
1
t = ln
71
1− x
1− y
2 × 71 × 124
ln 2
1 
1
1
= 332.226 sec.
t = ln
⇒ ln 2  −
= ln 4 ⇒ t =
 t = ln
 71 124 
1− x
124 − 71
124
1− y
ln
599.
Kc
E − Ec
= uc
K uc
RT
Now,
ln
⇒ ln 10 =
80 × 10 3 − E c
8.314 × 300
⇒ E c = 74.25 kJ
K c (127° C) 74.25 × 10 3  100 
=


8.314
K c (27° C )
 300 × 400 
K c (127° C)
t (27° C)
= 1706.6 = 1/ 2
K c (27° C)
t1/ 2 (127° C)
⇒
100 ln 2
=
t ⇒ t = 84.4 seconds
25 42.2
Xe (CF3 ) 2 ( g ) → Xe ( g ) + C 2 F4 ( g ) + F2 ( g )
⇒ t1/ 2 (127° C) = 42.2 sec. ⇒ K t = ln
600.
Total mole :
0.005 − x
After time t : n =
x
2.63 × 0.1
PV
=
= 0.0107 = 0.005 + 2x
RT 0.082 × 300
ln 2
0.005
t = ln
30
0.005 − 2.85 × 10 −3
⇒
K1
→ β
Ac
x
x = 2.85 × 10 −3
⇒
601.
x
K2
→ α
K1
[β]
=9=
K2
[α ]
⇒ t = 36.53 sec.
424
Problems in Chemistry
nα =
Also,
PV
= 6.5 × 10 −3
RT
nβ = 9nα = 58.5 × 10 −3
⇒ Moles of Ac decayed in 5.0 hours = nα + nβ = 0.065
ln 2
0.44
⇒ t1/ 2 = 21.68 years
⇒
× 5 = ln
t1/ 2
0.44 − 0.065
t
= 1.5
1/ 2
602. NO →
After 5.0 hour
x − y + 1.5 − x − z = 0.225
x− y
t
NO 2
1.5 − x − z
Also,
= 2.5
1/ 2
→
⇒
ln 2
x
× 5 = ln
⇒
1.5
x− y
ln 2
1.5 − x
× 5 = ln
2.5
1.5 − x − z
⇒
z = 1.275 − y
…(i)
x
=10
x− y
…(ii)
1.5 − x
=4
1.5 − x − z
…(iii)
Solving Eqs. (i), (ii) and (iii)
x = 1.0 g and
y = 0.5 g
2/ 3
K K 
2
K = 1 2
⇒ E = ( E1 + E 2 − E 3 ) = 153.34 kJ
3
 K3 
604. Since, absorbance of solution is directly proportional to the concentration of coloured species
A1 C1
A (absorbance) ∝ C ⇒
=
A2 C 2
603. Q
⇒
C1
0.600
=
0.200 4.00 × 10 −5
⇒ C1 = 1.2 × 10 −4 M
Also, decomposition of X follow first order kinetics:
C
A
0.600
Kt = ln 0 = ln 0 ⇒ K × 35 = ln
= ln 3
0.200
C
A
0.600
Kt = ln
= ln 80
0.0075
ln 80
t = 35
= 139.6 minutes.
⇒
ln 3
605. (a)
The decomposition reaction is following 1st order kinetics, since, for a 1st order reaction:
[ A ]0
Kt = ln
[ A]
where [ A ]0 is initial concentration and [ A ] is concentration of reactant left unreacted at any time
“t ”.
⇒
Kt = ln [ A ]0 − ln [ A ] ⇒ ln [ A ] = ln [ A ]0 − Kt
i.e., A plot of ln [ A ] vs time is a straight line with negative slope.
425
Solutions
Rate = K [H 2O 2 ]
Hence, rate law:
(b) From the graph, at t = 0
ln [ A ]0 = 0 ⇒ [ A ]0 =1.0
At t = 2800 min,
ln [ A ] = − 3
3
= 1.07 × 10 −3 min −1
⇒
−3 = 0 − 2800 K ⇒ K =
2800
3
At 2000 min,
ln [ A ] = −
× 2000 = − 2.142 ⇒ [ A] = 0.1173 M
2800
606. Let P0 be the partial pressure of each Cl 2O and N 2O 5 in the beginning. Therefore, the total
pressure after infinite time would be 2P0 + P0 / 2.
5
⇒
P0 = 750 and P0 = 300 mm of Hg.
3
Now, let us consider “ p” be the decrease in partial pressure of both of the reactants after time “t ”,
then:
2Cl 2O + 2N 2O 5 → 2NO 3Cl + 2NO 2Cl + O 2
300 − p
10 K = ln
At 10 min.
300 − p
300
300 − p
p
and 600 +
p
p
= 650 ⇒
2
p/ 2
p =100
 3
10 K = ln  
 2
⇒
30 K = ln
At 30 min.
…(i)
300
300 − p
…(ii)
Solving Eqs. (i) and (ii) yields p = 211.
Total P = 600 + p/ 2 = 705 mm of Hg.
607. The rate expression for a general n th order reaction is:
1  1
1 
Kt =
 n −1 −

n − 1  [ A ]
[ A ]0n − 1 

1 
1
1
1
1
⇒
(2 n − 1 − 1)
=
−
400 K =
1


n −1
n −1
−
n − 1  (0.5) n − 1
n
1
[ A ]0
 [ A ]0
964 K =
⇒

1 
1
1
1
1
( 4 n − 1 − 1)
=
−
1


n −1
n −1
n −1
−
n − 1  (025
n
1
. )
[ A ]0
 [ A ]0
964 K
4n − 1 − 1
= 2 n − 1 + 1 ⇒ 2 n − 1 = 141
. = (2)1/ 2
= 2.41 = n − 1
400 K
−1
2
Substituting, n =1.5 in Eq. (i) gives
400 K =
⇒
…(i)
2( 2 − 1)
0.6
K = 2.67 × 10 −3 sec –1 ( M ) − 1/ 2
⇒ n=
…(ii)
3
2
426
Problems in Chemistry
608. From the given graph, half-life is determined to be 20 sec.
ln 2
⇒
K=
20
0.4
⇒ At 10 sec.
K × 10 = ln
[ A]
[ A ] = Concentration of reactant left unreacted after 10 sec.
ln 2
0.4
and [ A ] = 0.28 M
× 10 = ln
⇒
20
[ A]
Rate = K [ A ] =
Hence, at 10 min:
ln 2
× 0.28 = 9.7 × 10 −3 mol L–1 sec –1 .
20
609. The Arrhenius equation is:
log K (s –1 ) = log A −
Ea
2.303 RT
⇒ From the given graph,
Ea
10 4
=
3
2.303 R
and slope
log A = 3
⇒ E a = 63.82 kJ
Therefore, at 500 K,
63.82 × 10 3
= − 3.66
2.303 × 8.314 × 500
1 100
⇒ t = ln
= 7485.75 sec = 2.08 hrs.
K
20
log K (S –1 ) = 3 −
⇒
K = 2.15 × 10 −4 S –1
610. For a 1st order reaction:
R = K [ A] ⇒
[ A ]0
[ A ]0
and 30 K = ln
[ A ]10
[ A ]30
[ A ]10
= 0.239 ⇒ K = 1.2 × 10 −2 min –1 .
20 K = ln
[ A ]30
10 K = ln
Also
⇒
[ A ]30 min
similarly,
⇒
611. For reaction:
R10 [ A ]10 min
=
= 1.27
R 30 [ A ]30 min
[ A ]2 hrs
= 2.94 and [ A ]30 min =
[ A ]2 hr = 0.237 M and
A
a − xe
R 30
= 0.6975 M
K
R 2 hr = 2.84 × 10 –3 mol L–1 min –1 .
P
xe
K =2=
xe
a − xe
⇒ xe =
2
a
3
Also, at any instant:
d[ P ] dx
=
= K f ( a − x ) − K b x = K b (2a − 3x )
dt
dt
…(i)
427
Solutions
x
∫0
⇒
Now,
dx
= Kb
2a − 3x
t
∫0
⇒ Kb t =
dt
1
2a
ln
3 2a − 3x
x e 2a
ln 2
= ; Kb t =
2
6
3
ln 2 2 ln 2 2
ln 2
t=
=
= ×
= 18.48 min.
3 Kb 3 K f
3 2.5 × 10 −2
x=
at
⇒
100
5
= ln
80
4
80
In presence of 1st catalyst:
= ln 2
30 KC = ln
1
40
40
In presence of 2nd catalyst:
10 KC = ln
= ln 4
2
10
EUC − EC
KC 1
KC 1
2 ln 2
1
=
= 6.21 ⇒ ln
⇒
= ln 6.21 =
KUC
RT
KUC ln(5 / 4)
612. In absence of catalyst:
60 KUC = ln
Solving, EC = 70.44 kJ; Similarly, EC = 70.975 kJ.
1
2
 −dN 
–1
613. (a) Activity
 = λN = 12.5 count min

 dt 
⇒
N =
12.5 12.5 t1/ 2 12.5 × 5730 × 365 × 24 × 60
=
=
= 5.4 × 1010
λ
ln 2
ln 2
Total number of C-atom in 1 g =
⇒
%
14
1
. × 10 22
× 6.023 × 10 23 = 502
12
C per gram of carbon =
5.4 × 1010
5.02 × 10 22
× 100 = 1.082 × 10 −10
 −dN 
(b) Initial activity 
 = λN 0 = 12.5
 dt  0
 −dN 
Activity after 50,000 years = 
 = λN
 dt 
Also
λt = ln
N0
N
⇒
⇒
 −dN 


 dt 
N
=
12.5
N0
N
ln 2
× 50,000 = ln 0
5730
N
N0
= 423.4 Substituting in equation (i):
N
1
−dN
. ×
= 125
= 2.95 × 10 −2 count per minute per gram.
423.4
dt
(c)
12.5 = λN 0
N0
N
1
5730
ln 1.785 = 4794 years
=1.785
⇒ t = ln 0 =
7 = λN ⇒
λ
ln 2
N
N
⇒
…(i)
428
Problems in Chemistry
COLLIGATIVE PROPERTIES
614. (a) − ∆T f = 0.6 = K f ⋅ m
∆Tb = K b ⋅ m ⇒
(b)
⇒
0.6 K f 1.86
=
=
∆Tb K b 0.52
− ∆T f = 0.6 = K f m = 1.86 ×
x1 =
n1
=
n1 + n2
⇒ ∆Tb = 0.167
n2
× 1000 ⇒
18n1
n2
= 5.8 × 10 −3
n1
1
= 0.9942
n
1+ 2
n1
⇒
P = P0 x1 = 23.75 × 0.9942 = 23.61 torr.
0.6
(c)
= 0.32 ≈ Molarity
m=
1.86
π = MRT = 0.32 × 0.082 × 293 = 7.688 bar.
60
615. Mole fraction of benzene = 78 = 0.638 ⇒ Mole fraction of toluene = 0.362
60 40
+
78 92
V. P. = 0.638 × 93.4 + 0.362 × 26.9 = 69.327 mm
n
616.
+ 0.744 = 1.86 ⋅ 2 × 1000 ⇒ n2 = 0.1
250
752.7 = 760 ⋅
Also,
n1
n1 + 0.1
⇒ n1 =10.31
⇒ Mass of ice formed = 250 − 10.31 × 18 = 64.42 g.
−∆ T f = 1.86 ×
617. V.P. of solution =
⇒
0.1
× 1000 = 1 ⇒ T f = − 1° C.
185.58
0.25 × 0.082 × 300 × 760
= 21.638 mm.
18 × 12
21.638 = 23.75 ⋅
n1
n1 + n2
⇒
n2
= 9.76 × 10 −2
n1
 n  1000
= 2.82
∆Tb = K b ⋅ m = 0.52 ×  2  ×
18
 n1 
618.
619.
x + 6 10 − x
+
0.77 180
60
=
⇒ x = 5.843 g
x 10 − x
0.58
+
180
60
∆Tb = iK b m ⇒ 0.15 = 3 × 0.5 × m ⇒ m = 0.1
429
Solutions
Now,
Pb(NO 3 ) 2 + 2NaCl
0.1
0
0.2
0
→ PbCl 2 + 2NaNO 3
0
0.1
0
0.2
Now, the solution contains two salts NaNO 3 and PbCl 2 .
− ∆T f = 0.83 = K f (2 × 0.2 + 3S ) where S is molar solubility of PbCl 2 .
S = 1.54 × 10 −2
⇒ K sp = 4S 3 = 1.46 × 10 −5 .
620. In mixture of A and B if mole fraction of A = x, then :
400x + 500(1 − x ) = 470 ⇒ x = 0.3
In final mixture, if combined mole fraction of A and B = y, then
470 y + (1 − y)600 = 496 ⇒ y = 0.8
Therefore, in the final mixture, mole fractions are : A = 0.24, B = 0.56, C = 0.2
⇒ Mole fraction in vapour phase are :
A=
x A ⋅ PA0 0.24 × 400
=
= 0.1935,
P
496
B=
x B PB0 0.56 × 500
=
= 0.5645
P
496
C = 1 − (0.1935 + 0.5645) = 0.242
Moles of solute =
621.
Mass of H 2O =
0.582 = 3 × 0.52 ×
⇒
⇒
38.2
202 + 18x
38.2 × 18x
+ 250
202 + 18x
38.2 × 1000

38.2 × 18x 
(202 + 18x )  250 +

202 + 18x 

x =10, formula = Na 2 B 4 O 7 ⋅ 10H 2O
10.2 − x 
x

RT
×4+
π = π (NaCl) + π (sucrose) ⇒ 7.32 = 2 ×
342 
58.5

622.
⇒
342 × 8x + 58.2(10.2 − x ) =
7.32 × 58.5 × 342
0.082 × 296
⇒ x = 2.03 g
m% NaCl = 19.9
623. In benzene solvent : 1.15 = 5.12 ×
In diethyl ether :
1.5
× 1000 ⇒
M × 36
1.55 = i × 2.02 ×
0.48 × 1000
185.5 × 10
M =185.5 amu.
⇒ i =3
Hence, per molecule of complex, three particles are produced on dissociation in diethyl ether.
Also, in one mole complex, mass : C = 185.5 × 0.6488 ≈ 120
H = 185.5 × 0.0545 ≈ 10 ⇒ Molecular formula = M (CH) 10
430
Problems in Chemistry
n2
× 1000
100
n
1 = K b ⋅ 2 × 1000
W1
0.26 = K b ⋅
624.
…(i)
…(ii)
Solving Eqs. (i) and (ii), W1 = 26 g
10
625.
Mole fraction of urea = 60 = 0.0322
10
+5
60
20
180
Mole fraction of glucose =
= 0.01
20
+ 10
180
Q mole fraction of glucose is less, vapour pressure above the glucose solution will be higher than
the vapour pressure above urea solution. To establish equilibrium, some H 2O( v ) molecule will
migrate from the glucose side to urea side in order to make the solutions of equal mole fraction.
10
20
60
180
=
⇒
10
20
+5+ x
+ 10 − x
60
180
Solving,
x=4
So, now in urea solution mass of water = 90 + 72 = 162 g
100
m% =
× 100 = 6.17
162
nH 2O
626. At equilibrium : 23.6 = 24 ×
⇒ nH 2O = 9.833 ⇒ m(H 2O) =177 g.
nH 2O + n urea
⇒ Mass of H 2O( v ) migrated from the glucose solution = 177 − 90 = 87 g.
n1
n2
⇒
23.6 = 24 ×
= 16.95 × 10 −3
⇒
n1 + n2
n1
Also,
18n1 + 180n2 = 113 ⇒ n1 + 10n2 = 6.277
Solving, Eqs. (i) and (ii), n1 = 5.367 and n2 = 0.09 ⇒ mass of glucose =16.2 g.
m% glucose in original solution = 8.1
64.6 = 70 ⋅
627.
⇒
628.
n1
n1 + n2
where
…(i)
…(ii)
n1 = moles of (methanol + ethanol)
n2
x 80 − x
= 0.0836 ⇒ n1 = 1.9938 =
+
n1
32
46
⇒ x = 26.78 g CH 3OH
1 0 1 0
PB + PT = 80 ⇒ PB0 + PT0 = 160
2
2
…(i)
431
Solutions
Mole fractions of benzene and toluene in the vapour phase :
PB0
P0
and xT = T
160
160
Now, these mole fractions will be the mole fractions of benzene and toluene in the liquid phase of
condensate.
xB =
PB0
P0
0
0
⋅ PB0 + T ⋅ PT0 = 100 ⇒ PB 2 + PT 2 = 160 × 100
160
160
Solving, Eqs. (i) and (ii)
⇒
PB0 = 120 torr and
…(ii)
PT0 = 40 torr.
 25 
 25 




629. V.P. of mixed solvents =  32  × 94 +  46  44 = 73.48 mm
 25 + 25 
 25 + 25 
 32 46 
 32 46 
n1
where, n1 = moles of solvent =1.324
Now,
58 = 73.48
n1 + n2
⇒
n2 = moles of solute = 0.35
2 × x 11.85 − x
n2 = 2 × moles of NaCl + moles of urea =
+
= 0.35 ⇒ x = 8.7 g
58.5
60
630. ln
P2 ∆H  T2 − T1 
=


P1
R  T1T2 
⇒
631. π = CRT
ln
and
V.P. of H 2O at 100°C is 760 mm

760 40.6 × 10 3 
50
=


P1
8.314  373 × 323 
⇒ P1 = 100.15 mm
⇒ C = 6.286 × 10 −2 M
⇒
P = 55.324 ×
Also,
2 = 1.86 ×
55.55
55.55 + 6.286 × 10 –2
6.286 × 10 −3 × 1000
w1
= 55.26 mm
⇒ w1 = 5.84
Initial amount of water = 100 − 2.15 = 97.85
Mass of ice formed = 97.85 − 5.84 = 92.01 g
2.3 = 5.1 ×
632.
1.3 = 5.1 ×
MA
1
1000
×
+ 2M B
20
…(i)
MA
1
1000
×
+ 4M B
20
…(ii)
Solving, Eqs. (i) and (ii)
M A = 25.58 and
M B = 42.64
432
Problems in Chemistry
633. Mol. wt. of acid :
⇒
0.2
= 15.1 × 0.1 × 10 −3
M
0.168 = (1 + α ) × 1.86 ×
0.558 = i × 1.86 × 0.1
634.
635. In benzene i =
In water :
1
2
⇒ − ∆T f = 2.16 =
⇒
⇒
31.78 = 31.82 ×
⇒ Formula = [Co(H 2O) 5 Cl]Cl 2
10
1000
×
10
118.5
⇒
M =118.5
⇒ α = 0.115
M 2+ + 2Cl −
55.55
55.55 + 3S
2S
S
⇒ S = 2.33 × 10 −2
K sp = 4S 3 = 5.06 × 10 −5
5.93 × 10 −3 = ( x + 1) × 0.52 ×
637.
⇒ α = 0.196
1
10 1000
× 5.12 ×
×
M 100
2
− ∆T f = 1.75 = (1 + α ) × 1.86 ×
Solubility = S
M =132.45
1
1000
×
132.45 100
⇒ i =3
MCl 2
636.
⇒
0.25 1000
×
10
M
M
= 23x
100
From Eqs. (i) and (ii)
x = 20.3 ≈ 20
Formula of protein = H 20 P
Mol. wt. = 2300 × 20 − 23 × 20 + 23 = 45563 amu
100
638.
⇒ M =123.46
754.5 = 760 ×
 100 5 
18 
+ 
 18 M 
Also,
⇒
754.5 = 760 ×
100
 100  α 
2 
78 
+ 1 −  ×

2  123.46 
 78 
α = 0.846 ⇒ % dimerization = 84.6
n
639.
3.31 = 5.12 × 2 × 1000 ⇒ n2 = 12.93 × 10 −3
20
Let the mixture contain x g cyclohexane.
x 1.32 − x
+
= 12.93 × 10 −3 ⇒ x = 0.64 g
84
128
1
640.
0.25 = K f ⋅
× 1000 ⇒ K f = 3
60 × 200
1.5
Also,
0.2 = 3 ×
× 1000 ⇒ M y =180
M y × 125
…(i)
…(ii)
433
Solutions
and 99 = 100 ×
n1
n1 + n2
⇒
⇒
K f =3=
M 1 RT f 2
∆H f
=
Also,
K b = 3.6 =
∆H v
=
119.5 × 10 −3 × 8.314 × (335) 2
∆H v
∆H v = 30.97 kJ mol –1
Now,
ln
⇒
ln
P2 ∆H v  T2 − T1 
=


P1
R  T1T2 

760 30.97 × 10 3 
12
=


P1
8.314
 335 × 323 
P1 = 502.8 mm of Hg.
2.5 1000
 α
− ∆T f = 2 = 1 −  × 14 ×
×
⇒ α = 0.925

2
94 100
Mass of solution =102.5 g ⇒ V =
⇒
643.
⇒ ∆H f = 34.1 kJ mol –1
0.4
× 1000 ⇒ K b = 3.6
128 × 25
M 1 ( RT f0 ) 2
⇒
642.
1.5
= 0.825
180
125
= 151.51 amu
0.825
151.51 × 10 −3 × 8.314 × (285) 2
∆H f
0.45 = K b ⋅
641.
⇒
⇒ n1 = 99n2 = 99 ×
M (Solvent) =
0
∴
n2 1
=
n1 99
m
=116.5 mL
f
2.5 1000
= 0.228 M
×
94 116.5
α
Kc =
= 360.6
2C (1 − α ) 2
Initial molarity =
n2
× 1000 ⇒ n2 = 0.041
300
Let mixture contain x g anthracene, then
0.7 = 5.12
6−x
x
+
= 0.041 ⇒ x = 2.677 g.
178 128
After cooling to 4.5°C, − ∆T f = 1 = 5.12 ×
⇒
644.
w1 = 209.92 g
0.041
× 1000
w1
⇒ mass of solid benzene produced = 90.08 g
1.9 = (1 + α ) × 1.86 ⇒ α = 0.0215
Molarity =1.098 ⇒ K a =
Cα 2
= 5.18 × 10 −4 .
1−α
434
Problems in Chemistry
SOLID STATE
645. Number of A per unit cell =1
1 7
=
8 8
Empirical formula = AB 7/ 8 = A8 B7
646. Number of A per unit cell =1
Number of B per unit cell = 7 ×
1
⇒ empirical formula = AB 5/ 2 = A2 B5
2
31
7
647. Number of atoms per unit cell = + 3 =
8
8
Also, in face centred cubic lattice, a = 2 2r
Number of B per unit cell = 5 ×
Packing fraction (φ ) =
31 × 4πr 3
8 × 3(2 2r ) 3
648. Volume occupied by atoms = 2 ×
= 0.717
4
8π 63 3
π (R 3 − r 3 ) =
R
×
3
3 64
Packing fraction (φ ) =
⇒
8π × 63R 3 × 3 3
3 × 64 × 64 × R 3
= 0.669
 1 1  16
649. Number of atoms per unit cell = 6 −  +  =
 6 2 3
Volume of unit cell = 24 2r 3
650. Density ( s) =
nM
Na
⇒ 2.7 =
3
⇒ Packing fraction =
16πr 3
3 × 24 2r
n × 27
6.023 × 10
8.27 =
651.
23
× ( 4.05 × 10 −8 ) 3
3
×
4
= 0.658
3
⇒ n = 4, FCC
4 × 128
6.023 × 10 23 a 3
⇒
a = 46.8 × 10 −9 cm = 4.68 Å = 2[ r (O 2− ) + r (Cd 2+ )]
⇒
r(Cd 2+ ) = 1.1 Å
652. Density =
2M
6.023 × 10
23
× ( 4.47) 2 × 10.89 × 10 −24
Formula of salt is Hg 2Cl 2 .
4M
653. 4.269 =
23
6.023 × 10 × (6.93) 2 × 10 −24
⇒
= 7.15 ⇒
M = 212
Solving x = 3
⇒ Mass of sulphur required =
96
× 1500 = 672.89 g
214
M = 468.52
Formula Co 2S x
435
Solutions
654. Given
h
2
r = 3.35 Å = 3.35 × 10 −8 cm.
=2
2
3
3
 3 1

Volume of unit cell = 24 2r = 24 2 
× × 3.35 × 10 −8  = 2.93 × 10 −22 cm 3
 2 2

3
density =
6 × 12
6.023 × 10
23
× 2.93 × 10 −22
= 0.4 g / cc
655. In HCP, number of atoms per unit cell = 6
Volume of unit cell = 24 2 r 3
Packing fraction =
656. h = 2r− and A = 6 3r−3
3 × 24 2r
3
=
π
3 2
= 0.74
⇒ V = 12 3r−3
Packing fraction =
657. Given a − 2r = 60 and
6 × 4π r 3
3 × 4π( r+3 + r−3 )
12 3r−3
= 0.64 ( r+ / r− = 0.414)
for BCC, 4r = 3a ⇒ a −
Density (ρ) =
3
a = 60 ⇒ a = 448 pm
2
2 × 48
6023
× 10
.
23
= 1.77 g/cc
× ( 4.48) 3 × 10 −24
658. For Fe 95O100 , if there are x Fe 2+ ,
(a)
2x + (95 − x )3 = 200 ⇒ x = 85
Fe 2+
⇒
Fe
3+
85 17
=
15
3
=
(b) 5% of cation are vacant.
659. (a) 5% of anion sites are vacant.
(b)
Density (ρ) =
(c)
Density (ρ) =
4 × 128
6.023 × 10
23
× ( 4.7) 3 × 10 −24
4 × 127.2
6.023 × 10 23 ( 4.7) 3 × 10 −24
= 8.81g/cc.
= 8.136 g/cc.
660. Number of cylindrical atoms per unit cell = 3
Packing fraction =
661. Edge not covered by atom = a − 2r
also in BCC, 4r = 3a
3πr 2 l
6 3r 2 l
= 0.906
436
Problems in Chemistry
Therefore, edge not covered = a −
2 − 3
3

a =a
2
 2 
2− 3
= 0.134
2
In FCC
2( r+ + r− ) = a = 4.8 Å ⇒ r+ + r− = 2.4 Å
3
 rB  
4 ( 4rA3 + rB3 )
π 
 4 +    = 0.753
Packing fraction = π
=
3 (2 2rA ) 3 12 2 
 rA  

2
2
πR + πr
π
Packing fraction =
= [1 + (0.414) 2 ] = 0.92
2
4
(2R )
4
Edge length of unit cell =
r
3
16
16
⇒ Area of a surface of unit cell = r 2 = × (96) 2 × 10 −20 cm 2 = 4.9152 × 10 −16 cm 2
3
3
6
⇒ Number of unit cell on surface =
= 1.22 × 1016
4.9152 × 10 −16
⇒ Fraction of edge not covered by atoms =
662.
663.
664.
665.
Volume of unit cell = a 3 =
64
3 3
(96) 3 × 10 −30 = 1.089 × 10 −23 cm 3
Volume of solid = 6 cm 3
⇒
Total number of Ca present =
4×6
1.089 × 10 −23
= 2.2 × 10 24
Number of Ca present on surface = 1.22 × 1016 × 2 = 2.44 × 1016
⇒ Fraction of Ca lying on the surface =
666. 2.32 =
667.
2.44 × 1016
2.2 × 10
24
= 11 × 10 −9
2× M
⇒ M =1268 ⇒ x = 47
6.023 × 10 (1.22) 3 × 10 −21
4 × 78
⇒ a = 54.6 × 10 −9 cm = 546 pm
3.18 =
6.023 × 10 23 a 3
23
668. Let us consider 100 cc of solid Ar : mass =159 g
Volume of 159 g liquid Ar =
159
= 112 mL
1.42
Q Packing fraction of solid Ar = 0.74, actual volume occupied by Ar = 74 mL.
⇒ % empty space in liquid phase =
112 − 74
× 100 = 33.92%
112
a a 
669. If the side length of unit cell is ‘ a’. Coordinate of centre of guest atom B will be  , , 0 .
2 4 
437
Solutions
1/ 2
 a  2  a  2 
a
5
⇒
R + r =   +    =




2
4
4


3a
Also, in BCC
R=
4
r
5
⇒ r = 0.29R
1+ =
⇒
R
3
3
4  3 r3  3 3
3π  1  r  
Packing fraction = π  2R +  ⋅
=
2 +    = 0.684
3 
2  64R 3
16  2  3  
670.
a − 2r = 34 pm
4r = 3a
Also,
a−
⇒
54 × 2
3
a = 54 ⇒ a =
= 395.6 pm
2
2− 3
Density (ρ) =
671. a = 2R
and
2 × 23
6.023 × 10
23
× (3.956) 3 × 10 −24
= 1.23 g/cc.
3a = 2R + 240 ⇒ ( 3 − 1) a = 240 ⇒ a =
ρ=
209
6.023 × 10
× (3.28) 3 × 10 −24
23
240
3 −1
= 9.83 g / cc
672. Volume of unit cell = 4.6 × 8 × 5.7 × 10 −24 cm 3 = 2.0976 × 10 −22 cm 3
⇒
2.93 =
n × 100
6.023 × 10
23
× 2.0976 × 10 −22
⇒ n = 3.7 ≈ 4
673. (a) Number of Na + = 3, Number of Cl − = 3
Packing fraction =
1
4π
(3r+3 + 3r−3 ) ×
= 0.59
3
16 2r−3
1 7
=
2 2
1 7
Number of Cl − = 4 − =
2 2
4π 7 3
1
⇒ Packing fraction =
× ( r+ + r−3 ) ×
= 0.64
3 2
16 2r−3
(b) Number of Na + = 4 −
 1 5
(c) Number of Na + = 4 − 1 +  = , Number of Cl − = 4
 2 2
Packing fraction =
4π
3
1
5 3
3
= 0.77
 r+ + 4r−  ×
2
 16 2r−3
= 327.84 pm.
…(i)
…(ii)
…(i)
438
Problems in Chemistry
SURFACE CHEMISTRY
1.36
× N A = 3.65 × 1019
22400
M
Molar volume (H 2 ) =
ρ
M
⇒ Volume of one H 2 molecule =
= 4.743 × 10 −23 cm 3
ρN A
674. Number of molecules of H 2 in 1.36 cc =
 3V 
radius =  
 4π 
1/ 3
= 2.246 × 10 −8 cm
A (H 2 ) = πr 2 = 1.583 × 10 −15 cm 2
Surface area covered on Cu per gram = 3.65 × 1019 × 1.583 × 10 −15 = 5.78 × 10 4 cm 2
675. Let V (mL) of solution is required
0.016 V
N (acid) =
× 6.023 × 10 23 = 9.637 × 1018 V
1000
Area covered = 9.637 × 1018 V × 0.2 × 10 −14 cm 2 = 19274 V = 500 cm 2
V = 25.94 × 10 –3 mL
676. Total A = 25000 m 2 = 25 × 10 7 cm 2 , r = 1.5 × 10 −8 cm
A ( NH 3 ) = πr 2 = 7.068 × 10 −16 cm 2
⇒
N =
25 × 10 7
−16
= 3.53 × 10 23
7.068 × 10
moles of NH 3 = 0.587
Vol. of NH 3 at STP required = 13153.5 mL
100
N =
× 6.023 × 10 23 = 2.69 × 10 21
22400
⇒
677.
A/molecule = 0.16 × 10 −14 cm 2
Total area covered by N 2 ( g )/gram of catalyst = 1.6 × 10 −15 × 2.69 × 10 21 = 43.04 × 10 5 cm 2
REACTION MECHANISM
682. (a) O 2 NCH 2CH •• < CH 2 •• < CH 3CH •• < C 6 H 5CH •• < (C 6 H 5 ) 2 C ••
(b) O2N—
Cl—
+
+
+
+
+
+
(c)
(aromatic)
H3C—
+
439
Solutions
+
+
(d)
O
O
+
O
O
+
O
+
O
+
O
O
O
O




−
−
−
(e) CH 2 —C—NH 2 < CH 2 —C—OCH 3 < CH 2 —C—CH 3 < CH 2 —C—H
−
O

< CH 2 —C—CH 2OCH 3
−
683. (a)
(b)
(c)
(d)
(e)
H 2O <1-butanol <1-butyne <1-butene.
Butanamide < aniline <1-butylamine.
Oxalic acid <1,3-propandioic acid < acetic acid.
Pyridine < N-methyl pyrrole < pyrrole.
Cyclohexancarboxylic acid < p-nitrophenol < phenol < cyclohexanol.
684. γ-H is acidic and conjugate base is stabilized by resonance
H γ —CH 2 —CH==CH—CHO
−
CH 2 —CH==CH—CHO + H + ←→
−
O
O



−
CH 2 ==CH—CH —C—H ←→ CH 2 == CH— CH== C

H
685. (a) 2 < 1 < 3 (b) 3 < 1 < 2 (c) 3 < 1 < 2 (d) 1 < 2 < 3
686. Due to high hydrocarbon (hydrophobic) proportion, both are insoluble in pure water. NH 2 of A is
basic in nature, forms salt with dilute H 2SO 4 , hence soluble. NH 2 of B is neutral due to resonance
delocalization of lone-pair of electron of nitrogen with carbonyl group hence does not form salt
with acid and insoluble. However, α-proton of amide B is quite acidic, dissolves in aqueous
solution of strong base :
O
H
O
O
H
O
NH2 + OH –
B
NH2 + H2O
H
(resonance stabilized base)
O
NH2
687. (a)
NH
OH
N—H
(b)
N
OH
N
N
(Aromatic)
OH
440
Problems in Chemistry
OH
(c)
CH2
C
O
O
CH2
C
CH3
C
CH3
OH
COOH
C
CH
O
OH
CH3
CH2
OH
C
CH
C
CH3
–
O
C
688.
COOH
O
C
H + H+
Phthalic acid
O
Stable conjugate base
I
O
O
C—OH
C—O–
+ H+
COOH
Isophthalic acid
COOH
Conjugate base is not stabilized
by intramolecular H-bonding.
II
Due to greater stability of I, phthalic acid has greater K a 1 than isophthalic acid. Removal of second
proton is difficult from I, since H-is bonded to two oxygen atom, while acidic proton of II is free
from H-bonding, can easily be deprotonated.
689. Conjugate base of benzene sulphonic acid has greater resonance stabilization than benzoate ion,
hence benzene sulphonic acid is stronger acid than benzoic acid.
O−
O
O



Ph—S —O − ←→ Ph — S == O ←→ Ph—S == O



O
O
O
−
Three equivalent resonance structure, more stable.
O−
O

 −
Ph—C—O ←→ Ph— C == O
Only two equivalent resonance structure, less stable.
••
−
690. Thiophenol is stronger acid since it produces a weak conjugate base C 6 H 5 S.
691. m-hydroxybenzoic acid > benzoic acid : Inductive effect.
o-hydroxy acid > benzoic acid : Ortho effect.
p-hydroxybenzoic acid < benzoic acid : +R effect from para-position.
441
Solutions
O
COOH
C
O–
H
OH
O
692.
+ H+
Stable conjugate base
(ortho effect)
O
COOH
C
O–
OCH3
OMe
+ H+
No ortho effect
693. A is stronger acid than B although electron donating tertiary butyl groups are closer to —COOH
functional group. It is due to steric inhibition to resonance of —NO 2 group by two bulky tertiary
butyl groups in B, while —NO 2 is free for resonance in A and increases acid strength markedly by
−R effect.
694. B is stronger base than A. Here basic strength is controlled by −R effect of —NO 2 . In B, due to
steric inhibition to resonance of —NO 2 by two bulky tertiary butyl group, it is not decreasing basic
strength by −R effect, hence B is stronger base than A.
695. (a)
+
H+
+
–H+
+
H
H

 •
disproportionation
(b) CH 3 —CH 2 —C • + H—C—CH 2 → CH 3CH 2CH 3 + CH 3CH== CH 2 .


