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1. What are the hybridisation states of each carbon atom in the following compounds ?
CH2=C=O, CH3CH=CH2 , (CH3)2CO, CH2=CHCN, C6H6
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•
Solution:
The hybridization state of each carbon in the given compounds, starting from left is given as:
(i) CH2=C=O sp2,sp2
(ii) CH3- CH =CH2 sp3,sp2, sp2
(iii) (CH3)2CO sp3, sp2
(iv) CH2=CHCN sp2sp2, sp
(v) C6H6 sp2.
2. Indicate the σ and π bonds in the following molecules :
C6H6, C6H12, CH2Cl2, CH2=C=CH2, CH3NO2, HCONHCH3.
•
Solution:
C6H6
C6H12
CH2Cl2
CH2=C=CH2
CH3NO2
HCONHCH3
12 σ (6 C-C,6C-H) and 3 π bonds
18 σ bonds (6 C-C,12C-H)
4 σ bonds
6 σ€(2C-C,4 C-H) bonds and 2 π bonds
6 σ (3C−H, 1C−Ν, 2N-O) bonds and 1 π bond
8 σ (4C-H, 1C-O,1N-H,2N-C) bonds and 1 π bond.
•
3. Give the IUPAC names of the following compounds
in
•
Solution:
(a) propyl benzene
(b) 3-methyl pentane nitrile
(c) 2,5- dimethyl heptane
(d) 2- bromo –3- chloroheptane
1
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(e) 3- chloro propanal
(f) 2,2- dichloroethanol
4. Which of the following represents the correct IUPAC name for the
compounds concerned ?
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(a) 2,2-Dimethylpentane or 2-Dimethylpentane
(b) 2,4,7- Trimethyloctane or 2,5,7-Trimethyloctane
(c) 2-Chloro-4-methylpentane or 4-Chloro-2-methyl pentane
(d) But-3-yn-1-ol or But-4-ol-1-yne.
•
Solution:
The correct IUPAC names from each pair is given below:
(a) 2,2, dimethyl pentane,
(b) 2,4,7-trimethyl octane,
(c) 2-chloro-4 methyl pentane.
(d) But-3-yn-1-ol
5. Write the formulas for the first five members of each homologous series beginning
with the following compounds. (a) H–COOH (b) CH3COCH3
•
Solution:
Carboxylic acid: H-COOH, CH3-COOH, CH3-CH2-COOH, CH3-CH2-CH2-COOH,
CH3-CH2-CH2-CH2-COOH
Ketones : CH3-CO-CH3, CH3-CH2-CO-CH3, CH3-CH2-CH2- CO-CH3, CH3-CH2-CH2-CH2-CO-CH3, CH3CH2-CH2-CH2-CH2-CO-CH3
6. Give condensed and bond line structural formulas and identify the functional group(s)
present, if any, for :
(b) 2-Hydroxy-1,2,3-propanetricarboxylic acid Hexanedial
© Hexanedial
•
in
(a) 2,2,4-Trimethylpentane
Solution:
BOND LINE FORMULAE
CONDENSED FORMULAE
FUNCTIONAL GROUP
It is Alkane
(a)
2
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Carboxylic acid group
and Hydroxyl group
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(b)
Aldehyde group
(c)
7. Identify the functional groups in the following compounds:
•
in
Solution:
a) 1) Aldehyde group
2) Ether group
3) Hydroxyl group
b) 1) Amino group
2) Ester group
3) Substituted Amino group
c) 1) Alkenes group
2) Nitro group
8. Which of the two: O2NCH2CH2O– or CH3CH2O– is expected to be more stable and why ?
3
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•
Solution:
O2 NCH2CH2O- is more stable than CH3CH2O-.
The stability of a molecule depends on the resonance stabilization energy which is in turn depends on
the more number of contributing structures which a molecule can provide. The molecule O2N – CH2 –
CH2O- has more contributing structures. Further O2NCH2CH2O- has more number of unpaired
electrons and more number of covalent bonds. Hence it is more stable.