H
CH 3
696. (a) Due to resonance in vinyl chloride, C—Cl bond acquire some double bond character while in
chloroethane, C—Cl bond is pure, single sigma covalent bond.
••
−
+
CH 2 == CH—Cl •• ←→ CH 2 —CH== Cl
••
Vinyl chloride
442
Problems in Chemistry
(b) Due to resonance in methyl vinyl ether, bond order of C—C is slightly reduced from two while
in ethene C—C bond order is exactly 2.
••
−
+
CH 2 ==CH—O— CH 3 ←→ CH 2 —CH== O—CH 3
••
Methyl-vinyl ether
(Bond order < 2)
(c) Larger size of sulphur compared to oxygen gives greater acidity to CH 3SH.
(d) Lone-pair of electron in vinyl amine is involved partly in resonance delocalization and less
available for donation to a Lewis acid, hence a weaker Lewis base.
••
−
+
CH 2 == CH—NH 2 ←→ CH 2 —CH== NH 2
Vinyl amine
697. In aqueous medium, Lewis base strength of an amine depends on two important factors — number
of electron donating groups bonded to nitrogen and hydration of conjugate acid by water molecule.
In general, increase in either of these increases basic strength. In case of trimethyl amine, three
methyl group increases electron density at nitrogen by +I effect but its conjugate acid is not
sufficiently hydrated by water due to lack of H-capable of forming hydrogen bonding with water.
On the other hand, methyl amine has only one electron-donating methyl group bonded to nitrogen
but conjugate acid is hydrated excessively due to three hydrogen available which are capable of
forming hydrogen bonding with water. Effect of hydration here exceeds the electron-donating
effect and hence CH 3 NH 2 is stronger base than trimethyl amine.
H
CH3—NH2 + H+
O
H
H
H
H—O
H
+
CH3—N—H---O
H
H
H---O
(Excessive hydration)
H
CH3
+
(CH3)3N + H
CH3—N—H---O
+
H
H
CH3
698. The compound must be 1,2-cyclobutan-dicarboxylic acid since all other constitutional isomers are
non-resolvable.
COOH
COOH
HOOC
H
(cis-isomer)
H
Polar but
non-resolvable
due to plane of
symmetry
H
H non-polar
but resolvable.
COOH
(trans-isomer)
699. A is thermodynamically favoured rearrangement since a less stable (highly strained) cyclopropane
ring is converted into a less strained (more stable) cyclobutyl ring. On the other hand, cyclohexane
is more stable ring than cycloheptane hence rearrangement B is thermodynamically not favoured.
443
Solutions
700. Catalytic hydrogenation involve adsorption of unsaturated compounds on the metal surface, which
is restricted by the bulky substituent at unsaturated carbons. In this regard, an alkyne has only two
substituent, poses less steric hindrance during adsorption, undergo easier hydrogenation while
alkene has four substituents, poses greater steric hindrance during adsorption, undergo
hydrogenation with difficulty.
On the other hand, electrophilic addition involve formation of a carbocation intermediate, stability
of which control the reactivity. Alkyne produces a less stable vinyl carbocation on addition of
electrophile, hence less reactive while an alkene produces a more stable alkyl carbocation and it is
more reactive.
–
–
+
701.
–
+
+
–
δ–
+
δ+
702. The order of acidic strength is: B(1) > A(2) > D(3) > C(4).
A, B, D are all more acidic than “C” because they are all phenols and the conjugate base that is
formed upon deprotonation can be stabilized by resonance. C is incapable to delocalize negative
charge via resonance.
B is most acidic, because in addition to all the resonance forms available in A and D, the charge of
anion can also be delocalized through the  NO 2 group as:
O–
O
O
O
–
–
N
O
+
O
–
N
O
+
O
–
N
O
+
O
–
O
–
N
O
–
+
O
–
(most stable)
O
–
N
O
–
+
O
N
O
–
+
O
444
Problems in Chemistry
D is less acidic than A because the two OCH 3 groups on ring make the system more electron
rich, which inturn destabilize the conjugate base relative to A.
O–
O
O
–
OH3C
H3CO
OCH3
CH3O
CH3O
O
O–
O–
OCH3
–
–
Others
+
CH3O
OCH3
–
OCH3
CH3O
OCH3
CH3O
(destablizing the conjugate base)
l
l
N
703.
l
Basic
l
N
Non-basic
H
Lone pair of the basic nitrogen is fully available with nitrogen, i.e., not involved in delocalization
while lone pair of non-basic nitrogen is part of aromatic delocalization as:
N
l
N
+
l
N
H
–
–
N
+
–
N
+
N
N
N
H
H
H
N
–
+
N
H
704. III < II < IV < I
O–
Negative charge is on electronegative atom and stabilized by resonance. This
is the most stable conjugate base of the four acids.
445
Solutions
O–
The negative charge on electronegative atom but not stabilized by resonance.
This anion is then less stable than phenoxide ion but more stable than the
anions produced from II and III.
CH2–
Negative charge is on electropositive carbon but stabilized by resonance.
CH2CH2–
Negative charge is on electropositive carbon, and not stabilized by
resonance, hence least stable.
705. (i)
(ii)
Equilibrium favour left-side since amide nitrogen is not basic because the lone pair on
nitrogen is delocalized through carbonyl groups via resonance. Therefore, the reaction will
not proceed in the forward direction.
Equilibrium favours left side. The positive charge will be better stabilized on pyridine
nitrogen as:
+
+
N
N
H
H
+
+
N
H
N
H
+
N
H
(more stabilized)
+
+
CH 3 —C ≡≡ N—H ←→ CH 3 —C== NH (less stabilized)
(iii) Equilibrium favours left side because the positive charge will be more stabilized on nitrogen
bonded to three ethyl groups. The inductively electron withdrawing nature of three trifluoro
methyl groups will destabilize the ammonium salt on the right side of reaction.
706. 2,6-ditertiarybutyl pyridine is more basic than pyridine since t-butyl group stabilized the
protonated ammonium salt by inductive effect. However, nucleophilic strength depends on its rate
at which it attack to an electrophile. In case of di-tertiarybutyl pyridine, the alkyl group adjacent to
the nitrogen, hinder the lone pair of nitrogen to attack at an electrophilic center. Therefore, it is
weaker nucleophile than pyridine.
R
N
l
l
R
O ⇒ Posing steric hindrance to the attack of lone pair.
446
Problems in Chemistry
py
py
H
xz-plane
707
H
pz
xy-plane
C–———C———C
Br
Cl
pz
Br  C  H and Cl  C  H are two perpendicular planes of the molecule.
708. I > III > IV > II
709. Equilibrium favours left side since on right side we have strong conjugate base HO – and strong
acid, they react to produce more stable species H 2O and a weak conjugate base
O
O–


CH 3O  C CH==C  OCH 3 . K eq = 10 −3 also predict equilibrium to favour left side.
710. Reaction (a) will proceed to right. For a spontaneous acid-base reaction, a strong acid must react
with a strong base to produce weak acid weak base.
+
s
In (a) PhCH 2 NH 3 ( pK a = 9.33) is stronger acid than CH 3CH 2 NO 2 ( pK a =10) and CH 3C HNO 2 is
stronger base than C 6 H 5CH 2 NH 2 . The opposite is case in reaction (b).
711.
(a)
(b) HO—
—NO2
N+
H
(c)
O
O
(d)
MeO
OMe
Conjugate base is resonance stabilized.
N+
H
COOH
(e)
(h) PhCOOH
(f)
(g)
(i) p-nitrophenol.
OH
NH3Cl
712. (a) O2N
COOH
(c)
(b)
COOH
447
Solutions
713. (a) p-methoxyaniline, (b) pyridine, (c) CH 3CH 2 NH 2 , (d) 2,4,6-trinitro-N-N-dimethylaniline.
(d) N,N-dimethylaniline,
(b) H 2 N  C— NH 2 ,
(c) PhNH 2 ,
714. (a) CH 3 NH 2 ,

NH
(e) m-nitroaniline.
715. (a) IV < II < I < III
(b)
+
CH 3CH 2 OH 2 + CH 3CH CH 3

S–
and 2HSO 4 –
CH 3CH 2OH + CH 3 CH CH 3

SH
H 2SO 4 + SO 2–
4
Major components at equilibrium are: CH 3CH 2OH, CH 3CH(SH)CH 3 , HSO 4 – .
716. (a) is better site of protonation, since it gives a resonance stabilized conjugate acid:
CH3
N
+
CH3
+
N—H
N
CH3
CH3
N
N—H
CH3
N—H
+
CH3
CH3
+
N
N—H
CH3
(Complete octet of all atoms)
717. Oxygen, since it gives resonance stabilized oxonium ion as:
H
+
O
O
O
OH
OH
+
H+
N
+
H+
N
N
+
N
N
H
no resonance
stabilization
(Resonance stabilization)
OH
+
N
Complete octet of
all the atoms.
448
Problems in Chemistry
718. (a) (i) A > D > B > C
(ii) C > A > B
(b) D > C > E > B > F > A
(c) B > E > C > A > D
(d) C > B > A > D
(e) B > C > D > A
(f) A > C > D > B
(g) B > A > E > D > C
(h) (i) C >D > B > A (ii) E > C > D > B > A
(i) B > E > D > C > A
(j) B > C > D > A > E
CH3
l l
N
CH3
CH3
CH3
CH3
l l
N
CH3
+
N
CH3
CH3
δ+
N
+
719. (i)
(ii)
δ–
O
(Resonance hybrid)
–
+
B
–
O
–
O
O
δ–
δ+
+
B
B
B
δ+
(Resonance hybrid)
O
O
–
l l
(iii)
l l
–
N
N
N
Oδ–
O
–
CH2
(major)
+
H2N
H2N
+
(i)
H2N
O–
O–
O
H2N
H2N
+
H2N
+
(ii)
–
δ–
(Resonance hybrid)
720. The resonance forms of (i), (ii) and (iii) are:
H2N
δ–
N
+
–
–
449
Solutions
+
H2 N
H2 N
–
(iii)
721.
722.
723.
724.
725.
726.
The resonance forms of three compounds reveal that compound (i) has maximum double-bond
character in C—N bond and compound (iii) has minimum double bond character in C—N bond.
Therefore, the order of C—N bond-lengths in three compounds is: (i) < (ii) < (iii).
Order of acid strength is: III > II > I.
I > II > III.
For carbocations, more stable the carbocation, weaker will be electrophile. Hence, order of
electrophilic strength is: II > I > III (aromatic).
(i) To left since ammonia is weaker acid than water.
(ii) To right since in aqueous medium, secondary amine is stronger base than tertiary.
(iii) To left since hydroxide ion is stronger base than methylamine.
(a) B > C > A, (b) A > C > B, (c) A > B > C.
Strength of a base depends on stability of its conjugate acid. For all the above bases, their
conjugate acids formed after protonation acquire very high stability from resonance as:
H
I
N
H+
N
H
N+
N
N
N
II
N
+
+
N
H+
N
+
N
N
N
H
H2N
NH2
NH2
NH
III
H
+
NH2 + H+
H2N
NH2
+
H2N
NH2
NH2
H2N
Here all three resonance structures are equivalent.
+
NH2
450
Problems in Chemistry
–
+
l l
N
727.
N
+
N—
–
+
N
+
N—
+
N
+
N—
+
N—
+
+
N
N
+
N—
+
N—
–
–
(ii) H a : Conjugate base is aromatic.
728. (i) H c > H a > H b .
(iii) H b : Electron withdrawing group increases acidity.
(iv) b: Methoxy group decreases electrophilic character of carbonyl carbon by resonance
effect.
729. (iii) > (ii) > (iv) > (i).
+
NH2
N
730. H2N
+
+
+
NH3
NH2
NH3
COOH
H2N
H
N
COO–
H
at pH < 2.18
pH between 2.18 – 9.09
+
NH2
N
H2N
COO–
H
NH2
NH
NH2
H2N
N
COO–
H
pH between 9.09 – 13.2
pH > 13.2
H
N
731. (a)
(b) p-nitroaniline > aniline.
O
N
H
451
Solutions
COOH
+
732. (a) H3N
NH3
NH3
+
+
COO–
+
(c) H3N
COO–
+
(b) H3N
COO–
(d) H2N
NH2
NH2
733. 3 < 4 < 1 < 2.
734. Conjugate base is resonance stabilized as:
CH 3  CH==CH  CHO
O–

CH 2  CH==CH  CHO ←→ CH 2 ==CH  CH==CH
735. The relative strengths of ammonium ions as shown above lies in stability of conjugate base.
Stronger acid produce more stable conjugate base and hence lower pK a value. Also, since
aromatic amines are weaker base than aliphatic amines, conjugate acid of former will be stronger
than the conjugate acid of later.
⇒ A > B: acid strength.
STEREOCHEMISTRY
736.
Cl
Cl
Cl
H
Cl
H
H
H
H
H
cis-1,3-dichlorocyclopentane
(1)
Cl
Cl
(2)
(3)
trans-1,3-dichlorocyclopentane
Here (2) and (3) are non-superimpossible mirror images of one another, hence they are
enantiomers. Structure (1) is meso isomer and it is diastereomers of both (2) and (3).
Br
Br
Br
Br
737.
H
Cl
H
Cl
H
H
H
H
H
H
Cl
Cl
(1)
(2)
(3)
(4)
1 and 2, 3 and 4 are enantiomers while 1 and 3,
1 and 4, 2 and 3, 2 and 4 are diastereomers.
738. (a) Enantiomers (b) Identical (c) Diastereomers (d) Identical
739.
13.9x − 13.9(100 − x ) = − 300 ⇒ x = 39.2,
Hence, 39.2% of the mixture is (+) 2-butanol and 60.8% is (–) 2-butanol.
452
Problems in Chemistry
H
740. (a)
Br
H
F
Cl
Br
H
Cl
Br
C==C
Cl
C==C
II
I
H
C==C
F
IV
C==C
V
F
H
Br
F
F
H
Cl
Br
Cl
C== C
Br
III
Cl
C== C
F
VI
I and II, III and IV, V and VI are pair of geometrical isomers.
Cl
Cl
(b)
Cl
H
C==C
Cl Cl
H C==C
HH
H
H
C==C
H C==C
HH
Cl
H
H
C==C
Cl Cl
(ii)
(i)
H C==C
HH
Cl
H
(iii)
II and III are enantiomers while I is optically inactive, meso form.
CH 3
CH 3 Cl
CH 3




(c) Six isomer : CH 3 —CH—CH 2CH 2Cl, CH 3 CH—CH—CH 3 CH 3 —C—CH 2CH 3 ,

Cl
(two stereo isomers),
CH 2Cl

CH 3 —CHCH 2 CH 3 (two stereo isomers).
CH 3

741. HC ≡≡C—CH—CH== CH 2
CH 3

CH 3CH 2 —CH—CH 2CH 3
B
A
Br
742.
A
743. CH 3 CHCH 2CH 3

OH
( A)
( B)
CH3CH2
CH3
B
C
CH 3 —C —CH 2CH 3

O
C==N
CH3
+
C2H5
CH3CH2
CH3
(C)
C
C==N
CH3
C2H5
H
CH3 NHCH3
(D)
H
C
CH3
NHCH3
453
Solutions
744. Addition of bromine at carbon-carbon double bond proceed via a cyclic bromonium ion
intermediate followed by nucleophilic attack of bromide ion as:
Br
CH3 C H
2 5
Br
+
CH3 C2H5
Br
+
Br –
Br
C 2H 5
C————C
CH3
CH3 C2H5
CH3 C2H5 CH3 C2H5
Br
Mixture of enantiomers obtained in equal amount giving racemic mixture.
Catalytic hydrogenation occurs at metal surface giving syn hydrogenation.
H
C2H5
CH3
C==C
C 2H 5
H2/Pt
CH3
H
CH3
C2H5
C2H5 CH3
COOH
Cl
745.
A
B
D
COOH
C
746. CH 3CH 2 —CH—CH==CH 2

CH 3
CH 3CH 2 CHCH 2CH 3 CH 3CH 2 CHCH 2CH 2Cl


CH 3
CH 3
B
A
C
Cl

CH 3CH 2CH CH CH 3

CH 3
Cl

CH 3CH 2 CCH 2 CH 3

CH 3
CH 3CH 2 CHCH 2 CH 3

CH 2Cl
CH 3CH 2 C== CHCH 3

CH 3
D
E
F
G
CH 3CH 2 CCH 2 CH 3

CH 2
H
747. A: Achiral, B: Achiral, C: Meso, D: Meso, E: Meso, F: Meso, G: Chiral,
H: Chiral, I: Achiral.
748. (a) Distereomers, (b) Identical and meso.
749. Compound I is optically inactive since it contain a plane of symmetry.
Compound II is enantiomeric since it does not contain plane of symmetry, hence chiral. Also
compound I is polar while II is non polar.
454
Problems in Chemistry
πz
πy
750. In allene, the two pi-bonds are in perpendicular
Cl
planes, hence the two terminal H  C  H planes are
also perpendicular. Therefore, if allene is substituted xz
property it becomes chiral as:
C–———C—–——C
The two terminal Cl  C  H planes are
perpendicular to one-another and Cl and H at one H
terminal are equidistant from either Cl or H from
other carbon. This is non superimposable on its mirror image, hence optically active.
751. (a) II and IV are chiral, hence optically active.
(b) I and III are achiral, possess plane of symmetry, hence optically inactive.
(c) There is no enantiomer pair, both II and IV are identical structure.
(d) I and II, II and III, I and III are pair of distereomers.
752. (a) Both I and II are optically, but they are not mirror image of one another, hence, they
distereomers.
(b) Distereomers.
(c) Enantiomers.
(d) I and V are enantiomers.
(e) IV and VI are distereomers.
CH3
CH3
CH3
753.
+ HCl
xy
H
are
H
Cl
+
H
Cl
H
H
Chiral
(two stereoisomers)
Cl
Chiral
(two stereoisomers)
Therefore, in the above addition reaction four stereoisomers (two pair of enantiomers) will be
produced.
CH3
CH3
CH3
CH 3 CH 3
H
CH3
CH3


H
754. H 3C  C 2  C 3  CH 3
60°
60°


CH3
H
H
H
CH3
H
H
CH3
CH3
H
CH3
CH3
2,3-dimethylbutane
CH
3
I
II
III
60°
CH3
H
H3C
60 °
60 °
60 °
I ← II ← III ←
H
CH3
IV
CH3
455
Solutions
I
I
III
III
Energy : I > III > II > IV
II
II
E
IV
0
60
120
180
240
300
360
Angle
are
755. I: CH 3  C ≡≡C  CH 2  COOH and CH 2 ==CH  CH==CH  COOH
isomers.
CH 3  C ≡≡C  CH 2COOH and HC ≡≡C  CH 2CH 2COOH are positional isomers.
H
functional
H
HC ≡≡C
C ≡≡CH stereoisomers.
COOH + HOOC
CH 3
CH 3
O

II: CH 3CH 2CH 2CH 2  C  NH 2
O

and CH 3  CH  CH 2CH 2  C  H

NH 2
functional isomers.
O

CH 3CH 2CH 2CH 2  C  NH 2
O

and CH 3CH 2CH 2  C  NHCH 3 are positional isomers.
CH 3
CH 3
H
and
NH 2
CH 2CH 2CHO
H2 N
H
CH 2CH 2CHO
are stereoisomers.
756. Only two stereoisomers, since H 2O will attack to the intermediate carbocation from a side anti to
the methyl group on adjacent carbon.
H3 C
CH3
CH3
H3 C
H3 C
CH3
l l
H+
+
OH2
(bottom side attack)
OH
Enantiomeric (two stereomers)
456
Problems in Chemistry
757. The stereocenters are labeled by stars:
H
N
*
*
O
O
O
I
The distereomer pairs are:
H
N
H
N
O
O
O
O
and
O
O
I
I
758. In acid catalyzed hydration of alkene, a planar carbocation is generated in the 1st step. The
nucleophile (H 2O) attack in the next subsequent step from a side where the steric hindrance is
minimum.
H+
I
H2O
–H+
+
CH3
CH3
OH
CH3
(Pair of enantiomers)
H+
+
H2O
II
–H+
CH3
CH3
OH
CH3
(Pair of enantiomers)
457
Solutions
H+
H2O
III
OH
–H+
+
CH3
CH3
CH3
(Pair of enantiomers)
⊕
H+
IV
CH3
CH3
(No chiral carbon in generated on protonation)
759.
+
••
CH 2 ==CH  N HCH 3 ←→ CH 2  CH== NHCH 3
N-methylethenamine
H
+
CH3
+
C== N
–
C2H5
CH3
60°
760.
H
H
CH3
CH3
I
C== N
CH 3
CH 2
(These are geometrical isomers)
C2H5
C2H5
H
CH
3
H
60°
H
CH3
II
CH3
H
+
–
H
H 2C
H
H
60°
CH3
CH3
CH3
C2H5
H
H3C
H
CH3
III
CH3
IV
60°
CH3
H
C2H5
H
CH3
CH3
V
Energy: I > V > III > II > IV
Stability: I < V < III < II < IV
761. (a) This compound contain a symmetry plane, hence optically inactive.
(b) This compound contain ‘one’ chiral carbon, hence enantiomeric.
458
Problems in Chemistry
Chiral carbon
Enantiomers
(c) This compound contain a symmetry plane, hence, optically inactive.
(d) This compound contain ‘one’ chiral carbon, hence, enantiomeric.
Chiral carbon
O
OH
O
HO
Enantiomers
CH 2OH
H
HOH 2C
762.
OH
HO
H
HO
H
HO
H
≡≡
CHO
HO
H
HO
H
HO
H
HO
H
CHO
CHO
180°
→
CHO
H
OH
HO
H
H
OH
HO
H
H
OH
HO
H
H
OH
CH 2OH
HO
H
CH 2OH
II
V
Enantiomers
Also II → D-sugar.
V → L-sugar.
Sugar I, III and IV are all L-sugar.
Any pairs of two structures other than II and V are distereomer pair.
763. This compound has two chiral carbon, and a double bond capable of showing geometrical
isomerism giving rise to total eight different configurations possible for the molecule as shown
below:
O
O
H
HO
H
H
Enantiomeric
COOH
H
H
OH
H
Enantiomeric
COOH
459
Solutions
O
O
H
H
COOH
H
OH
COOH
H
H
Enantiomeric
O
O
764. (a)
OH
–
+
NH2
NH2
no stereo-isomerism
O
–
O
(b)
N
CH3
+
N
CH3
CH3
CH3
no stereoisomerism
–
O
O
(c)
+
N
N
H
H
(exhibit geometrical isomerism)
O
O
(d)
H
Enantiomeric
N
H
H
–
+
N
H
H
(exhibit geometrical isomerism)
765. (a) Both are similar structures.
(c) They are enantiomers.
(b) They are positional isomers.
(d) They are constitutional isomers.
O
OH
OH
: All have chiral carbon
766. I
I
keto
enol
460
Problems in Chemistry
OH
O
OH
II
II
keto
enol
All have chiral carbon.
O
OH
OH
III
III
keto
enol
All have chiral carbon.
767. (a) I and II are identical.
(b) I and II are identical.
(c) I and II are distereomers.
(d) I and II, I and III are pairs of distereomers. II and III are enantiomer pair.
768. Both has nitrogen as their chiral center. Ion II recemizes in aqueous solution due to the existence of
following equilibrium:
H
CH
+
N
H2O +
H3C
CH2
l l
H3O+ +
C2H5
N
H3C
CH
C2H5
CH2
(achiral)
769. It has one chiral center and can exhibit enantiomerism as:
C2H5
H
C2H5
and
O
O
O
O
are enantiomers.
H
770. The cis and trans forms of 3-methoxy cyclohexanol are:
OH
OH
OCH3
H
H
H
cis
H
trans
OCH3
O–
OCH3
OCH3
OCH3
cis-3-methoxycyclohexanol
CH3I
SN2
base
H
H
H
H
meso form (optically inactive)
461
Solutions
O–
H
trans-3-methoxycyclohexanol
base
H
H
O–


OCH3 
 CH3I

OCH3 



OCH3
H
H
OCH3
+
H
OCH3
OCH3
H
Enantiomers (Racemic mixture)
H
771. (a) Tautomers (Isomer), (b) Conformers, (c) Resonance structures, (d) Positional isomers.
HYDROCARBONS
O

772. Ph—CH 2 —C ≡≡C—CH 3 PhCH 2CH 2 CCH 3
A
B
PhCH 2COOH PhCH 3
D
E
CH3
O

PhCH 2 CCH 2CH 3
C
PhCH 2Cl
F
CH3
CH3
Cl
Cl2
AlCl3
+
Cl
E
OH
773.
C
CH
A
B
OH
COOH
OH
OH
C—CH3
CH—CH3
O
OH
C
(i) KCN
(ii) H3O+
O
D
O
O
COOH
774.
O
CHO
A
B
COOH
C
D
O
462
Problems in Chemistry
CH3
775.
•
CHCH3
A
Labelled carbons are chiral
CH3
Et
•
H
Et
H
CH3
Enantiomeric (two structures)
Enantiomeric (two structures)
CH3
CH3
CH
C
CH2—C—CH3
CH3
776.
Br
A
B
CH3
COOH
CH—CH—CH3
Br
777.
E
D
C
—CH2—C
—CH2—CH2—CH2—
C—
O
A
CH3
—CH2COOH
CH2Br
—COOH
O
C
D
E
F
CH3
778.
—CH2OCH2CH2CH—CH==CH2
A
O
B
CH3
—CH2OCH2CH2CHCH2CH2OH
B
463
Solutions
CH3
—CH2OH
OH
HO
O
D
C
E
RO•
SO 2Cl 2 → ROSO 2Cl + Cl •
784.
RH + Cl • → R • + ClH
R • + SO 2Cl 2 → R  Cl + SO 2Cl •
Br
Br
Br
785.
B
A
D
C
F
E
Cl
Cl
786.
A
B
C
787.
A
B
Cl
CH3 CH3


788. (a) CH 3  C  C H  CH 2I

Cl
I

(c) C 6 H 5  C H  C HCH 3

Br
C
Cl
D
Cl
D
Cl
Cl
E
Cl
E
Br

(b) C 6 H 5  C H  C H  CH 3

Cl
CH3 Br
 
(d) C 6 H 5  C  C (C 6 H 5 ) 2

I
G
F
F
G
464
Problems in Chemistry
Cl
789.
(a)
(b)
+
ring expansion
+ H+
Cl
+
+
ring expansion
+ H+
Cl
Cl
+
CH3
790.
CH3
C H2SO4
CH3
B2H6
OH
H2O2/NaOH
OH
791. (a)
(c)
792. (a)
(b)
(c)
(b) CH 3CH 2CH 2OH
(CH 3CH 2CH 2 ) 3 B
2-methyl-3-pentanol
(d) 2-methylcyclohexanol.
1-butene + BH 3 / H 2O 2 / NaOH → 1-butanol
1-butene + dil. H 2SO 4 /heat → 2-butanol
2-bromo-2,4-dimethylpentane + alcoholic KOH → 2,4-dimethyl-2-pentene
2,4-dimethyl-2-pentene + BH 3 /H 2O 2 /NaOh → product.
CHO
O3
H2-Pt
(A)
Zn-H2O
2 CH3CHO +
(D)
(E)
CHO
H2-Pt
(C)
(B)
H+/KMnO4
CH3CH2CH2COOH
(F) + CH COOH
3
793. Ozonolysis products shows that A is a symmetrical diene:
794. Cold, dilute, alkaline KMnO4 solution adds on C==C producing cis-diols. One of the stereomer A
produces meso cis-diol therefore it is confirmed that the starting compound is a symmetrical
alkene.
Also oxidative ozonolysis of the starting compound gives no acid as product, it indicates that there
is no hydrogen attached to doubly bonded carbon. From these finding it is now concluded that the
starting compound has following structural formula-3,4-dimethyl-3-hexene
Also: A + Cold, dil-KMnO4 → C(Meso) therefore, A is cis isomer and B is thus trans one. Thus
the various compounds are: A: cis-3,4-dimethyl-3-hexene. B: trans-3,4-dimethyl-3-hexene. C:
Meso-3,4-dimethyl butan-3,4-diol.
D: 2-Pentanone.
Cl
795.
(A)
(B)
(C)
465
Solutions
796. (a)
OH
H2SO4
B2H6/H2O2
heat
NaOH
B2H6/H2O2
(b)
OH
NaOH
OH
KMnO4
+
H
Br2
(c)
heat
CH3
Br2
KOH
Br Alcohol
KOH
Br
Alcohol
Br
Ph
Ph
Br
Br2
(d)
Ph
C
+
Na
hν
Ph
Pd/H2/BaSO4
H
(e)
H
Na/liq NH3
(1) Na/heat
NBS
H
Br
H
(2) CH3Br
H
H
Mg/ether
D2O
D
H
H
Na/liq NH3
(f)
KCN
(g)
Br
Br
BrH
+
(CH2=CH)2Cu
Br2
H
OH2
Br+
HOOC
Br
–H+
O
HOOC
O
466
Problems in Chemistry
Br2
(h)
Product
OH2
797. (a)
+
+
H
+
–H
••
OH
••
OH
O
+
–H
Me-shift
••
OH
••
OH
+
••
OH
••
H 2O
+
–H
(b) IV < II < III < V < I
B : Et 2 NCHCl 2
798. A : Et 3 N + CCl –2
Ph
Ph
Ph
Ph
+
+
H
799. (a)
C : Et 2 N  CHO
+
Cl
Cl
+
H
(b)
••
O
••
O
Cl
+
resonance stabilized
(c) HOCl
C6H5CH
HO– + Cl
CHCH3
••
O
••
••
O
+
Cl
+
Cl
+
OH
+
C6H5CH
CHCH3
Cl
stabler carbocation
HO–
C6H5CH
CHCH3
Cl
••
OH
••
467
Solutions
CH2Cl
Cl
Cl
800.
A
C
B
E CH2Cl
D
Cl
CHO
801.
H 3C
H O
G
F
J
CH3 H
H
O
H 3C
CH3
H 3C
I OH
CH3
CH3COOH
OH OH
B
A
H 3C
C
CH3
D
CH3 H
H 3C
CH3
Cl
Cl
F
E
802.
A: 5-methyl-2,4-heptadiene;
B: CH3CHO
E: 2-hydroxyethanoic acid.
803. A: 3-methyl-1-butene; B: 3-methyl-2-butanol;
C: 2-butanone;
D: Glyoxal
C:3-methyl-2-butanone
CH3
H2
H
804. CH3
C
C
C
C
C
CH3 H
H
CH3
Pd/BaSO4
(A)
CH3
Pt
CH3
H 3C
H
Na/liq NH3
H
H
H
( Optically Inactive C )
H
H 3C
O3
H
CH3
H
H 3C
H
( O A) D
H
H
HOOC
C
CH3
CHO
E(OA)
H
468
Problems in Chemistry
CH2Cl
805.
A
Cl
B
Cl
H
D
C
H H
H
Cl
OH OH
E
F
G
C2H5
806.
C2H5 HOOC
COOH
B
A
C
D
O
807.
A
B
C Cu
Cu(I)
Br
A
+
B
808.
HO
Br
E=A
C
809. (A)
(D)
OH
(B)
F
ONa (C)
OMe
MgBr (E)
OH
Br
(F)
(G)
Br
(H)
(I)
Br
+
H
D=A
469
Solutions
OH
(J)
C Na
+
(K)
Br
(L)
Br
Br
CH2Br
810.
Br
Br B
A
811. A
Br C Br
Br E H
D
D
C
B
CH2Cl
Cl
E
Cl
OH
Cl
812. (a)
(b)
(c)
Cl
H
813.
B
A
CHO
O C O
814.
A
Br Br
B
OH Br
C
Br
D
O
815.
816.
A
B
C
OH
817. (a)
+
H
–H2O
+
Me-shift
+
–H+
E
OH
OH
470
Problems in Chemistry
Cl
Br
Br2
(b)
Cl
Br
Br
+
Br
Br
Br
OH2
OH
CH3
CH3
HOOC
818.
H
CHO H3C
CHO
H 3C
C
B
A
H
CH3
CH3
H
D
H 3C
COOH
I
E
COOH
O
Br
OH
819.
B
A
C
D
E
820. (a)
+
+ H+
(b) (i)
+
–H
+
+
Br
(ii)
H
Bu–
C
COOH
471
Solutions
Cl
(iii)
Cl
Cl
N
–
OC(CH3)3
N
H
H
821. (a)
B
A
H
(b) HOCH2
Br
Br
Br
Br
CH2OH
Y
X
Z
Br
822.
A
C
B
COOH
COOH
D
O
O
COOH
OH
COOH
E
F
H
H
Br + Br
HO
OH
(b) H
H
823. (a)
CH3
CH3
CH3
O
O
O
(e)
(d)
(c)
H
CH3
CH3
H
H
H
OH
(f)
OH
H
Br
(g)
H
Br
OCH3
(h)
H
Cl
OH
472
Problems in Chemistry
H
CH3
—OCH3
824.
H
NaOH
syn hydration
CH3
trans
—OCH3
OH
BH3/H2O2
C
C
CH3
H
CH3
H
CH3
CH3
CH3
NaOH
syn hydration H3C
—OCH3
H
—OCH3
OH
BH3/H2O2
C
C
H
O
—C
825. CH3—
COOH
CH3—C—CH2CH2—CH—CH2COOH
C—CH3
B
A
COONa
COOH
CH—CH2COONa
CHCH2COOH
CH2CH2COONa
CH2CH2COOH
D
C
H
826. (a) Br
CH 3
H
Br
H
Br
Br
H
H
CH 3
(enantiomeric)
H
Br
CH 3
Br
Br
Br
H
Br
Br
(enantiomeric)
CH 3
(meso)
Br
H
H
(enantiomeric)
CH 3
Br
Br
Br
Br
H
H
(d)
H
H
CH 3
CH 3
CH 3
CH 3
H
H
Br
H
Br
Br
CH 3
(meso)
CH 3
H
Br
H
H
Br
Br
H
(b) H
H
(enantiomeric)
CH 3
(c)
H
Br
Br
CH 3
CH 3
(meso)
Br
H
CH 3
Br
H
H
Br
H
Br
Br
H
CH 3
(meso)
473
Solutions
827.
(CH2)10C
OMe
O
B
A
C
CH3
HO
H
D
Ph
H
CH3
E
Br
+
Br
828. Starting compound
+
Br2
H—O
HO
–H+
Product
l
l
829. (a) (i) HgSO 4 /H 2SO 4 (ii) NaBH 4
(b) (i) H 2SO 4 /heat (ii) Br 2 /MeOH
(c) (i) C 2 H 5ONa/C 2 H 5OH (ii) HOOC  CH==CH  COOH
830. (a)
CH3
CH
———CH
(b)
3
3
CH3
—CH3
831. If the number of double-bonds are same, heat of hydrogenation depends on degree of substitution
at carbon-carbon double bonds. More the number of alkyl substituents at carbon-carbon double
bond, more stable will be the alkene towards hydrogenation, hence smaller will be the heat of
hydrogenation. On that basis :
C—least substituted, highest heat of hydrogenation.
A—most substituted, least heat of hydrogenation.
H+
OH –H2O
832.
+
H2C
H
–H+
I
+
474
Problems in Chemistry
+
–H+
H+
H+
H
+
+
+
H–-Shift
–H+
II
OMe
833. (a)
(b)
(c)
Br
Ph
(racemic)
OCH3
HO (racemic) OH
Ph
(d)
(e)
+
H
H
Br
834.
C2H5ONa
RCO3H
O
C2H5OH
HC
CNa
O–
CH3I
C
Cl
835. (a)
(c)
(b)
D
CH
HgSO4
H2SO4
Product
475
Solutions
+
+ H+
836.
+
I
I
Cl
Cl–
Cl
+
II
Cl–
II
+
Cl
Cl
837.
A
B
H
H
COOH
C
H
H
D
CN
E
F
O
G
OH
C2H5
C2H5
A
839. CH 3  C ≡≡C  CH 2  COOH
(A)
O

H  C ≡≡C  CH 2  C  OCH 3
C
CH3
840. (a)
Br2
heat
COOH
HOOC
838.
O
B
O
C
CH 3

H  C ≡≡C CH  COOH
B
CH 3C ≡≡CCH 3
D
CH3
Br
C2H5ONa
CH3
CH3
MCPBA
O
C2H5OH
OH
CH3
(i) CH3COONa
(ii) H3O+
OAc
476
Problems in Chemistry
Cl
Cl2
hν
(b)
C2H5ONa
(i) CHCl3
C2H5OH
(ii) NaOH
Product
O
CH3I
Na/∆

Na
∆
(c) H  C ≡≡C  H → → CH 3C ≡≡CH → CH 3C ≡≡C – Na + →
H 3O +
→ Product
CH 2


Na
→ CH 3  C  C  CH 3
Heat

CH 2
HBr
(d) CH 3  C ≡≡CH → CH 3  C==CH 2

Br
H
H
O
l
l
O+
CH3
O
H
l
l
+
CH3
+
O
CH3
H+
841. (a)
O
–H+
(b)
O
O
+
+ NH3
H3N
H
+
HO
O
N
–
HO
H
–
OH
N
HO
H
O
NH2
O
H
+
HO
N
O
–
(c)
OH
OH
H
OH
N
HO
OH
H
O
O+
Br2
Br
+
–H+
Br
O
Br
477
Solutions
H+
842.
+
H
+
HSO4–
H
+ H2SO4
H
(3° carbocation)
H
+
OCH3
l l
Cl2
843. (a)
+
Cl
O
H
CH3
–H+
H
Product
Cl
H
(b)
Br
l
l O—H
+ Br2
H
HO–
H
+
O
+
H+
(c)
O
H
O
OH
H
O—H
H
H
O
Br
Br
+
O
–H+
l l
O—
OH
OH
O
OH
OH
H
–H+
+
H+
l l
O
O
CH3
OH
CH
OH
Br
CH3
CH3
OH
CH2—C—CH3
CH—CH—CH3
C—CH3
O
O
H
844.
Br
B
A
OH
CH3
CH2—C
OH
C
CH3
CH2—C—CH2OH
CH2
O
H
D
E
F
478
Problems in Chemistry
+
CH3
+ H+
845.
+
I
H+
I–
A
I
+
I–
+
B
+
Br–
+H+
846.
+
Br
H– shift
Br
Br–
+
+
Br–
Br
847. This compound has a chiral carbon. Therefore can be resolved into enantiomers. Treatment of this
compound with concentrated sulphuric acid bring about isomerization into a more stable
compound but without any chiral carbon, hence optical activity is lost as:
+
H+
H– shift
2° carbocation
CH3
CH3
+
3° carbocation
–H+
No chiral carbon
CH3
OH
848.
A
C
B
+
849.
+
+ H+
+
Protonation at any other side will produce a less resonance stabilized carbocation.
479
Solutions
ALKYL HALIDES
850. (a) n-iodoheptane since iodide ion is a better leaving group.
(b) Sodium ethoxide since it is a stronger nucleophile than ethanol
(c) 1-butyltosylate, due to less steric hinderence.
(d) triarylbutoxide with ethyl bromide.
(e) with-2-bromopentane.
851. Both the reactions are examples of S N 2 reaction. Since aniline is a weaker nucleophile than
cyclohexylamine, former undergoes slower substitution reaction.
CH3
Cl
852.
CH3
CH3
H3C
CH3
H3C
CH3
H3C
CH3
B
A
C
CH3
CH3
CH3
H2C
CH3CH2CH—COOH
E
D
Cl
853.
H
H
H
H3C
CH3
CH3 H3C
B
A
H
H
H
H
D
C
H
CH3
CH3
CH3
CH3
H3C
CH3
E
854. (a) Methylbromide undergo halogen exchange with NaI forming methyl iodide. Iodide being a
better leaving group then bromide, rate of hydrolysis is increased.
(b) The predominant reaction in this case is elimination (E-2) which depends on concentration of
both alkyl halide and base. Increasing concentration of base increases the rate of E-2 reaction
decreasing concentration of alkyl halide.
(c) NaCN is ionized completely giving ambident (CN − ) nucleophile. Since carbon is better
electron donor than nitrogen, bonding occurs from carbon side. AgCN being covalent, CN − is
not totally free for attack and attack primarily occurs from nitrogen side giving isocyanide.
480
Problems in Chemistry
855. (a)
OH
H+
–H2O
CH2+
Cl–
Cl
CH2+
CH2Cl
Cl–
(b)
856. (a)
(b)
(c)
857. (a)
(b)
858. (a)
H+
OH
–H2O
+
Cl–
Cl
(3° carbocation)
C—F bond is the strongest carbon-halogen bond, undergo very difficult heterolysis.
Latter forms a more stable carbocation.
t-butylchloride form a more stable carbocation.
AgNO 3 forms insoluble AgX driving reaction in forward direction.
More acidic solvent increases the rate of heterolysis facilitating formation of carbocation.
CH2CH2CH2OCH3
(b)
CH 3 CHCH 3 ( A ) CH 3 CHCH 3 ( B )


OTs
N3
C
H
CH3S
CH3
859. III < II < I
860. (a) Ethanol being a very weak nucleophile can’t activate S N 2 reaction, and CH 3 Br being very less
reactive in S N 1reaction, neither S N 1nor S N 2 take place. Adding excess of C 2 H 5ONa involve
S N 2 reaction and ether is formed.
(c) Et 3 P is a better nucleophile than Et 3 N.
N3
CH2
OCH2CH
861. (a)
(b)
I
Br
(c)
(d) MeO—
N—H
(e)
Br
862. Tertiary butyl being a bulky base abstract H-from less hindered β-carbon giving this alternative
product.
481
Solutions
OTs
OMe
863.
;
CH3ONa
SN2
OTs
OTs
OTs
OTs
CH3ONa
E-2
OTs
1,5-hexadiene,
864.
1,4-pentadiene
N
N
H
CH3
2-methyl-1,4-pentadiene,
3-methyl-1,4-pentadiene
N
N
H
H
H
H
(b) CH3——OH
865. (a) HO——CH3
(c)
C4H9
C4H9
CH3
CH3
H
H
(racemic)
CH3
CH3——H
(d)
CH3——H
C2H5
866. (a) First
867. (a)
CH3
CH3O——H
(e)
H——CH3
C2H5
(b) Second (c) First
Br
+
Heat
Br
+
OH
H2O
–H+
CH3
(b)
H2O
Heat
+
2H
+
H 3C H
+
OH
CH3
H3C H
OH
482
Problems in Chemistry
868. (a) Though neopentylbromide is primary, bulky tertiary butyl group poses very large steric
hindrance to the attack of bulky nucleophile N −3 .
(b)
Br
H——CH3 + N3–
N3
CH3——H
S N2
D
D
(c) Rate will double (d) Rate will double (e) not related
(f) Racemization occur through carbocation intermediate.
869. (a) n-iodoheptane (b) C 2 H 5ONa (c) n-butyltosylate
CH3I
Product
Excess
NaOH
—OH
870. (a) HO—
(d) isopropoxide with C 2 H 5 Br.
NH
TsCl
3
(b) C 6 H 5CH 2OH →
→
Product.
CH CH CH SNa
3
2
2
TsCl
(c) C 6 H 5OH →
→
Product.
(d)
B2H6
Cl
KOH
Product.
H2O2/OH–
871. Allyl bromide produces a resonance stabilized carbocation justifying its greater reactivity in S N 1
reaction. Also carbon β to Br has an empty p-orbital which share some of the shared electron pairs
of pentavalent transition state formed in S N 2 reaction, increasing its stability and hence reactivity.
872. (a) CN − being weaker nucleophile than H − .
(b) 3° carbocation is more stable than 1°.
(c) Geminal diols are unstable.
+
873. (a)
–H+
+
••
OH
••
Product
O
H
(b)
—Br
+ Br –
O
–
O
874. (a) CF3– < CH 3O – < CH 3S – < F –
(b) CH 3COO – < CH 3SO 3– < CF3SO 3–
Br
Br
Br
875.
A
B
Br
Br
C
D
E
483
Solutions
CH2Br
H——CH3
H3C——H
CH2Br
CH2Br
H——CH3
H——CH3
CH2Br
F
G
COOH
876.
Br
A
B
(CH3)2CHOH
877.
O
A
C
H
CH3—CH—CH2CH3
B
CH3
C
D
HOOC
C
CH3
H
OH
C
CH3
C
C
H
CH3
D
H
E
Cl
C—CH3
CH3CH
878.
CH3
A
B
Br
C
O
O
879.
COOH
A
E
B
C
D
484
880.
Problems in Chemistry
O
I
A
I
NC
I
I
OH
D
C
B
Br
Br
881.
OH
A
C
B
C H SNa
TsCl
2 5
882. (a) →
→
TsCl
NaI
(b) →
→
PhCH CH I
CH C ≡≡ CNa
NaH
2
2
(c) →
→
3
(d) n-butylbromide →
NH
3
TsCl
(e) →
→
O
883.
S
O
O
CH3
D
H2N—CH— CH2OH

CH==CH2
→
(f)
NaI
I
Rate = 1
mesylate
O
S
O
O
CF3
NaI
I
Rate = 5000
triflate
In both mesylate and triflate, the leaving group anion is stabilized by resonance. In the case of
triflate, the three fluorides provide additional stabilization to the anion via electron withdrawing
inductive effect.
O–
O
O
O




–
Mesylate: O  S  CH 3 ←→ O ==S  CH 3 ←→ O ==S  CH 3 ≡≡ O  S  CH 3




O
O
O
O
–
O–
O
O F
O


 

–
Triflate: O  S  CF3 ←→ O ==S  CF3 ←→ O ==S  CF3 ≡≡ O  S  C  F


 

O
F
O
O
O
–
(more stable)
485
Solutions
O
O
884.
1.
O
HOCH3
Ag+
+
OCH3 +
–H+
O
O
O
Ph
Ph
OCH3
Ph
Planar carbocation
O
2.
Ag+
O
O
OCH3
HOCH3
+
O
O
O
+
O
–H+
O
O
Ph
Ph
Ph
a resonance stabilized
acyl oxonium ion
(c) SN–1, CH 2 ==CH  CH 2 Br.
885. (a) SN–2, CH 3CH 2 Br, (b) S N 1, (CH 3 ) 3 CCH 2 Br,
H
886. (a)
n-Bu
(c)
CH3—CH—CH2—
Et
Ph
Cl
(b)
C==C
CH3
CH3
Et
iPr
(d)
CH3
(e)
CH(CH3)2
(f)
H
OH
Et
H
COOCH3
Et
CH3
(g)
(Retention of configuration)
(Racemic mixture)
CH3
CH 3CH 2
887. (a)
(b)
CH3
OH
(c)
C
CH  CH 2CH 3

CH 3
H 3C
CH3
(Enantiomers)
SCH(CH3)2
(d)
Ph
(e)
CH3
ClCH 2
Ph
Et
H
(Retention of configuration)
(f) CH 2 == C
C(CH 3 ) 3
486
Problems in Chemistry
(g)
CH 3
CH 3
Cl



+
heat
H– -shift
888. CH 2 == C  CH 2  CH  CH 3 → CH 2 ==C  CH 2  CH  CH 3 ←→
CH 3
CH 3
 +

+
CH 2 == C  CH  CH 2CH 3 ←→ CH 2  C== CH  CH 2CH 3
X
Y
CH 3
CH 3


+
CH3OH
X + CH 3OH → CH 2 == C  CH  CH 2CH 3 → CH 2 ==C  CH CH 2CH 3 + CH 3OH 2


H 3C  O  H
OCH 3
+
H
CH 3


CH3OH
Y + CH 3OH → CH 3  O  CH 2 C== CHCH 2CH 3 →
+
CH 3

+
CH 3OCH 2  C== CHCH 2CH 3 + CH 3OH 2
KOH
I
889.
NBS
—Br
EtOH
NBS
KOH
EtOH
—Br
KCN(aq)
Product
890. Since, the tosylate is tertiary, E–2 and S N 1 are equally probable:
F
F
PhNH2
E-2
H3C
OTs
CH3
F
SN1
H3C
F
PhHN
+
: Net result is racemic mixture.
NHPh
Enantiomeric
CH3
Enantiomeric
487
Solutions
C3H7 H
891.
H
Br
Br
C4 H 9
H
H
C3H7
H
C==C
C4H9
H
C4H9
H
H
C3 H 7
C2H5OK
Major alkene
More stable conformer
H
C3 H 7
H
Br
Br
H
C3H7
H
Repel each other
C4 H 9
H
C3H7
C4H9
H
H
H
C2H5OK
C==C
C4 H 9
Minor alkene
Less stable conformer
CH2CH3
892. (a)
H
CH3CH2CH2O
C
CH3
OCH3
(c)
(b)
CH3
+
CH3
OCH3
CH3
OCH3
(Racemic mixture)
Br
893.
O
B
A
D
C
OMe
894. (a)
(b) no reaction
H
Cl
(c)
OEt
Ph
(d) CH3CH2CH
C—CH3
CN
(e)
(f)
ONa + CH4
(cis + trans)
Br
Br
895.
+
l l
H—N
N
–
H
N
488
Problems in Chemistry
Br
H
N
H
+
N
l
l
– H+
N
896. Esters are usually water insoluble, if acetic acid is used, it gives homogeneous medium driving
reaction in forward direction.
897. (a) 2-chlorobutane,
(b) 2-chloro-2,3-dimethylbutane,
(c) 1-methylchlorocyclopentane,
(d) 1-ethyl-2-methyl chlorocyclohexane.
D
898. (a)
D
(b)
C
NC
CH3
D
C
NC
H
H
899. (a)
OH
(b)
OH
(c)
OH
(d)
OH
(d)
OH
C
+
CN
H3 C
TsCl
KI
TsCl
CH3CH2NH2
CH3
H
CH3CH2COOH
H+/heat
TsCl
NaHS
TsCl
CH3ONa
900. Only II can be used for successful synthesis of Grignard reagent, rest all contain acidic proton and
will react with R – (from Grignard reagent) forming alkane.
Cl
O–
O
O
–
O
O
+ OCH3
901. (a)
+ Cl–
Cl
(b) Br
–
+ OCH3
Br
OCH3
O
O–
O
–Br–
OCH3
489
Solutions
O
–
+ OH
(c)
O
O–
HO
O
HO
O–
O
H2O
Product
O
HO
O–
O
—Br
902. (a)
Mg
—Br
s (MgBr)+
SN2
H
(b)
+
H–-shift
H+
—OH
H
+
–H2O
H– – shift
+
Br–
Product
+
ring
expansion
903. (a) (CH 3 ) 3 CI is more reactive since I – is a better leaving group than Cl – .
(b) I is more reactive since β H is less sterically hindered compared to II.
CH(CH3)2
CH(CH3)2
CH(CH3)2
Cl
Cl
CH3O–
CH3O–
H
H
E-2
904. (a)
E-2
H
H
(achiral)
Enantiomers
(b) As shown in (a), the predominant elimination product is achiral.
(CH3)2HC
(c)
H
Cl
H
CH(CH3)2
OCH3
CH3
O–
SN2
CH(CH3)2
OCH3
H
H
H
H
Enantiomers
CH3O–
H
Cl
H
CH(CH3)2
490
Problems in Chemistry
CH3
905.
I
(CH3)3COK
II
(CH3)3COH
(CH3)3COK
(CH3)3COH
II is more reactive towards elimination reaction since the bulky base will abstract a H from side of
least steric crowding giving more stable alkene.
OCH3
+
OCH3
Cl
OCH3
+
Resonance stabilization
906. (a)
(b)
(c)
(d)
(e)
907.
First is more reactive since it produces a more stable carbocation:
Second is more reactive since it produces a resonance stabilized allylic carbocation.
First is more reactive since it produces a more stable benzylic carbocation.
Second is more reactive due to the same reason as with (b).
Second is more reactive due to the same reason as with (a).
I
–I+
+
Cl–
CH3OH
–H+
I
III
Me– -shift
Cl–
+
CH3OH
–H+
II
IV
ALCOHOLS AND ETHERS
908. Ethylene oxide being a three membered ring, suffers from very large angle strain and therefore
highly unstable whereas THF is a five membered ring, it is stable.
O
CH3ONa
H2SO4
909. Ph—CH —CH 2OCH 3 ←
Ph—CH—CH 2OH
Ph  →
CH
OH
CH
3
3OH


OH
OCH 3
491
Solutions
O
CH2OH
910. (a)
CH2OH
N2H4
H2SO4
O3
Heat
Zn-H2O
NaOH/Heat
LiAlH4
OH
Br
(b)
aqueous
Heat
+ Br2
KOH
+
H+
–H+
Heat
+
O
O
O3
+
Zn/H2O
Y
X
X
Conc. H2SO4
LiAlH4
Y
Product.
Heat
H2CO
NaOH
CH2
O
Heat
O3
O
O
Zn-H2O
CH3
A
B
CH3
—CH2
E
—CH2D
—CH2MgBr
—CH2Br
911.
C
CH3
C
—CH2—C—OH
CH3
D
492
Problems in Chemistry
912. At higher temperature, intramolecular dehydration is entropy favoured.
913. 2-butanol
2-butanone
2-bromobutane
B
C
A
OMgBr
D
CH 3
CH 3


Ph—C—CH 2CH 3 ( B ) Ph—C== CHCH 3 (C ) CH 3COOH( D ).

Br
914. Ph—CH —CH —CH 3 ( A )


OH OH
O
NBS
915. (a) Ph—CH2—
Ph—CH—
CCl4
KOH
O3
Ethanol
Zn–H2O
+ PhCHO
Br
H 3O
(b)
C2H5ONa
Ag2O
Br2/Heat
O
C2H5OH
O
CN
Br
NaOH
KMnO4
CaO Heat
OO
O
H
H2SO4
O
COOH
O
Br
Ag2O
Br2 Heat
O
O
C2H5ONa
KMnO4
C2H5OH
H2SO4
COOH
O
O3
N2H4/OH–
Zn-H2O
Heat
CH2
O
(CH3)3COK
Ag2O
(CH3)3COH
Br2 Heat
493
Solutions
H3O+
(c)
O
CH3COCl
OH
H2O
OH
O
O
O3
H2SO4
Zn-H2O
Heat
NaBH4
OCOMe
OCOCH3
O
OCOCH3
H
SOCl2
H3O+
CH3CH2CHO +
O
Zn
Heat
OCOCH3
PBr3
H
NaBH4
Br
O
OH
OH
Br
916.
O
A
C
B
OH
E
D
F
H
G
917. (a)
OH
H+
+
–H+
–H2O
(b)
OH
••
OH
OH
+ H+
+
O
–H+
+
(c)
+
OH + H
+
CH2
–H2O
H-shift
–H+
+
CH3
–H+
+
–H
494
Problems in Chemistry
OCH3
OH
H+
(d)
+ CH3OH
H2O
NaBD4
H2O
+ (CH3COO)2Hg
918. (a)
O
OH
D
OH
(b)
SOCl2
Mg
Cl
A
Ether
H
PCC
CrO3
H3O+
A
H2SO4
O
O
OH

H3
PhMgBr
(c) CH 3COCl →
→
C6 H 5
CH
—C—
3
excess

C6 H 5
O+
O
O
O
MeMgBr
(d)
O
H3O+
OH
C2H5ONa
SOCl2
C2H5OH
OH
OH
OH
B2H6
CH3MgBr
H2O2/OH–
OH
H3O+
Cl
NaCl
TsCl
919. (a)
SN2
(b) Ethylbromide being a primary halide undergo predominantly S N 2 reaction while cyclohexyl
bromide being a secondary halide, undergo E-2 reaction.
O
OH
920. (a)
A
OH
O
(b)
H
H
B
A
495
Solutions
O
(c)
+ CH2O
OH
(d) (i)
O
(ii)
OH
COCH3
OH
921.
A
922.
H
H
CH3
Br
B
H3O+
H
CH3
C
OH
CH3
H
OH
H
CH3
CH3
O
OH
CH3
H
H3O+
H
(Racemic mixture)
CH3
H
OH
CH3
(meso diol)
923. Chlorination of alcohols with HCl proceeds by S N 1 mechanism, rearranged products are also
obtained. Chlorination of alcohol with SOCl 2 proceed by S N 2 mechanism, no rearranged product
is obtained.
924. (a)
ONa +
(b)
ONa + PhCH2Br
(c)
—ONa + CH3CH2Br
Br
Product.
Product.
Product.
(d) CH 3CH 2ONa + CH 2 == CH—CH 2 Br → Product.
925. The major product is 2-chloro-2-methylbutane.
926. (a) (CH 3 ) 2 C==CHCH 2Cl gives a resonance stabilized tertiary carbocation as:
+
+
Allyl chloride gives primary carbocation and therefore less stable than the above mentioned
carbocation.
496
Problems in Chemistry
(b) In acidic medium alcohol undergo isomerization via carbocation intermediate as:
H 2O
–H+
H+
–H2O
OH
+
+
HO
Ph
927. (a)
Ph
H+
Ph
+
+
–H2O
OH OH
Ph
Ph
••
O—H
••
Ph
+
OH Ph
–H+
Ph
O
OH
+
H+
(b)
+
–H+
–H2O
H+
(c)
+
+
–H2O
OH OH
—— OH
••
OH
H+
(d)
+O
O
–H+
+
O
–H+
HO
HO
H
928. PhCH(OH)CH 2 Ph
PhCOCH 2 Ph
Y
X
R
929. (a)
R
HBr
Peroxide
(CH3)3COH
Br
(b)
OH
O
R
(CH3)3COK
H+
–H2O
–H+
+
O
CH2
+
497
Solutions
OH
CHO
(c) (i)
CHO
OH
A
(ii)
OH
CH3—CH—CH2SPh
(iii)
OH
C
B
OH
D
OH
(iv)
E
930. C 6 H 5 —CH 2 —CH —CH 3

OH
C 6 H 5 —CH—CH 2OH

CH 3
( A)
CH 3

931. CH 3 —CH— CH —CH 3

OH
( B)
CH 3
CH 3


CH 3 —CH CH 2CH 2OH CH 3 —C—CH 2CH 3
( B)

OH
( A)
(C )
CH 3
CH 3


CH 3 —CH CH 2COOH CH 3 —C—CH 2CH 3
(D )

Br
CH 3

CH 3 —CH—COOH
(F )
(E )
OH
932.
A
CH 3

933. CH 3 —CH —C—CH 2CH 3

CH 2
( A)
O
C
B
CH 3

CH 3 —CH —CH—CH 2CH 3

CH 2OH
CH 3

CH 3 —CH —CH—CH 2CH 3

COOH
( B)
(C )
CH 3

CH 3 —CH— C —C 2 H 5 ;

O
CH 3

CH 3 —CH— CH —CH 2CH 3

OH
(D )
(E )
CH 3
CH 3
C== CHCH 2CH 3
(F )
498
Problems in Chemistry
936. 1,3-butandiol (A) 3-ketobutanoic acid (B).
+
Methyl shift
H+
934.
H2O
–H+
+
OH
(Major)
Cl
935.
*
Cl
–
OC2H5
O
I
937. (a)
*
CH2—CH—CH2—OC2H5
*
CH2—CH—CH2—OC2H5
O
O
–
—OH + HCl
D2O
Mg
—Cl
—D
ether
O
H2SO4
(b) CH 2 == CH 2 + Cold, dilute, alkaline KMnO 4 → CH 2 —CH 2 →
Heat


OH OH
O
CH 3

Heat
938. (a) CH 3 —CH—CH 3 + Br 2 → (CH 3 ) 3 CBr
CH 3

(b) (CH 3 ) 3 CBr (from a) + C 2 H 5ONa → CH 3 —C== CH 2
C2H5OH
E-2
Peroxide
(c) (CH 3 ) 2 C==CH 2 + HBr →
(CH 3 ) 2 CHCH 2 Br
CH ONa
SN 2
3
(d) (CH 3 ) 2 CHCH 2 Br →
(CH 3 ) 2 CHCH 2OCH 3
(e) (CH 3 ) 2 CHCH 2 Br + CH 3COONa → (CH 3 ) 2 CHCH 2OCOCH 3
(f) (CH 3 ) 2 CHCH 2 Br + KCN( aq ) → (CH 3 ) 2 CHCH 2CN
O
939.
OH
Cl
COOH
—O—C—
A
B
D
C
CH2OH
Cl
F
COOH
G
H
I
E
499
Solutions
OH
OH
940.
A
B
C
I
O
941.
A
O
B
I
C
H O
2 2
942. (a) 2-methyl propene + B2 H 6 →
Product
NaOH
CH COOT
3
(b) 2-methyl propene + B2 D 6 →
Product.
CH COOT
3
(c) 2-methyl propene + B2 H 6 →
Product.
C2H5Br
NaH
NaOH
(d) 2-methylpropene + B2 H 6 + H 2O 2 →
(CH 3 ) 2 CHCH 2OH → → Product.
OH
943.
H+
–H2O
+
CH2
–H+
+
CH3O
CH3O
944. HO
CH
CH—CH2OH
CH—CH—CH2OH,
HO
Br
X
Br
A
O
HO—CH2—CHO
CH3O
O
C6H5—C—O
CH
B
C
CH—CH2O—C—C6H5,
500
Problems in Chemistry
CH3O
I
HO
CH—CH2Br,
CH
HO
CHCH2OH
CH
D
E
CH3O
CH—CH2OH
CH
CH3O
F
945.
O
O
A
C
Cl
Cl
B
CH3
CH3
946.
C—CH2CH3
C
C
CH3
CH3
and
H
OH
A
CH3
B
O
OH
—C—CH3
CHCH3
—CH—CH3
C
E
D
CH 3
CH 3


947. CH 3CH 2 —CH—CH== C—CH—CH 2CH 3

CH 3
CH 3

CH 3CH 2 — CH— C—CH 3

O
A
CH 3
CH 3


CH 3CH 2 —CH— C—H CH 3 —CH 2 —CH—COOH

O
C
H
C
C
D
B
CH 3

CH 3CH 2 —CH—CH —CH 3

OH
E
501
Solutions
CH 3

CH 3 —CH 2 —C== CHCH 3
CH 3CHO
CH 3 —C —CH 2CH 3

O
F
G
H
CH3
CH3
—CH—CH2OH
948.
CH3
—CH—COOH
A
C
B
OH
O
C—CH3
D
H
H
O
O
E
F
OH
CHO
OH
949. (a)
I
(b)
(enantiomers)
(c)
(d)
OCH3
HO
OH
(g) H
(f)
(e)
CH2CH3
SCH3
(h) HO
OH
OH
OH
(j)
(enantiomers)
(i)
OCH2CH3
502
Problems in Chemistry
950. (a) PCC, (b) NaBH 4 , H 3O + , (c) Pd/C, (d) Mg, Ethyleneoxide, H 3O + , PBr 3 ,
(e) TsCl, CH 3ONa, (f) SOCl 2 , Mg/ether, CH 2O, H 3O + ,
(g) Ethylene glycol/ H + , LiAlH 4 , H 3O + , (h) CH 3 MgBr, H 3O + .
Br
Mg
951.
C6H5MgBr
A
O
O
H
H
OH
PCC
A
H3O+
CH3MgBr
OH
Product
CH3CH2ONa
OCH2 CH3
952. (a)
H3O+
NaH
Br
(b)
+
Br + CH3 Br
OH
l
+
(c)
S
Br–
CH3
l
S
O
(d)
(e)
O
enantiomers
OH
OH
CH3
(f)
OCH3
(g)
O
OH
SCH2 CH3
O
(i) O
—S—
O
(h)
(j)
OH
CH3
CH3
503
Solutions
Si
O
953.
O
Si
+ CF3CO3H
F–
O
H3O+
O + (CH3)3SiF
O
O–K+
+
954. A : CH 3 MgBr/Ether, H 3O ;
D:
B:
OH
(A)
+
NaOCH 3 , H 3O , NaH/CH 3CH 2 Br;
Hg(OAc) 2 , CH 3OH, NaBH 4 ; E : CH 3CH 2SNa/H 3O +
F:
C:
mCPBA;
SO2
MeO
955. In case of reaction 1, the mechanism is a series of protonation and S N 2 reactions as:
H
O
+
CH3
O
I
CH3
H+
+ CH3OH
I–
+
H+
CH3—OH2
I–
CH3I + H2O
In the case of reaction 2, the mechanism is also a series of protonation and S N 2 reaction. However,
phenol is not converted to iodobenzene because of protonated phenol can’t be substituted by
iodide ion (I – ) due to strong C  O bond in phenol:
H
O+
O
CH3
OH
CH3
H+
I–
+
others
H
O
H
+ CH3I
O+
–
–
Carbon-oxygen bond acquire partial double bond character,
can’t be substituted under SN2 conditions.
504
Problems in Chemistry
O
OH CH2N2
OH K2Cr2O7
956. (a)
O
OCH3
H3O+
(1) CH3CH2MgBr (2eq)
Product
(2) H3O+
OH
(b)
N3
NaN3
TsCl
NH2
H2/Pd
O
Cl
OH
(c)
CH2
PCC
PPh3
OH
TsCl
C–Na+
CH
PCC
CH2Cl2
(d)
Product
O
OH
Product
C
H 3O +
C
CH
CH
NaBH4
Pd
BaSO4
mCPBA
Product
O
957. (a)
—D
(b)
OH
H
H
(c)
(d)
SCH3
(e)
CH2CH3 +
(enantiomers)
OH
CH3
OH
CH2CH3
CH3
505
Solutions
OH
OH
I
(g)
(f)
HO
OH
HO
OCH2CH3
(h)
O
958. (a) (1) PCC, (2) CH 3 Li, (3) H 3O + ; (b) HI/Red P; (c) (1) SOCl 2 , (2) Mg/ether; (3)  ,
(4) H 3O + ; (d) (1) PCC, (2) NaBH 4 , (3) H 3O + .
H
959.
H
CH3
CrO4
OH
H+
2–
B–
O
O
O—Cr—OH
Slow
O
C—H bond cleavage occur in slow rate determing step. Therefore, replacement of H by D will
slow down the rate of reaction.
O
H
D
C:
OH
PCC
CH2Cl2
D
C  H bond will be broken in preference to C  D bond.
O
O
960.
O
H
l
l
+
O —S—
+ Cl—S—
H
O
— + Cl–
O
H
O
Pyridine
O —S—
O
+
N Cl–
—+
506
Problems in Chemistry
Cl– H
O
N
O—S—
+
+
Cl
60°C
O
H
O
+
N
–
+ O—S—
+
O
Higher temperature promotes SN–2 reaction of tosylate by chloride ion generated in the first step
of reaction.
Ot-Bu
O
961. (a)
(b)
SCH2CH3
(c)
S
OCH3
O
OCH3
O
O
(d)
Br
(f)
(e)
+
O
OH
N3
O
OCH3
(g)
O
(h)
(i)
OH
O
OPh
O
(j)
(k)
O
O
OH
507
Solutions
HO
Br
O
N3
NaN3
NaOH
962. (a)
HO
CH3I
NaH
Product
H3O+
O
HO
CH2
(b)
CHLi
mCPBA
NaH
CH3I
LiAlH4
NaH
CH3I
Product
from a
OH
O
H
963.
H 3O +
Br
PCC
Mg/ether
O
H 3O +
O
CH3CH2ONa
HO
OCH2CH3
CH2N2
HO
OCH2CH3
mCPBA
O
964. I > II > III > IV.
OCH2CH3
965. (a)
D
(b)
(c)
OH
O
Br
Ph
(d)
(e) H
OH
CH3
CH2CH3
508
Problems in Chemistry
OCH2CH3
OH
HO
OH
966.
(a)
(b)
(c)
(enantiomers)
(Major)
(Major)
OH
O
Br
(d)
(e)
OH
OH
(b)
PCC
CH2Cl2
CH2N2
NaH
EtONa
(c)
H3O+
CH3Li
PCC
CH2Cl2
967. (a)
NaH
CH3I
Product
Product
CH3I
O
OH PBr3
(d)
PhSNa
mCPBA
Product
968. There are two problems with the route outlined above. First, because the above alcohol is tertiary,
it will be difficult to make the tosylate, which is a large, sterically demanding group. If you can
make it, the second step is also difficult as it require SN–2 reaction of a tertiary tosylate by H. This
system is very hindered and will not readily undergo susbtitution reaction. Elimination will be a
serious competition reaction. A better method would be as follows:
S
OH
O—C—SCH3
NaH
CS2
CH3I
Bu3SnH
509
Solutions
OH
Cl
K2Cr2O7
969. (a)
SOCl2
NH3
H2SO4
Br
(b)
NaOH
O
POCl3
COOH
K2Cr2O7
Product
Na/NH3
H3O+
EtOH
H2SO4
O
OH
OH
(c)
TsCl
NaCN
CH2N2
NH2
LiAlH4
CH3COCl
OCH3
970. (a)
—Cl
(b)
Product
O
H
Product
CH3
H
(c)
H
(d)
971. (a)
CH3
(e)
Br
NaOH
EtOH
mCPBA
O
NaCN
H 2O
Product
510
Problems in Chemistry
(b)
—OH
C. H2SO4
NBS/hν
CCl4
—Br
Mg
ether
—OH
PCC
CH2Cl2
—MgBr
O
OH
NaBH4
TsCl
OH
CH3
Cl
973. O
(c)
OH
CH3
O
PhS–
OH
OH
Ph
(b) CH3—C—CH2CH3
972. (a)
(d)
I
O–
O
O
O
H2O
Product
OH
O
O
O
O
–
SPh
Product
OH
C H ONa
Hg(OAc)2
C2H5OH
H 2O
NaBH
2 5
4
975. (a) PhCH 2CH 2CH 2 Br →
PhCH 2  CH==CH 2 → →
O
OH



PCC
Ph  CH 2  CH  CH 3 →
Ph—CH 2 —C—CH 3
CH2Cl 2
I
OH

H 3O +
I
Na
Pd
CH 3C ≡≡CH → CH 3C ≡≡C – Na + → →
CH 3C ≡≡C  C  CH 3 → Product
heat
CaCO3 /PbO

CH 2 Ph
O
(b)
B2H6
—CH2OH
H2O2/NaOH
Cl2
PCC
CH2Cl2
—C—H
I
Cl
Mg
ether
AlCl3
H 3O +
I
OH
—CH—
OH
Cl2
AlCl3
—CH—
—Cl
PCC
CH2Cl2
Product
511
Solutions
O
SOCl 2
C2H5ONa
mCPBA
(c) CH 3CH 2CH 2CH 2OH → → → CH 2CH 2  CH  CH 2
C2H5OH
OH

PCC
→ → CH 2CH 2 CH  CH 2CN → Product
NaCN
H 2O
CH2Cl 2
NBS/hν
Mg
CCl 4
THF
976. CH 2 ==CH  CH 3 → CH 2 ==CH  CH 2 Br → CH 2 ==CH  CH 2 MgBr
OH

→ → CH 2 ==CH  CH 2  CHCH 2CH 3
H 3O +
Propanal
CH3I
NaH
Product ← ←
Conc. H2SO4
977. (a)
mCPBA
PhO–
H 2O
Product
heat
CH3
O
(b)
CH3
OH
HO–
H3C
O
Br NaNH2
OH PBr3
H 2O
I
OH
C
NH3
Br
CH
OH
Na
heat
I
C2H5—C
+ H+
978.
O
C—CH2—CH—C2H5
PCC
CH2Cl2
Product
HOCH2CH2—CH—CH2—CH3
O+
CH—CH3
+
H
H– Shift
Br
HOCH2CH2—C—CH2CH3
CH2CH3
Br–
+
HOCH2CH2—C—CH2CH3
CH2CH3
512
Problems in Chemistry
C H ONa
Hg(OAc)
C2H5OH
H 2O
NaBH
2 5
2
4
979. (a) CH 3CH 2CH 2CH 2 Br →
CH 3CH 2CH==CH 2 →

OH

CH 3CH 2  C  CH 3

CH 2CH 3
(b) PhCH 2CH 3
O
OH


PCC
← CH 3CH 2  C  CH 3 ← CH 3CH 2  CH  CH 3
(i) C2H5MgBr
CH2Cl 2
(ii) H2O
Br

C2H5ONa
mCPBA
→ Ph  CH  CH 3 → Ph  CH==CH 2 →
NBS/hν
C2H5OH
CCl 4
O
HBr
Ph   → Product
B2H6
H2CrO4
SOCl 2
(c) (CH 3 ) 3 C  CH==CH 2 → (CH 3 ) 3 C  CH 2CH 2OH → →
H2O2 /NaOH
O

→ (CH 3 ) 3 C  CH 2  C 
AlCl 3
PhI
—I
H2SO4
→ Product
SO3
PCC
CH2Cl 2
(d) CH 3CH 2OH → CH 3CHO(I)
Br
OH
O
NaOH
OH
PBr3
H 2O
CH3—CH—CH2Br
NaNH2
I
CH3—C
H 3O +
C–Na+
Product
CH 3
CH 3
O


C2H5ONa
PBr3
Mg

(e) CH 3 —CH—CH 2 Br → →
CH 3 —CH—CH 2CH 2CH 2OH →
→
Et 2O
C2H5OH
CH 3

(i) KCN
mCPBA
→
CH 3 —CH—CH 2 —CH== CH 2 →
Product
+
(ii) H3O
CH 3CH 3
 
980. (a) CH 3  C  CH  CH 2CH 3
 *
Br
(Racemic)
+ CH3C
(b)
O
–
OMgBr
CH
513
Solutions
CH3 SEt
OH H
(c)
CH3 SEt
OH H
+
CH 3

(d) CH 3  CH  C  CH 3


NHPh CH 3
(Racemic)
O
S
(e)
(f)
Br
OEt
Br
981. (a)
OCH3
O
Mg
Et2O
OH
H2O
OH
NBS/hν
CCl4
Br
ONa
Product
Br
(b)
Cl
(CH3)3COK
(CH3)3COH
HO
KOH
NBS/hν
CCl4
∆
Br
NaOH
NBS/hν
CCl4
H2O
OH
OTs
TsCl
982. (a)
CH3
CH3
OTs
CH3
Cl
NaCl
Et2O, SN2
SN2
(from a)
CH3
I
NaI
(b)
Cl
NaCl
SN2
CH3
CH3
514
Problems in Chemistry
Cl
Cl
NBS/hν
CCl4
983. (a)
Cl
CH3ONa
OCH3
Br
CH3
O
Mg/Et2O
Br
Product
Br
C2H5OK
NBS/hν
CCl4
(b)
H 2O
NBS/hν
CCl4
C2H5OH
PhS–
Product
Cl
(c)
C2H5ONa
CHO
Br Mg
NBS/hν
CCl4
Et2O
CH3
TsCl
(d)
CHO
C2H5OH
I
H 3O +
CHO
I
KCN
mCPBA
O
EtOH
OH
984. (a)
OH
PCC
CH2Cl2
O
Product
OH
Br
CH3
EtOK
Br
NBS/hν
CCl4
Product
Ph
H3O+
Ph
PhMgBr
H+
EtOH
OH
OMgBr
OH
(b)
Ph
OH
(as in a)
OH
H2SO4
heat
Ph
O
(c)
—Ph
O3
H2O2
O
HO–, H2O2
O
Ph–C—CH2CH2CH2—COOH
from (b)
B2H6
OH
Ph
EG/H+
O
Ph—C—CH2CH2CH2COOH
O
H2SO4
Ph
Ph—C—(CH2)3—CH2OH
LiAlH4
H3O+
515
Solutions
Ph
Ph
Ph
–H2O
H+
985.
OH2
OH
—CH3
(a planar carbocation)
+
(+)-2-phenyl-2-butanol
CH3OH
Ph
+
H3C—C
attack from
both side giving
racemic product
(±)
OCH3
Br

986. CH 2 ==CH  CH 2CH 2OH ( A ) CH 2  CH  CH 2  CH 2OH ( B )

Br
Br
O
D
O
C
O
E
OH
987.
H
Br
988.
PPh3
OTs
TsCl
H
NaOH
SN2
Product
H3O+
mCPBA
Acetone
Product
H
989. (a)
O
Cl
Cl2
Et2O
hν
OH
Mg
H 3O
CN
TsCl
O
(b)
mCPBA
KCN
CH3
CH3MgBr
H 3O +
CH3
H2SO4
heat
OH
O
Na

(c) CH 3CH 2C ≡≡ CH → CH 3CH 2C ≡≡C – Na + → CH 3CH 2  C ≡≡C  CH 2CH 2OH
heat
H 2O
Na liq. NH 3
H 3O +
Product
516
Problems in Chemistry
CH3
H
H
990. HO
HOOC
COOH
CH3
O
A
B
CHO
CHO
OHC
CH3
OH
D
C
CHO
H2/Pt
∆
E
l l
OH
+
H
991.
OH
O—Ts
–H2O
–H+
OH
+
l l
O
O
H
992. (a)
(b)
993. (a)
PBr 3 (SN – 2) → C 2 H 5SNa(SN–2) :
TsCl → NaCN
H2SO4
OH
heat
Two inversion will finally lead to retention.
B2H6
OH
TsCl
OTs
H2O2/NaOH
Me3N
Br2
(b)
NaNH2
CCl4
994. (a)
(b)
Br +
ONa
ONa
+ CH3CH2Br
(1° halide)
Product
Br
C
Product
Product
C–Na+
Product
517
Solutions
O–
Br
–Br–
(c)
995. CH3OH
PBr3
mg
ether
Product
CH3MgBr
CH3CH2CHO
PCC
CH2Cl2
PCC/CH2Cl2
OH
CH3CH2CH2OH
O
OH
—MgBr
Mg Et2O
OH
Br
PBr3
O
PBr3
996.
H3O+
Mg/Et2O
OH
Product
PCC/CH2Cl2
MgBr
OH
997. (a) HO
NH2
(b) Br
(c) CH3O
OH
(f)
Cl
Br
OH
(e)
(d)
S
OH
(g) D
(h)
OH
—CH2CH2OH
HO
518
Problems in Chemistry
O
998. (a)
O
Br
OCH3
(b)
(c)
Br
OH
N3
Ph
O
OEt
(d)
D
(e)
D
(f)
O
HO
OH
OH
999. In acid catalyzed hydrolysis, protonation activate the reaction and stability of carbocation
intermediate decides the reactivity:
H
O
HO
O
–H+
+ H+
OH2
3°-carbocation
fast
OH2
l l
+
H
OH
O+
O
+ H+
Slow
OH2
+
l l
OH2
1°-carbocation
In alkaline medium, the first step is nucleophilic attack of hydroxide ion,which preferably occur at
less hindered carbon:
O
O
less hindrance is offered
from adjacent carbon.
more hindrance is
offered from adjacent carbon
O–
O
–
CH2
1000. HO
N
+
+
l l
N
l l
N
HO
O
O
–
HO–
HO
N –N
2
O
O
H 2O
OH
519
Solutions
OH
OH
O
l l
O
–H2O*
*
OH2
H+
OH
–H+
O
O
+
OH
HO
H
+
O
OH
O
H+
Ph
Ph
Ph
Ph
O
*
OH
1001.
+
+
O+—H –H
H2O
*
l l
OH
*
O*
Ph
Ph
Ph
1002. (a)
OTs
A
CN
O
O
O
H
H
A
D
O
O
O
H
O
C
B
CHO
O
(b)
CN
H
D
C
B
1003.(a) mCPBA → H + /MeOH,
(b) mCPBA → NaN 3 /H 2O,
(c) (CH 3 ) 3 SBr.
–H2O
1004.
H+
HO
l
l
O
O
H2N—NH2
l l
+
H—NH
H—N +
O
NH2
NH2
O
–H+
H 2O
O
H—N
N
N
O–
l
l
NH2
O
520
Problems in Chemistry
NaCN
TsCl
1005. (a)
CH3Li
H 3O +
Ph3P
PPh3
(CH3)2C
O
CH2
Product
OH
H
LiAlH4
(b)
PBr3
Product
BuLi
COOH
O
OH
Cl
CN
NaOH
H 2O
(c)
C—OH
H 3O +
l
l
2I
H+/heat
CN
C—OH
O
I
OH
+
1006.
+ CH3COOH
O—H
N
N—H
N
–
OAc
Product
OH
H 3O +
+ N2
–
OAc
N—H
H
O
1007.
+
H+
OH
O
+
H
H
+
–H+
OH
Product
521
Solutions
OH
CH3
1008. (a) H
OH
OH
and its enantiomer. (b)
H
H
Ph
OH (c) CH3
CH3
CH3
H
Ph
OCOCH3
D
Br
OCH3
(d)
(e)
OH
CH3
1009. (a)
Mg
Et2O
PBr3
—OH
Oxirane
H 3O +
CH2CH2OH
O
Cr2O7
(b)
OH
O
PBr3
OH
OMe
H+
Br
(1)
OMe
O
2
MeO–
MeOH
O
Br
(2)
O
(1)
Product
C2H5ONa
MeOH
OH
HO
C2H5OH
O
2–
H+
C2H5ONa
SOCl2
H 3O +
heat
COOMe
COOMe
O
O
O
HO–/heat
aldol
O
1010. Br
O
CHO Br
MgBr
A
B
O
O
C
O
OH
O
O
O
H
H
D
E
F
522
Problems in Chemistry
H
OH
Ph
(b)
1011. (a)
H
(Racemic)
OH
Ph
CH3
O
OH
(c)
(d)
H
CH3
CH3
(Phenyl attack from back side, i.e.,
from H-side to steric reason).
O–
CH3
OCH3
–
O
O
1012. (a)
OCH3
H 2O
KH
Product
–
OCH3
OCH3
+
OH
(b)
CH3
CH2
H+
–H2O
H3CO
+
H 2O
+
OCH3
O
+
CH3
l l
+
H 2O
CH3OH2 +
O
ALDEHYDES AND KETONES
NMgBr




H 3O +
NaCN
 → C 2 H 5 —C —C 2 H 5 →
1013.
3-pentanone.

Mg
C 2 H 5 Br → C 2 H 5 MgBr 
ether

aq
C 2 H 5 Br → C 2 H 5CN
1014.(a) Br —
( CH 2 —
)5 Br
aq
Mg
NaCN
(1.0 equivalent)
Ether
)5 CN →
→ Br —( CH 2 —
NMgBr
O
H3O+
523
Solutions
or Br —
( CH 2 —
)5 Br
aq
Sn
HCl
dilute
→ →
NaCN
(Excess)
NaOH
(Aldol)
CHO CHO
H2CrO4
Heat
CHO
OH
CH3
(b)
Br
O
CH2Br
NBS
CCl4
aq
C
NaCN
Br
Br
N
Mg
Ether
H 3O +
O
C2 H 5
1015.
C==C
H
C2 H 5
and
CH 3
C2 H 5
C==C
H
C2 H 5
A
OH

CH 3CH 2CH 2OH CH 3CH 2 CCH 3
D

H
B
C
F
O

1016.(a) Cl 3CCHO (b) CH 3CCl 2 —C—CH 3
G
(c) C 6 H 5CHO
NO2
O
—C—
CH 3CH 2CHO
O

CH 3CH 2 CCH 3
OH
OH


CH 3CH 2 —C—COOH CH 3CH 2 —C—COOH


H
CH 3
E
(d) O2N—
CH 3
—NO2
(e)
O
—C—
NO2
524
Problems in Chemistry
O
H
H
1017.
O
CHO
B
A
OH
OH
1018. (a)
CH2OH
D
F
COOH
E
C
O
O
A
O
B
C
OH
CHO
(b)
CHO
OH
A
B
C
O
O
(c)
O
A
B
—COO– + O2N—
1019. (a) CH3CH—COO– (b) H2N—
OH
(c)
CHO
COO–
CHO
CH2OH
B
A
Ph
1020. (a)
CHO
(b)
A
B
OH
—CH2OH
525
Solutions
O
OH
OH
O
O
O
1021. (a)
(b)
CHO
O
1022. (a)
O
OH
N
O
C
H+
NH
Beckmann’s
rearrangement
(Oxime)
A
O
(Caprolactum)
B
(b)
Ph
H3C
C==N
Ph
C==N
H3C
OH
C
O
OH
CH3—C—NHPh
CH3COO –
E
F
D
O
PhCOO –
Ph—C—NHCH3
G
H
H
I
H
C==N
N
CH3NH2
N
O
C==N
OH
(c)
H—C—NH—
N
HCOO–
M
L
K
J
O
C—NH2
N
N
COO–
+ NH3
P
N
O
O
1023. Ph—C—CH
A
Ph
CH3
C2 H 5
C2H5—CH
CH3
O
CH 3


Ph—C —NH—CH—C 2 H 5
D
C==N
OH
Ph
C2H5—CH
CH3
B
C==N
OH
C
C2 H 5
CH 3 O



C 2 H 5 —CH—C —NHPh CH 3 —CH— NH 2
E
F
526
Problems in Chemistry
O
O
HOOC
Cl
1024.
OH–
H
–
OH–
COOH
–
not
COO–
1025. (a)
OH
TsCl
NaCN
CN
SOCl2
Mg
MgCl
Ether
H3O+
O
Mg
(b) C2H5Br
C2H5MgBr
H3O+
aq KOH
C2H5OH
PCC
CH3CHO
CH
O
C—CHO
CH2CH—CHO
CH3
C
O
O
Br
CH3—C
CH
CH– +
EtO—C
O
O
CH3
B
A
1027.
OH
CHO
1026.
H2CrO4
C—CH3
–
C
EtO–
COOEt
COOEt
Br
Br
C—CH3
Br
Product
527
Solutions
O
Cl
O
1028.
A
D
C
B
OH
O
O



1029.(a) Ph—CH== CH—C—CH 3 + CH 3 —C—CH 2 —C—CH 3

CH 3
O
O


(b) C 6 H 5CH== CH—C—C 6 H 5 + C 6 H 5 —C—CH== C—C 6 H 5

CH 3
O
(c) C 6 H 5CH== CH—NO 2
(d)
OH
O
H
(e)
(f)
O
O
O
O
1030. (a)
O
CH3
C6H5CHO
OH–
O
(b)
CH
Aldol
O
OH–
C6H5CHO
Aldol
C6H5CH
CHC6H5
E
528
Problems in Chemistry
NO 2

C6H5CHO
→
C 6 H 5CH==C—CH 3
OH−
(c) CH 3CH 2 NO 2 → CH 3CH—NO 2
H 2O
NH 2
NH 2


BH3
LiAlH4
CH3COOH
→
C 6 H 5 —CH==C—CH 3 →
→
C 6 H 5CH 2 CH—CH 3
O
O
C
OH–
(d)
CH3
Aldol
NaBH4
Product
O
O
O
(e)
CHC6H5
C6H5CHO
OH–
Aldol
CHC6H5
N2H4
OH–/Heat
CH 3
CH 3


1031.CH 3 —C== CH—CH —CH 3
A
CH 3
CH 3


CH 3 —CH —COOH CH 3 —C—COOH
B

Br
CH 3

CH 3 —C—COOH

OH
C
D
CH 3
CH 3
CH 3



1032.CH 3 —CH—C ≡≡CH CH 3 —CH—C ≡≡C—CH 2CH 2CH 3 CH 3 —CH— C—CH 2CH 2CH 2CH 3
A
B

O
C
CH 3

CH 3 —CH—COOH CH 3CH 2CH 2COOH
D
E
Cl
CHO
1033.
A
C
Cl
B
D
529
Solutions
CH 3OH
 
1034.CH 3 —C— CH —CH 2CH 3

CH 3
CH 3

CH 3 —C— C—CH 2CH 3
 
CH 3O
B
A
C6H5
1035.C 6 H 5CHO
H
A
OH
C==N
C6H5
H
B
C 6 H 5COOH
F
C==N
OH
O
O


C 6 H 5 —C—NH 2 H—C—NHC 6 H 5
D
E
C
C 6 H 5 NH 2
G
O
O
O
O
CHO
COOH
NNHPh
1036.
O
A
C
B
OH O
COOH
CHO
O
F
D=E
CHO
MgBr
G
H
J
CH 3 OH CH 3
H
CH 3





1037.CH 3CH 2 —C—CH 2OH CH 3CH 2 —CH—CHO CH 3CH 2 CH— CH —C—CH 2CH 3
B


CHO
CH 3
C
A
CH 3
CH 3


CH 3CH 2 —C== CH—C—CH 2CH 3

CHO
CHO

CH 3 —C—CH 2CH 3

CHO
CH 3CH 2 CCH 3

O
D
E
F
1038.(a) O 3 / Zn-H 2O followed by Aldol condensation.
(b) O 3 / Zn-H 2O followed by Aldol condensation.
(c) X = KMnO 4 / NaOH (dilute) Y = HIO 4 , Z = dilute NaOH (Aldol)
530
Problems in Chemistry
CH3
(d) X = CH3OCH2Cl + Ph3P
BuLi
C—
: Y = CH3OCH
—OMe :
CH3
Z = H—C—CH—
—OH
O
1039. (a)
–H +
+ H+
O
O
OH
•O
•
+
H
R
O
OH
H+
(b)
OR
+
OH
OH
–H+
OH
O
••
OH
••
O
••
H
+
–H2O
+
–H+
OH2
O—H
–H+
O
O
OH
O
LiAlH4
1040. (a)
OR
H
OH
O
+
O
—CH—CH3
H2SO4
Heat
CHCH3
O3/Zn-H2O
H 3O +
CH3
CH3MgBr
Heat
OH
H 2O
HBr Peroxide
aq KCN
Br
H 3O +
COOH
O+CH3CHO
Product
531
Solutions
H O+
PhMgBr
MnO
3
2
(b) PhCHO → →
Ph —CH —Ph →
Ph—C—Ph →
Ph 3P == CH2