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9. Explain why alkyl groups act as electron donors when attached to a π system.
•
Solution:
Carbon is slightly more electronegative than hydrogen. Thus, carbon atom in an alkyl (say methyl)
group has higher electron density around it as compared with an H atom. As a result, alkyl group are
able to donate electrons inductively when attached to a π system.
10. Draw the resonance structures for the following compounds. Show the electron shift
using curved-arrow notation.
in
4
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•
Solution:
•
in
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11. What are electrophiles and nucleophiles ? Explain with examples.
Solution:
Nucleophiles
A nucleophile is a reagent containing an atom having unshared or lone pair of electrons. A
nucleophile is electron rich and supplies electrons to electron deficient sites, i.e., (nucleus loving).
According to Lewis concepts of acid and base, nucleophiles behave as Lewis bases. Nucleophiles are
electron rich species. They carry a negative charge and they are nucleus loving. They attack electron
deficient centers. Cl-, I-, OH-, CN-, NO-2 CH3O-etc. are examples of nucleophiles.
5
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Neutral molecules like H2O; R3N etc . are also nucleophiles due to the presence of lone pair of
electrons.
NH3, H2O, ROH, ROR (Neutral species).
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Electrophiles
A positively charged on neutral species which are deficient of electrons and can accept a pair of
electrons are called electrophiles. These are also called electron loving (philic) species for examples:
H+, H3O+, Cl+, CH3, NO2+(positively charged)
AlCl3, BF3, SO3 (Neutral species)
Both aluminium and boron have total of six electrons i.e., two less than octet. Therefore, they try to
complete their octet and act as electrophile.
According to Lewis concepts of acids and bases they are called Lewis acids.
12. Identify the reagents shown in bold in the following equations as nucleophiles or
electrophiles:
•
Solution:
(a) CH3COOH + HONegative nucleophile
CH3-COO- + H2O
(b) CH3 - CO - CH3 + NCCN- is a negative nucleophile
CH3 - C (CN) - OH CH3
13. Classify the following reactions in one of the reaction type studied.
in
€€€€€€ ⊕
CH3CO is a positive electrophile
(a) CH3 – CH2 – Br + HS- CH3CH2 – SH + Br(b) (CH3)2C = CH2 + HCl (CH3)2 – Cl – C – CH3
(c) (CH3)3 – C – CH2 – OH + HBr (CH3)2 – CBr . CH2 – CH3
(d) CH3 CH2 – Br + HO- CH2 = CH2
6
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•
Solution:
(a) Nucleophilic substitution
(b) Addition reaction
(c) Nucleophilic substitution followed by rearrangement reaction.
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(d) Elimination reaction.
14. What is the relationship between the members of following pairs of structures ? Are
they structural or geometrical isomers or resonance contributors?
•
Solution:
a) They are positional isomers.
b) They are geometrical isomers.
c) Resonance contributors.
in
d) Geometrical isomers
15. For the following bond cleavages, use curved-arrows to show the electron flow and
classify each as homolysis or heterolysis. Identify reactive intermediate produced as free
radical, carbocation and carbanion.
•
Solution:
7
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16. Explain the terms Inductive and Electromeric effects. Which electron displacement
effect explains the following correct orders of acidity of the carboxylic acids?
•
Solution:
Inductive effect
Inductive effect involves electrons in bonds. It occurs in saturated compounds. The electron pair is
only slightly displaced and there are only partial charges. It is transmitted only a quite short
distance.
Mesomeric effect
Mesomeric effect occurs only in unsaturated compounds. It involves electrons in bonds. The
electrons pair is transferred completely with the result full positive and negative charges are created.
It is transmitted from one end to the other of quite large molecules.
In the first case the inductive effect in di and tri halogen substituted acids is more marked with the
result these acids are progressively more stronger than the corresponding mono halogeno
substituted acid.
In the second case, the decreasing acidic strength is due to increase in +I effect due to alkyl groups.