O
OH
Ph
CH==CH 2 .
Ph
O
(c)
CHCH2CH2MgBr
+ CH2
H 3O +
H2SO4
Heat
HBr(excess) Peroxide
Br
Br2
Na
Heat
Heat
Br
Br
OH
Alc. KOH
OH
H2CrO4
B 2H 6
H2O2/OH–
O
–
O
–
CH
(CH2)9O
O (CH2)9
–
O
C
O
CH
O
A
C
O
C
C
O
B
OH
O
O (CH2)9
O–
CH
1042. (a)
O
(CH2)9O
C—CH(CH2)5—CH3
C–MgBr
H2O
Product.
532
Problems in Chemistry
H
H
C
O
O
C
CH—
( CH2—
)5 CH3
D
OH
—CH2CHO
—CH2CH2OH
—MgBr
(b)
E
F
G
O
O
O
O
O
(c)
H
OH
OH
I
J
O
O
1043. (a)
—CH
O
CH—C—CH3
(b) Ph—CH
CH—C—CH
CH—Ph
1041.Via aldol condensation :
(c) Ph—CH 2 —CH==C— Ph

CHO
OH
O
O
—–



H
O
Mg
PCC
2
1044.(a) PhBr → PhMgBr → →
Ph—CH 2 —CH —CH 3 →
PhCH 2 —C—CH 3
ether
O
H 2O
Pd-CaCO3
(b) CH 3CH 2C≡≡ CNa + ——— → CH 3CH 2C ≡≡CCH 2CH 2OH →
Product.
H2
Mg
O
——
H O
PBr
2
3
(c) PhBr  –→ PhMgBr → →
PhCH 2CH 2OH →
PhCH 2CH 2 Br
ether
O


Mg
H3
CH3CH2CN
(d) PhBr → PhMgBr →
→
Ph—C —CH 2CH 3
O+
ether
N H /OH–
2 4
(e) Product from (d) →
PhCH 2CH 2CH 3
Heat
533
Solutions
Na
liq. NH3
Na
PhBr
(f) H—C ≡≡ C—H →
Ph—C ≡≡ C—Ph →
→
Heat
H
Ph
C
Ph
C
O
RCO3H
H
H
Ph
Ph H
CH 3
O


1045. CH 3CH 2CH 2 C—O—CH—CH 2CH 3 CH 3CH 2 CHCH 3 CH 3CH 2CH 2CH 2OH
A
C

OH
B
O

CH 3 —C—CH 2CH 3
D
O

CH 3CH 2 —C== CH—C —CH 2CH 3

CH 3
CH 3CH 2 C== CH—CH 2CH 2CH 3

CH 3
E
F
O
OH


Mcpba
CH 3CH 2 CH— CH —CH 2CH 2CH 3 CH 3CH 2 CH— C—CH 2CH 2CH 3 → A
Bayer villiger oxidation


CH 3
CH 3
G
H
O
OH
O
1046.
O
OCH3
A
OCH3
B
C
O
COOH
O
E
O
E
dil. NaOH
Heat
reflux
O
O
I
D
534
Problems in Chemistry
1047. (a) Ph—C —CH2CH2CH2OTs
O
CH2Ph
Ph—C—CH2CH2CH2—N
O
O
O
CH2Ph
B
A
CH 2 Ph
Ph—C—CH 2CH 2CH 2 —N

O
CH 2 Ph
C
OTs
OH
O
(b)
A
C
B
D
(c)
OH
A
O
B
Ph
—CH—Ph
(d)
—C—Ph
OH
A
B
O
—C
NNHPh
C
C H MgBr
2 5
(e) H—C ≡≡CNa C 6 H 5CH 2C ≡≡CH →
C 2 H 6 + C 6 H 5CH 2C ≡≡CMgBr
E
D
NHNH2
(f)
KOH
EG
OH
O


H 3O +
1048. (a) Ph—C—OMe + MeMgBr(excess) → Ph— C—CH 3

CH 3
O
O


LiAlH4
H 3O +
PCC
(b) Ph—C—OMe  –→ Ph—C—OH →
PhCH 2OH →
PhCHO
535
Solutions
OH
CH 3
+


–H2O
H 3O
(c) Ph—C —OMe →
Ph—C—CH 3 →
Ph—C—CH 3



O
OH
Br
Unstable
O
O
CHO
CH
HOCH2CH2OH
1049. (a)
CH
KCN
O
O
H+
CH2Br
CH2Br
CH2CN
OH
O
CHO
O
O
CH2CH2OH
(Hemiacetal)
CH2CHO
H
Et
O
(b)
CH
LiAlH4
H+
O
H+
O
+OH
O
Et
+
Sn/HCl
2
••
••
H2O
O
OH
OH
O
–H+
OH
Me-Shift
H+
1050. (a)
–H+
–H2O
+
OH
OHOH
••
H—O
••
+
O
H+
(b)
+
O
••
H—O
••
O
••
CH
+
HO
+
CH3
(Enol)
CH2
O
O
+
CH
+
CH3
(Keto)
536
Problems in Chemistry
–H+
CH3
H+
+
O
O
CH3
H–-Shift
+
O
–H+
O
O
CH2OH
(c)
O
CH2+
O
H+
+
O
+
–H2O
–H+
O
COOMe
COOMe
O
CH
–
1051. (a) O
–
CH
H2O
+
[O]
O
O
O
O
COOMe
COOMe
COOH COOH
OH
O
H2O
O
COOMe
O
OH
COOH COOH
O
OH
COOMe COOMe
OH
O
∆
COOMe
OH
OH
COOMe COOMe
O
O
–
CH—COOMe
O
CH—COOMe
–
537
Solutions
OH
H+
O
(c)
+ HN—Me
••
–H2O
H
–H2O
CH
••
CH—N—OH
+
N—Me
HO
Me
OH
Product
–H+
+
O—H
N
Me
O
1052. (a)
H
CH—CH2CH2—C—CH3
H+
O
O
O
HOCH2CH2OH
O
O
CH—CH2CH2—C
CH3CH2CH2CH
CH—(CH2)2—CH3
PPh3
O
CH3
H3O+
H
N2H4
H2/Ni
OH–/Heat
Product
O
H
C
C
Br
H
CH—CH—CH3
Br2
(b)
Br
1 equivalent
H
(c) PhCH2OH
C
C
SOCl2
CH3
KOH
C
Na/liq NH3
Heat
C—CH3
H
Mg
Ether
O
PhCH2MgCl
PhCH2CH2CH2OMgCl
H 2O
Product
538
Problems in Chemistry
Cl
H3O+
(d)
+
HBr
–H+
Heat
Peroxide
Br
(CH3)2CuLi
Br
Br2
(e)
C2H5ONa
Heat
HBr
X
O
Peroxide
Br
OH
H2CrO4
aq KOH
X
Mg Ether
H+
Product
OH
Br
OH
1053.
CH—CH
CH2
C—CH2CH3
CH3
CH3
A
B
HO
OH CH2
O
CH2
CH
CH3
D
MgBr
X
E
OH
C
H3C
C
CHCH3
539
Solutions
O
O
C
CH
C
CH
CH3
1054.
CH3
O
B
A
D
C
(b) RLi → CH 3I → Product.
1055. (a) N 2 H 4 / NaOH/ Heat
(c)
OH
O
H HCN
(d)
H 3O +
OH
COOH
COOCH3
CH3OH
H+/Heat
NH2
Product
O
COOCH3
Hydrolyse
O
O
O
NH3
O–
O
(e)
H
–
+ CH3–
H+ –H2O
O
(b)
1056. (a)
O
O
Ph
SOCl2
540
Problems in Chemistry
CH 3
O— BF3


Ph—CH—CH 2 —C—CH 3
+−
1057. Ph—CH== O + BF3 → Ph—CH== OBF3 →
===
+
CH 3
O


→ Ph—C —CH 2 —CH—CH 3
OH

H2CrO4
PhC ≡≡ CNa H3O+
1058. CH 3CHO → → Ph—CH —CH 3 → → →
H 3O +
PhMgBr
OH
OH
H



Na
Ph—C—– C == C —Ph
Ph—C—C ≡≡ C—Ph →
liq. NH3



CH 3
CH 3 H
1059. (a)
+ CH2
+
O
O
O–
CH2
+
H
OCH3
(b)
O
O—CH2
–
H
HO
HO
O
X
1060. (a)
O + CH2—COOEt
Br
(b)
O + PhCH
PPh3
O
(c)
Zn
Reformatsky
reaction
H2O
CH3
Product.
O
O
+ Cl2
∆
CH—COOEt
CH3—C
Cl
aq KCN
CN
RCO3H
Product.
541
Solutions
(d)
CH2—OH
O+
H+
Product.
CH2—OH
O
OH
(e)
H3
+ PhMgBr
O+
H+
Ph—C—CH3
CH2
Ph—C
∆
CH3
CH3
O
(f)
Aldol
+ OH– (dilute)
H+
∆
Product.
O
O–MgCl
1061.
OH
H 3O +
MgCl
Reverse
O
MgCl +
+
N
N
OH
Ag2O
1062. (a)
A
+ N2
O
O
O
CH3OH
O
O
(b) H2C
C
(Ketene)
B
O
O
C
542
Problems in Chemistry
O
(c)
O–
+
N—N
–
H 2O
O
N–
N
LiAlH4
NH
NH
D
+
H+
(d)
+
O
HO
HO
Me-Shift
+
–H+
HO
HO
OH
H+
(e)
Enolization
(3 molecule)
OH
HO
OH
+
HO
HO
–3H2O
(f)
+
O
CH2
CH2
+
+
HO
HO
O
H 2O
–H+
••
HO—
O
O–
H
CH2
543
Solutions
O
O
CH3
CH3 CH2N2
1063.
CH3CH2ONa
H 3O +
∆
OH
alternatively
CH3OH/HCl
OCH3
CH3CH2Br
CH3I
NaH
OCH2CH3
NaH
OMe
OH
OMe
OEt
OEt
O
O
O
O
O
COOH
O
1064.
A
Alternative
OH
B
O
O
O
CH2CH2OH
D
C
D
—CH—CH2OH
COOH
A
B
O
COOH
O
1065.
H
COOH
HOOC
A
O
B
C
H
O
OH
O
OH
D
O
E
O
544
Problems in Chemistry
OH
l
1066.
C
N
l
C
N
OH
O
l
l
A
C
B
O
C
NHCH3
N
NH2
H
O
D
E
F
COOH
G
H
1067. (a)
CH3
Ph
H
C
N
l
Ph
l
C
N
l
(b)
H
l
OCH3
OCH3
CH3
O
N
(c)
H
(e)
(d)
O
O
O
O
(f)
NNH2
X
OH
+
O
O
Y
1068. (a) N 2 H 4 /NaOH/heat, H 3O + , Ph 3 P==CH 2 ,
(b) N 2 H 4 /NaOH/heat, H 3O + , HNCH 3 /H 2 –Pd/C
(c) Ph 3 P==CH 2 , H 2 N  OCH 3 ,
(d) CH 3 Li/HCl/H 2O, Ph 3 P==CH 2 .
1069. III < II < I.
Z
OH
545
Solutions
1070.The nucleophilic addition of cyanide ion will occur at carbonyl carbon. Therefore, the stable
conformation at α-carbon will be:
O
OH
H O
H
OH
OH
H
2
–
CN
–
H
NC
H
==R
1
R
As shown above, attack from side 1 will produce B and attack from side 2 will produce C. Since,
attack of CN – from side 2 will suffer less steric hindrance, C will be formed in greater amount than
B.
H
H
1071. (a)
CH3CH2MgBr
PCC
OH
O
CH3
CH3
Planar arrangement
H
H
+
CH2CH3
CH3
OH
OH
CH3
Major product
(nucleophilic addition of
CH3CH2– occur from side opposite
to —CH3 group)
–
H
H3O
H
Ph
(attack occur from
less-hindered side)
CH2I
1072.
Ph3P
BuLi
Ph H
+
Ph
C6H5CH2CHO
Products
CH2CH3
(Minor product)
(b) In the most stable conform shown below:
O
H3C OH
CH3 H
H3C
H
CH3
H3O+
H
OH +
CH3
H
CH3
(Major)
Ph H
CH3
OH
CH3
(Minor)
546
Problems in Chemistry
OH
OH
O
HCN
(b)
CN
H
H3O+
COOH
H2O
H2SO4
O
O
(c)
LiAlH4
H+
H
COOH
Heat
O
TsCl
NaCN
H
OH
OH
O
O
(1) CH3Li
CN
O
O
CH3Li
excess
K2Cr2O7
O
H
Product
O
O
(d)
(2) HCl/H2O
H2SO4
O
N2H4
OH–/heat
O
1073.A will proceed more efficiently. The electronegative chlorine, via inductive effect, increases the
electrophilicity of the carbonyl carbon, making it more susceptible to attack by cyanide ion.
O
O
EG
H+
1074. (a)
(CH3)2S
HCl/H2O
EG
H+
O
Product
O
O
O
(b)
(CH3)2NH
O3
O
O3
DMS
N 2H 4
O
NaOH/∆
HCl
H 2O
N 2H 4
Product
547
Solutions
O
(c)
H
O
O
CH3Li
H3
H3 C
O+
O
H+
H 2O
O
CH3
OH CHO
O
O
NaBH4
O
O
H+
CH3
CH3I
NaH
Product
l l
O
O–
OH
O
OH
+
O
OH2
H+
1075.
l l
HO—N—H
HO—N +
H
H
HO
O
N
–H2O
O
N
O
O
N
O
–
N
+
HO
O
Cl–
●●
O
N
Cl–
H
OCH3
OCH3
1076. (a)
CH3
CH3 +
CH3
CH3
(Major)
H OH
H OH
(c)
+
(Major)
O
(d)
N
H
HO
H
H
+
H+
O
HO
H
H
l
l
N
+
N
H
548
Problems in Chemistry
O
O
HOCH2CH2OH
1077.
O
O
O3
O
H
(CH3)2S
H+
O
H
C2H5NH2
O
O
H2/Pd
O
1078.
O
OPh
CH3O
CH3COCl
N
O
OPh
l l
CH3O
CH3O
O+
O
l l
+
O–Ph
HO
H
H+
H
O
–H+
Product
O
F
HCN
CN
CH3O
OH
l l
O+
H
O
CH3
HO
1079. (a) F
HO
H2/Pd
F
NH2
Pd/H2
O
HCN
CHO
O
EG
H+
B2H6
O
O
H3
O+
H 2O
Product
Pd
H2
O
NH3
∆
PCC
NaOH/H2O2
O
H
H2N
Product
OH
SnCl2
HCl
O
(c)
CH3
OPh
O
(b)
Product
H
O
O
549
Solutions
1080.The bond angle required at carbonyl carbon is 120°. In the case of cyclohexanone, the six
membered ring can accommodate this bond angle easily and thus, the driving force towards
hemiacetal formation, where the bond angle is 109° is small. On the other hand, in
cyclopropanone, the system is highly strained, and hemiacetal formation enables some of the
strain to be alleviated by allowing the carbonyl carbon to be sp 3 hybridized.
H
H+
O + CH3OH
OCH3
ring is less strained
than cyclopropanone
1081. (a) CH 3COCH 2 Br, (b) CH 3CHO, (c) CH 3COCF3 , (d) CH 3CHO.
1082. A : CH 3ONa/CH 3OH,
B : (CH 3 ) 2 NH,
C : CH 3CH==CH 2 MgBr; H 3O + ,
D : CH 3OH/H 2SO 4 .
CH3
O
O
PPh3
CH3CH
(CH3)2CuLi
1083. (a)
H2O
(Conjugate addition)
CH3
CH3
CH3
O
mCPBA
CH3
O
O
O
O
PBr3
EG
H+
(b)
Mg
ether
O
O
H 2O
OH
OH
OH
PCC
H3
O+
Product
H 2O
Cl
Cl
H
1084.
H
OH
trans
OH–
H
H
SN2
H
O
H
–
[—O– and Cl are anti, substitution is preferred]
O
550
Problems in Chemistry
Cl
H
OH
Tautomerism
–OH
O
OH
[H and Cl are anti, elimination is preferred]
H
cis
+
H
O
O


1085. HO  CH 2CH 2CH 2  C  H + H + → HOCH 2CH 2CH 2  C  H →
H
O+
OH
O
–H+
+
OH2
O
OH
H+
CH3OH
Product
–H+
CN
1086. (a)
(b)
NH2
O
Ph
O
CHCH2
(c)
N
+ CH3CH2—CH
cis-trans
–
CH3O H
O
C—CHO
(e)
(d)
O
CH3
O
O–
O
–
1087.
H
A
H
H
H
O
O
CH3OH
+
H
more stable
H
H
less stable
551
Solutions
O
O–
–
+ C
1088. (a)
H 2O
CCH3
H2/Pd
CCH3
C
Product
BaSO4
OH
H2SO4
heat
(b)
mCPBA
CH2N2
(c)
O
NaCN
H 2O
O
O
O
CH3MgBr
H 3O +
Product
Product
OH
(d)
CH3MgBr
H 2O
NaH
CH3CH2I
ether
1089. (a)
Product
OCH3
CH3ONa
mCPBA
O
CH3OH
OCH3
PCC
OH
Ph3P
C
CH3
CH3
Product
O
OH
O
(b)
H
CH3CH2MgBr
PCC
H3O+
H2O
H
N
N
552
Problems in Chemistry
OH

1090. CH 3  CH 2  CH  CHO
A
CH 3  CH==CH  CHO
B
CH 3  CH  CH 2CHO

Cl
C
CH 3  CH  CH  CHO


Cl
Cl
D
CH 3CH  CH 2  CH 2OH

Cl
CH 3 CH  CH 2CH 2OH

OH
E
F
O

CH 3  C  CH 3
NaNH 2
H 3O +
1091. (a) CH 3CH 2  C== CH 2 → CH 3CH 2C ≡≡CNa → →
H 2O
NH 3

Br
KMnO4
CH 3  CH  CH 3 

OH
OH

PBr3
KOH
CH 3CH 2C ≡≡C  C  CH 3 → → Product
EtOH

CH 3
CH 3
O

PBr3
KOH
mCPBA
(b) CH 3  C  OH → → → 
EtOH

CH 3
HBr
HNO2
→ → Product
CH 3
CH 3


Hg(OAc)2
NaBH4 H2CrO4
(c) CH 3CH 2CH  CH==CH 2 → → → CH 3  CH 2  CH  C  CH 3
H 2O

O
H 3O +
→ → Product
HC ≡≡ CMgBr
Br2
(i) NaNH2 /NH3
CH 2 ==CH 2 → CH 2  CH 2 
(ii) H3O+


(iii) Mg/Et 2O
Br
Br
CH 2CH 3
CH 2CH 3


SOCl 2
B2H6
H2CrO4
(d) CH 3CH 2  C==CH 2 → → CH 3CH 2  CH  COOH →
H2O2 /NaOH
PhOH
Cl 2
AlCl 3
AlCl 3
→ → Product
553
Solutions
O
O
O


 s
CH3O–
CH2 == CH  C  CH3
1092. C 6 H 5  CH 2  C  CH 2C 6 H 5 → C 6 H 5  CH 2  C  CH  C 6 H 5 
O
O




CH3OH
CH3 O
C 6 H 5 —CH—C—CH 2 — C 6 H 5 ←
C 6 H 5 —CH—C—CH 2 —C 6 H 5

O
O–


CH 2 —CH 2 —C—CH 3
CH 2 —CH==C—CH 3
−
O

C 6 H 5  CH  C  CH  C 6 H 5
s

CH 2  CH 2  C  CH 3

O
O
Ph
Ph
Ph
O
–
Ph
O
CH3OH
CH3
CH3
Ph
CH3OH
H
CH3O–
O
Ph
OH
CH3
O
CH3
1093. (a) H3C—
O
(b)
—C—SCH2C6H5
—CH
O
CH3
CH—C—CH3
O
CHC6H5
(c)
O
O
Br
1094. (a)
(CH3)3CO–
O
–
Br
554
Problems in Chemistry
O
O
O O
Ph
Ph—C—C—Ph
Ph
O
(b) Ph
Ph
KOH
Ph
Ph
s
O
O
Ph
OH–
s
EtOH
Ph
Ph
H
Ph
Ph
O–
O
Ph
s
Ph
EtOH
–H2O
Ph
Ph
O
Ph
O
O
O
Ph
Ph
Ph
Ph
–H2O
H
HO
Ph
Ph
Ph
Ph
1095.Difference in reactivity towards hydrolysis reaction lies in the stability of carbocation intermediate
as:
H
H
O
O
H+
H
O+
O
+
H
O
+
H
H
H2C
+
O
O
OH
OH
Allylic group provide extrastability due to resonance
H
H
O
O
H+
O
+
O
H
H
O
+
not stabilized by
allylic resonance
OH
OH
555
Solutions
O
H
1096. H3C
H+
H
CH3—C
–H+
+
OH
CH2
CH—OH
+
–
CH2—CH—O—H
Product
H—O—–CH—CH2–
+
1097. II < III < I.
OH
H+
1098.
O
O
O
O+
O
l l
–H+
l l
OH2
+
H
OH
O
HO
H2O
OH
OH
+
OH
l l
–H+
Product
O
1099.
H
Br
EG
H+
Mg
O
H2O
CH3CHO
H
CH3
H
Ether
O
O
OH
l l
OH
H3O+
Product
CH3
O
1100. I
I
–
H + OCH3
O–
–I–
O
OCH3
A
OCH3
H3O+
HO
H
O
556
Problems in Chemistry
H2O
1101. Compound + EtO–
–
O
O–
CH3
O
O
CH3
H
CH3
H
CH3
50%
Planar (achiral)
H 3O +
MeMgBr
+
50%
PCC
CH2Cl 2
1102. (a) PhCHO → → → Product
(b)
CH3
OH
mCPBA
OH
O
(c)
O
Mg
H 2O
O
OH
O
O
1103.
H
OH
C
B
E
D
O
F
O
O
OH
Product
Br
O
A
H 3O +
CH3CHO
Et2O
H+
Br
Product
OH
O
OH
CH3I
NaH
O
H+
O
O
OH
1104.
B
A
CH 3
CH 3


1105. CH 3  C   C== C  CH 3


CH 3 CH  CH 3

CH 3
A
D
C
O

CH 3  C  CH 3
B
CH 3 O CH 3
 

CH 3  C  C  CH  CH 3

CH 3
C
557
Solutions
CH 3
CH 3


CH 3  C  CH CH  CH 3


CH 3 OH
CH 3
CH 3


CH 3  C  CH==C  CH 3

CH 3
D
E
Br

B : C 6 H 5CHCH 3 ;
1106. A : C 6 H 5CH 2CH 3 ;
OH

E : C 6 H 5 CH CH 3 ;
D : C 6 H 5CHO;
1107. ClCH 2 
C : C 6 H 5CH==CH 2 ;
H
(CH 3 ) 2 CH—
1108. (a)
Ph
O
EG
CHO
Ph3P
CH
H+
O
O

F : C 6 H 5  C  CH 3 .
CH2
H 3O +
H 2O
Product
O
O
O
O
(b)
(i) H+
H2 N
H2/Pd
H (ii) H2/Pd
HO
NH
N
H+
H
1109. (a)
OH
(b)
(Racemic)
(CH3)2HC
O
H
(c)
H
(d)
CH3
(e) HO
H
Ph
CD2OH
CH3
CH2Ph
OH
OH
(f) Ph—CH—C(CH3)2 (g)
H
OH
CH3
OH
H
O
1110. (a)
N
H
N
(b)
NH—C—CH3
558
Problems in Chemistry
(c) PhCH 2 MgBr/Et 2O, H 3O + , PCC/CH 2Cl 2
(d)
O
O
CH3
(b)
Fe/HCl
H+
H2O
H3O+
(CH3)2S
H+
O
MgBr
O3
EG
1111. (a)
Product
CH3
KI
NaNO2
NH
(e)
CrO2Cl2
Ac2O
HCl/0°C
I
NO2
CHO
CH3MgBr
H3O+
CH2Cl2
H2O
PCC
Product
I
(c)
1112.
NH2
OH
O3
(CH3)2S
TsCl
H+
BH3
OTs
Product
(CH3)3COK
(CH3)3COH
B2H6
H2SO4
NaOH/H2O2
heat
Major product
(not required)
PCC
CH2Cl2
Product
559
Solutions
OH
1113. (a)
,
OH
OH
O
OH
,
,
,
OH
OH
O
O
OH
O
OH
,
OH
OH
O
O
(b) (i)
D
D
D
D
O
Br
(ii)
O
D
(iii)
O
O
O
CHO
1114. (a)
OMe
(b)
OH
(c) Ph—CH2—CH
NHCH3
O
OH
CHO
(e)
(d)
OH
O
O
D
CHO
1115. (a) D
D
OH
(d)
O
CHO
(b)
Br
(c)
CHO
CH
OH
OH
H
C—Ph
560
Problems in Chemistry
1116.A has enolizable hydrogen bonded to chiral carbon, therefore, it acquire planarity at chiral carbon
when converted to its enol form. Due to planarity at chiral carbon in the enol form, it forms both
enantiomers in equal amount when converted back to keto form. No such enolization is possible
with B since it has no α – H at chiral carbon and hence, its optical activity is retained even in
aqueous alkaline solution.
EtOH
EtO–
A
H
+
O–
O
Planar structure
CH2—H
1117.
O
H
Racemic mixture
O
O
O
H
O
–
O
–
CH2
O
O
–
+ OH
O
H2O
∆
O
O
1118. (a)
OH
(b)
(c)
CH—COOH
O
O
O
(d)
(e)
OH
(f)
Br
H
(g) Cl
1119.
H
O
O
Cl
(h)
N
H
OH PCC
HO–
CH2
∆ (Aldol)
CH2Cl2
H
O
O
PPh3
561
Solutions
O
O
1120. (a)
(b) H
O
O
D
D
CHO
H (c)
O
(d)
OH
CN
(e)
N
(f)
(g)
N
H
+
O
OH
O ll
1121. (a) HO
H
H
+ H+
–H+
O
l l
O
H
H
+
OH2
+
O CH3
l l
H+
O
OH
C
C
H
l l
(c)
H H+
CH3
–H+
O
Product
NO2
+
O
O
—NO2
H2N—O—
H
NO2
OH
Ph—CH—N—O—
—NO2
–H+
+
NO2
–H2O
+
H
H
l l
Ph—CH—N—O—
—NO2
—CH—N—O—
H
H+
NO2
OH
NO2
—NO2
Ph—HC
N—O—
H
—NO2
A
NO2
–
PhCN + O—
—NO2
–
Ph—C
N—O—
NO2
—NO2
CH3O–
562
Problems in Chemistry
1122. High proportion of enol in the equilibrium mixture of compound II can be rationalized by two
factors:
(a) It is conjugated system.
(b) It forms a six membered cyclic ring via intermolecular H-bonding as:
O
H
O
H Intermolecular H-bonding.
O
O
–
O N
OH OH–
1123.
C
OMe
N
H
–
O
N
N–
H
N
O
H 2O
O
OMe
N
O
H
N
OMe
OH
O3
1124. (a)
(CH3)2S
NaOH
heat
OH
LiAlH4
Product (b)
O
RCO3H
O
PCC
CH2Cl2
Product
OH
O
O
(c)
OMe
N
O
–H2O
H
OH
H3O+
COOH
CH3OH
H+
Product
563
Solutions
anti to t-butyl, shift
OH
HNO2
1125. (a)
H
OH
OH
H+
+
OH
NH2
OH2
l l
Product
–H+
H–-Shift
OH
OH
OH
H+
HNO2
NH2
OH2
OH
Product
O
+
–H+
+
l l
CH—OH
O
O
–
MeO–
(b)
OH
+
H 2O
Product
O
OH
O
–
Ph
O
Ph
Ph
O
O –
MeO–
OH
Ph
O
H 2O
O
–
O
Ph
Ph
O
Ph
564
Problems in Chemistry
1126. (a)
O
H+
OH
OH
H 2O
+
OH
–H+
H+
–H2O
OH
OH
–H+
O
O
CH3
OH
(i) CH3MgBr
(b)
+
H+
–H2O
(ii) H2O
O
HO–
O
O
CH2–H
HO–
O–
–
O
O
O
H
H2O
OH
Br
Br
1127.
O
OH
O
SN2
LDA
O
+ Br–
O
O–
Br
Br
LDA
O
CH3
+ Br–
O
CH2–
O
+
565
Solutions
1128.
OH
B
A
C
O
NOH
O
E
F
D
G
CH 3 C 2 H 5


1129. CH 3  CH C==CHCH 3
A
CH3 O


CH 3  CH  C  CH 2CH 3
D
CH 3 C 2 H 5


CH 3  CH  C  C 2 H 5

OH
CH 3 C 2 H 5


CH 3  CH  C  CH  CH 3
 
H OH
B
C
CH 3

CH 3CH  CH  CH 2CH 3

OH
CH 3

CH 3  CH  CH==CH  CH 3
F
E
CH 3

CH 3  C== CH  CH 2CH 3
G
CARBOXYLIC ACID AND ITS DERIVATIVES
1130.Extraction of a solution of carboxylic acid and phenol in CCl 4 with aqueous solution of NaHCO 3
will bring carboxylic acid in aqueous phase as RCOONa leaving phenol in organic phase.
1131. A : 2-methylmalonic acid, B : Propanoic acid.
OH
O
Br2
1132. (a)
–H+
O
Br
O
• OH
•
O
+
Br
+
(b)
CH3COOH
+ H+
–H+
+
OCOCH3
566
Problems in Chemistry
OH
(c)
+
H+
–H2O
NHCOCH3
NHCOCH3
•• NHCOCH
3
–H+
+
Product
O
O
1133. (a) PhCONHCH3
O
COOH
(b)
O
(c)
CH3
O
OH
O
O
(d)
OCH3
COOH
dilute
H2SO4
HBr
NaCN
1134. (a) →
→
→
HBr
Peroxide
dilute
H2SO4
NaCN
(b) → →
→
O
Cl
SOCl2
COOH
O
(c)
I
Ag2O
Br2/Heat
1135. (a) CH 3OH + Na 2CO 3
Br
Mg
H3O+
ether
COOEt
(b)
COOH
(d) CH 3CH 2CH(COOEt) 2
I
(c) Ph—CH—COOEt

COPh
567
Solutions
O
—CH2NH2
(ii) H3C—
—COOH
(e) (i) H3C—
(iii) H3C—
—C—Ph
O
O


1136. (a) HOOC—
( CH 2 —
) 4 COOH →
C 2 H 5O—C —
( CH 2 —
)4 C— OC 2 H 5
+
C2H5OH
H
O
COOEt
EtO–
EtOH
I
O
O
I
LDA
–
COOEt Cl
P
COOEt
aqueous
NaCN
COOEt
COOEt
H O+
4
3
(b) CH 3COOH →
ICH 2COOH → → HOOC—CH 2 —COOH
I2
H 2O
C 2 H 5OH/ H + / Heat
tEOOC
COOEt
SOCl 2
NaCN
1137. (a) PhCH 2OH →
PhCH 2Cl →
PhCH 2CN
CH 3

H3
SOCl 2
CH3MgCl
(b) PhCOOH →
→
PhCOCl →
Ph—C—CH 3
Excess

OH
O+
EtONa
(c)
EtOOC
CH2
CH—CH2Br
COOEt
COOEt
COOEt
H 3O + H 2 O
C2H5OH
COOH
SOCl2
COOH
Heat
COOH
COOEt
568
Problems in Chemistry
O
1138. (a)
OH
Cold, dilute
Cl
Cl
O
Cl
OH
Cl2
aqueous
hν
O
O
NaOH
O
O
H2CrO4
CH3
(c)
O
alkaline/KMnO4
OH
(b)
O
RCO3H
CH3
CH3
CH3Cl
H2SO4
CH3Cl
AlCl3
Heat
AlCl3
CH3
SO3H
O
SO3H
COOH
P 2O 5
O
O
Heat
KMnO4/H3O+
dilute
NaOH/Heat
H2SO4
COOH
O
O
COOH
(d)
P 2O 5
O
Heat
COOH
O
OH
O
H+
COOH
O
NH2
1139. (a) 6-hepten-1-ol
PCl5
NH3
NH2
O3
H2O2
COOH
Heat
NH
O
NaBH4
NaCN
(b) PhCH 2 Br →
→ PhCH 2CH 2 NH 2
569
Solutions
H O+
H2SO4
HBr
3
NaCN
(c) PhCHO + CH 3 MgBr →
Ph CH CH 3 → PhCH==CH 2 → →
Heat
Peroxide

OH
PhCH 2CH 2CN
O
1140. (a)
P4/I2
COOH
H 3O
Na
COOH
COOH
∆
I
COOH
P 2O 5
O
Heat
O
O
(b) C2H5O—C—(CH2—
)4 COOC2H5
O
COOEt
E t ONa
H 3O +
E t OH
He a t
O
L DA
CH3
O
HCOOH
1141. (a)
Heat
NH2
H—C—NH—CH—CH2CH3
C2H5MgBr
(b) CH3COOEt
C H 3I
H 3O +
Excess
OH
NH2
(c)
CH3COOH
NHCOCH3
Heat
OMgCl
Br
(d)
Mg
Et2O
HCOCl
H3O+
Product.
CH3
570
Problems in Chemistry
O
COOH
COOH
O
1142.
OH
COOH
C
A
O
D
B
OH
O
O
1143.
O
A
B
COOH
COOH
C
O
COOH
D
E
CH3
CH3
1144. CH3—CH—CH2COOH
CH3CH—CH—COOH
A
Br
B
O
COOH
O
H
E
D
C
H
F
CH 3
CH 3
CH 3



1145. CH 3CH 2 CHC H 2COOH CH 3CH 2 CH—CH—COOH CH 3CH 2 C== CHCOOH
A
C

Br
B
CH 3

CH 3CH 2 —C== CH 2
D
O
O
1146.
NH
A
B
O
OH
NH2 HOOC
HOOC
C
D
571
Solutions
O CH 3 O

 
1147. CH 3CH 2 —C —CH— C—OC 3 H 7
O

CH 3CH 2 —C —CH—CH 3

COOH
A
CH 3CH 2CH 2OH
C
B
OH

CH 3CH 2CHO CH 3CH 2 CH —CH—CH 3
D

CHO
E
O
COOCH3
COOH
O
1148.
COOH
COOH
O
C
B
A
1149. C2H5OOC—CH—COOC2H5
HOOC
COOC2H5
A
O
COOH
B
O
O
O
COOEt
COOEt
EtO
C
OEt
D
E
CH3
COOC2H5
1150.
COOH
A
CH3
B
O
OH
1151.
A
B
C2H5OH
C
D
572
Problems in Chemistry
O
1152. (a) MeO—
—NHCOCH3
CH2COOH
(b) Ph—C—NHOH
(c)
C
O
COOH
O
C
CH3
O
OH
O
(d)
(e)
O
Y
X
Cl
OH
1153. (a)
OCH3
OCH3
O
A
CN

(b) Ph—C—CF3

OH
C
1154. TsO—CH 2C ≡≡CH
A
B
O
CN

Ph—C—CF3

OCH 3
COOH

Ph—C—CF3

OCH 3
D
E
TsO—CH 2C ≡≡CLi TsOCH 2C ≡≡ C—CH 2CH 2Cl
B
NC—CH 2C ≡≡CCH 2CH 2CN
D
C
HOOC —CH 2 —C ≡≡C —CH 2CH 2COOH
E
O–
O–
+
1155.
OCOCH3
O
O
Y
X
+
CH2
O
A
CH3COCl
CH3COCl
O
O
O
Cl
O
Cl
573
Solutions
H
1156. (a)
H
H
NaCN
CH2Cl
H
H
H 3O +
H
CH2CN
CH2COOH
LiAlH4
NaBr
Product.
H
TsCl
H
OH
CH3
H
(b)
H
H
(CH3—CH—CH2)2AlH
H
COOEt
CHO
O
(c)
COOH
H 3O +
O
LiAlH4
OH
OH
OH
O
O
(d)
H3O+
O
NH3
OH
OH
Heat
Heat
COOH
(e)
BaSO4
CONH2
CONH2
H 3O +
Pd
H2SO4
Conc. H2SO4
Et
COOH
OH
Et
Heat
O
O
CH2OCOPh
COOH
LiAlH4
PhCOCl
O
1157.
CD3
CD3
CD2OH
LiAlD4
H2O
CD2OCOPh
PhCOCl
O
CD3
O
CD3
574
Problems in Chemistry
CH 3

1158. Ph—CH —CH 2COOH Ph—CH—CH 2CH 3
A

COOH
CH 3

PhCH 2 CH—COOH
PhCH 2CH 2CH 3
C
D
B
PhCH(CH 3 ) 2
E
O
1159.
A
COOH
COOEt
COOH
COOEt
COOEt
C
B
O
O
COOEt
D
E
O
O
O


1160. (a) Ph—CH 2 —C—CH—C—OCH 3

Ph
(b)
18
1161. (a) HOOC—CH 2CH 2CH 2 —OH (b) H 2 N—CH 2COO − (c) HOOC—CH 2CH 2CH 2CONH 2
O
O
COONa
COOCH3
1162. (a)
O
CH2CH2CH3
Y
X
Z
(b) CH 3 —( CH 2 —
)10 COOH
O
1163. (a)
O
O
–
CH2
+ LDA
CH2O
H 2O
OH
PCC
O
EtONa
(b)
—Br
H 3O +
O
COOH
O
Heat
O
H
O
O
O
575
Solutions
O
O


C2H5ONa
Cl 2
(c) Ph—C—CH 2CH 3 → Ph—C—CH CH 3 →
heat
C2H5OH

Cl
O
O



H+
HBr
Ph—C —CH== CH 2 →
Ph—C—CH 2CH 2 Br →
CH2 —CH2


OH
OH
O
Ag2O
Product
O
Mg
Ph—C—CH2CH2CH2CH2OH
H2O
O
LDA
LDA
OMe
CH3I
O
O
O
O
OC2H5
(e) C H O
2 5
O
Ph—C—CH2CH2Br
ether
H3O+
O
O
(d)
O
O
COOEt
EtONa
LDA
O
O
O
Heat
COOEt
H 3O +
CH3CH2CH2Br
O
O



N 2H 4
PhC ≡≡ CNa Ph —C —C ≡≡C— Ph 
1164. (a) PhCOOH → Ph— C —Cl →
OH– /Heat
SOCl 2
Pd
BaSO4 /H2
Product ← Ph—CH 2 —C ≡≡C—Ph
O
(b)
Ph—C—OMe +
OMgBr
MgBr
Ph—C—OMe
O
H 3O +
Heat
Ph—C—
LiAlH4
Ph—CH
H2SO4
Heat
576
Problems in Chemistry
(c)
PhCH2CH2CH2Br
C2H5ONa
COOCH3
Ph
COOCH3
+
O
OH
OH
H
1165. (a)
H
••
OH
••
H+
OH
O
+
O
COOEt
Product
O
HO
(b)
–H+
O—H
EtO
O
O
O
HCl
O
O
HO
O
+
O
1166.
O
O
EtO–
EtO
OEt
O–
EtOOC
(EtOOC)2CH–
COOEt
O
H 3O +
HOOC
Heat
O
O
EtO–
1167. (a)
OC2H5
OEt
CH3CH2COOC2H5
s
CH3—CH—C—OEt
O
s
H 3O +
COOEt
Heat
O
O
O
O
O
CH3CHCOOEt
577
Solutions
OEt
(b)
CH3MgBr
O
O
OEt
(c)
Ph
PhMgBr
O
O
O
O
[from (a)]
O
O
O
CH3MgBr
(d)
HCN
CN
OC2H5
H3O+
OH
B2H6
COOH
OEt
(e)
COOH
OH
PhNHNH2
CH3COOH
H2SO4
Heat
Product.
O
O
[from (a)]
O
O
EtO–
1168. (a)
OEt
OEt
O
(b)
O
OH
OH
O
O
CH2—CH2
O
OEt CH2
H+
OEt
O
O
CHMgBr
O
O
O
H3
O+
O
O
N 2H 4
OH–/Heat
578
Problems in Chemistry
H
O
O
1170. K2Cr2O7 + H2SO4
H 2O
l l
OH2
COOH
HO—Cr—OH
O—Cr—OH
heat
O
O
H
H
OH
OH
O
H2CrO4
OH
O
H2O
l l
C
CH
OCr—OH
+
O
O—H
C
C
OH
OH
+
C—O—CH3
l l
HOCH3
H+
OH
OH H
–H+
+
OH
OH2
C
+
O
l l
H
OCH3
C—O—H
–H2O
OCH3
C—OH
H+
OCH3
H 2O
O
OCH3
O
1171. (a)
O
Br
NaN3
H2/Pd/C
HCl
Product
O
H 2O
NH2
CH3COCl
N
HCl
N—H
H2
Ni
579
Solutions
CHO
(b)
CHO
CHO
NaNO2
CuCN
HCl/0°C
HCN
N2+ Cl–
NH2
CN
CN
Product
OH
H2/Pd
CN
PBr3
Mg
1172. CH 3OH → CH 3 Br → CH 3 MgBr
ether
OH
O
O
CH3MgBr
H3O+
HO
OH
excess
O
K2Cr2O7
H2SO4
O
C
1173. (a)
–
OLi+
O
OH
C
+ CH3Li
Product
C
O–Li+
–
CH3Li+
CH3
OH
Product
–H2O
C
CH3
OH
H3O+
O–Li+
580
Problems in Chemistry
H
H

O+
(b) R —C ≡≡ N •• +
H
O

H 2O
OH2
R —C ≡≡ N—H →
R —C== N—H 
••
+
H
H
+
OH
O


H 2O
R —C— NH 2
R —C — NH 2
− H+

OH
H
H
+
H+
O

Tautomerism
R —C—NH 2 ←→ R —C== NH

OH
+
OH
 +
R —C— NH 3

• OH
•
O

S
O

R —C—OH
+
••
Cl
OH
 +
R —C== O—H + NH 3
O

R —C—OH + NH +4
O H Cl
O H Cl
   −
  
→ R —C—O —S—O → R —C —O—S== O 
+
+

Cl
Cl –
Cl
O


Cl – + SO 2 + R —C—Cl ←→
Cl
R
C
–
O
O

O—S —Cl
Cl
O


Cl −
← R —C—O—S== O
O−
O

+
 −
+
H 2O
(c) R —C—N—N ≡≡ N ←→ R —C== N—N ≡≡ N → R —N ==C==O + N 2 
H 2O
H 2O + CO 2 + R —NH 2 ←
O

C
R
N

H
OH ←
581
Solutions
Cl–
H
H
CH3
N
l l
CH3
N
+
Cl
1174.
O
CH3O
Cl
P
Cl
O
P
CH3O
Cl
O
Cl
O
+
CH3
N
+
N
H3CO
CH3O
H
Cl–
Product
O
LiAlH4
CH3O
1175. (a)
OH
CH3O
NaH
O
OH
CH3CH2I
COOH
(b)
1177. (a)
COOH
OH
LiAlH4
H3O+
Product
1176. (a) II > III > IV > I,
Product
O
NaCN
(b) I > II > III > IV,
SOCl2
Mg
ether
H3O+
OH
TsCl
(c) III > II > I.
PBr3
LiAlH4
OH
H 2O
OCH3
H2SO4
heat, CH3OH
O
Br
H 3O +
NaCN
582
Problems in Chemistry
(b)
H3O+
O
COOH
OH
O
H2SO4
CH3
∆
HBr
ROOR
CH2
COOH
O
O
aq. NaOH
H2SO4
COONa
∆
1178. (a) A will hydrolyse first,
(d) Rate will decrease.
(b) Rate will increase,
OH
(c) Rate will decrease,
O
O–
O
Ph
Ph
OCH3
1179.
Ph
OCH3
–
H
–
OCH3
H3O+
Ph
Ph
CH3O– + Ph
O
O–
O
O
OCH3
CH3
Ph
OCH3
OCH3
O
Ph
OCH3
CH3OH
∆H = ( BE ) Reactants − ( BE ) Products = 91 + 99 − (85 + 102) = 3 kcal
O–
O
O–
O
O
O
+
*
O
1180. H3C
CF3
CH3
*
O
l l
H
O
CH3
H
CF3
+
OCH3
O
CH3OH
CH3
+ CH3OH2
*
O
CF3
OCH3
O
O
+
CF3—C—OCH3 + CH3—C—O*–
Necleophile attack at more electrophilic. C
CH3OH2
*
CH3—C—OH
583
Solutions
O
OH
OH
O
H3C
1181. H3C
H3C
COOH
OH
O
A
H
B
O
O CH3 O
CH3
H3C
COOH
CH3—C—C—–C—OC2H5
X
CH3
CH3
E
COOH
D
1182. A : C 6 H 5COOCH 2CH 3 ; B : CH 3CH 2OH;
C : CH 3CHO;
1183. C 6 H 5CH==CH  COOH C 6 H 5CHBrCHBrCOOH
A
C 6 H 5COCOCONH 2
E
F
HO
O
O
C
C 6 H 5COCOCOCl
D
D : C 6 H 5COOH.
C 6 H 5CH(OH)CH(OH)COOH
B
C 6 H 5COCOCOOH
O
C
O
H3C
–
–O
1184. (a)
COOH
O
O
COOH
–
HO
H 2O
O
O
HO
N
H
TsO
COOH
HO
O–
–
TsCl
(b)
O
OH
N
H
TsO
Product
N
584
Problems in Chemistry
O
N—OTs
s
BuLi
(c)
O
O
O
BuLi
N
C
–
C—N
O
OTs
O
O
H2O
s
N
O
Product
C
O
O
O
O
2–
1185. (a)
OEt
+ CO3
OEt
s
H
CH2CH2Br
Br
O
O
O
O
CO32–
Product
–
Br
OEt
H
OEt
CH2CH2Br
585
Solutions
COOEt
COOEt
OEt
COOEt
CH2CH2—COOEt
s
Na
(b)
–
O
COOEt
COOEt
COOEt
EtOOC
COOEt
O
O
COOEt
COOEt
Na/∆
s
COOEt
EtOOC
EtOOC
OEt
O
O
O
COOEt
COOEt
EtOOC
EtOOC
OEt
O
–
O
EtOOC
EtOOC
O
t-BuOK
1186.
N
R
O
–
N
CH3
O
EtO
O
N
CH2–
R
R
AcO
O
O
O
HO
Ac2O
O
Tautomerize
N
N
N
R
R
R
AcO
MeMgBr
HO
OH
H2O
Me
N
R
OH
NaOH
H2O
Me
N
R
586
Problems in Chemistry
O
CH3
OH
–H2O
N
R
R
–
COOEt
COOEt
COOEt
COOEt
Tautomerize
Me
N
EtO–
1187. (a)
O
O
–
OEt
COOEt
EtO– +
O
(b) (i)
COOEt
EG
H+
O
COOEt
O
CH3MgBr
O
O
O
O
(i) N2H4
(ii) HO–/heat
(iii) H3O+
O
Br
(ii)
COOEt
H2SO4
Br2/H+
heat
H 2O
O
O
O
Br
EG
H+
O
O
X
2X + Na
O
Heat
O
O
O
H2SO4
H2O
Product.
587
Solutions
H2SO4
(iii)
COOEt
KMnO4
Heat
H 2O
COOH
H+/heat
O
O
CH3Li
excess
NaBH4
TsCl
O
NaCN
OH
CN
H2SO4
H 2O
COOH
R
O
(iv)
N
COOEt RNHNH2
H+
R
N
+
H
COOEt
N
1188.
EG
H+
O
EtO–
N
H
O
N 2H 4
HO–/heat
H 2O
O
COOCH3
N+
PCC
LiAlH4
O
OEt
O
–
R
Product
N—H
COOCH3
Product
H 3O +
H 2O
O
O
1189. (a)
—Br KCN
—CN
H3O+
H2O
—COOH
Br2/P
Br
(HVZ)
COOH
588
Problems in Chemistry
H O+
LiAlH4
3
(b) CH 3CH  CH 2CN →
→ Product
H
2O

C6 H 5
1190.
O
O
H+
H+/H2O
H
COOH
OH
O
H—O
HO
HO
O
–H2O
–H+
OH
HO
O
CH2MgBr
CH2MgBr
O
O
OH
+
O
H
–
BrMgCH2
O+
H 2O
OH
O
1191. (a)
OH
l
l
OMgBr
–
O
CH2MgBr
BrMgO
OMgBr
BrMg—O
O
H 3O +
OH
HO
S–
(b)
S
O
l l
NH2
NH2
1192. (a)
(b) CH 3I (excess); Ag 2O/heat,
HS
O–
S
NH2
+
O
N
H
+
H
N
H
O
589
Solutions
(c) CH 3ONa; CH 3CH 2CH 2 Br; NaOH/H 2O; heat,
(d) Mg/Et 2O; CO 2 /H 3O + ; CH 3OH/H + .
O

LiAlH4
1193. PhBr → → PhCOOH → → Ph  C  NH 2 →
SOCl 2
Mg
CO2
Et 2O
H 3O +
NH3
O
Product
H+
Br
O

Br HOOC  CH 2  C  CH 3
1194.
C
A
HOOC  CH 2  CH  CH 3

OH
B
D
COOH
1195 CH  COOCH 3

CH  COOCH  C 6 H 5

CH 3
H
C 6 H 5  CH  CH 3

OH
C==C
COOH
H
B
C
A
COOCH 3
COOH
H
H
H
H
Br
Br
Br
Br
COOCH(CH 3 )C 6 H 5
COOH
(E and F )
D
1196. All the three reaction involves SN–1 reaction of tertiary halide as:
CH3
A : CH3O—
—C—Br
CH3
∆
CH3O—
CH3
—C+ + Br–
CH3CN
CH3
Here CH3O— from para position stabilize
the carbocation increasing reactivity.
CH3O—
CH3
+
—C—N
C—CH3
CH3
CH3
CH3O—
+
—C—N
CH3
CH3
C—CH3
OH
2
l l
–H+
CH3O—
—C—N
CH3
C—CH3
OH
tautomerize
Product
590
Problems in Chemistry
The above mechanism reveal that electron donating group will increase the reactivity and electron
withdrawing group will decrease the reactivity. Therefore, reaction A will be the fastest and
reaction B will be the slowest.
OH
OH
O
1197.
LiAlH4
OH
NH4Cl
H–
TsCl
OEt
H
O
O
O
OH–
OTs
l
OH
O
O
l
HO
+
N
(b)
1198. (a)
OH
(c)
OH
O
O
O
O
O
CH3
I
N
(e)
(d)
N—CH3
(f)
H
O
O
O
O
C
C
OH
1199.
H2N
OH
–H+
l
Ph
l
O
Ph
HN
l
l
O
O
–H+
Cl
O
O
O
O
O
Ph
–H+ Ph
Ph
O
N
H—N
O
+
Ph
O
HN
Ph
Ph
l
l
O
O
591
Solutions
Cl
O
SEt
1200. (a)
OMe
OEt
(c)
(b) no reaction
CCl3
O
COOH
(d)
OLi
(f)
(e)
+
OEt
OHC
O
1201. (a) Ag 2O/H 2O, (b) {(CH 3 ) 3 CO}3 Al, (c) CH 3 MgCl; PCC, (d) NaBH 4 .
O–
O
OEt
1202. (a)
C
O
C
N
N–Li+
+ CH3Li
NaOEt
1203.
Product
H 2O
Product
O
O
OEt
O
–
O
(b)
O
OEt
CH3MgI
Br
Br
NaOEt
EtO
EtO
Diethyl malonate
O
EtO
Product
O
1204. (a)
NH3
SOCl2
H 3O +
∆
O
O
(b)
OEt
O
OEt
592
Problems in Chemistry
O
O
O
(d)
(c)
COOEt
O
O
O
Br
CH3
1205.
A
CN
CN
C
B
OH
OH
D
O
OC2H5
E
O
O
O
OC2H5 C2H5OK
O
C 2H 5I
OC2H5
C2H5OH
C2H5
F
O
H+
∆
1-Phenyl-1-butanone
AMINES
SOCl
H
KCN
2
2
1206. (a) R —OH →
R —Cl →
aqueous R —CN → RCH 2 NH 2
Ni
(b)
—OH
H2CrO4
==O
NH2OH
==N
OH Na
Hg
O

LiAlH4
(c) C 6 H 5 NHCH 3 → C 6 H 5 —N— C —CH 3 →
C 6 H 5 —N—CH 2CH 3


CH 3
CH 3
CH3COCl
—NH2
593
Solutions
NaNO
LiAlH
CuCN
CH COCl
4
2
3
(d) C 6 H 5 NH 2 →
C 6 H 5 N 2Cl → → C 6 H 5CH 2 NH 2 →
HCl 0°C
LiAlH
4
→
C 6 H 5CH 2 NHCH 2CH 3
CH 3
CH 3
CH 3



1207. CH 3CH 2 —CH—CH 2 NH 2 , CH 3CH 2 CH—CH 2OH, CH 3CH 2 CH—CHO, CH 3CH 2 CH(CH 3 ) 2
B
A
D
C
—CH3
1208. H2N—
O
O
1209.
OH
COOH
COOH
—CN
A
C
B
O
O
OH
NH2
D
E
CH3
N—CH3
1210.
+
+
N
CH3 I–
N
CH3 OH– CH3
CH3
C
B
A
CH3
CH3
+
N—CH3
OH–
CH3
+
N—CH3
I–
CH3
F
E
D
N(CH3)2HBr
N(CH3)2
N(CH3)3I
N(CH3)3OH
N(CH3)2HBr
N(CH3)2
N(CH3)3I
N(CH3)3OH
A
B
C
D
1211. (a)
E
594
Problems in Chemistry
NOH
(b)
NH2
Ph
Ph
N(CH3)3I
Ph
Ph
Ph
Ph
A
CH3
C
B
CH3
N –
Br
Ph
Ph
D
+
H
1212.
+ N(CH3)3
H
A
H 3C
N
B
CH3
O
N+
H 3C
C
CH3
CH3
O
O
H H
1213.
N3
H2
Pb/CaCO3
+
NH2
O
N
O
H
H H
–
+
N
N
1214.Hinsberg’s method : Add benzene sulphonyl chloride to the mixture of butylamine and
dibutylamine and filter off the precipitate containing sulphonamides. Now dissolve the precipitate
in aqueous KOH solution. Sulphonamide of butylamine will come into solution as their potassium
salt leaving sulphonamide of di-butylamine insoluble. Filter the solution and treat both residue and
filtrate separately with dilute acid.

H O

 
CH3 CH2 CH2 CH2 —N— S —C6 H5 



O
CH3 CH2 CH2 CH2 NH2  C6 H5 SO2 Cl
 Insoluble
 →
(CH3 CH2 CH2 CH2 ) 2 NH

O


(CH3 CH2 CH2 CH2 ) 2 N— S —C6 H5 



O
↓ KOH
K+
−
(CH 3CH 2CH 2CH 2 ) 2 N—SO 2C 6 H 5 + CH 3CH 2CH 2CH 2 —N — SO 2C 6 H 5
Insoluble
Soluble
dilute H+
dilute H+
+
(CH 3CH 2CH 2CH 2 ) 2 NH 2
+
CH 3CH 2CH 2CH 2 NH 3
595
Solutions
H
H
NO
N
N
N
CH3
CH3
+
–
N I
1215.
NH2
NH2
OH
B
A
–
N+ I
H 3C
CH3
CH3
D
C
O
H
CH2CH2COOEt
1216. (a)
(b)
N—
D
N
A
N
CH2CH2COOEt
C
B
(c) H2N—CH2CH2CH2CH2—NH2(E)
O
O



NH3
CHCl 3
Br2
H3
1217. (a) R —CN → R —C —OH → R —C—NH 2 →
R —NH 2 → RNC
O+
Heat
H 2O
H O+
NaOH
KOH
HNO
TsCl
2
3
KCN
(b) R — NC →
→
R —NH 2 →
R —OH →
R —CN
H 2O
O

1218. (i) PhCH 2 NH—C—CH==CH 2 (ii) PhCH 2COCH 2Cl (iii) ClCOOEt (iv) BrCH 2CH 2COOH
1219. (a) (CH 3CH 2CH 2CH 2 ) 3 B
CH 3CH 2CH 2CH 2 NH 2
A
(b) PhNHCOOH
C
(c) C 2 H 5 NHCN
F
B
PhNH 2
PhNHCONHPh
E
D
C 2 H 5 NHCH 2 NH 2
G
O−
O
O

+

−
1220. H 3C—C
+ CH 2 —N 2 → CH 3 —C—CH 2 —N 2 → CH 3 —C  CH 2 + N 2



CH 2Cl
CH 2Cl
CH 2Cl( B )
O–
O

+

Also, ClCH 2 —C—CH 2 —N2 → ClCH 2 —C—CH 2CH 3 + N 2
A

CH 3
596
Problems in Chemistry
CH 3 CH 3


CH 3 —C—–N + —CH 2CH 3OH –


CH 3 CH 3
CH 3

1221. CH 3 —C— NHCH 2CH 3

CH 3
CH 3

CH 3 —C== CH 2
C
B
A
CH 3CH 2 N(CH 3 ) 2
(CH 3 ) 3 COH.
D
E
NO 2
1222. (a) MeC ——CH 2OH
CH 2OH
(b) RCH 2 NMe 2
OH

CH2 == CO
(c) CH 2 == C—CN →
CH 2 == C—CN

OCOCH 3
Br
−
2
1223. (CH 3 ) 2 CH—C— NH 2 + OH – → (CH 3 ) 2 CH—C— NH →
(CH 3 ) 2 CH—C— NHBr
–Br –



O
O
O
–
OH−
••
–Br −
→ (CH 3 ) 2 CH—C— N—Br → (CH 3 ) 2 CH—C—N •• → (CH 3 ) 2 CH—N ==C==O
••


O
O
H O
2
→
(CH 3 ) 2 CHNH 2 + CO 2
O
O
O
1224. (a)
O + NH3
O
N—H
KOH
PhCH2Br
N—CH2Ph
O
O
–
+ KBr OH PhCH2NH2
H2O
KMnO
NH
Heat
LiAlH
PO
4
3
2 5
4
(b) PhCH 2 Br →
PhCOOH →
PhCONH 2 →
PhCN → PhCH 2 NH 2 .
–
OH /Heat
O
1225. (a)
+ Cl2
Heat
O
Cl
aqueous
KOH
O
PCC
O
H
H2N—CH2CH2—NH2
∆
N
N
H2
Ni
N
N
H
597
Solutions
H
O
O
(b)
NH
O
+ NH3(excess)
[form (a)]
NH
O
N
H2
Ni
N
H
AROMATIC COMPOUNDS
1226. C2H5—
—CH—CH3 C2H5—
—C—CH3 C2H5—
OH
O
A
C
O
—COOH
O
—
[ C—
D
O
Cl
B
HOOC—
—C—CH3
]n
—C—OCH2CH2O —
E
COOH
O
1227. (a)
—CH
O
Cl
CH—C—CH3
(b)
O2N
Br
Br
OH
(c)
COOH
COOH
(d)
S
O
(e)
OCH3
(f)
NO2
(g) X : HNO 3 / H 2SO 4 followed by reduction with N 2 H 4 /OH – / Heat.
Y : Reduction with N 2 H 4 /OH – / Heat followed by nitration.
598
Problems in Chemistry
Ph –H+
+
OH2
1228.
+
–H+
Product
+
Br
NBS
1229.
aq. KOH
CCl4
OH
H+
+
+
CH3 CH3
Ph
Ph
CH3
Ph
+
C—CH3
PhCH
–H+
CH3
CHCH3
—C
1230. CH3O—
—C—CH3
CH3O—
O
B
A
Cl
O
C
O
—O—C—CH3
—C—CH3
CH3O—
D
O
E
O
1231. (a)
—C—CH3
HO—
PhCH2COCl
AlCl3
O
Ph
Ph
Cl2
AlCl3
Cl
Ph
N 2H 4
OH–/Heat
Cl
599
Solutions
OH
OH
CHO
CH
+
H 3O
(b)
CH3MgBr
CH3
CH—CH3
CH3Cl
AlCl3
PCC
(c)
H2SO4
KMnO4
Heat
NaOH
COOH
H3O+
HO3S
SO3H
(d)
+ CH3CH2CH2Cl
AlCl3
HNO3
H2SO4
NO2
1232.
H3C—CH
CH2OH
H3C—CH
OC2H5
CH2I
I
B
A
H3C—CH
CH2OH
CH3
CHO
O
OH
D
C
Cl
CH3
CHO
O
E
Product
600
Problems in Chemistry
OH
1233. (a) PhCH2CH2CH2Br
C2H5ONa
C2H5OH
CH2
PhCH2CH
(CH3COO)2Hg
NaBH4
H2O
PhCH2—CH—CH3
OH
Product
Pd/BaSO4
H2
C
Cl
(b)
H3O+
Ph—CH2—C—CH3
CH3C
CNa
PhCH2—C—CH3
C—CH3
PCC
O
Cl
O
HNO3
Mg
ether
H2SO4
O2N—
—CH2CH2OH
NBS
CCl4
Product.
NO2
O
(c)
+ Cl2
AlCl3
O
—C—Cl
Cl—
AlCl3
AlCl
Mg
ether
—C—
Cl
Oxirane
PCC
3
(d) C 6 H 6 + Cl 2 →
C 6 H 5Cl → C 6 H 5 MgCl →
C 6 H 5CH 2CH 2OH →
+
H 3O
O
OH


C6H5 MgCl
H 3O +
PCC
C 6 H 6CH 2 —C—C 6 H 5 ← C 6 H 5CH 2 CH —C 6 H 5 ←
←
I
(e)
+ I2
HNO3
I
AlCl3
(CH3)3CCH2COCl
C
SOCl2
(CH3)3CCH2COOH
B2H6/H2O2
NaOH
H3O+/Na2CrO4
(CH3)3CCH2CH2OH
O
CH2
601
Solutions
O
O
1234. (a) Ph
Ph
NO2
X
O
–
(b) Ph—C—CH2
Ph—C—CH2—CH2—C—Ph
CN
Ph
Ph
CHO
X
Ph
O
Z
Ph
O
Z
Y
AlCl3
CrO2Cl2
CH3Cl
O
—CH3
—CHO
Ac2O
Product
O
(b) Ph
Z
O
Ph
O
CH3COOMe
C2H5ONa
Heat
O
Br2
AlBr3
—CH
AcOH
HOCH2CH2OH
Ph
CN
H+/Heat
H3PO4
Product
H 3O +
Ph
O
O2 N
HNO3
CH3COCl
H2SO4
AlCl3
S
O
Ph
O
S
CH—COOMe
O
O
+ KCN
(c)
O
O
O
1235. (a)
O
Y
O
(c)
Ph
NaBH4
Product
CH3MgBr
O
CN
602
Problems in Chemistry
NO2
NH2
HNO3
1236. (a)
F
Zn-HCl
HNO2
HBF4
0°C
∆
H2SO4
CH3COCl
NHOC—CH3
NHCOCH3
HNO2
H 3O +
0°C
HNO3
CH3Cl
H2SO4
AlCl3
NH2
H 3O +
NO2
F
CH3
F
F
HBF4
KMnO4
HNO2
HBF4
∆
NaOH
0°C
∆
NO2
CH3
COOH
–
(b) O2N—
O
—F
O
–
F
+
N
H
+ •• N—Et
+
Et
(Positive charge at p-carbon activate
the nucleophilic attack)
–
O
O
–
+
N
F
H
Et
O2N—
—N
Et
O
O
(c) (i)
+
O
AlCl3
O
OH
CH3OH
H+/Heat
O
+ HF
Et
N—Et
+
Product
603
Solutions
O
(ii)
R
+
—COCl
R
AlCl3
R′
R′
O
O
C—Cl
AlCl3
(iii)
+
(iv)
CH3Cl
AlCl3
HNO3
CH3—
COOC2H5
H2SO4
CH3—
—NO2
Zn-HCl
CH3—
—NH2
COOH
CH3CH2OH
Ag2O
OHC—
H+
NH2
—NH2
CrO2Cl2
H 3O +
Ac2O
NH2
O
OH
1237. (a)
H2SO4
KOH
∆
∆
H3O+
OCOPh
O—C—Ph
HNO3
PhCOCl
H2SO4
NO2
NO2
(b)
HNO3
Br2
H2SO4
AlBr3
N2Cl
Zn-HCl
NaNO2
H 2O
HCl/0°C
boil
Br
Br
O
O—C—Ph
PhCOCl
Br
604
Problems in Chemistry
CH3
(c)
COOH
CH3Cl
HNO3
KMnO4
AlCl3
H2SO4
NaOH
PhOH
H 3O +
NO2
NO2
NHCOCH3
NO2
HNO3
(d)
Product.
H+/Heat
H 3O +
HNO3
CH3COCl
Zn-HCl
H2SO4
H2SO4
NO2
NaNO2
NaNO2
HCl/0°C
Heat
Product
NH2
NO2
HNO3
(e)
CH3COCl
Zn-HCl
H 3O +
HNO3
H2SO4
H2SO4
OH
Product
aq. KCN
NaNO2
NO2
Zn-HCl
HCl/0°C
H 2O
NaNO2
boil
HCl/0°C
NO2
OH
(f)
OH
Conc. H2SO4
OH
Br2
OH
Br
Br
FeBr3
SO3H
NO2
NO2
+ HNO3/H2SO4
Br
Br
reflux
SO3H
1238. (a)
dilute
H2SO4
Cl2
NaNO2
Zn-HCl
FeCl3
HCl/0°C
Cl
aq
KI
Product
605
Solutions
Cl
(b)
Cl
HNO3
+ Cl2/FeCl3
Zn-HCl
H2SO4
NaNO2
aq
HCl/0°C
KI
Product
NO2
NO2
NO2
Cl2
(c)
Zn-HCl
FeCl3(excess)
NaNO2
aq
HCl/0°C
KI
Product
Cl
Cl
NHOAc
NHOAc
HNO3
1239. (a)
Zn-HCl
CH3COCl
HNO3
Blocking
H2SO4
H2SO4
N2+Cl–
Ph
NO2
NaNO2
NO2
NH2
Br2
Zn-HCl
Br
Br
Br
aqueous
HCl/0°C
H3PO2
Br
NH2
NH2
NaOH
Br
NaNO2
Br
COOH
(c)
H 3O +
Reflux
HCl/0°C
NO2
(b)
SO3H
NH2
NO2
PhLi
LiCl + N2 +
NO2
Conc. H2SO4
NH2
NH2
CaO/Heat
NH2
NaNO2(excess)
H 2O
HCl-0°C
Boil
NH2
O
O
O
OH
HO
Tautomer
OH
606
Problems in Chemistry
NH2
H
H
CH3
N:
1240.
C
MeO
NH2
+
N
CH3
_
O
O
MeO
CH2CH2COOMe
H 2C
H 2O
–H2O
CH2COOMe
H
H
NH2
+
_ N
+ CH
3
MeO
CH3
MeO
H
CH2COOMe
H
HC
CH2COOMe
H
H
N
CH3
N
–H+
MeO
MeO
H O+
CH O
+
CH3
H
CH2COOMe
CH2COOMe
Mg
ether
NH2
+
N
PCC
2
3
1241.(a) PhBr → →
→
PhCH 2OH → Product
CrO Cl
2 2
(b) PhCH 3 →
Product.
Ac2O
Etard’s reaction
OH
(c)
+ CHCl3
H 3O +
NaOH
Reimer-Tieman
reaction
OH
OH
Conc H2SO4
(d)
Product
H 3O +
AlCl3(excess)
Reflux
CH3COCl
Product.
SO3H
OH
(e)
OH
H2
Ni
O
H2CrO4
Cl2
C2H5ONa
∆
C2H5OH
Product
– NH3
607
Solutions
O
(f) PhCHO
H3
PhMgBr
O+
PCC
Ph—C—Ph
O
(g)
Product
O
O
+
H 3O +
HCN
AlCl3
H 3O +
Zn(Hg)
HCl
HOOC
COOH
O
HF
O
CH3
CH3
HNO3
1242. (a)
CH3
Zn-HCl
H2SO4
NaNO2
H2O
HCl/0°C
Boil
HCl
HCN/AlCl3
NO2
OH
CH3
H3O+
CHO
OH
SO3H
SO3H
HNO3
(b)
OH
OH
KOH
LiAlH4
H3O+
H2SO4
NO2
OH
(c)
NO2
OH
Br2
Br
Br
Br
H 2O
Zn-powder
Heat
Br
Br
Br
NH2
608
Problems in Chemistry
OH
OCOCH3
OCOCH3
CH3COCl
(d)
AlCl3
Protection
H 3O +
CH3CH2CH2Cl
OH
Conc H2SO4
(e)
∆
OH
KOH
OCH3
OCH3
H 3O +
∆
(i) NaOH
Conc H2SO4
(ii) CH3Cl
Blocking
SO3H
OCH3
KOH
aq
Product
CH3
NBS
H 3O +
CH3Cl
Reflux
AlCl3
1243.PhCH 2 Br + CHCl 3 .
CH3
CH3
HNO3
Conc H2SO4
1244. (a)
CH3
NO2
H 3O +
Zn-HCl
H2SO4
Product
SO3H
CH3
(b)
CHO
CH3
CHO
CrO2Cl2
HNO3
N2H4
Ac2O
H2SO4
OH–
Product
NO2
H2O
NaNO2
Boil
HCl/0°C
CH3
(c)
Conc H2SO4
KOH
Heat
H3O+
Product
NO2
Zn-HCl
NaNO2
HCl/0°C
H 2O
Boil
609
Solutions
NO2
HNO3
(d)
CuCl
NaNO2
Zn-HCl
H2SO4
(Excess)
Product
HCl/0°C
NO2
NO2
Zn-HCl
(e)
CuCN
NaNO2
Product
HCl/0°C
[from (d)]
NO2
NO2
NO2
NO2
NaNO2
Na2S
(f)
Zn-HCl
KI
HCl/0°C
[from (d)]
NO2
NO2
Product.
NaNO2
NH2
CuBr
KCN
NaNO2
Zn-HCl
HCl/0°C
HCl/0°C
Br
CH3
(h) Toluene
Boil
NO2
HNO3
Zn-HCl
H2SO4
(excess)
NaNO2
Br
CuBr
HCl/0°C
Br
Product.
NO2
HNO3
H2SO4
Product.
CH3
NO2
(i)
HCl/0°C
H 2O
NO2
(g)
[from (f)]
I
NH2
NaNO2
CH3Cl
AlCl3
NO2
Br2(excess)
Zn-HCl
FeBr3
Br
Br
Product
610
Problems in Chemistry
NH2
HNO3
(j)
NHCOCH3
Conc H2SO4
CH3COCl
Zn-HCl
H2SO4
Br2
H3O+
Protection
FeBr3
SO3 H
Br
NH2
NH2
Br
Br
CuBr
NaNO2
Br
Br
Br
Br
KOH
H3O+
fusion
HCl/0°C
OH
SO3 H
OH
NH2
(k)
CH3COCl
Zn-HCl
HNO3
H3O+
HNO3
Br2
H2SO4
H2SO4
FeBr3(excess)
NO2
Br
Br
Br
NaNO2
CuBr
Zn-HCl
KCN
Product
CH3
Br2(excess)
Br
Br
FeBr3
H 3O +
reflux
H 3O +
SO3H
SO3H
NH2
I
Br
Br
(m)
SO3H
[from (j)]
Br
NO2
NO2
CH3
Conc H2SO4
(l)
Br
NaNO2
HCl/0°C
HCl/0°C
CH3
NH2
H 3O +
NaNO2
Reflux
HCl/0°C
KI
aq
Br
Br
KMnO4
NaOH
Product.
611
Solutions
CH3
CH3
NO2
NO2
Br2
(n)
FeBr3
[from (a)]
Br
CH3
CH3
NO2
NO2
SnCl2
(o)
CH3
NO2
[from (h)]
NaNO2
H2O
HCl/0°C
Boil
NH2
CH3
OH
CH3
CH3
NO2
NaNO2
(p)
NO2
CuBr
Zn-HCl
HCl/0°C
(r) H3C—
NaNO2
HCl/0°C
NH2
[from (o)]
(q) Toluene
NO2
Br
Br
HNO3
Zn-HCl
NaNO2
HCl/0°C
H2SO4
—N2+Cl–
p-cresol
CN
KCN
aq
H3C—
—N2+Cl–
Phenol
Product
Product.
[from (q)]
NH2
NO2
Na2S
1245. (a)
NaNO2
H3PO2
CH3COCl
Zn-HCl
HNO3
H2SO4
HCl/0°C
NO2
NH2
H3O+
CF3CO3H
NO2
Product.
612
Problems in Chemistry
CHO
N2H4
(b)
Cl2
OH–/Heat
AlCl3
CrO2Cl2
—Cl
H3C—
Product.
Ac2O
NO2
Na2S
(c)
NaNO2
CuCl
Product
HCl/0°C
NO2
NO2
(d)
KCN
∆
H 3O +
NO2
NaNO2
Zn-HCl
HCl/0°C
KCN
aq
Product
CN
Cl
Br
—CH3
(e) H2N—
CF3CO3H
—CH3
O2N—
Br2
—CH3
O2N—
FeBr3
Br
H2N—
CH3
Zn-HCl
—CH3
CH3
NaNO2
(f)
Br
H3PO2
HCl/0°C
Br
NH2
NO2
NO2
CHO
(g)
H+
HOCH2CH2OH
O
CH
O
Zn-HCl
NaNO2
HCl/0°C
KCN
aq
LiAlH4
CH2NH2
H3O+
Boil
CHO
613
Solutions
Zn-HCl
(h)
NHCOCH3
NHCOCH3
NO2
H 3O +
CH3Cl
CH3COCl
AlCl3
CH3
NO2
KMnO4
NaOH/Heat
COOH
CH3
CH3
HNO3
(i)
CH3
Zn-HCl
CH3COCl
CH3
HNO3
H2SO4
H 3O +
H2SO4
NO2
NO2
NHCOCH3
NH2
CH3
CF3CO3H
NO2
NO2
NH2
(k)
NH2
Br2-H2O
Br
Br
NaNO2
H3PO2
Br
Br
HCl/0°C
Br
Br
614
Problems in Chemistry
CH3
1246. (a)
CH3
CH3Cl
HNO3
AlCl3
H2SO4
COOH
NO2
NO2
KMnO4
NaOH/Heat
NO2
NO2
CONH2
NO2
NH3
∆
NO2
NO2
(b)
HNO3
NH2
Zn-HCl
H2SO4
Br2
Br
Br
CF3CO3H
H2O
NO2
Br
Br
Br
Br
NO2
(c)
NO2
HNO3
Cl2
H2SO4
FeCl3
NH2
Zn-Hg
CH3COCl
Cl
Cl
NH2
Br2
H 3O +
FeBr3
Cl
Br
615
Solutions
Cl
(d)
Cl
AlCl3
+ Cl2
NO2
HNO3
Excess
H2SO4
Cl
C2H5
C2H5
Cl
C2H5
C2H5
C2H5
SO3H
COOH
CN
1247.
A
B
SO3H
C
E
D
O
COOH
O
COOH
F
G
1248. (a) H3C—
Mg
—Cl
O
CH2O
H 3O +
Ether
H3C—
—CH2OH
OMe
Br
(b)
KNH2
OMe
NH2
CHO
(c)
+ CO + HCl
AlCl3
CH3
CHO
Cl2
N2H4/OH–
AlCl3
Heat
Cl
Cl
616
Problems in Chemistry
NO2
NH2
HNO3
(d)
NHCOCH3
Zn-HCl
Conc H2SO4
CH3COCl
H2SO4
NH2
Na2S
(e)
H2SO4
Warm
NO2
Br
Br
Br
Br2
NaNO2
CuBr
Br
Br
HCl/0°C
FeBr3
NO2
NO2
NO2
H 3O +
NH2
NH2
NO2
HNO3
NO2
NO2
(f) H3C—
—NO2
Zn-HCl
CH3 COCl
HNO3
H3C—
H2 SO4
—NHCOCH3
H3 O+
NO2
NaNO2
H3 PO2
HCl/0°C
OH
(g)
CHO
H3C—
CHO
Zn
CO/HCl
HNO3
Heat
H2SO4
H2SO4
NO2
CH3
CH3
N 2H 4
Zn-HCl
OH–/Heat
NO2
NaNO2
H 2O
HCl/0°C
boil
OH
617
Solutions
O
O
O



1249. (a) Ph—C—C 2 H 5 (b) C 2 H 5 —C—OEt (c) C 6 H 5 —C—CH 3 (d) C 6 H 5CH==CH—COOEt
O
COCl AlCl3
1250. (a) C6H6 +
Zn(Hg)
HCl
O
AlCl3
(b) C6H6 +
Zn(Hg)
Cl
HCl
O
O
(c)
+
O
HF
O + AlCl3
HO—
O
O
O
Zn(Hg)
HCl
(d)
+ C2H5Cl
(e)
+ Cl2
AlCl3
NBS
CCl4
Cl
AlCl3
C2H5ONa
C2H5OH
OH
O
H 3O +
Mg
Ether
O
(f)
+ CH3COCl
OMe
NaBH4
AlCl3
NHCOCH3
N
OMe
Cl
Br
(b)
1251. (a)
OH
Br
(d)
(c)
COCH3
NO2
Br
COOCH3
618
Problems in Chemistry
CH3
CH3
CH3
HNO3
Conc. H2SO4
1252. (a)
CH3
NO2
H2SO4
H2SO4
(Heat)
SO3H
SO3H
NO2
(b)
NO2
dilute
NO2
HNO3
Br2
H2SO4
FeBr3
NH2
Zn-HCl
Br
CH2CH3
C2H5Cl
(c)
AlCl3
Br
CH2CH3
Br—CHCH3
HNO3
Br2
H2SO4
hν
NO2
NO2
COCH3
aq
PCC
KOH
NO2
COCH3
(d)
CH3COCl
HNO3
AlCl3
H2SO4
NO2
NH2
Zn-HCl
(e)
[from (d)]
CH2CH3
COCH3
N2H4/OH–
Heat
NO2
NO2
619
Solutions
NO2
(f)
HNO3
Cl2
H2SO4
AlCl3
Cl
CH3
CH3
H+
—N
1253.
—N—CH3
NO2+
—N—CH3
+
+
H
H
O2 N
(meta directing)
NH3
O
O
1254. (a)
+
Product.
AlCl3
Cl
NaBH4
H2SO4
Heat
B2H6
PCC
OH–/H2O2
NO2
NO2
(b)
NO2
Na2S
HNO3
H2SO4
Heat
NH2
NO2
FeSO4
H 2O 2
CH3
(c)
CH3Cl (excess)
Product.
NO
COOH
KMnO4
AlCl3
NaOH/Heat
CH3
COOH
CH3
CHO
CHO
O
(d)
CO/HCl
Cl2
AlCl3
AlCl3
Zn(Hg)
Cl
HCl
Mg
Cl
Ether
620
Problems in Chemistry
CH3
O
H 3O +
Mg
SOCl2
H 3O +
Ether
OH
CH3
OH
NO2
(e)
HNO3
Br2
H2SO4
FeBr3
Zn-HCl
Br
CH3
CH3
(f)
NH2
Br
CH2Cl
CH3Cl
Br2
AlCl3
Cl2
FeBr3
hν
Br
O
(g)
+ CH3COCl
Br
O
AlCl3
HNO3
H2SO4
NO2
CH2 CH3
(h)
+ CH3 CH2 Cl
AlCl3
CH2 CH3
HNO 3
Conc. H2SO4
H2SO4
SO3 H
Product
C2H5ONa
Cl2
H3O+
C2H5OH
hν
Heat
CH2 CH3
NO2
SO3 H
621
Solutions
Br
(i)
Br
FeBr3
+ Br2
Conc. H2SO4
SO3H
NO2
NH2
HNO 3
(j)
Zn-HCl
CH3COCl
H3O+
CH3Cl
H2SO4
AlCl3
CH3
N
CH3I
CH3
O
O
(b)
1255. (a)
(c)
X
X
N
X
OH
O
(d)
(e)
X
X
CH3
O
Br
NO2
1256. (a)
(b)
Cl
(c)
CH3
(e)
(d)
Br
NO2
OCOPh
Cl
622
Problems in Chemistry
OD
OH
H 2O
1257.
D
D
NO2
NO2
COOH
1258. (a)
CH3Cl
KMnO4
HNO3
AlCl3
NaOH/Heat
H2SO4
Heat
HNO 3
(b) H3 C—
Zn-HCl
H2SO4
NO2
NaNO 2
NO2
CuCN
HCl/0°C
H3 C—
—CN
H 3O +
Product
CH3
(c)
AlCl3
NO2
HNO3
CH3Cl
H2SO4 (Heat)
NO2
O
O
(d)
AlCl3
+
Cl
OH
NaBH4
(e)
+ CH3 Cl
AlCl3
CH3
COOH
KMnO 4
NaOH/Heat
HNO 3
H2SO4
COOH
NH 3
Heat
NO2
Product
623
Solutions
(f)
+
OH
—Cl
O
H2
AlCl3
Pt
O
O
OH
1259. (a)
OH
(b)
(c)
O
1260. (a)
+ CO/HCl
CHO
AlCl3
Product
C2H5C
COCH3
CNa
H3O+
OH
C—C2H5
C
Pd-BaSO4
C—C2H5
C
PCC
OH–/Heat
Product
O
O
(b)
N2H4
+
O
Zn(Hg)
HCl
AlCl3
HOOC
HF
HOOC
O
O
CH3Cl
AlCl3
CH3
O
O
Cl AlCl
3
(c) PhH +
O
Ph—C—CH2CH3
CH3MgBr
H3O+
Product.
624
Problems in Chemistry
NH2
O
OMe
OH
CH3
(b)
1261. (a)
(c)
(d)
HOOC
SO3H
Cl
NO2
COOH
1262. (a)
COOH
Br2
CH3Cl
KMnO4
AlCl3
NaOH/Heat
FeBr3
Br
O
+
O
O
O
(b)
SOCl2
(CH3)2CHOH
Product
AlCl3
HF
HOOC
O
O
N 2H 4
Product
OH –/Heat
CH3
(c)
CH3Cl
Cl2
AlCl3
(excess)
hν
CH3 —
—CH2 Cl
Mg
ether
O
H 3O +
CH3
Product
NO2
O
(d)
HNO3
Cl2
H2SO4
AlCl3
Mg
ether
Cl
NO2
NO2
H3O+
Ag2O
OH
CH2COOH
625
Solutions
O
(e)
+ CH3 Cl
AlCl3
AlCl3
Zn(Hg)
HF
O
HCl
HOOC
O
O
Product
NaBH4
H+
Heat
O
H
+
+ I+
1263.
Cl–
OH
1264.
I
I
+ HF
CH2
CH—CH2
+
H+
C 6H 6
–H+
Allyl alcohol
+
C 6H 6
Product
–H+
COOH
OH
CH3
1265.
H3C—C
CH3
Cl
1266. (a)
+ Cl2
HNO3
AlCl3
Zn-HCl
H2SO4
CHO
(b)
+ CO + HCl
AlCl3
NaNO2
CuCN
HCl/0°C
Product.
CH3
CHO
HNO 3
N 2H 4
H2SO4
OH –/Heat
NO2
NO2
626
Problems in Chemistry
NO2
HNO 3
(c)
H2SO4
Cl2
KMnO 4
HNO 3
H2SO4
Product.
COOH
CH3
CH3Cl
AlCl3
CuBr
HCl/0°C
Cl
CH3
(d)
NaNO 2
Zn-HCl
AlCl3
Sn/HCl
NaNO 2
HCl/0°C
NaOH/Heat
H 2O
NO2
NO2
Boil
Product.
CH3
Sn-HCl
(e)
NaNO2
H 3O +
CuCN
Product
HCl/0°C
NO2 [from (d)]
HNO 3
(f)
NHCOCH3
NHCOCH3
NO2
Zn-HCl
NO2
HNO 3
CH3COCl
H2SO4
H2SO4
Heat
NO2
Product
NO2
Heat
HCl/0°C
NH2
HNO3
(g)
HBF 4
NaNO 2
NaNO2
Na2S
H2SO4
Heat
HCl/0°C
NO2
NO2
OH
OMe
LiAlH4
NH2
CH3Cl
H2O
KOH
Boil
NO2
H 3O +
627
Solutions
I
(h)
I
HNO 3
+ I2
HNO 3
NaNO 2
Sn-HCl
HCl/0°C
H2SO4
I
NO2
H2O
Boil
OH
NO2
1267. (a)
NO2
(b)
N
C6H5S
NO2
NO2
NO2
NO2
NHPh
Ph
N—Ph
(d)
(c)
O2N
CH2OH
NO2
O2N
NO2
CH2OH
OMe
CHO
OMe
1268. (a)
A
OMe
(b)
OMe
O
O
B
OMe
OMe
OMe
(c)
HO
HO
O
O
E
F
G
H
OH
I
628
Problems in Chemistry
CH2 OH
(d)
CH2 O–
OH
OMe
OMe
O–
OMe
J
OMe
K
OMe
L
NO2
–
(e) CH(COOEt)2
Br
CH(COOEt)2
M
NO2
N
(f) NaH/CH 3I(excess)
LiAlH 4
P
O
OMe
Br 2 / FeBr 3
Q
OMe
OMe
OMe
NO2
NH2
N2Cl
B
C
D
1269. (a)
A
(b) O2N
N
O
O2N
E
I
I
(c)
NO2
H2 N
I
O2 N
I
I
H
G
Br
N2 Cl CH3 CH2 O
NH2 CH3 CH2 O
(d) CH3 CH2 O
I
Br
Br
J
K
629
Solutions
(e) HO
HO
N2Cl
I
M
L
(f) H2N
CH3
CH3
CH3
CH3
N
CH3
O
CH3
CN
CH3
N2Cl
COOH
CH3
P
Q
CHO
CHO
CHO
CHO
1270. (a)
D
C
B
A
Cl
N2Cl
NH2
NO2
N2HSO4
OH
(b)
MeO
OMe
MeO
OMe
F
E
CH3
NH2
H2N
CH3
CH3
N2Cl
N2Cl
HO
OH
(c)
I
H
G
COOH
(d)
N
N
N
N2 Cl
J
COOH
K
(e) No reaction (L)
630
Problems in Chemistry
CH3
N
(f)
N
N
H3C
+
CH3
M
CH3
P
CH3
O
Br
CH3
CH3
COOH
CH3
1271.
Br
CH3
CH3
A
C
B
COOH
CH3
D
O
COOH
N
N
COOH
Br
E
G
O
F
OMe
OMe
OMe
1272.
Br
A
O
B
O
OH
C
O
O
OH
OH
1273.
A
O
B
OH
C
OH
CHO
OH
OH
D
E
O
1274. H3C—
CH3
—C—O—CH—CH2CH3
A
OH
B
O
C
H3C—
—COOH
D
631
Solutions
COOH
1275.
Br
A
COOH
C
COOCOCH3
B
COOH
COOH
O
1276.
OH
OMe
C
B
A
COOH
COOH
CH3
O
COOH
Br
Br
OH
OH
OH
D
F
Br
E
O
C—OC2H5
CH3CH2OH
1277.
CH3CHO
C6H5COOH
C
D
B
A
1278.(a) Aromatic, cyclic and fully conjugated with six π-electrons.
(b) Antiaromatic, cyclic and fully conjugated with 4π-electrons.
(c) Non-aromatic, cyclic but not fully conjugated.
(d) Antiaromatic, cyclic and fully conjugated with 4π-electrons.
1279.Conjugate base of A invoke aromatic character in the middle ring therefore, deprotonation of
benzylic H in A is favoured. On the other hand conjugate base of B does not bring any stability by
aromaticity deprotonation of benzylic H is not as favoured as in case of A.
CH3
H3C
H
s
–H+
A
Aromatic anion
632
Problems in Chemistry
CH3
CH3
H
–
–H+
B
1280. (a)
non-aromatic anion
+
Br
Br
OCH3
OCH3
OCH3
(b)
+
+
NH2
NH2
NH2
O
H
H
(c)
H
(d)
+
O
O
O
O
NO2
NO2
(f)
(e)
Cl
CH3O
NO2
O
O
(g)
(h)
O2N
CH3
633
Solutions
CHO
CHO
(i)
(j)
O
I
1281.Nitration of benzene proceeds by the following mechanism:
O–
O

N
+
O

+ H—O—S —OH
••
O—H

O
O
−
O

N
+
+
OH 2
+ HSO –4
+
O== N ==O
nitronium ion
–
O—SO3H
O
+ N+
O
H
Slow
Rate determining
step
NO2
+
+
H
NO2
H
NO2
+
Resonance stabilized σ-complex
NO2
fast
H
NO2
Energy
+
Reaction energy diagram:
The reaction proceeds at same
rate for both benzene and
hexadeutrobenzene since C—H
bond cleavage is not occurring
in rate determining step.
C6H6
NO2
HNO3
H2SO4
Reaction coordinate
634
Problems in Chemistry
1282.Sulphonation of benzene proceeds by the following mechanism:
O
O
S—O–
S—OH
+
S
O
+
O
O
O
HO
O
S+
H+
–H2O
+
O
O
O
H
S
S
HSO4–
C6H6
+ O
O
(diphenyl sulphone)
O
C
CH3COCl
1283.
O
CH3 Na/NH
3
AlCl3
O3
t-BuOH
(CH3)2S
Products
1284.(a) CH 3COCl/AlCl 3 , Br 2 /FeBr 3 , (b) PCC/CH 2Cl 2 , NBS/hν,
(c) H 2 /Rh/Pressure, NaIO 4 , (d) CH 3COCl/AlCl 3 , HNO 3 /H 2SO 4 ,
(e) (CH 3 ) 2 CHCOCl/AlCl 3 , Na/NH 3 /EtOH, Zn(Hg) /HCl,
(f) NBS/hν, Mg/ether, CH 2O, PCC( g ) Br 2 /FeBr 3 , HNO 3 /H 2SO 4 .
1285.The reaction will be slower with acetophenone than with benzene as the acyl group is a
deactivating group in electrophilic aromatic substitution.
O
O–
O–
O–
+
+
+
Acetophenone
As shown above, electron withdrawing (  R) effect of acyl group decreases the electron density
from the aromatic ring making it less susceptible for attack to an electrophile (Br + in bromination
reaction).
Also, acyl group, by resonance, brings positive charge on ortho and para positions, hence,
meta-positions are relatively richer in electrons compared to the ortho/para positions. Hence,
electrophile, if attacks at all, it attacks at meta positions only.
1286.The mechanism of substitution reaction is:
N
Cl
–
OCH3
–
N
l l
OCH3
Cl
+ Cl–
N
OCH3
635
Solutions
On the basis of above mechanism, it can be concluded that the order of reactivity will be:
A < C < B.
The reason for the above order are : For the pyridinium salt, one of the resonance forms bear no
charge. This will be a stable, neutral species and the activation energy is, accordingly, low relative
to others A and C.
N+
OCH3
OCH3
N
Cl
Cl–
–
–
–
+ OCH3
CH3
Cl
+
+
N
Cl
CH3
N
Cl
CH3
CH3
stable, neutral
species
OCH3
+
N
OCH3
CH3
In the case of 3-chloropyridine, the negative charge on intermediate can’t be delocalized,
therefore, this reaction would be expected to be slower than 4-chloropyridine where the negative
charge is delocalized.
–
Cl
–
+ OCH3
Cl
OCH3
+
N
N
Cl
Cl
OCH3
–
N
N
–
OCH3
Cl
OCH3
Cl
Cl
OCH3
–
–
OCH3
N
l l
N
–
N
(more stable)
H
+
1287.
O
l
l
+
C6H6
+
H+
O—H
OH
+
HO
HSO4–
+
–
HSO4
H+
–H2O
+
H
OH
636
Problems in Chemistry
O
+ CH3COCl
1288.
HO
AlCl3
CN
NaH
CH3CH2Br
HCN
O
NH
—NH2
O
O
CN
O
LiAlH4
CH3COCl
H
CH3O
N
l l
1289.
O
H
CH3
CH3O
CH3O
C
H
CH3O
H
CH3
CH3O
+
NH
CH3O
HN
CH3O
H
CH3
+
OH2
l
l
H
CH3
CH3O
+
–H+
H
1290.(a)
(b)
(c)
(d)
Product
NH
CH3O
CH3
Antiaromatic, fully conjugated, planar with 4π-electrons.
Aromatic, planar, fully conjugated, 6π-electrons (two from ring oxygen).
Non-aromatic, planar fully conjugated, 7π-electrons (one from radical).
Non-aromatic, not fully conjugated.
Cl
Cl
+
1291. (a)
O2N
OH
HN
CH3O
(b)
NO2
OCH3
H+
637
Solutions
O
(d)
(c)
HO3S
O
O2N
NO2
OCH3
(f)
(e)
O
COOH
(b) Br
1292. (a)
—NO2
O
COOH
O2N
O
NO2
NO2
NO2
Br
NH2
(d)
(c)
+
+
NH2
OCH3
SO3H
NH2
NO2
(f)
(e)
OCH3
NO2
1293.The σ-complexes formed by attack at ortho, para and meta positions are as follows:
CH3
CH3
HNO3
CH3
+
NO2
H2SO4
+
2°
+
NO2
H
NO2
H
2°
CH3
H
3°
638
Problems in Chemistry
CH3
CH3
CH3
+
para
attack
+
+
2°
H
H
NO2
CH3
meta
H
NO2
3°
NO2
CH3
CH3
+
+
NO2
attack
2°
NO2
H
NO2
H
2°
H
+
2°
2°
1294.(a) CH 3Cl/AlCl 3 , NBS/hν, CH 3ONa,
(b) CH 3COCl/AlCl 3 , Fe(Hg) /HCl, Na/NH 3 /EtOH,
(c) H 2SO 4 /H 2O – /heat, KMnO 4 /NaOH/heat,
(d) CH 3 (CH 2 ) 2 COCl/AlCl 3 , NBS/hν, CH 3ONa, Rh/H 2 /∆.
1295.  B(OH) 2 is a meta directing group due to following resonance phenomena:
–
OH
HO
HO
B
–
B
OH
HO
–
B
OH
HO
+
–
B
OH
+
+
The above resonance structures reveals that  B(OH) 2 withdraw electrons from ortho/para
positions deactivating the ring for attack of electrophile. Even, if the compound is subjected to
electrophilic attack, it occurs at meta position.
CH3
1296. (a)
Cl2
CH3Cl
AlCl3
AlCl3
CH2Br
NaN3
SN2
NBS
hν
Cl
Product
Cl
Br
SO3H
(b)
H2SO4/H2O
Br2
CH3COCl
disulphonation
FeBr3
AlCl3
Product
639
Solutions
H
–
S
OH
E2
C6H5—C—O—S—
1297. C6H5—CH2—Br
+
C6H5CHO +
CH3
H
O
S
CH3
+
1298.
+ H+
+
–H+
Product
+
CH3
CH3
H
CH3
CH3
+
1299.
–H+
+ D+
+
H
D
H
3°
CH3
D+
D
D
CH3
CH3
H
D
–H+
+
D
D
+
D
D
D+
CH3
D
D
CH3
D
D
D
–H+
H
+
D
D
H
640
Problems in Chemistry
1300.Benzoic acid.
The mechanism involves addition of electrons to form a radical anion in the first step as:
COOH
COOH
COOH
COOH
l
l
Na
EtOH
+e
l
–
–
l l
COOH
COOH
–
l l
EtOH
Stabilized by
electron withdrawing group
O
O
1301.
+
O
AlCl3
O
Br2
OH AlCl
3
H2O
O
O
O
O
C
OH
OH
Zn(Hg)
HCl
CH3Li(2 mol)
H2O
Br
Br
O
Br
Br
NH2OH
Product
Br
NNH2
1302. (a)
(b)
(c)
(d)
Br
CN
O
OH
enantiomers
Cl
641
Solutions
Br
Br
NO2
Br2
1303. (a)
Br
HNO3
H2SO4
CH3COCl
FeBr3
AlCl3
O
O
Br
Br
Br
N2+Cl–
OH
NO2
NaNO2
H2O/∆
H+
HCl/0°C
H3
O+
EG
H+
Fe
HCl
O
O
O
O
O
O
EG
H+
(b)
HNO3
H2SO4
(o/p directing)
(Alternatively)
Zn(Hg)/HCl
H2SO4
H 3O +
O2N
O
Ac2O
O2N
O2N
O
O2N
(b)
O
CrO2Cl2
HNO3
1304. (a)
O
O
+
(c)
(Etards reaction)
(d)
N
642
Problems in Chemistry
NO2
Cl
Br
1305. (a)
NO2
(b)
O2N
NO2
OCH3
(c)
(d)
COOH
(e)
OH
Ph
1306. (a)
D
D
NH2
D
(b)
OH
HO
OH
D
D
(c)
D
OH
OH
CF3
O
(e)
(d)
CHO
CHO
N
+ CH3OH
(f)
OCH3
1307.(a) (CH 3 ) 2 CHCOCl/AlCl 3 , Ph 3 P==CH 2 ,
(b) O 3 /Zn–H 2O, Zn(Hg) /HCl,
(c) Br 2 /FeBr 3 , HNO 3 /H 2SO 4 , Mg/ether/oxirane/H 3O + , PCC/CH 2Cl 2 ,
(d) Alternative : CH 3Cl/AlCl 3 , H 2SO 4 /HNO 3 , NBS/hν, Mg/ether/CH 2O, PCC/CH 2Cl 2 .
1308.CH 3OCH== PPh 3 C 6 H 5CH==CHOCH 3 C 6 H 5CH 2CHO
A
B
C
1309.Only B, D and G are formed. In order to rationalize, we need to see the sigma complex in each case.
In sigma complex leaging to C, E, F, H, one of the resonance form places the positive charge on C1 ,
which is the carbon bearing nitro group. These products are then all disfavoured.
643
Solutions
NO2
NO2
NO2
σ-complex of D
σ-complex of G
NO2
+
+
O2N
σ-complex of B
NO2
+
In none of the resonance forms of the above σ-complex it is possible to place positive charge (+) on
carbon bearing  NO 2 group.
Sigma complex of remaining four places positive charge on C-bearing  NO 2 in one of their
resonance form:
NO2
NO2
+
O2 N
NO2
O2 N
O2 N
+
+
σ-complex of C
NO2
O2 N
+
not favoured resonance form
O
O
O
O
1310. (a)
H
O
H
EG
H+
NH3
Pd/C
H2
∆
NH2
H
Product
CF3
CF3
(b)
H3O+
O
O
CF3
HNO3
Fe/HCl; NaNO2
H2SO4
HCl/0°C
NO2
C6H5N(CH3)2
N2+Cl–
Product
644
Problems in Chemistry
1311.Br 2 + FeBr 3 → Br + + FeBr 4–
OH
OH
OH
CH3
H3C
CH3
H3C
H3C
CH3
+ Br+
+
Br
SO3H
O
S
l
l
O
OH
Br
+ HBr + SO3
Br–
1312.(a) I, (b) I, (c) I, (d) I, they are all aromatic.
1313.(a) III < II < I, (b) II < I < III, (c) III < II < I.
NO
NO
+ E+
1314.
meta
attack
+
+
+
H
E
+
NO
E+
NO
NO
l
l
H
H
E
E
O
N
+N
O
+
para
attack
+
E
H
H
E
H
E
σ-complex is stabilized by +R effect
of —NO functional group.
1315. (a)
NBS
HNO3
+ CH3CH2Cl
H2SO4
hν
NO2
C2H5ONa
C2H5OH
(b)
Cl AlCl
3
+
N 2H 4
NaOH/∆
O
O
H2SO4
Product
Product
645
Solutions
CH3
(c)
CH3
NO2
H2SO4
Cl2
FeCl3
AlCl3
+ CH3Cl
CH3
HNO3
Cl
Cl
COOH
NO2
KMnO4
NaOH/∆
Cl
OH
CH3
OH
CN
1316. (a)
(b)
CH3
OH
(c)
O
OCH3
(e) OH—
(d)
H2SO4
CH3Cl
AlCl3
1317. (a)
(b)
NaOH
fusion
SO3/∆
HNO3
Br2
H2SO4
FeBr3
N
N
H3O+
Product
SnCl2
HCl
O
(c)
+
Cl
AlCl3
N2H4
NaOH/∆
Br2
FeBr3
Br
NBS/hν
C2H5ONa
C2H5OH
Product
646
Problems in Chemistry
3
1318.
2
O
+
E+
H
attack at C-2
E
O
1
H
H
+
E
O
E+
attack at C-3
E
E
H
H
E
O
+
Three resonance structures
+
O
O
+
Two resonance structures,
less probable.
Therefore, electrophile is most likely to attack at C–2 position.
Zn-HCl
1319. (a)
N2+Cl–
NHCOCH3
NO2
CH3COCl
H 3O +
CH3Cl
NaNO2
HCl/0°C
AlCl3
H 2O
∆
Product
O
HCl
H 3O +
Na/NH3
EtOH
(b)
O
OCH3
OCH3
CH3
CH3Cl
Na/∆
Cl
Product
CH2
O
PPh3
OH
NBS
hν
(c)
Product
Mg
ether
CH3I (excess)
CH2O
H 3O +
PCC
CH2Cl2
NH2
H
NH3
Pd-C/H2
O
647
Solutions
1320.(a) II, (b) I.
1321.(a) III < I < II, (b) I < II < III, (c) I < II < III.
Cl
1322. (a)
+
AlCl3
Zn(Hg)
HCl
O
HNO3
(b)
Product
CH3COCl
H3O+
H2SO4
NBS/hν
CH3CH2ONa
CH3CH2OH
CHO
H2O
—
H2N—
CHO
Cl
CrO3
Ac2O
1323.
Product
AlCl3
NHCOCH3
Zn
HCl
H2SO4
CH3Cl
EG
O
H+
or AlCl3
O
O
O
Product
CH
H 3O +
Zn
H2/Pd
(CH3)2NH
HCl
N
CH3
CH3
CH3
N
O
CH3
CN
CN
1324.
O
O
CHO
CN
+
E+
ortho attack
+
H
H
E
E
648
Problems in Chemistry
CN
+
CN
H
H
E
+
E
destabilized by CN
CN
CN
+
E+
meta
attack
+
H
E
H
E
+
Meta attack of E avoides the electron withdrawing effect of CN group as positive charge can’t
interact directly with CN group. Therefore, meta attack is favoured over ortho/para attack of
electrophile.
1325. (a)
(b)
N
A
H
B
MeO
COOMe
OMe
COOMe
O
1326.
+
l l
NH
H
N
O–
NH2
NH2
OMe
OMe
COOMe
N
COOMe
H+
–H2O
N
NH2
NH2
H
OMe
OMe
H+
COOMe
COOMe –NH3
N
+
NH3
OH
+
N
649
Solutions
OMe
OMe
COOMe
H
COOMe
H+
–H+
N
N+
H
OMe
COOMe
+
+ NH4
NH
3
l l
N
H
CH2O
Mg
PCC
1327.(a) PhBr → → → Product
Et 2O
CH2Cl 2
NBS/hν
CCl 4
aq. NaOH
PCC
(b) Ph  CH 3 → → → Product
or
CH2Cl 2
CrO3
Ph  CH 3 → Ph  CHO
Ac2O
(Etard’s reaction)
OH
(c) Ph—OH
CHCl3/NaOH
Reimer-Tiemann Reaction
CHO
OH
OH
(d)
H3O+
Reflux
CH3COCl
H2SO4
HO—
O
SO3H
OCH3
OCH3
(e)
CH3
H2/Ni
Heat
O
OH
H3O+
Reflux
PCC
CH2Cl2
O
Cl2
EtONa
EtOH
heat
Cl
Product
650
Problems in Chemistry
OH
OH


H2SO4
PhMgBr
H3
PCC
HCN
(f) Ph—CHO → →
→
→
Ph—C— Ph
Ph—C— Ph →
H 2O
CH2Cl 2


COOH
CN
O+
O
O
(g)
+
AlCl3
O
Zn(Hg)
HCl
COOH
O
COOH
HF
O
(h)
H+
+
NHNH2
Heat
N
O
H
COOH
1328. (a)
(b)
(c)
Br
(e)
(d)
(f)
Br
O
(g)
COOH
CN
Br
(h)
SO3H
651
Solutions
O
O
O
1329. (a)
AlCl3
+
Cl2
Cl
AlCl3
Cl
Product
HNO3, H2SO4
N 2H 4
OH–/heat
Cl
OH
Br
Mg
Et2O
Br2
(b)
FeBr3
(i) CH3CH2CHO
H2SO4
O+
heat
(ii) H3
O
H3O+
Product
NaBr
mCPBA
Br
+
NH3
1330. (a)
(b)
(c)
+
Br
NO2
Br
NO2
1331.
HNO3
H2SO4
Fe
HCl
Ac2O
HNO3
H2SO4
H 3O +
NaNO2
HCl/0°C
—N2+Cl–
O2N—
I
H2SO4
SO3
NaOH
fusion
H 3O +
OH
CCl4
OH
H 3O +
NaOH
COOH
(I)
Product
652
Problems in Chemistry
O
1332. (a)
O
Br2
CrO2Cl2
Ac2O
(Etard’s Reaction)
MeO
FeBr3
MeO
MeO
Br
(only product)
Ph3P
CH2
MeO
Br
O
Ph3P
(b)
CH2
Product
1 equiv.
O
N
(c)
(d)
C6H6
CH3COCl
OH
LiAlH4
H2NOH
AlCl3
NBS
hν
NaOH
PCC
CH2Cl2
CH3CH2CH
PPh3
Product
Product
1333. (a) Non-aromatic, (b) Aromatic, (c) Aromatic, (d) Nonaromatic, (e) Aromatic, (f) Aromatic.
1334. Since, both rings are aromatic with the following resonance structure:
⊕
s
both the rings are now aromatic hence double bond character between the
rings are very less, therefore less barrier to rotation,
1335. (a) Deactivated, (b) Deactivated, (c) Activated, (d) Activated, (e) Deactivated, (f) Activated.
1336. (a)
Cl2
AlCl3
H2SO4
HNO3
Product
653
Solutions
Br
O
Br2
(b)
Zn(Hg)
Cl
FeBr3
NBS/hν
CCl4
HCl
AlCl3
Product
O
H2SO4
SO3
Cl
(c)
AlCl3
Zn(Hg)
HCl
Product
O
O
1337. (a)
(b)
(c)
O2N
Br
O
O
O
COOH
(e)
(d)
(f)
CH3
NO2
CHO
NO2
NH2
OH
O
1338. (a)
(c)
(b)
NO2
O2N
Br
O
NHCH2CH3
O
N
O—C—Ph
NO2
O2N
(d)
(f)
(e)
Cl
NO2
CF3
654
Problems in Chemistry
Cl
(h)
(g)
HO—
OCH3
1339. O2N
O
l
l
NHCH3
–
+ O
N
O
O2N
O
CH3
+N
+
N
O–
NO2
N—CH3
O
H O–
–H+
+
O2N
N
O–
O–
NO2
NO2
O–
N
O2N
NO2
H2O
Product
NO2
1340.(a) Non-aromatic, (b) Non-aromatic, (c) Antiaromatic,
(d) Antiaromatic, (e) Antiaromatic, (f) Aromatic,
(g) Aromatic, (h) Aromatic.
1341.Since, double bond is in resonance and involved in aromaticity as:
s
1342. (a) C6H6
⊕
CH3Cl (excess)
AlCl3
Here, both the rings are aromatic, therefore, this resonance form is
a major contributor. Due to this resonance structure the double
bond between the two rings has acquired significant single bond
character giving low energy barrier to rotation.
CH3—
—CH3
KMnO4
NaOH/Heat
H3O+
Product
655
Solutions
NO2
HNO3
(b)
NHCOCH3
NH2
CH3COCl
excess
Zn
HCl
H2SO4/heat
NHCOCH3
NO2
H3O+
CF3CO3H
NaNO2
HBF4
heat
NHCOCH3
NO2
HNO3
Sn
HCl
H2SO4
F
F
CHO
HNO3
H2SO4
CO
HCl/AlCl3
(c)
NHCOCH3
NH2
NO2
NaNO2
HCl
Fe/HCl
NO2
NHCOCH3
HNO3
H2SO4
(d)
CH3
Fe
HCl
Product
NH2
NHCOCH3
H2SO4
CH3COCl
H2O
heat
HNO3
H2SO4
SO3
SO3H
N2+Cl–
OH
NO2
NH2
NO2
H 2O
boil
NHCOCH3
NO2
NaNO2
NO2
dil. HCl
HCl/0°C
OH
SO3H
NH2
Sn
HCl
NHCOCH3
(e)
HNO3
H2SO4
Fe/HCl
CH3COCl
NHCOCH3
H2SO4
SO3
dil. HCl
CF3CO3H
Product
SO3H
656
Problems in Chemistry
O
CHO
(CH3)2C
HF
CO/HCl
AlCl3
(f)
CH2
O
CH
HNO3
H2SO4
EG/H+
O
N 2H 4
O
CH
H 2O
boil
dil. HCl
NaOH/heat
SO3H
(b)
(c)
Br
O
O2N
O
NO2
O
Cl
NO2
NO2
Cl
(e)
+
CH3
(f)
SO3H
CH3
Cl
O2N
CH3
NHCOCH3
(h) CH3CO
Br
CH3
HCl/0°C
N
1343. (a)
(g)
NaNO2
OH
OH
(d)
Fe/HCl
COCH3
657
Solutions
O
O
O
EG
Cl2
Cl
1344.
H+
AlCl3
AlCl3
Cl
O
H3
O
O
H3O+
O+
O
O
Mg
Et2O
H
Cl
HO
O
N2H4
PCC
CH2Cl2
NaOH
heat
Product
HO
O
Cl
1345. (a)
CH3Cl
AlCl3
O
AlCl3
HNO3
Zn
HCl
H2SO4
NO2
NH2
NaNO2
H2O
HCl/0°C
boil
OH
CH3
(b)
COOH
CH3Cl
HNO3
KMnO4
AlCl3
H2SO4
NaOH/heat
NO2
H3O+
NH4HS
NO2
Product
658
Problems in Chemistry
O
O
COOH
1346.
(a)
A
C
B
D
NO2
NO2
Co(OAc)2
(b)
AcOH/HBr/O2
NO2
No reaction since there is no α—H
to phenyl ring.
E
NO2
F
O
Br
1347. (a)
+ Br2
FeBr3
Mg
Et2O
H3O+
OH
PCC
CH2Cl2
Product
O
(b)
(CH3)2CHCOCl
AlCl3
HNO3
Zn
H2SO4
HCl
NH2
Product
NaNO2
CuCN
LiAlH4
HCl/0°C
CN
Br
Br
(c)
Br2
FeBr3
H2SO4
HNO3
SO3
H2SO4
SO3H
NO2
dil. HCl
Fe/HCl
NaNO2
HCl/0°C
SO3
NaBF4
∆
Product
659
Solutions
NHCOCH3
HNO3
(d)
H2SO4
Fe
HCl
CH3COCl
HNO3
Cl
Zn/HCl
O
AlCl3
H2SO4
NO2
NO2
NHCOCH3
NO2
CF3CO3H
Sn/HCl
NO2
dil.
HCl
NH2
NO2
1348.(a) Non-aromatic since, it has a sp 3 carbon.
(b) Aromatic.
OCH3
NO2
F
(b) CH3O—
1349. (a)
O
C
—CH3
(c)
NO2
CH3
Br
(d)
—OH
(e)
F
1350. (a)
(b)
O
OMe
O
OMe
O
OMe
OMe
(c)
CH2Ph
(Minor)
O
(Major)
CH2Ph
660
Problems in Chemistry
CH3
1351. (a)
(b)
CH3
Br
CH3Cl
H2SO4
Br2
AlCl3
SO3
FeBr3
H2SO4
NaOH
SO3
fusion
dil. HCl
HNO3
H2SO4
SO3H
SO3H
ONa
OCH3
CH3I
Br2
FeBr3
Product
Product
NHCOCH3
HNO3
(c)
Fe/HCl
H2SO4
CH3COCl
H2SO4
dil. HCl
SO3
CF3CO3H
SO3H
SO3
1352. (a)
H2SO4
H 3O +
KMnO4
NaOH
(b)
t-BuCl
AlCl3
KMnO4
(c)
t-BuCl
AlCl3
CH3COCl
Product
H 3O +
Product
NaOH
AlCl3
Product
OMe
(d)
OMe
t-BuCl
AlCl3
HNO3
H2SO4
Product
Cl
NH3
(e)
H2S
NO2
NaNO2
HCl/0°C
CuCN
H2SO4
CH3OH
H2O
H+/heat
Product
Product
661
Solutions
(f)
CH3Br
(g)
HNO3
NaNO2
Fe/HCl
Heat
Product
HBF4/0°C
AlCl3
H2SO4
Br2
HNO3
Cl2
NH3
NaNO2
H 2O
H2SO4
AlCl3
H 2S
HCl/0°C
boil
FeBr3
Product
O
(h)
+
Cl
AlCl3
HNO3
N 2H 4
H2SO4
NaOH/heat
Product
O
(i)
+
Cl
HNO3
AlCl3
Zn-HCl
H2SO4
NaNO2
H 2O
HCl/0°C
boil
CH3I
Br
Br
1353. (a)
Br2
H2SO4
FeBr3
SO3
Br
O
NaOH
Product
O
Cl
AlCl3
SO3H
SO3H
dil HCl
Zn(Hg)
HCl
Product
O
Cl
(b)
Br2 (excess)
Zn(Hg)
FeBr3
AlCl3
Product
HCl
Cl
O
(c)
Cl2
AlCl3
Cl
SO3
H2SO4
dil. HCl
AlCl3
Zn(Hg)
HCl
(d)
H2SO4
HNO3
Fe/HCl
O
Ac2O
HNO3
H2SO4
dil HCl
Mg
D 2O
Ether
NaNO2
HCl/0°C
CuCN
Product
Product
662
Problems in Chemistry
NO2
HNO3
(e)
Ac2O
Te/HCl
H2SO4
CH3I
(f)
AlCl3
Br2
(g)
Fe
(h)
H2SO4
SO3
NaOH
HNO3
H2SO4
H 2S
CH3I
Heat
Ac2O
NH3
CH3I
SO3
Br2(excess)
AlCl3
H2SO4
FeBr3
CH3CH2COCl
AlCl3
dil HCl
NaNO2
HCl/0°C
HBF4
Heat
NBS
CH3—
—OCH3
Cl2
FeCl3
dil HCl CF3CO3H
HCl(dil)
hν/CCl4
NaOH
Product
aqueous
Product
Product
Cl
CN
1354. (a)
(c)
(b)
Br
NO2
NO2
Cl
OEt
NH2
(f)
(e)
(d)
O
(like Fries rearrangement)
O
H
A:
1355.
O
Cl
O
and AlCl 3 ; B:
O
D:
Ag 2O/H 2O;
E:
O
Ph
MeMgBr/H 2O; C:
Product
H 2SO 4 /H 2O;
663
Solutions
NHOAc
CH2CH3
1356. (a)
(b)
O
CH3
C(CH3)3
Cl
OH
(c)
(e) O2N—
(d)
Br
CH3
CH(CH3)2
NO2
Cl
HNO3
1358. (a)
COOH
Cl
Li/Et2O
NH3/H2S
CO2/H2O
NaNO2
HCl/0°C
H2SO4
CuCN
NO2
NO2
H3O+
H2O
Product
Cl
Mg/Et2O
(b)
CO2
CH3I
H2O
AlCl3
KMnO4
NaOH/heat
OCH3
Cl
HNO3
(c)
H2SO4
Product
CH3ONa
CH3OH
Fe/HCl
O2N
OCH3
H2O
boil
HO
F
(d)
HNO3
H2SO4
OCH3
CH3ONa
CH3OH
NaNO2
HCl/0°C
Fe/HCl
NaNO2
HCl/0°C
O2N
CuCN
OCH3
NaOH
H2O
H3O+
H2O
HOOC
664
Problems in Chemistry
Br
O
1359. (a)
Br
O
(b)
—OCH3
—NO2
OCH3
(c)
—OCH3
O2N
F
Cl
NH2
(c)
(b)
1360. (a)
MeO
NO2
Br
SO3H
OMe
NO2
OMe
OMe
(d)
(e)
O
I
1362.(a)
(b)
(c)
(d)
1363.A:
Aromatic,
Not aromatic, it has a sp 3 carbon which prevent delocalization,
Aromatic,
Not aromatic.
KMnO 4 /NaOH/heat; B: SOCl 2 ; C: Et 2 NCH 2CH 2OH; D: Na 2S.
OCH3
OCH3
O
O
1364.
HO
O
HO
A
O
H
OCH3
O
H3CO
B
O
OH
O
C
OH
COOH
O
H3CO
D
HO
OH
E
O
COOH
HO
F
COOH
665
Solutions
—NO2
1365. (a) Cl—
(b) O2N—
Cl
(d) O2N—
—CH3
(c) H3CO—
NO2
Cl
Cl
Cl
Cl
Cl
1366.
Cl2
AlCl3
N2H4/OH–
O
AlCl3
C
O
Cl
HNO3
Cl
NH4HS
H2SO4
NO2
NH2
666
Problems in Chemistry
CARBOHYDRATES, AMINO ACIDS AND POLYMERS
1367. (a) A reducing sugar reduces Tollen’s reagent, Fehling solution as well as Benedict’s solution
while non-reducing sugar donot give positive test with any one of these reagents.
(b) α-D-glucopyranose has hemiacetal structure and reduces Tollen’s reagent, Fehling solution,
Benedict solution while methyl-α-D-glucopyranoside has acetal structure and are
non-reducing, therefore fail in the above tests.
—C—O
—C—O
O—H
O—R′
C
—C
Alkyl group
C
—C
R(=H or CH2OH)
α-D-glycopyranose
give +ve Tollen’s test
R
If R′ = CH3, methyl α-D-glycopyranoside
do not gives Tollen’s test
1368. Maltose and sucrose are the two most common disaccharide.
H+
Maltose + H 2O →
2 moles of glucose (C 6 H12O 6 )
H+
Sucrose + H 2O → Glucose + Fructose
1369. Cellulose and starch are the most common polysaccharides. Both on hydrolysis produces many
molecules of glucose, i.e., both are glucose polymers.
1370. Lactose is a disaccharide. Lactose is a reducing sugar and it reduces Tollen’s reagent, Fehling’s
solution and Benedict solution. Lactose on hydrolysis gives D-glucose and D-galactose.
1371. Epimers are the diastereomers that differs in configuration at only one stereocenter. The above
two diastereomers differ in configuration only at C-2 and this can be confirmed by their reaction
with phenyl hydrazine.
CH==NNHPh
C==NNHPh
HO
H
H
OH
H
OH
D-glucose + C6H5NHNH2
C6H5NHNH2 + D-mannose.
CH2OH
Same phenyl osazone formation confirm that only C-2
has different configuration in D-glucose and D-mannose.
1372.
HO
H
H
H
CHO
H
OH
OH
OH
CH2OH
H2
Ni
HO
H
H
H
CH2OH
H
OH
OH
OH
CH2OH
H2
Ni
HO
HO
HO
H
CHO
H
H
H
OH
CH2OH
667
Solutions
C6H5NHNH2
CH==NNHC6H5
C6H5NHNH2
CH==NNHC6H5
C==NNHC6H5
C==NNHC6H5
H
H
H
OH
OH
OH
Different osazone
HO
HO
H
CH2OH
H
HO
HO
H
CHO
OH
H
H
OH
CH2OH
H
H
OH
CH2OH
CH==NNHC6H5
C6H5NHNH2
HO
HO
H
C==NNHC6H5
H
H
OH
CH2OH
Same osazone as obtained from B.
C
1373. Amino acid exist as dipolar ion as :
 +
—C— NH 3

COO –
(Dipolar ion)
(Zwitter ion)
Due to its existence in the form of dipolar ion, there exist a very strong intermolecular attraction
which is responsible for its high melting point.
1374. (a) pH = 2 indicate highly acid solution and cationic form will predominate :
+
H 2 N—CH—COO –
H 2 N—CH—COOH
H 3 N—CH—COOH
pH = 12
pH = 2

←

→

H 3C—CH—CH 3
CH 3 —CH—CH 3
CH 3 —CH—CH 3
(b) At pH =12, anionic form predominate.
1375. (a) In strongly acidic solution —NH 2 gets protonated and cationic form will predominate :
HOOC—CH 2CH 2 —CH—COOH

NH 3
+
(b) In strongly basic solution —COOH will get deprotonated and anionic form will predominate:
−
OOC—CH 2CH 2 —CH—COO –

NH 2
668
Problems in Chemistry
(c) At isoelectric pH, zwitterion will predominate :
HOOC—CH 2CH 2 —CH—COO – .

NH 3
+
1376. Isoelectric point is the pH at which zwitter ion has its maximum concentration. Glutamic acid is a
dicarboxylic acid, its zwitter ion undergo deprotonation as:
HOOC—CH 2 —CH 2 —CH —COO –

NH 3
+
−
OOC — CH 2CH 2 —CH — COO – + H +

NH 3
+
In order to suppress the above ionization so that zwitter ion predominate, higher [H + ]will have to
be maintained and therefore isoelectric pH will be lower. Glutamic is a monocarboxylic acid no
extra H + is required to suppress above type of ionization and hence isoelectric pH is higher.
+
N H 3Cl –
NH 2


1377. A : H —C— COOH B : H —C— COOCH 3


CH 2OH
CH 2OH
−
+
NH 3 Cl

C : H—C— COOCH 3

CH 2Cl
NH 2

D : H—C— COOH

CH2Cl
NH2

Na(Hg)
D → H—C—COOH
dilute H+

CH 3
O

H3O+ /Heat
1378. CH 3 —C —H + NH 3 + HCN → CH 3 —CH—CN →
CH 3 —CH—COO –
H
O
2


NH 2
NH 3
+
Alanine
1379. (a) A step growth polymerization results from reaction of two functional group releasing out
some small molecules as H 2O, HCl, CH 3OH as biproduct.
O
(b) CH3O—C—
O
—C—OCH3 + HO—CH2—CH2—OH
Dimethylterphthalate
Ethylene glycol
O
O
Heat
—OCH2CH2O—C—
—C—
Poly(ethyleneterphthalate)
Decron
+ 2CH3OH
669
Solutions
1380. Teflon is a polymer of tetrafluoroethylene and synthesized as follows :
O
O
O




Heat
Ph—C—O—O—C—Ph → 2Ph—C—O • ( ≡≡RO • )
Benzoyl peroxide
Heat
RO • + CF2 == CF2 → RO—CF2 —CF2• (Initiation)
ROCF2 —CF2• + nCF2 == CF2 → RO —
[ CF2 —CF2 —CF
]n 2 —CF2• (Propagation)
RO—
[ CF2 — CF2 —
]n CF2 —CF2• + RO • → RO —[ CF2 —CF2 ]—
n CF2CF2 —OR
RO—[CF2 —CF2 —
]n CF2 — CF2• + ROCF2 —CF2• → RO—[CF2 —CF2 —
]n +2 OR
2RO—
[ CF2 —CF2 —
]n CF2 —CF2• → RO—
[ CF2 —CF2 —
]2n +2OR
O
O


Heat
1381. (a) n HO—C —( CH 2—
) 4 C — OH + nH 2 N—( CH 2 )—
6 NH 2 →
Adipic acid
O
O


—[ C —(CH 2 —
) 4 C—NH —
( CH 2 —
)6 NH —
] n + H 2O
Nylon 6,6
(b) Due to extensive intermolecular H-bonding among the polymer chains, nylon possess very
high melting point.
MISCELLANEOUS
O
O
1382. (a)
O
NH2
(b) H
(c) CH3CH2OH + H—C—H
PhNH
OH
OH
(d)
(e)
(f)
CN
O
Br
1383. (a)
O
OH
Cl
(c) CH3COCH3
(b)
O
(d) CH3
CH3
670
Problems in Chemistry
O
COOH
N
(e)
O
(g)
—N
OH
+
(f)
CH3
H
N—
—OH (h)
H
NH2
O
Br
1384. A :
—C
NH2
B:
Br
C:
N
D:
O
H—N
O
O
1385. (A)
N
O
Ph3P
(B)
O
O
O
(C)
CHCH3
OH–
Aldol
OH
H+
Heat
671
Solutions
H3C
H3C
N
COOCH3
1386.
O
H3C
N
N
COOH
C6H5
COOH
O
OH
O
(C)
(B)
Cocaine (A)
O


CH2N2
PCC
PCC
1387. C 6 H 5CH 2OH → C 6 H 5CHO → →
C 6 H 5  C  CH 3 →
(i) CH3MgBr
(benzyl alcohol)
CCl 4
∆
CCl 4
(ii) H3O+
O
CH3
X
OH
X
H 3O +
CH3(CH2)3CH2MgBr
O
OH
LiAlH4
1388.
PCl3
H2O
MeO
Mg
ether
MeO
—OMe
OH
—OMe
H3O+
(MgCl)+
O
MeO
MeO
—OMe
H+/Heat
MeO
(Dimestrol)
672
Problems in Chemistry
OH
O
O
1389.
A
C
OH
B
O
O
O
O
Br
D
E
F
OH
Br
G
O
CHO
O3
+
Zn-H2O
H
O
O
I
Mg
O
(i) 
ether
O+
PCl
C H ONa
3
2 5
1390. (a) PhCH 2 Br → PhCH 2 MgBr → Ph—CH 2CH 2CH 2OH →

(ii) H3
NaBH4
OCH3
(CH3COO)2Hg, CH3OH
C2H5OH
Ph—CH2 —CH
CH2
673
Solutions
Br
(b)
HC
C
C–Na+
CH
B 2H 6
NaOH/H2O2
OH
CHO
(i) CH3MgBr
PCC
CCl4
Product
(ii) H3O+
OH
CH2CH3
OH
(b)
1391. (a)
(c)
O
OH
OH
O
CH3
Br
C2H5O
1392.
A
O
OC2H5
Ph
C
B
D
COOH
E
Br
1393. (a)
OH
(b)
CN
OH
H
H
H
(Enantiomeric)
Ph
OH
Minor
Major
Ph
(c)
+
CH3
(d)
OEt
OH
CH3
(e)
O
674
Problems in Chemistry
1394.
OH
H+/∆
OH
B2H6/H2O2
–H2O
NaOH
1
3
Cold, dilute, alkaline
KMnO4
OH
Regioselective hydroboration
followed by oxidation.
OH
2
O
O
LiAlH4
EG/H+
1395.
O
OEt
O
O
O
O
O
H 2O
OEt
O
PBr3
Br
OH
O
Product
HCl
O
PhONa
O
1396. Enol 2 is thermodynamically more stable than 3 because the alkene in this tautomer is conjugated
with aromatic ring as well as with carbonyl group. The C==Cdouble bond in enol 3 is isolated and
the stability afforded to 2-through resonance is not available to 3.
675
Solutions
O
O
H
O
O
O
H 3O +
CH3
OH
CH2
H 2O
2 (thermodynamically more stable)
1
H+ H
+
O
–H3O
O
3 (less stable)
O
OH
+
H
2
1397.
X
O
H
H
OH–
O–
O–
most stable
Y
l l
This enol is highly disfavoured-the individual 2pz orbital of the
double bond are orthogonal and consequently, the pπ-bond can’t form
(Berdt’s rule)
OH–
O–
most stable enolate anion because the double
bond is in conjugation with aromatic ring.
O–
Z
O–
OH–
Less stable
Most stable, because the double bond is tetrasubstituted
and consequently thermodynamically more stable than
the alternative, trisubstituted enolate anion.
1398. Here, the acidity is due to COOH groups and not due to α  H. Also, COOH is a strong
electron withdrawing functional group, distance of one COOH group from other determine
acidic strength. Larger the distance, weaker is the acid. Thus, the order of acid strength is:
C>B>D>A
676
Problems in Chemistry
COOH
1399. A : X
1. KOH, H2O
COCl
SOCl2
or (COCl)2
or PCl3 or PCl5
2. H3O+
CH3CH2COONa
Br
Product
O
H 3O +
EtONa
B: X +
Product (Through nucleophilic
addition of enolate ion)
O
N
Heat
C: X +
N
H
O
1400.
O
CH3
CH3O–
–
CH2
+ CH3O–
MeOH
O
OH
O
O
O
O
O
O
–
MeOH
O
O
–O
OH
OH
O
Heat
–H2O
O
OH
677
Solutions
CHO
CH2
CH
O
1401.
OH
OH
O
OH
O
CF3CO3H
O
CH3
N
C
A
CH3
D
B
H2SO4/H2O
E
O
O
Br OH
O
Br
OH
1403.
O
O
CN
KCN
H+/∆
O
CN
H+
O
O
1404. EtO
OEt
EtO–
EtO
s
+ EtO
EtO
O–
O
(CH3COO)2Hg
H2O
Styrene
NaBH4
(not formed)
OEt
–EtO–
B
1405. (a)
OEt
s
O
O
OEt
O
O
more acidic
less acidic
OEt
H
H
H
O
O
PCC
CCl4
O
678
Problems in Chemistry
OH
H
B2H6–H2O2
(b)
PCC
CCl4
NaOH
Br
KCN
HBr
(c)
Peroxide
O
H 3O +
H2SO4
O
OH
OH
LiAlH4
O
1406. (a)
(b)
—C
—C
CH
CH
HgSO4
—C—CH3
H2SO4/H2O
H2/Pd/BaSO4
B2H6–H2O2
NaOH
Br2
AlCl3
Product
Product
O
(c)
—C—CH3
PhMgBr
H3O+
ether
H2O
Product
(from a)
QUALITATIVE ANALYSIS
1407. A : PbCO 3 , B : PbO, C : Pb 3O 4 , D : Pb(NO 3 ) 2 , E : PbO 2 , F : PbCl 2 ,
G : PbI 2
2+
1408. A : Cu(NO 3 ) 2 ; B : NO 2 , C : Cu(OH) 2 , D : CuO, E : [Cu(NH 3 ) 4 (H 2O) 2 ]
F : [CuCl 4 ]2– .
B : 2NO 2 + H 2O → HNO 3 + HNO 2
A : ZnCO 3 ⋅ Zn(OH) 2 , B : ZnO, C : ZnSO 4 , D : Zn(OH) 2 , E : [Zn(NH 3 ) 4 ]2+
A : Fe, B : FeSO 4 , C : Fe 2O 3 , D : Fe 3O 4 , E : FeCl 3 , F : Fe(OH) 2 , G : Fe(OH) 3
SOFCl.
A : BaCO 3 , B : BaO
A : CrCl 3 , B : Na 2CrO 4 , C : Na 2Cr 2O 7 , D : NaCl, E : CrO 2Cl 2 , F : (NH 4 ) 2 Cr 2O 7
G : Cr 2O 3 .
1414. A : ZnCO 3 ⋅ Zn(OH) 2 B : CO 2 , C : ZnO, X : K 2 Zn 3 [Fe(CN) 6 ]2 , D : ZnS,
E : Zn(OH) 2 .
1416. A : CaCO 3 , B : CaO, C : CaC 2O 4 .
1409.
1410.
1411.
1412.
1413.
679
Solutions
1417.
1418.
1419.
1420.
1421.
1422.
1423.
1424.
1425.
1426.
1427.
1428.
1429.
1430.
1431.
1432.
1433.
1434.
1435.
1436.
A:
A:
A:
A:
A:
A:
A:
A:
A:
A:
A:
A:
A:
A:
A:
A:
A:
A:
A:
A:
Bi 2O 3 + PbCrO 4 , B : Bi(OH) 3 + Pb(OH) 2 , C : Bi(OH) 3 , D : PbO 2 , E : PbCl 2 , F : BiI 3
CuCl 2 , CrCl 3 , B : CuS, C : CrO 2–
4 , D : CrO 5
Cr 2O 3 , B : Al 2O 3 , C : PbCrO 4 , D : Al(OH) 3
Fe 2O 3 , B : MnCl 2 ⋅ 2H 2O, C : Fe(OH) 3 , D : MnS, E : CrO 2Cl 2 .
Pb 3O 4 , B : Fe(CrO 4 ) 2 , C : PbO 2 , D : PbS, E : Fe 4 [Fe(CN) 6 ]3 , F : PbCl 2 .
HgI 2 , B : AlI 3 , C : K 2 [HgI 4 ], D : HgO ⋅ Hg(NH 2 )I, E : Al(OH) 3
ZnO, B : Ca(NO 3 ) 2 , C : Zn(OH) 2 , D : [Zn(NH 3 ) 4 ]2+ , E : CaCO 3 .
Ba(NO 3 ) 2 , B : CaCl 2 ⋅ 6H 2O, C : CaC 2O 4 , D : BaCrO 4 , E : BaCO 3 .
Ag 2CrO 4 , B : Ag 2CO 3 , C : Ag 2O, D : CrO 5 .
Mg, B : Mg 3 N 2 , C : Mg(OH) 2 , D : NH 3 , E : HgO ⋅ Hg(NH 2 )I.
FeSO 4 ⋅ 7H 2O, B : Fe 2O 3 , C : SO 2 , D : SO 3 , E : FeCl 3 .
HgCl 2 , B : Hg 2Cl 2 , C : Hg, D : HgI 2
Bi 2S 3 , B : H 2S, C : Bi(OH) 3 , D : BiO –3 , E : BiI 3
ZnCl 2 , B : MgCl 2 , C : Mg(OH) 2 , D : ZnS, E : HCl.
Fe 2O 3 , B : MgSO 4 ⋅ 7H 2O, C : FeS, D : Mg(OH) 2 , E : S, F : BaSO 4 .
CuBr 2 , B : ZnSO 3 , C : CuS, D : Zn(OH) 2 , E : BaSO 3 , F : SO 2
Fe 3O 4 , B : AgI, C : Fe 4 [Fe(CN) 6 ]3
Ag 2CO 3 , B : CuCl 2 , C : AgCl, D : CuS, E : [Cu(CN) 4 ]3–
FeCl 3 , B : ZnSO 4 , C : Fe(OH) 3 , D : Zn(OH) 2 , E : HgO ⋅ HgSO 4
AlCl 3 , B : BaSO 4 , C : BaCrO 4 , D : Al(OH) 3 , E : CoOAl 2O 3 .
COORDINATION COMPOUNDS
1437. (i) hexaamminecobalt (III) chloride.
(ii) pentaamminechlorocobalt (III) ion.
(iii) tetraamminesulphatocobalt (III) nitrate.
(iv) potassium pentachloronitridoosmate (VI).
(v) sodium dithiosulphatoargentate (I).
(vi) potassium amminedicyanodioxoperoxochromate (VI).
(vii) pentaamminenitrito irridium (III) chloride.
(viii) potassium tetrachloroplumbate (II).
(ix) copper (II) potassium hexacyanocobaltate (III).
(x) potassium hexacyanoaurate (III).
(xi) lithium tetrahydrido aluminate (III).
(xii) sodium tetrahydridoborate (III).
(xiii) sodium hexafluoroaluminate (III).
(xiv) tetraaquadichlorocobalt (III) chloride dihydrate.
1438. (i) hexaammine chromium (III) hexaisothiocyanatochromate (III).
(ii) tetraammine copper (II) tetrachloroplatinate (II).
(iii) tetraammine platinum (II) tetrachloroplatinate (II).
(iv) hexaammine cobalt (III) tetracyanonickelate (II).
(v) tetraammine dichloroplatinum (IV) tetrachloroplatinate (II).
(vi) hexaammine cobalt (III) hexanitrocobaltate (II).
(vii) pentaamminenitrochromium (II) hexanitrochromate (II).
680
Problems in Chemistry
(viii) tetrapyridylplatinum (II) tetrachloroplatinate (II).
(ix) hexaammine nickel (II) hexanitrocobaltate (III).
(x) hexammine cobalt (II) diamminetetrachlorochromate (II).
(xi) tetraammine dibromocobalt (III) tetrachlorozincate (II).
(xii) pentaammine carbonatocobalt (III) tetrachlorocuperate (II).
(xiii) diammine silver (I) hexacyanoferrate (II).
(xiv) dichlorobis(ethylenediamine) chromium (III) tetrachloro palladate (II).
1439. (i) µ-amidobis (pentaammine cobalt (III)) nitrate.
(ii) µ-amido-µ-sulphurdioxide bis (tetraammine cobalt (II)) nitrate.
(iii) tri-µ-carbonyl bis (tricarbonyl iron (0)).
(iv) µ-amido-µ-hydroxotetraammine cobalt (III) bis(ethylenediamine) cobalt (III) chloride.
(v) µ-amido-µ-hydroxo dibis(ethylenediamine) cobalt (III) bromide.
(vi) µ-amido-µ-nitro dibis (ethylenediamine) (cobalt (III)) chloride.
1440. (i) [Co(NH 3 ) 5 Br]SO 4 (ii) [Pt(en) 2 Cl 2 ]SO 4 (iii) [Pt(Py) 4 ][PtCl 4 ]
(iv) K 3 [Fe(CN) 5 (CO)] (v) Cs[TeF5 ] (vi) [Pt(NH 3 ) 3 Br]NO 3
(vii) [Co(en) 2 Cl 2 ]Cl ⋅ H 2O (viii) [Cr(H 2O) 6 Br 2 ]Cl (ix) (NH 4 ) 3 [ZrF7 ]
(x) [Ni(NH 3 ) 6 ]3 [Co(NO 2 ) 6 ]2 (xi) [Cr(en) 2 Cl 2 ]2 [PtCl 4 ] (xii) Al[AuCl 4 ]3
(xiii) Fe 4 [Fe(CN) 6 ]3
1441. A : [Co(NH 3 ) 3 (H 2O)ClBr]Br ⋅ H 2O; B : [Co(NH 3 ) 3 (H 2O) 2 Br]Cl ⋅ Br
1442. A : [Co(NH 3 ) 5 Br]SO 4 , B : [Co(NH 3 ) 5 SO 4 ]Br, Ionization isomerism.
1443. A : [Pt(NH 3 ) 4 Cl 2 ]Br B : [Pt(NH 3 ) 4 Br 2 ]Cl 2
1444. A : [Cr(H 2O) 6 ]Cl 3 , B : [Cr(H 2O) 5 Cl]Cl 2 ⋅ H 2O, C : [Cr(H 2O) 4 Cl 2 ]Cl ⋅ 2H 2O.
1445. [Ti(H 2O) 6 ]Cl 4
1446. [CoCl 6 ]4− , octahedral, sp 3 d 2 -hybridization, an outer orbital complex.
1447. In case of [Co(NH 3 ) 6 ][Cr(NO 2 ) 6 ], cobalt will be discharged at cathode while with
[Co(NO 2 ) 6 ][Cr(NH 3 ) 6 ], chromium will be discharged at cathode.
1448. [Cr(NH 3 ) 6 ][Cr(NO 2 ) 6 ] will have lower conductivity due to larger hydrated radius.
1449. Conducting reagent : dimer 482, non-conducting reagent : monomer 241.
1450. [Pt(NH 3 ) 4 ][Pt(NO 2 ) 4 ] and [Pt(NH 3 ) 3 (NO 2 )][Pt(NO 2 ) 3 NH 3 ] only.
1451. A : [Cr(NH 3 ) 4 BrCl]Cl B : [Cr(NH 3 ) 4 Cl 2 ]Br
1452. A : [Cr(NH 3 ) 4 Br 2 ]Cl ⋅ H 2O B : [Cr(NH 3 ) 4 BrCl] ⋅ Br ⋅ H 2O C : [Cr(NH 3 ) 4 (H 2O)Cl]Br 2 .
1453. A : [Co(NH 3 ) 3 (H 2O) 2 SO 4 ]NO 3 ⋅ H 2O,
B : [Co(NH 3 ) 3 (H 2O) 2 NO 3 ]SO 4 ⋅ H 2O
C : [Co(NH 3 ) 3 (H 2O)(NO 3 )(SO 4 )] ⋅ 2H 2O D : [Co(NH 3 ) 3 (H 2O) 3 ]SO 4 ⋅ NO 3 .
1454. A : [Pt(NH 3 ) 4 Cl 2 ]Br 2 B : [Pt(NH 3 ) 4 ClBr]BrCl C : [Pt(NH 3 ) 4 Br 2 ]Cl 2
1455. A : [Co(NH 3 ) 3 (H 2O)Br 2 ]ClH 2O, B : [Co(NH 3 ) 3 (H 2O) 2 Cl]Br 2
1456. [Cr(H 2O) 5 Cl]Cl 2 ⋅ H 2O
1457. (a) Red is complement colour of green wavelength and vice-versa. Hence, ligand B is producing
smaller crystal field splitting.
(b) [CoA 6 ]3+
1459. [Cr(en) 3 ]3+
1458. [Cu(NH 3 ) 4 ]2+
1460. Due to lack of d-d-electronic transition.
−
1461. (a) CN − is a strong ligand, d 8 (Ni 2+ ) has all paired electrons in Ni(CN) 2–
4 , while Cl is a weak
ligand Ni 2+ ( d 8 ) has two unpaired electrons.
681
Solutions
1462.
1463.
1464.
1465.
1466.
1470.
(b) Ni has d 10 -configuration in Ni(CO) 4 and complex is tetrahedral.
(c) Ni in [Ni(NH 3 ) 6 ]2+ is sp 3 d 2 -hybridized.
Iodide ion being a strong reducing agent, reduces Cu 2+ to Cu + .
Weak ligand field of H 2O gives five unpaired electrons in d-orbitals of Fe in Fe(H 2O) 3+
6 and no
electronic transition from one d-orbital to other d-orbital can occur.
On the otherhand, CN − being a very strong ligand, gives a very large crystal field splitting and
electronic transition from one d-level to other require photons of high energy and absorption
occur in UV-region.
K 3 [Mn(CN) 6 ] : µ = 2.82 Bm. Octahedral, inner orbital complex.
K 2 [MnBr 4 ] : µ = 5.92 Bm, tetrahedral.
(a), (b), (d), (e)
F − is a weak ligand, forms high spin complex and CoF63− has four unpaired electrons in d-orbital.
CN − is a strong ligand and all six electrons in the d-orbital of Co 3+ in Co(CN) 3–
6 are paired.
The trans-isomer of [CoCl 2 (en) 2 ] possess a plane of symmetry and hence, does not show optical
isomerism while cis form is devoid of any symmetry, it shows optical isomerism.
Cl
NH2
H2N
Cl
NH2
NH2
Co
Co
H2N
Cl
NH2
NH2
Cl
H2N
trans [CoCl2(en)2]
(has plane of symmetry)
Cis [CoCl2(en)2]
1471. Complex [ZnA2 B 2 ]2+ will be superimposable on its mirror image due to presence of plane of
symmetry while [ZnABCD ]2+ is optically active.
A
A
2+
Zn
2+
2+
B
Zn
A
D
B
B
Possess plane of Symmetry
C
Devoid of Symmetry
(Optically active)
1472. [Pt(NH 3 ) 2 Cl 2 ] is a square planar complex, therefore, show cis-trans isomerism as shown below :
Cl
NH3
Cl
Pt
Cl
Pt
NH3
Cis
NH3
H3N
Cl
Trans
682
Problems in Chemistry
On the other hand [Zn(NH 3 ) 2 Cl 2 ] is tetrahedral and all the positions around metal are equivalent,
therefore, no cis-trans isomerism is observed.
1473. In octahedral complex [CrL 6 ]3+ , Cr 3+ has 3d 3 configuration. The splitting pattern and electron
filling in d-orbital of Cr 3+ can be shown as:
∆0
3d3
Cr3+
The nature of ligand will affect the ( ∆ 0 ) value but not the electronic configuration in the lower
level of d-orbital. Therefore, magnetic properties, which depends on number of unpaired
electrons with central metal, will not be affected by identity of ligand.
1474. “a ” and “c” are identical and both are trans isomers. “b” and “d ” are pair of enantiomers and both
are cis-isomer.
1475. The complex is [CoCl 6 ](C 2 H10 N 2 ) 2 and it is octahedral.
Cl
Cl
4–
Cl
Co2+
Cl
Cl
Cl
1476. The complex has formula [Ti(H 2 NCONH 2 ) 4 I 2 ]I.
1477. (a) Cr 3+ hydrolyses in water according to the following reaction:
Cr 3+ + H 2O
Cr(OH) 2+ + H +
In the above reaction, H + is produced which impart acidic nature to the solution.
(b) As OH – is added, a gelatinous precipitate of Cr(OH) 3 is produced at first instant which
dissolves on adding excess of OH – due to formation of complex [Cr(OH) 6 ]3– as:
3OH–
Cr 3+ + 3OH – → Cr(OH) 3↓ → [Cr(OH) 6 ]3– ( aq )
1478. (a) tetraoxalatozirconate (IV) ion
(b) diaquatetrachlorocupperate (II) ion
(c) amminetrichloroplatinate (II) ion (d) tetracyanotetrahydroxomolybedate (IV) ion
1479. Electronic configuration of Fe(II) is
2+
Fe
:
3d
6
4s
0
4p
0
4d
0
683
Solutions
Here Fe 2+ is readily converted into Fe 3+ (3d 5 ) due to the reason that Fe 2+ , after losing an extra
electron yields a stable electronic configuration : 3d 5 where the 3d-orbital is completely
half-filled.
With Ni(II) (3d 8 ) and Co(II) (3d 7 ), no such stable electronic configuration is formed after
removing an extra electron from the 3d-orbitals, therefore, formation of Ni(III) and Co(III) from
Ni(II) and Co(II) respectively are very difficult.
1480. (a) An aqueous solution of [Ni(SO 4 )(en) 2 ]Cl 2 which produce precipitate with AgNO 3 solution
and it will not give any precipitate with aqueous BaCl 2 solution. However, aqueous solution
of [NiCl 2 (en) 2 ]SO 4 will form a precipitate with BaCl 2 .
(b) Aqueous solution of [Ni(en) 2 I 2 ]Cl 2 will form a white precipitate with aqueous AgNO 3
solution which will be soluble in excess of ammonia solution. However, aqueous solution of
[NiCl 2 (en) 2 ]I 2 will form an yellow precipitate of AgI which will be insoluble in
concentrated ammonia solution.
1481. (a) In [MnCl 6 ]4– , Mn 2+ has 3d 5 configuration. Since, Cl – is a weak ligand, there will be smaller
crystal-field splitting. In [Mn(CN) 6 ]4– , Mn 2+ has again 3d 5 configuration, but now CN – is a
strong ligand, there will be larger crystal-field splitting as:
3d5
3d5
[MnCl6]
4–
[Mn(CN)6]4–
(b) Since, strong ligand field absorb shorter wavelength light, therefore, [Mn(CN) 6 ]4– will
transmits longer wavelengths.
1482. (a) Since, the complex is emitting long wavelength (yellow) light, short wavelength visible light
would have been absorbed. This is also evident from the high strength ligand field given by
strong ligands-cyanide ion.
(b) In [Co(CN) 6 ]3– , Co is in 3d 6 state and ligand is strong, there is no unpaired electrons as:
3d6
Strong ligand
field
No unpaired electron, complex is diamagnetic
(c) If NH 3 molecules are substituted for cyanide ions, complex will absorb long wavelength
visible light and the shift in absorbance will be towards red region of the spectrum. This is
due to the fact that NH 3 is a weaker ligand than CN – .
684
Problems in Chemistry
1483. In high spin complex of Mn 2+ (3d 5 ), all the electrons are occupying singly the five-d-orbitals as:
Mn2+
As evident from the above orbital energy diagram for Mn 2+ , electron transition is not allowed
from lower level to higher level since, any electronic transition from low energy level to high
energy level will violate “Pauli Exclusion Principle”. Therefore, complex of Mn 2+ is very faintly
coloured.
1484. For a spontaneous reaction ∆G ° < 0, i. e., E ° > 0. In the above example, both the reduction
reactions are spontaneous. Also reduction potential of reaction (II) is more positive,
corresponding ∆G° will be more negative, reaction will be more spontaneous. Therefore,
[Co(NH 3 ) 6 ]3+ is more stable and [Co(H 2O) 6 ]3+ is more reactive.
1485. (a) [Cr(H 2O) 4 Cl 2 ]Cl (b) [Cr(H 2O) 6 ]Cl 3 .
1486. (a) Coordination, (b) Ionization, (c) Linkage, (d) Hydrate.
O
O
OH
OH2
OH
N
N
OH2
1487.
Co3+
Co3+
OH
OH2
O
O
O
O
H
Trans isomer
Nitrilotriacetic acid [ ••N(CH2COOH)3 ]
NH
H2N
1488. (a)
NH2
Co3+
H2N
NH2
HN
OH
OH2
O
H
Cis isomer
O
685
Solutions
1489. Weaker the ligand field strength, smaller will be the crystal field splitting and longer will be the
wavelength of absorption. Hence :
[Cr(CN) 6 ]3– < [Cr(en) 3 ]3+ < [Cr(NH 3 ) 6 ]3+ < [CrCl 6 ]3– < [Cr(H 2O) 6 ]3+
1490. Stronger the ligand field, greater will be the crystal field splitting and hence, higher will be the
frequency of light to be absorbed. Hence, order of frequency of light absorbed by these complexes
is :
[Cr(H 2O) 6 ]3+ < [Cr(en) 3 ]3+ < [Cr(CN) 6 ]3–
1491. (a) Stronger the ligand field, greater will be crystal field splitting, smaller will be the
wavelength of light absorbed.
λ :[Fe(CN) 6 ]4– < [Fe(H 2O) 6 ]2+
⇒
(b) Higher the oxidation state of central metal, greater will be the crystal field splitting. Hence,
there will be greater crystal field splitting of d-orbitals in Mn(III) complex compared to that
in Mn(II) complex. Hence, [Mn(CN) 6 ]4– will absorb at longer wavelength (low energy
photons) than [Mn(CN) 6 ]3– .
Br
H2O
1492.
(a)
H2O
Br
Pt
Pt
H2O
Br
Br
OH2
Cis
Trans
Py
Cl
(b)
Pt
(c) Zn
Cl
Pt
H3N
2+
Br
Cl
Br
H3N
Pt
Py
Py
has tetrahedral arrangement of ligands around it :
Cl
Zn2+
H3N
NH3
NH3
Cl
H2O
(d)
Cl
Cl
Ni
H2O
Ni
OH2
OH2
Cis
OH2
H2O
H2O
OH2
Cl
trans
Br
NH3
686
Problems in Chemistry
O
O
Cl
O
Cl
O
O
O
(e)
O
Co
O
Co
Cl
O
O
O
O
O
O
Cl
O
trans
O
Cis
(Optically active)
Cl
en
Cl
1493. (a)
en
Co
en
Co
NO2
en
NO2
(trans)
Cl
(enantiomeric)
en
Cl
(b)
en
Co
en
Co
Cl
en
Cl Cl
(trans)
OH2
(cis)
H 3N
(c)
OH2
H 3N
Cr
(d)
OH2
Cl
H 3N
Cl
NH3
NH3
Cl
Cl
Pt
NH3
H 2O
Cl
en
NH3
(e)
Pt
Cl
OH2
Co
OH2
Cl
Cl
H 3N
Cr
H3N
Cl
Cl
Cr
Cl
Cl
en
NH3
Cl
en
687
Solutions
1494. Co(III) = 3d 6 , weak ligand, magnetic moment = 4 × 6 = 24 BM = 4.9 BM.
(b) Co 3+ : 3d 6 , magnetic moment = 0; en is a strong ligand.
(c) Mn 3+ (3d 4 ) and CN – strong ligand, two unpaired electron, µ = 8 BM = 2.82 BM.
(d) Mn 2+ (3d 5 ) and Cl – is a weak ligand, µ = 5 × 7 BM = 5.916 BM
1495. NiCl 2 + (CH 3 ) 3 P → Compound
From elemental composition, empirical/molecular formula = [NiCl 2 ((CH 3 ) 3 P) 2 ] . Since, it has
no isomeric forms, complex must be tetrahedral :
Cl
Ni
(CH3)3P
Cl
P(CH3)3
bis (trimethyl phosphine) dichloronickel (II)
Magnetic moment : Ni 2+ (3d 8 ) and sp 3 hybridization indicate presence of two unpaired
electrons, hence, µ = 8 = 3.82 BM.
1496. Zero magnetic moment indicate that there is no unpaired electron and complex is square planar
(dsp 2 hybridized Ni).
CN
Br
2–
CN
Br
Ni2+
CN
Ni2+
Br
Br
CN
Cis
NH3
Cl
1497. (a)
NH3
Cl
–
NH3
Cl
Co3+
Co3+
Cl
trans
Cl
Cl
NH3
Cl
Cl
trans
Cl
Cl
Cl
(b)
NH3
Co3+
Co3+
Cl
NH3
NH3
(Facial)
NH3
Cl
NH3
H3N
Cl
(Meridional)
–
2–
688
Problems in Chemistry
1498. (a) No isomers (b) No isomers.
O
O
O
O
O
O
Co3+
(c)
O
O
3–
O
O
O
O
3–
Co3+
Enantiomers
O
O
O
O
O
O
O
O
O
O
8
1499. Cyanide ion is a strong ligand and in 3d ( Ni
2+
) it gives square planar geometry:
Energy
dx2–y2
dxy
dz2
dxz
2–
[Ni(CN)4 ]
O
dyz
crystal field splitting.
Energy
There is no unpaired electron, complex is diamagnetic.
Chloride ion is weak ligand, in 3d 8 ( Ni 2+ ), it produces tetrahedral ligand field.
[NiCl 4 ]2– crystal field splitting.
There are two unpaired electrons and complex is paramagnetic.
µ = 8 BM = 2.82 BM.
O
689
Solutions
1500. When a substance absorbs a particular colour when bathed in white light, the perceived colour of
the reflected or transmitted light is the complementary colour. Also, wavelengths of absorbed and
transmitted lights are inversely related.
Cyanide ion being stronger ligand than H 2O, former absorbs light of shorter wavelength than later
and wavelengths of transmitted light is longer for cyanide ion (yellow) and shorter for H 2O (blue
or green).
1501. The crystal field splitting energy is equal to the energy of photon causing absorption maximum.
∆=
⇒
hc 6.625 × 10 –34 × 3 × 10 8
=
= 4.23 × 10 –19 J
λ
470 × 10 –9
1502. Stronger the ligand field, shorter the wavelengths of photon absorbed.
(a) [Co(H 2O) 6 ]3+
(b) [FeF6 ]3–
(c) [CuCl 4 ]2–
0.875 1000
1503. − ∆T f = 0.56 = i × 1.86 ×
×
; i = vant Hoff’s factor
233.5
25
⇒ i = 2. Hence, formula of complex is [Co(NH 3 ) 4 Cl 2 ]Cl.
1504. Rust consists of mixture of Fe(OH) 3 and Fe 2O 3⋅nH 2O. Oxalic acid, being a bidentate, chelating,
strong ligand, forms stable complex with iron, removing rust as:
Fe 3+ + 3C 2O 2–
→ [Fe(C 2O 4 ) 3 ]3–
4
CN–
2–
1505. Cu 2+ + 2CN – → Cu(CN) 2 ↓ →
excess Cu(CN) 4
Soluble
White
The complex Cu(CN) 2–
4 is highly stable and it ionizes very feebly as:
2–
2+
Cu(CN) 4
Cu + 4CN – : K d is very small and equilibrium lies predominantly to left side.
If H 2S gas is bubbled at this time, no precipitate of CuS is formed due to presence of very small
amount of Cu 2+ ion in the solution (less than required to exceed the solubility limit).
1506. The initial green colouration is due to presence of [CuCl 4 ]2– in concentrated solution. On
dilution, solution turned blue due to conversion of [CuCl 4 ]2– into [Cu(H 2O) 6 ]2+ .
1507. K f (Formation constant)
=
[Fe(H 2O) 5 NCS]2+
[Fe(H 2O) 6 ]3+ [SCN – ]
=
7.3 × 10 –5
(0.2) (10 –3 − 7.3 × 10 –5 )
= 0.393
1508. Electrical conductance of electrolyte depends on number of ions available. By comparison with
the conductance of standard electrolyte, nature of unknown electrolyte can be established. In the
given experiment, conductance of unknown will match as :
[Co(NH 3 ) 6 ]Cl 3 = FeCl 3 , [Co(NH 3 ) 5 Cl]Cl 2 = MgCl 2 and
[Co(NH 3 ) 4 Cl 2 ] Cl = NaCl.
1509. CoCl 2 at low concentration appear pink coloured due to its existence as [Co(H 2O) 6 ]Cl 2 . On
adding HCl, following reaction occurs :
[Co(H 2O) 6 ]Cl 2 + HCl → [Co(Cl) 6 ]4–
blue-coloured
Adding HgCl 2 , pink colour is restored due to the following reaction:
690
Problems in Chemistry
[CoCl 6 ]4– + HgCl 2 + H 2O → [Co(H 2O) 6 ]2+ + [HgCl 4 ]2–
pink
colourless
1510. The complex is square planar and exhibit cis-trans isomerism in which cis-form is polar while
trans-form is non-polar. Dipole moment measurement can allow distinction between these two
forms :
Polar
Cl
Cl
NH3
Cl
Pt
Pt
NH3
H3 N
Cis
H3 N
Cl
trans - Non - polar
1511. Both solutions contain hydrated iron [Fe(H 2O) 6 ] and [Fe(H 2O) 6 ]3+ ions. Higher the oxidation
state of transition metal, greater will be the crystal field splitting and shorter will be the absorption
wavelength. The wavelength of emitted light is inversely related to absorption wavelength. On
this basis, transmitted light from [Fe(H 2O) 6 ]3+ will have longer wavelength-will appear yellow,
while [Fe(H 2O) 6 ]2+ solution will be brown. This can also be explained on the basis of electronic
configurations :
Fe 3+ (3d 5 ) and weak ligand field, completely half-filled d-orbital. Difficult d-d-transitions.
Fe 2+ (3d 6 ) and weak ligand field, less than full-filled d-orbital, easier d-d-transition.
2+
REPRESENTATIVE ELEMENTS
1512. (a) B2O 3 + 3Mg → 3MgO + 2B
(b) Al 2O 3 will come in solution as Na[AlO 2 ].
(c) CO 2 + 2Na[Al(OH) 4 ] → 2Al(OH) 3 + Na 2CO 3 + H 2O
H
H
H
1513.
B
B
H
H
H
H
 s
1514. (a) BF3 + EtOH → Et —O → BF3
⊕
+
s
(b) BCl 3 + PhNH 2 → PhH 2 N •• → BCl 3
(c) BF3 + KF → K[BF4 ]
1515. (a) HF is a weak electrolyte, but when mixed with KF, forms a strong electrolyte KHF2 which
ionizes as :
KHF2 → K + + H + + 2F –
Fluoride ions makes a complex AlF4− with AlF3 , making it soluble. Adding BF3 , causes AlF3 to
reprecipitate as :
BF3 + AlF4– → BF4– + AlF3 .
691
Solutions
1516. (a) In trisilyl amine, pπ - dπ back bonding occurs making nitrogen sp 2 -hybridized.
H3Si—N—SiH3
SiH3
No such back bonding is possible in trimethylamine and nitrogen is in
sp 3 -state.
(b) Due to presence of vacant d-orbital with silicon, it forms a network structure as:
O
O==Si==O
Si==O
O==Si==O
No such bonding in possible with CO 2 since carbon does not possess vacant d-orbital. Therefore,
CO 2 remains as isolated gaseous molecule.
1517. (a) 6NaOH + SiCl 4 → Na 2SiO 3 + 4NaCl + 3H 2O
(b) H 3SiCl + H 2O → H 3Si(OH) → H 3Si—O—SiH 3 + H 2O
(c) SiF4 + 4H 2O → Si(OH) 4 + 4HF
1518. (a) Sn + 2NaOH → Na 2SnO 2 + H 2
(b) 2SO 2 + PbO 2 → Pb + 2SO 3
(c) SiH 2Cl 2 + 2H 2O → 2HCl + SiH 2 (OH) 2 → —( SiH 2O —
)n
Polymer
(d) 4ClCH 2SiCl 3 + 3Li[AlH 4 ] → 4ClCH 2SiH 3 + 3Li[AlCl 4 ]
1519.
H
H
Be
H
H
H
Be
H
Be
H
H
A part of the chain structure of polymeric BeH2
Here each Be is tetrahedrally coordinated and each hydrogen bridging two Be atoms. The
Be—H—Be bond is a 3 centred 2 electron bond.
1520. (a) NaH + H 2O → NaOH + H 2
(b) 2NaN 3 + Heat → 2Na + 3N 2
(c) 4KO 2 + 2H 2O → 4KOH + 3O 2
(d) NaF + BF3 → Na[BF4 ]
∆
1521. (a) [NH 4 ]2 [BeF4 ] → BeF2 + 2NH 4 F
(b) 2NaCl + BeCl 2 → 2Na 2 [BeCl 4 ]
Water
(c) BeF2 ( s) → [Be(H 2O) 4 ]2+ + 2F −
1522. (a) BeCl 2 monomer is linear and Be atom has two vacant 2 p-orbitals, following diagram shows
one of these vacant orbitals :
Cl—Be—Cl
Vacant 2-p-orbital
Cl—Be—Cl
Cl—Be
Cl—Be—Cl
Cl
Cl
Dimer
Cl—Be
692
Problems in Chemistry
Dimerization occurs by Cl lone pair donation into vacant 2 p-orbital of Be. Hybridization of
Be in dimer is sp 2 and each Be atom still has one vacant 2 p-orbital.
(b) Diethyl ether is a Lewis base and BeCl 2 is a Lewis acid, adduct formation occurs.
OEt2
Be
Et2O
Cl
Cl
Be is tetrahedrally coordinated and sp3-hybridized.
1523. Anhydrous CaCl 2 is hygroscopic and forms a hydrate, probably CaCl 2 ⋅ 2H 2O. In the presence of
a lot of water, hydrate becomes liquid since it is deliquescent. Therefore, action as a drying agent
depends on hygroscopic nature of CaCl 2 .
On the otherhand, CaH 2 reacts with H 2O liberating H 2 :
CaH 2 + 2H 2O → Ca(OH) 2 + 2H 2
Therefore, CaH 2 is used to remove H 2O from a solvent.
1524. NaF and NaCl are ionic species. The lattice energy and hence, melting temperature, of NaF is the
greater of the two because of its smaller internuclear distance. CF4 and CCl 4 are covalent
compounds, and so do not adhere to the ionic model and their melting temperature are determined
by van der Waal’s force which depends on molar masses.
1525. n/ p ratio for 4 Be 7 and 6 C11 are 0.75 and 0.83, less than one, i.e., their nuclei contain fewer
neutrons than protons, therefore positron emitted as :
7
→
3 Li
7
+
+1 e
11
6C
→
11
5B
+
+1 e
4 Be
0
0
n/ p ratio for 6 C14 is 1.33, greater than one, i.e., its nucleus contain large number of neutrons and
therefore some neutrons are converted into protons.
0n
1
→
1
1H
+
−1 e
0
, i.e.,
14
6C
→
7N
14
+
−1 e
0
1526. He consists of two protons, two neutrons and two electrons and sum of the masses of these
nucleons is 4.033 amu. Therefore, ∆m = 0.304 amu per atom.
∆E = ∆mc 2 = 2.736 × 1012 J/ mol = 2.835 × 10 7 eV.
1527. Ionic mobility and conductivity in solution depends on hydrated radius of the ion. Larger the
hydrated radius, lower will be the ionic mobility and conductivity in aqueous solution. Size of
hydrated ion depends on extent of hydration which consequently depends on positive charge
density of cation. Smaller size and higher charge gives a greater charge density and cations are
heavily hydrated. Li + is most hydrated in its group and hence, has lowest ionic mobility and
conductivity.
1528. Li has most negative reduction potential suggest that ∆G° should be most negative for the
reaction:
Li → Li + + e
E ° = + 3.05 V, greater than E ° for Cs.
On the basis of above observation, Li should be most reactive. The actual reason for higher
reactivity of Cs lies on kinetics. On descending down the group, melting point of metal (Group I)
decreases and the energy produced by reaction is sufficient to melt the metal, which spread metals
693
Solutions
throughout the solution allowing larger surface area for reaction. Hence, kinetics favour greater
reactivity of alkali metals down the group.
1529. Dilute solution of alkali metals in liquid ammonia are dark blue due to presence of solvated metal
ions and solvated electrons:
M + NH 3 → M + NH 3 + e − NH 3
1530.
1531.
1532.
1533.
1534.
1535.
If the solution is allowed to stand, the colour slowly fades until it disappears owing to the
formation of metal amide. At concentration above 3M, solutions are copper-bronze coloured
because of formation of metal ions clusters.
A : Li, B : Li 2O, C : LiH, D : H 2 , E : LiOH.
Mg 2+ is more heavily hydrated than Li + due to greater charge/size ratio in the former case.
In salt, Be is divalent ion Be 2+ .Since Be belongs to IInd period of periodic table the only available
orbitals for coordination with H 2O are 2sand 2 p and a maximum of four lone pair of electrons can
be accommodated.
A : Ca, B : Ca(OH) 2 , C : CaCO 3 , D : CaO, E : CaC 2 .
A : BaCO 3 , B : CO 2 , C : CaCO 3 , D : BaO.
Due to inert pair effect, the stable oxidation state is +2 and not +4 while for Sn both Sn 2+ and Sn 4+
are possible and it acts as a reducing agent.
Sn 2+ → Sn 4+ + 2e.
1536. In non-polar solvent, dimeric structure is retained while in aqueous medium, high hydration
energy is sufficient to break the covalent dimer into [Al(H 2O) 6 ]3+ and Cl − as:
Cl
Cl
Al
Cl ••
•
• Cl
Al
Cl
Cl
→
–
Al(H 2O) 3+
6 + Cl
In polar (aqueous) solvent
In non-polar solvent
Low
T
1537. (a) B2 H 6 + NH 3 (excess) → B2 H 6 ⋅ 2NH 3
High
T
(b) B2 H 6 + NH 3 (excess) → (BN) x [ Boron nitride].
1: 2
High T
(c) B2 H 6 + NH 3 → B3 N 3 H 6 (Borazine).
1538. (a) B3 N 3 H 6 + 3HCl → B3 N 3 H 9Cl 3
(b) B2 H 6 + 6H 2O → 2H 3 BO 3 + 6H 2
(c) B3 N 3 H 6 + 9H 2O → 3NH 3 + 3H 3 BO 3 + 3H 2
(d) B2 H 6 + 6EtOH → 2B(OEt) 3 + 6H 2
(e) B2 H 6 + 2LiH → 2Li[BH 4 ]
(f) B2 H 6 + HCl → B2 H 5Cl + H 2
(g) B2 H 6 + 3Cl 2 → 2BCl 3 + 6HCl
1539. A : Al 2Cl 6 , B : Al 3+ ( aq ) + Cl − ( aq ), C : AgCl,
D : Al(OH) 3 ,
E : Li[AlH 4 ]
694
Problems in Chemistry
Heat
1540. (a) SiO 2 + 2C → Si + CO
Si + 2Cl 2 → SiCl 4
(b) SiO 2 + 2C → Si + 2CO
2200°C
Si + C → SiC
1541. (a) Al 4C 3 + 12H 2O → 3CH 4 + 4Al(OH) 3
(b) CaC 2 + 2H 2O → C 2 H 2 + Ca(OH 2 )
1100°C
(c) CaC 2 + N 2 → CaNCN + C
(Nitrolium)
O
2
(d) SiC + 2NaOH + O 2 →
Na 2SiO 3 + CO 2 + H 2O
(e) Mg 2C 3 + 4H 2O → 2Mg(OH) 2 + CH 3C ≡≡CH
1542. 2C 2 H 5 MgCl + SiCl 4 → (C 2 H 5 ) 2 SiCl 2 + 2MgCl 2
C2 H 5
C2 H 5


Heat
(C 2 H 5 ) 2 SiCl 2 + 2H 2O → HO—Si—OH →
)n
—Si
( —O—


C2 H 5
C2 H 5
1543. R 3SiCl is monofunctional and act as chain terminator, hence stop the growth of propagating
polymer chain as:
R
R
R
R




( Si—O) n—O— Si— R
—Si—O
(
—
) n H + HO—Si— R → —




R
R
R
R
(From R3SiCl)
(growing chain)
1544. (a)
CH3 CH3
Si
O
O
H 3C
Si
H 3C
Si
O
Chain terminated
–
(b)
CH3
–
–
CH3
_
_
_
_
_
.
.
_
_
_
.
_.
(d)
_
.
.
_.
(c)
_
_
_
.
.
_
_
.
_
_
_
695
Solutions
_
_
_
_.
.
_
.
_
.
_.
(e)
_
_
.
.
_
.
.
_
_
.
_
_
H O
2
1545. (a) P4 + 3NaOH →
PH 3 + 3NaH 2 PO 2
(b) CaNCN + 3H 2O → 2NH 3 + CaCO 3
(c) N 2 H 4 + 2I 2 → N 2 + 4HI
(d) N 2O + 2NaNH 2 → NaN 3 + NH 3 + NaOH
1546. Superphosphate :
3[Ca 3 (PO 4 ) 2 ⋅ CaF2 ] + 7H 2SO 4 → 3Ca(H 2 PO 4 ) 2 + 7CaSO 4 + 2HF
14444244443
Superphosphate
Triple superphosphate:
[Ca 3 (PO 4 ) 2 ⋅ CaF2 ] + 6H 3 PO 4 →
4Ca(H 2 PO 4 ) 2
+ 2HF
(Triple superphosphate)
1547. (a) N 2O 5 + NaCl → NaNO 3 + NO 2Cl
(b) N 2O 5 + Na → NaNO 3 + NO 2
(c) N 2O 4 + H 2O → HNO 2 + HNO 3
(d) 2NO 2 + Cl 2 → 2NO 2Cl
(e) 2NO 2 + HCl → 2NOCl + Cl 2 + 2H 2O
1548. In N 2 , due to smaller size of 2 p-orbital, effective pπ - pπ overlap takes place giving very high
“bond-energy” to N 2 molecule. In P2 , bond energy is just half of the N—N bond energy in N 2
due to poor pπ - pπ overlap between the larger 3 p-orbitals. On the other hand in P4 , each
phosphorus is tetrahedrally bonded to three other phosphorus atoms giving three σ-bond to each
phosphorus, greater bond energy than P2 , hence more stable P4 is formed.
P
N
N
better pπ-pπ overlap
P
P
poor pπ-pπ overlap
P
P
P
more stable structure
1549. In NF3 , N—F bond is strongest and it behaves like other inert molecules CF4 , etc.
696
Problems in Chemistry
O
..
P
O
P
O
O
1550. : P
O
O
O
..
P
P 4O 6
O
P:
O
O
P
O
P
O
O
O
P
O
O
P4O10
1551. (a) Ostwald method : In Ostwald process, NH 3 is oxidised first into NO in presence of
platinum, rhodium catalyst at high temperature and pressure. NO is then further oxidised into
NO 2 and absorbed in water.
Platinum/Rhodium
5 atmosphere, 850°C
4NH 3 ( g ) + 5O 2 ( g ) → 4NO( g ) + 6H 2O( g )
→ 2NO( g ) + O 2 ( g )
2NO 2 ( g )
2NO 2 ( g ) + H 2O( l) → HNO 3 + HNO 2
2HNO 2 → H 2O + NO 2 + NO
3NO 2 + H 2O → 2HNO 3 + NO
2HNO 2 + 2FeO → 2NO + H 2O + Fe 2O 3 Oxidising action
5HNO 2 + 2KMnO 4 + 3H 2SO 4 → 2K 2SO 4 + 2MnSO 4 + 3H 2O Reducing action
1552. (a) Ligand oxygen has two unpaired electrons in its molecular orbitals, giving paramagnetic
character while N 2 has all paired electrons in molecular orbital, and diamagnetic.
(b) NO + has bond order of three while bond order of NO is 2.5.
1553. Preparation :
2H 2SO 4 → 2H + + 2HSO –4
(b)
at anode
2HSO –4 → H 2S 2O 8 + 2e −
H 2S 2O 8 + 2H 2O → 2H 2SO 4 + H 2O 2 .
O
Structure :
Open book like structure.
H
(a)
(b)
(c)
(d)
H
97°
O
94°
2KMnO 4 + 3H 2SO 4 + 5H 2O 2 → K 2SO 4 + 2MnSO 4 + 8H 2O + 5O 2
K 2Cr 2O 7 + 4H 2O 2 + H 2SO 4 → 2CrO 5 + K 2SO 4 + 5H 2O
2KI + H 2O 2 → I 2 + 2H 2O
2K 4 [Fe(CN) 6 ] + H 2O 2 + H 2SO 4 → 2K 3 [Fe(CN) 6 ] + K 2SO 4 + 2H 2O
697
Solutions
O

1554. (a) HO—S —OH
+ IV
O O
 
(b) HO—S —S —OH

O
O O
 
(c) HO—S —S —OH
+ III
O

(d) HO—S —OH

S
+V and + III
O O
 
(e) HO—S —S — OH
 
O O
(+V)
+IV and –II
O

(f) HO—S —O—OH

O
O
O


(g) HO—S —O—O—S —OH


O
O
(+VI)
(+VI)
1555. In contact process, SO 2 is oxidized by air into SO 3 in presence of V2O 5 catalyst:
V2O5
1
SO 2 + O 2 → SO 3
2
SO 3 is then passed through 98% H 2SO 4 , forming H 2S 2O 7 (oleum). This solution is then diluted
to obtain sulphuric acid of desired strength.
1556.
boiling
water
Na 2SO 3 + S → Na 2S 2O 3
Heat in
air
2Na 2S 3 + 3O 2 → 2Na 2S 2O 3 + 2S
Thiosulphuric acid can’t be prepared by adding acid to the aqueous solution of thiosulphate salt,
since the free acid decomposes in water into mixture of S, H 2S, SO 2 , etc. It can be made in the
absence of water and at low temperature.
1557. (a) 2F2 + 2H 2O → 4HF + O 2
(b)
(c)
(d)
(e)
1558. (a)
Cl 2 + CO → COCl 2
3Cl 2 + 8NH 3 → N 2 + 6NH 4Cl
3Cl 2 + 6NaOH → 5NaCl + NaClO 3 + 3H 2O
SiO 2 + 6HF → H 2SiF6 + 2H 2O
Cl 2O + 2NaOH → 2NaOCl + H 2O
(b) 4HClO 4 + P4O10 → 2Cl 2O 7 + 4HPO 3
(c) 2ClO 2 + 2NaOH → NaClO 2 + NaClO 3 + H 2O
(d) KBrO 3 + F2 + 2KOH → KBrO 4 + 2KF + H 2O
(e) Cl 2O 6 + 2NaOH → NaClO 3 + NaClO 4 + H 2O
(f) Cl 2O 6 + HF → FClO 2 + HClO 4
1559. First, water must be absent in preparation of F2 by any method, otherwise F2 , being the strongest
oxidising agent, oxidises water into O 2 converting itself into HF. Secondly, HF in anhydrous
state, is only slightly ionized and is therefore a poor conductor of electricity. Therefore,
anhydrous HF is difficult to electrolyse. For the purpose of producing F2 by electrolytic method, a
mixture of HF and KF is electrolysed. HF first combines with KF forming a conducting K[HF2 ],
enabling electrolysis easier.
698
Problems in Chemistry
1560. (a) F
O
103°
F
O
F Cl
111°
Cl
O—O
O==I—O—I==O
F
O
O
(b) Repulsion between the lone pair reduces the bond angle in OF2 while in Cl 2O, bond angle is
greater than tetrahedral bond angle due to steric crowding of the larger halogen atoms.
1561. (a) XeF2 + 2HCl → Xe + 2HF + Cl 2
(b) XeF4 + 4KI → Xe + 4KF + 2I 2
(c) XeF4 + 2SF4 → Xe + 2SF6
(d) 2XeF2 + 2H 2O → 2Xe + 4HF + O 2
3
(e) 3XeF4 + 6H 2O → 2Xe + XeO 3 +12HF + 2 O 2
(f) XeF6 + 3H 2O → XeO 3 + 6HF
XeF6 + H 2O → XeOF4 + 2HF
(g) 2XeF6 + XeO 3 → 3XeOF4
(h) XeO 3 + XeOF4 → 2XeO 2 F2
1562. Fe 3+ ( aq. ) is a stronger oxidising agent than Br 2 but weaker than Cl 2 .
1563. In ClF3 , Cl has ten electrons in its valence shell, seven of its own and three shared electrons from
three fluorine. Since, chlorine has vacant d-orbitals (3s 2 3 p 5 3d 0 ), it can extend its valency
beyond eight, i.e., it can violate octet rule. On the other hand, in FCl 3 , the central fluorine atom
has ten-valence electrons which is not possible since, fluorine has no vacant d-orbitals and it
cannot violate octet rule (2s 2 2 p 5 ).
1564. Due to very small size of nitrogen, their p-orbitals makes effective overlap forming strong
pi-bonds and nitrogen acquire octet configuration by forming a sigma and two pi-bonds between
two nitrogen atoms as:
N
N
has complete octet at both nitrogen.
On the other hand, due to large size of phosphorus atom, their p-orbitals do not overlaps
efficiently to form pi-bonds. Due to this reasons, phosphorus atoms forms preferably sigma bonds
and at room temperature it exist as P4 containing sigma-bonds only. If heated at very high
temperature, it decomposes partially into less stable P2 molecules.
1565. Sulphur contains vacant d-orbitals, can extend its valency beyond eight while oxygen lacks
vacant d-orbital, can’t violate octet rule
1566.
CuF2 + H 2SO 4 → CuSO 4 + 2HF
2NaCl + H 2SO 4 → Na 2SO 4 + 2HCl
HBr and HI cannot be prepared by the above method because they are oxidised into elemental
bromine and iodine as:
2NaBr + 2H 2SO 4 → Na 2SO 4 + SO 2 + 2H 2O + Br 2 .
699
Solutions
1567. Boric acid B(OH) 3 , only partially react with water to form H 3O + and B(OH) 4 – , it behaves as a
weak monobasic acid
B(OH) 4 – + H 3O +
B(OH) 3 + 2H 2O
Thus, boric acid can’t be titrated satisfactorily with NaOH as a sharp end point is not obtained.
Adding cis-diol or triol (glycerol) enhances acidic strength of boric acid by forming a stable
complex as:
B(OH) –4 + H 3O +
B(OH) 3 + 2H 2O

 

C O
 C  OH –2H2O 

B(OH) –4 +
→


 
 C  OH
  C O

 
B

OH



OH

…(i)
–
diol
–2H2O
→
 
  C O

 
 
  C O
 
B
(X)

 
O  C

 
 
O  C 
 
s
The formation of complex of B(OH) –4 with diols, as shown above, removes B(OH) –4 from
solution, driving reaction (i) in forward direction and all B(OH) 3 reacts quantitatively with water
forming B(OH) –4 required for formation of stable complex (X). This increases acid strength and
boric acid becomes strong acid in presence of such organic compounds.
1568. In CCl 4 , carbon has no vacant orbital where water can attack and therefore, it is inert towards
water. On the other hand, Si in SiCl 4 has vacant 3d-orbitals, available for attack by water, which
makes it highly reactive with water.
Cl
Cl
Si
Cl
Cl
OH2
vacant d-orbitals are the site of
attack by water molecule
If sufficient energy is supplied by using superheated steam, CCl 4 hydrolyses forming phosgene
gas:
Superheated
CCl 4 + H 2O( g ) → COCl 2 + 2HCl
1574. Initially CaCO 3 is formed as white precipitate as:
CO 2 + Ca(OH) 2 → CaCO 3 ↓ + H 2O
700
Problems in Chemistry
On further passage of CO 2 , solid CaCO 3 comes in solution by forming soluble bicarbonate as:
CaCO 3 + H 2O + CO 2 → Ca(HCO 3 ) 2 ( aq )
1575. Phosphoric acid is proton donor and gives three proton on complete ionisation, exhibiting its
tribasic character:
+
PO 3–
4 + 3H
O== P(OH) 3
On the other hand, boric acid is electron deficient and boron has only six shared electrons in
valence shell, i. e., two electrons short of stable octet configuration. Hence, H 3 BO 3 behaves as
Lewis acid. Also, it can accept only one lone pair, it is monobasic:
s
⊕
(a vacant orbital) + •• NH 3 → (HO) 3 B ← NH 3
(HO) 3 B
1577. 3NaHCO 3 + NaAl(SO 4 ) 2 → 3CO 2 + 2Na 2SO 4 + Al(OH) 3
1578. Inert pair effect. As we descend down a group in the periodic table, inert pair effect becomes more
and more pronaunced and oxidation number decreases by two unit.
1579. Aluminium being strongly electropositive, reacts with acid components of food.
1580. In between pH = 5 to pH = 8, Al 3+ precipitate as Al(OH) 3 . As it settles, the gelatinous Al(OH) 3
traps and removes suspended solids from the water.
1581. (a) N 2 H 4 ( aq ) + HCl → N 2 H 5Cl
(b) 2NO( g ) + 32 O 2 ( g ) → N 2O 5 ( g )
(c) 2NO 2 ( g ) + H 2O( l) → HNO 2 + HNO 3
(d) 2NH 3 ( g ) + 4O 2 ( g ) → N 2O 5 + 3H 2O
200°C
(e) NH 4 NO 3 ( l) → N 2O + 2H 2O
Electric spark
1582. (a) 2Al + Fe 2O 3 → Al 2O 3 + 2Fe ∆H < 0
(b) Na 2O 2 + H 2O → 2NaOH + 12 O 2 ( g )
(c) Pb 2+ + O 3 + H 2O → PbO 2 + O 2 + 2H +
1583. (a) 2KO 2 + H 2O → 2KOH + 32 O 2
(b) 3O 3 + Cl – → 3O 2 + ClO –3
1584. Borax on treatment with CaF2 in concentrated sulphuric acid solution gives BF3 as:
Na 2 B4O 7 + 6CaF2 + 8H 2SO 4 → 6CaSO 4 + 4BF3 + 7H 2O + 2NaHSO 4
1585.
S + O 2 → SO 2
Na + H 2O → NaOH( aq ) + 12 H 2 ( g )
…(i)
…(ii)
From Eqs. (i) and (ii),
2NaOH + SO 2 → Na 2SO 3 + H 2O
SO 2–
3
–
+ SO 2–
4
…(iii)
+
+ H 2O + Cl 2 → 2Cl
+ 2H
{from Eq. (iii)}
Boiling elemental sulphur in alkaline solution of Na 2SO 3 gives thiosulphate as:
SO 2–
3 ( aq ) + S ( s) →
S 2O 32–
(thiosulphate ion)
1586. The NF3 has little tendency to act as donor molecule. The highly electronegative F atoms attract
electrons and these moments partially cancel the moment from the lone pair which reduces its
701
Solutions
donor power. On the other hand, in PF3 , P  F bond length is larger compared to N  F, bond,
electron withdrawing effect of fluorine does not transmit effectively onto the lone pair on
phosphorus in PF3 . Hence, PF3 act as a good donor molecule.
Ni[CO]4 + 4PF3 → Ni(PF3 ) 4 + 4CO
1587. A : NH 3 ; B: NO; C : NO 2 ; D : N 2O 3 ; E : HNO 2 ; F : N 2 ; G :CaNCN.
1588. In POCl 3 , there is pπ − d π back bonding between oxygen and phosphorus as:
O
pπ – dπ
back bonding
Due to this pπ − d π back bonding, P—O bond acquire double
bond character and hence, P—O bond is shorter than the
expected bond-length.
P
Cl
Cl
Cl
1589. To attain the pentavalent state, d-orbitals must be used. Hydrogen is not sufficiently
electronegative to make the d-orbitals contract sufficiently, hence, PH 5 is not formed.
1590. HCl( g ) is prepared from NaCl and H 2SO 4 as:
∆
2NaCl + H 2SO 4 → Na 2SO 4 + 2HCl( g )
HCl( g ) does not possess sufficient reducing power required for reducing H 2SO 4 , hence, the
above reaction is successful in preparing HCl( g ).
Contrary to HCl, HBr and HI are oxidised by H 2SO 4 as:
2HBr + H 2SO 4 → Br 2 + SO 2 + 2H 2O
2HI + H 2SO 4 → I 2 + SO 2 + 2H 2O
Hence, HBr and HI can’t be prepared by the reaction of NaBr and NaI with H 2SO 4 .
1591. (a) H  C ≡≡ P ••
F
(e)
H

P
H



H


+
F
P
F
F
1592. (a)
(b)
(c)
(d)
(e)


(b) 
H


P4
P4
P4
P4
P4
F
+ 4Al → 4AlP
+10F2 → 4PF5
+12Na → 4Na 3 P
+ Se → P4Se 6 + P4Se 10
+ Cl 2 → PCl 3 + PCl 5
2O
2
(f) P4 + 3O 2 → P4O 6 →
P4O10
••
••
(c) H  P  P  H


H
H
(d)
–
O

P
O–
O
O
–
702
Problems in Chemistry
1593. Double bond is formed by side-wise (lateral) overlaping of p-orbitals holding electrons of
opposite spin. For efficient pi-bonding (lateral overlap), the p-orbitals must be close-enough in
order to allow them to overlap up to some extent and this will be possible if atoms are small
enough in size. Oxygen, being the smallest in its group, its p-orbitals are close enough and
formation of pi-bond become possible. On the other hand, sulphur is too large in size that their
p-orbitals don’t come closer enough to allow the formation of pi-bonds, hence, sulphur exist as S 8
in which all bonds are sigma covalent bonds :
S
O2-molecule
showing π-bond
formation
S
S-atoms are
much distant and
no effective π-bond
are formed
S8 with
all single, sigma
covalent bonds
1594. As we descend down to hologen group in periodic table, molar mass increases rapidly and hence,
intermolecular force of attraction (van der Waals force of attraction). Due to this increase in force
of attraction, volatility decreases from top to bottom and their boiling/melting point increases.
1595. (a) Xe in XeO 2 F2 is sp 3 d hybridized and shape of the molecule is
F
O
Xe
O
F
(b) Xe in XeF + is sp 3 hybridized :
•• +
•
• Xe  F
••
Linear
(c) Xe in XeF3+ is sp 3 hybridized :
F
+
Xe
F
F
3 2
(d) Xe in XeO 4–
6 is sp d hybridized :
O
O
O
Xe
O
O
O
4–
703
Solutions
(e) Xe in XeO 3 F2 is sp 3 d hybridized :
O
F
O
Xe
O
F
1596. NaOH reacted with atmospheric CO 2 forming a white layer of NaHCO 3 .
1597. White phosphorus is highly reactive and it catches fire even at room temperature. It owes
reactivity from its structure in which a highly strained ring is available.
P
Here the P—P—P bond angles are 60° while the required bond angle is
109°. Due to this angle strain, P4 is highly unstable and reactive.
60°
P
P
P
Red hot
1598. (a) 2NH 3 + 3CuO → N 2 + 3H 2O + 3Cu
heat
(b) (NH 4 ) 2 Cr 2O 7 → N 2 + Cr 2O 3 + 4H 2O
1599. CuSO 4 + 2KF → K 2SO 4 + CuF2 ↓ (Insoluble, precipitate as green solid)
CuSO 4 + 2KCl → K 2SO 4 + CuCl 2 ( aq ) (Soluble, remains in solution giving green colour)
1600. (a) 4H 3 PO 3 → 3H 3 PO 4 + PH 3
(b) Li 4C + 4HCl → 4LiCl + CH 4
(c) 2HI + 2HNO 2 → I 2 + 2NO + 2H 2O
(d) H 2S + 2Cl 2 → 2HCl + SCl 2
1601. (a) PCl 5 < CCl 4 < SiCl 4 < BCl 3
(b) F – < O 2– < N 3– < C 4–
(c) I < Br < Cl < F
(d) SiO 2 < H 2O < CO 2
METALLURGY
1602. Mg is extracted from sea water exploiting the fact that Mg(OH) 2 is less soluble than Ca(OH) 2 and
precipitated out on treatment of sea water with lime.
Sea water
 Ca(OH)2
 → Mg(OH) 2 ↓
2+
contains Mg salt
The precipitate is filtered off, dried and dissolved in small volume of dilute HCl as:
Mg(OH) 2 + 2HCl → MgCl 2 + 2H 2O
The above solution is heated to dryness and anhydrous salt is electrolysed to obtain the pure metal
as:
Mg 2+ + 2e − → Mg
2Cl
–
→ Cl 2 + 2e
at cathode
−
at anode.
704
Problems in Chemistry
1603. Dolomite ore consists CaCO 3 ⋅ MgCO 3 . Extraction involves:
Heat
(i) Calcination: CaCO 3 ⋅ MgCO 3 → CaO ⋅ MgO + 2CO 2
(ii) Dissolution in dilute HCl :
CaO ⋅ MgO + 4HCl → CaCl 2 (aq) + MgCl 2 (aq) + 2H 2O
The above solution is then treated with excess of calcined ore and CO 2 is bubbled through it.
CaCl 2 ⋅ MgCl 2 + CaO ⋅ MgO + 2CO 2 → 2MgCl 2 + 2CaCO 3 ↓
Precipitated
The aqueous solution is then heated to dryness and anhydrous MgCl 2 is electrolyzed in molten
state to obtain pure metal.
1604. Mg itself being one of the strongest reducing agent, reduction of Mg 2+ by chemical method is not
possible.
1605. Due to smaller size and high charge, Mg 2+ is heavily hydrated in aqueous solution which is
responsible for its low ionic mobility and conductivity.
1606. (a) Epsom salt MgSO 4 ⋅ 7H 2O, occurs in nature as epsomite in certain gypsum deposite. It is also
obtained by hydration of its monohydrate MgSO 4 ⋅ H 2O, which exist in nature as kieserite.
Kieserite is allowed to remain in contact with water, when it slowly hydrated and sets to a
solid mass of crystals of heptahydrate:
MgSO 4 ⋅ H 2O + 6H 2O → MgSO 4 ⋅ 7H 2O
Epsom salt
Epsom salt is used in medicine as a purgative and as a catalyst in preparation of H 2SO 4 .
(b)
4Mg +10HNO 3 → 4Mg(NO 3 ) 2 + 3H 2O + NH 4 NO 3
1607. The net reaction of Solvay process is:
2NaCl + CaCO 3 → Na 2CO 3 + CaCl 2 .
Steps involved in the process are:
(i)
NH 3 + H 2O + CO 2 → NH 4 HCO 3
(ii)
NH 4 HCO 3 + NaCl → NaHCO 3 + NH 4Cl
(iii)
Heat
2NaHCO 3 → Na 2CO 3 + H 2O + CO 2 .
NaHCO 3 being less soluble, can preferentially be crystallized leaving NH 4Cl in solution, and this
forms basis of manufacture. On the otherhand, KHCO 3 being more soluble, not separated by
fractional crystallization and therefore this method fails for K 2CO 3 .
1608. Baeyer’s process : Ore is roasted first to oxidise any Fe 2+ into Fe 3+ and then digested with
NaOH at 130°C. Al 2O 3 dissolves, comes in solution leaving others in residue:
Al 2O 3 ⋅ 2H 2O + 2NaOH → 2NaAlO 2 + 3H 2O
Filtrate is treated with some Al(OH) 3 , where all aluminate is precipitated as Al(OH) 3 .
Hall’s Process : Ore is first fused with Na 2CO 3 , dissolved in water and CO 2 is bubbled.
CO
2
Al 2O 3 ⋅ 2H 2O + Na 2CO 3 → 2NaAlO 2 + CO 2 + H 2O →
2Al(OH) 3 ↓
Serpek’s Process :
Used when SiO 2 is the chief impurity:
705
Solutions
1800°C
Al 2O 3 + N 2 + 3C → 2AlN + 3CO;
AlN + H 2O → Al(OH) 3 + NH 3 .
1609. Al is extracted from bauxite ore by electrolytic reduction method. Some cryolite (Na 3 AlF6 ) and
CaF2 are added to the ore in order to lower the melting temperature and increase the electrical
conductivity facilitating electrolysis easier:
Al 3+ + 3e − → Al at graphite cathode.
Thermite welding: Thermite welding is used for commercial welding of iron. It exploit strong
reducing power of Al metal. Thermite is 3 :1 mixtrue of Fe 2O 3 and Al.
Fe 2O 3 + 2Al → 2Fe + Al 2O 3
∆H < 0
The heat evolved in the above reaction is sufficient to keep iron in molten state which serve the
welding purpose.
3
1610.
Al + 3HCl → AlCl 3 + H 2
2
2Al + 3H 2SO 4 → Al 2 (SO 4 ) 3 + 3H 2
3
H2
2
Concentrated nitric acid, being an oxidising agent, forms a protective layer of Al 2O 3 on the
surface of metal, which prevent it from further reaction.
1611. Aluminium take up hydrogen in presence of Zeigler catalyst (trialkyl aluminium) as:
3
Al + H 2 + 2Et 3 Al → 3Et 2 AlH
2
Then alkene is inserted 1st between Al—H bond as:
H 2O
Al + NaOH → NaAlO 2 +
Et 2 AlH + CH 2 == CH 2 → Et 2 Al—CH 2 —CH 3
Finally, applying high pressure, alkene molecules are inserted between Al—C bond as:
Et 2 Al—CH 2CH 2 + CH 2CH 3 → Et 2 Al—CH 2CH 2CH 2CH 3
n-times
→ Et 2 Al —
( CH 2 —CH 2 —
) n CH 2CH 3
Hydrolysis of this long chain growing aluminium alkyls gives straight chain hydrocarbon called
polyethylene. A modified catalyst TiCl 4 (Nata catalyst) causes polymerization to occur at faster
rate and even a much higher molecular weight polymer is produced.
1612. Iron is extracted from its oxide ore Haematite (Fe 2O 3 ) or Magnetite (Fe 3O 4 ) by carbon reduction
method, in a blast furnace. Blast furnace is charged with ore, lime-stone and coke. The reactions
occurring in blast furnace are:
(i) Coke is oxidised at the bottom of furnace as:
2C + O 2 → 2CO [CO 2 is not stable at this temperature]
(ii) Reduction of iron oxide at the top of the furnace:
Fe 2O 3 + 3CO → 2Fe + 3CO 2
Fe 3O 4 + 4CO → 3Fe + 4CO 2
706
Problems in Chemistry
Limestone decomposes giving CO 2 and CaO. CO 2 combines with coke providing CO as :
CO 2 + C → 2CO.
The mass impurity SiO 2 is removed as silicate (slag):
CaO + SiO 2 → CaSiO 3 (slag).
Iron obtained here is called cast iron.
1613. In presence of moisture and CO 2 , iron gets rusted with a brown deposition on the surface:
Fe + H 2O + O 2 → Fe(OH) 3 + Fe 2O 3 ⋅ nH 2O
14444244443
Rust
Rusting of iron is prevented by :
(i) Coating with more electropositive metals like Mg, Ni, Zn, Cr, etc.
(ii) Galvanization : Galvanization involves coating iron with Zn metal. It gives physical
protection by forming a thin, hard film of ZnCO 3 on the surface in due course of time. At the
same time, it also gives cathodic protection.
1614. (a) 3Fe + 4H 2O → Fe 3O 4 + 4H 2 ( g )
(b) Fe + 2HCl → FeCl 2 + H 2 ( g )
(c) Fe + 2H 2SO 4 → FeSO 4 + SO 2 + 2H 2O
2FeSO 4 + 2H 2SO 4 → Fe 2 (SO 4 ) 3 + 2H 2O + SO 2
(d) 2Fe ( s) + 3Cl 2 ( g ) → 2FeCl 3
(e) 4Fe +10HNO 3 → 4Fe(NO 3 ) 2 + NH 4 NO 3 + 3H 2O
(f) 4Fe +18HNO 3 → 4Fe(NO 3 ) 3 + 3NO + 3NO 2 + 9H 2O
(g) Fe + S → FeS
1615. Mohr’s salt FeSO 4 ⋅ (NH 4 ) 2 SO 4 ⋅ 6H 2O is prepared by making saturated solution of pure ferrous
sulphate and pure ammonium sulphate separately, in air free distilled water at 40°C. Mixing the
above solutions, and allowing to cool, the salt is obtained as pale green crystals.
Mohr’s salt has the advantage over ferrous sulphate in being not efflorescent and less readily
oxidized in solid state.
1616. Anhydrous FeCl 3 is prepared by direct combination of metal with dry chlorine gas at high
temperature:
2Fe ( s) + 3Cl 2 ( g ) → 2FeCl 3 ( s).
Anhydrous FeCl 3 is yellow coloured solid, sublime below 300°C. When heated above 500°C, salt
decomposes into FeCl 2 and Cl 2 .
Salt is soluble in water and its aqueous solution is brown coloured due to presence of colloidal
Fe(OH) 3 .
FeCl 3 + 3H 2O
Fe(OH) 3 + 3HCl.
On adding excess of HCl to the aqueous solution of salt, yellow coloured is restored.
Hexahydrated FeCl 3 ⋅ 6H 2O is prepared by dissolving metal in dilute HCl. The resulting solution
containing FeCl 2 is treated with chlorine gas to oxidise FeCl 2 to FeCl 3 . Finally solution is
evaporated to crystallization on a steam bath.
1617. Tin is extracted from cassiterite ore which contain small amount of SnO 2 .The extraction involves
roasting of ore followed by smelting with coke in blast furnace:
SnO 2 + 2C → Sn + 2CO
707
Solutions
The tin obtained here is always accompanied with some iron and tungeston as impurity which are
removed by liquation.
1618. (a) Sn + 2Cl 2 → SnCl 4
(b) 4Sn +10HNO 3 → 4Sn(NO 3 ) 2 + NH 4 NO 3 + 3H 2O
(c) Sn + 4HNO 3 →
H 2SnO 3
+ 4NO 2 + H 2O
(Metastannic acid)
(d) Sn + 2H 2SO 4 → SnO 2 + 2SO 2 + 2H 2O
(e) Sn + 2NaOH → Na 2SnO 2 + H 2
(stannite)
1619. Extraction of Cu from copper pyrite involves the following steps:
(i) Roasting : Pyrite is converted into cuprous sulphide and iron sulphide as:
2CuFeS 2 + O 2 → Cu 2S + 2FeS + SO 2
2FeS + 3O 2 → 2FeO + 2SO 2 .
(ii) Smelting : Roasted ore is mixed with coke and sand and smelted in blast furnace where FeO
combines with SiO 2 forming easily fusible slag
FeO + SiO 2 → FeSiO 3 (slag)
(iii) Self-reduction (Bessemerization) : To the molten Cu 2S, blast of hot air is passed in a
controlled manner so that Cu 2S is converted partially into Cu 2O and supply of air is cut-off
after some time. Now, more roasted Cu 2S is added and heated in absence of oxygen where the
following self reduction reaction occur:
2Cu 2O + Cu 2S → 6Cu + SO 2
Copper obtained by this method have blistered surface due to evolution of SO 2 , and thus named as
blistered copper.
1620. Moist air containing some CO 2 attacks metal slowly forming a green basic carbonate as:
2Cu + CO 2 + O 2 + H 2O → Cu(OH) 2 ⋅ CuCO 3
(Malachite green)
1621. Copper being less electropositive, below to hydrogen in electrochemical series, can’t displace H +
ion from solution.
1622. When copper is heated in oxygen atmosphere, at temperature up to 1100°C, a black cupric oxide
CuO is formed. If temperature is maintained above 1100°C, red cuprous oxide Cu 2O is produced.
T < 1100°C
O
2
Cu + O 2 → CuO(black) →
Cu 2O(red)
T > 1100°C
1623. (a) 3Cu + 8HNO 3 → 3Cu(NO 3 ) 2 + 2NO + 4H 2O
1
(b) Cu + 2HCl + 2 O 2 → CuCl 2 + H 2O
O
2
(c) 2Cu + 2H 2SO 4 →
2CuSO 4 + 2H 2O
(d) Cu + 4HNO 3 → Cu(NO 3 ) 2 + 2NO 2 + 2H 2O
(e) Fe 2 (SO 4 ) 3 + Cu → 2FeSO 4 + CuSO 4
1624. Ore containing PbS is partially oxidized by heating in limited supply of air. The supply of air is
cut off after some time and excess of roasted ore is added where the following self-reduction
occurs:
PbS + 2PbO → 3Pb + SO 2 .
708
Problems in Chemistry
1625. (i) Pb is unaffected by dry air, but in presence of moisture, it get tarnished due to formation of
Pb(OH) 2 first and finally into PbCO 3 which protect it from further reaction:
CO
2
Pb + air → Pb(OH) 2 →
PbCO 3 .
(ii) At temperature below 400°C, Pb combine with oxygen forming litharge (PbO). If the
temperature above 400°C is maintained, red Pb 3O 4 is produced.
O
Heat
2
2Pb + O 2 →
2PbO →
Pb 3O 4
Litharge
450°C
Red-lead
Heat
1626. Pb + 2H 2SO 4 → PbSO 4 + SO 2 + H 2O
PbSO 4 forms an insoluble layer on the metal surface, which causes reaction to stop after some
time.
1627. (a) 3Pb + 8HNO 3 → 3Pb(NO 3 ) 2 + 2NO + 4H 4O
(b) Pb + 4HNO 3 → Pb(NO 3 ) 2 + 2NO 2 + 2H 2O
1
(c) Pb + 2NaOH + 2 O 2 → Na 2 PbO 2 + H 2O
1
(d) Pb + 2CH 3COOH + 2 O 2 → (CH 3COO) 2 Pb + H 2O
1628. Silver is extracted from its sulphide ore by cyanide process. Ore containing Ag 2S is first leached
with an aqueous solution containing NaCN, where insoluble silver sulphide comes in solution by
forming soluble argentocyanide complex.
Ag 2S + 4NaCN → 2Na[Ag(CN) 2 ] + Na 2S
Soluble
The aqueous solution is then treated with a more electropositive metal Zn where Ag is produced :
Zn + 2Na[Ag(CN) 2 ] → 2Ag + Na[Zn(CN) 4 ]
1629. AgBr act as a light sensitive material and used for making photographic film. Light rays reduces
Ag + into Ag. Unreacted AgBr, where light has not fallen, is finally washed away by Na 2S 2O 3
solution:
AgBr + 2Na 2S 2O 3 → Na 3 [Ag(S 2O 3 ) 2 ] + NaBr
Fixer
1630. Ag reacts slowly with trace of H 2S in air, forming a black layer of Ag 2S on surface:
2Ag + H 2S → Ag 2S + H 2
Black
1631. (a) Ag + 2HNO 3 → AgNO 3 + H 2O + NO 2
(b) Ag + 2H 2SO 4 → Ag 2SO 4 + SO 2 + H 2O
(c) AgBr + 2NaCN → Na[Ag(CN) 2 ] + NaBr
(d) AgCl + 2NH 4OH → [Ag(NH 3 ) 2 ]Cl + 2H 2O
1632. Gold (Au), being a noble metal, exist in nature in free state. Powdered ore containing Au is
leached with aqueous solution of KCN in presence of oxygen where Au comes in solution by
forming soluble complex [Au(CN) 2 ]− :
4Au + 8KCN + 2H 2O + O 2 → 4K[Au(CN) 2 ] + 4KOH
709
Solutions
The complex is then treated with electropositive metal Zn and Au comes out of solution in a
redox-displacement reaction as:
Zn + 2K[Au(CN) 2 ] → K 2 [Zn(CN) 4 ] + 2Au.
1633. Au reacts with aqua-regia forming chloro auric acid as:
2Au +11HCl + 3HNO 3 → 3NOCl + 2HAuCl 4 + 6H 2O
1634. Both group I metals and copper group metals have one ‘ s’ electron in their outer orbital, but copper
group metals have ten d-electrons in the penultimate orbit, giving very poor screening effect. The
poor screening gives elements of copper group very small size and high density, high ionization
energies, etc. Also, in Cu-group, d-electrons are involved in metallic bonding. Thus, the melting
points and enthalpy of sublimation are much higher than for alkali metals. All this add up in order
to make elements of Cu-group noble.
1635. MnO 2 is first fused with KOH in presence of oxygen, where it is converted into potassium
manganate K 2 MnO 4 as:
1
MnO 2 + 2KOH + 2 O 2 → K 2 MnO 4 + H 2O
K 2 MnO 4 is then converted into KMnO 4 by one of the following technique:
(i) Passing CO 2 ( g ) :
3K 2 MnO 4 + 2CO 2 → 2K 2CO 3 + 2KMnO 4 + MnO 2
(ii) By the action of dilute H 2SO 4 :
3K 2 MnO 4 + 2H 2SO 4 → 2K 2SO 4 + 2KMnO 4 + MnO 2 + 2H 2O
(iii) By the action of ozone:
2K 2 MnO 4 + O 3 + H 2O → 2KMnO 4 + 2KOH + O 2 .
1636. Chrome-iron is first heated with Na 2CO 3 to produce Na 2CrO 4 as:
7
2FeO ⋅ Cr 2O 3 + 4Na 2CO 3 + 2 O 2 → 4Na 2CrO 4 + Fe 2O 3 + 4CO 2
Roasted ore is extracted with water where Na 2CrO 4 dissolve leaving impurities behind.
Na 2CrO 4 solution is then treated with concentrated sulphuric acid where Na 2SO 4 ⋅ 10H 2O
separate out first followed by crystals of Na 2Cr 2O 7 . Finally aqueous solution of Na 2Cr 2O 7 is
treated with KCl where K 2Cr 2O 7 separate out due to lower solubility than Na 2Cr 2O 7 :
Na 2Cr 2O 7 + 2KCl → K 2Cr 2O 7 + 2NaCl.
Heat
1637. (a) 4K 2Cr 2O 7 → 4K 2CrO 4 + 2Cr 2O 3 + 3O 2
(b) K 2Cr 2O 7 + 2KOH → 2K 2CrO 4 + H 2O
(c) K 2Cr 2O 7 + 4KCl + 3H 2SO 4 →
2CrO 2Cl 2 ↑
(chromyl chloride)
+ 3H 2O + 3K 2SO 4
1638. (a) 2KMnO 4 + 5K 2C 2O 4 + 8H 2SO 4 → 2MnSO 4 + 6K 2SO 4 +10CO 2 + 8H 2O
(b) 2KMnO 4 + 5O 3 + 3H 2SO 4 → 2MnSO 4 + K 2SO 4 +10O 2 + 3H 2O
(c) 6KMnO 4 +10FeC 2O 4 + 24H 2SO 4 → 5Fe 2 (SO 4 ) 3 + 6MnSO 4 + 20CO 2
+ 3K 2SO 4 + 24H 2O
(d) 2KMnO 4 +16HCl → 2MnCl 2 + 5Cl 2 + 2KCl + 8H 2O
710
Problems in Chemistry
1639. (a) In gaseous state, FeCl 3 remains in dimeric state as:
Cl
Fe
Cl
Cl
Cl
Cl
Fe
Cl
(b) In ether, FeCl 3 remains in solvated monomeric form as:
Et
Et
O •• → FeCl 3
(c) In water, it forms a hexahydrated complex as:
FeCl 3 + H 2O → [Fe(H 2O) 6 ]3+ + 3Cl – .
1640. Iodide ion (I − ) being a strong reducing agent, reduces Fe 3+ into Fe 2+ as
Fe 3+ + I − → Fe 2+ + 2 I 2
1
1641. Freshly prepared solutions of K 2SO 4 and Al 2 (SO 4 ) 3 are mixed together where potash alum
crystallizes as:
K 2SO 4 ( aq ) + Al 2 (SO 4 ) 3 ( aq ) → K 2SO 4 ⋅ Al 2 (SO 4 ) 3 ⋅ 24H 2O
Potash olum
Potash alum is mainly used as a mordant in dyeing.