17. Give a brief description of the principles of the following techniques taking an
example in each case.
(a) Crystallisation
(c) Chromatography
•
in
(b) Distillation
Solution:
Crystallisation
This is the most common method for the purification of solid organic compounds. It is based on the
fact that certain organic compounds are partly soluble in a solvent at room temperature and
solubility increases with increase in temperature. Example: Separation of sugar from a mixture of
sugar and common salt by using C2H5OH.
8
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Distillation
This method is based on the principle that at constant pressure every pure liquid boils at a definite
temperature called its boiling point. The method is used for the purification of those liquids which boil
without decomposition provided the impurities are non- volatile. The method is applied for the
purification when the two liquids differ in their boiling points by 30-50 K. Example: a mixture of
hexane (b.p.342K) and (b.p.384K) is separated by distillation.
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Chromatography
The technique depends on the distribution of the mixture between two phases, one fixed and the
other mobile (moving). The fixed phase may bean absorbent column (a solid chemical compound) or
a paper strip. The moving phase may be a liquid or gas. The mixture to be separated is dissolved in
the moving phase and passed over the fixed phase. There are three types of chromatography.
Adsorption Chromatography
The fixed phase is a solid, e.g., alumina, magnesium oxide, silica gel, etc. and mobile phase is a
liquid. Different constituents are adsorbed in different parts of the adsorbent, the component which
has maximum adsorption affinity for the absorbent is adsorbed at the nearest starting point and the
component having minimum affinity is adsorbed at the farthest point of the adsorbent. Thus different
bonds, zones or chromatograms are formed at different parts of the column. The adsorbed
components are then extracted (eluted) with the help of a suitable solvent.
Partition Chromatography
Here the fixed phase may be solid or liquid strongly adsorbed on a solid support while the mobile
phase is liquid. Paper chromatography is a special case of adsorption chromatography in which
adsorbent column is a paper strip (solid).
Gas Chromatography
Here the fixed phase may be solid or liquid while the mobile phase is a mixture of gases.
18. Describe the method, which can be used to separate two compounds with different
solubilities in a solvent S.
•
Solution:
The two compounds with different solubilities in a solvent S can be separated by the method of
crystallisation.
Crystallization is used to separate two compounds with different solubilities in a solvent ‘S’. Less
soluble component will crystallize first and more soluble component will crystallize on heating again
and then cooling it.
in
19. What is the difference between distillation, distillation under reduced pressure and
steam distillation?
•
Solution:
Distillation
This is used to separate either
(a) Volatile liquid from nonvolatile liquid or solid separately
or
Distillation under reduced
Steam distillation
pressure
This is used to purify liquids This is used for purifying substances
which decompose at or below which are steam volatile and
their boiling points.
immiscible with water.
9
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(b) Two volatile liquids having
their boiling points differing
appreciably (greater than
200C).
a
a
20. Discuss the chemistry of Lassaigne’s test.
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•
Solution:
Preparation of Lassaigne's filtrate or sodium fusion extract: This step is common in all the Lassigne's
tests,i.e. for nitrogen, sulphur and halogens. A small piece (pea-size) of dry metallic sodium is
heated in an ignition (fusion) tube till it melts. A small quantity of the compound is now added to
molten metal and the tube is again heated very strongly till it becomes red hot. The red hot tube is
broken into pieces by immersing it in cold distilled water in a china dish (if necessary, break the tube
with the help of tongs). The contents of the china dish are boiled and then filtered. The filtrate
obtained is known as Lassaigne's extract and can be used for testing the presence of N, S and
halogens in the given compound.
Chemical reactions involved in the formation of Lassaigne's extract.
If N is present:
Na + C+N
NaCN
Metallic fusion from organic compound Sodium cyanide
If S is present :
fusion
2Na + S
Na2S
(sod. Sulphide)
If both N and S are present:
fusion
Na + C + N+ S
NaCNS
From organic compound Sod. Sulphocyanide
If X is present:
where
NaX (Sod. Halide)
X = Cl, Br or I
in
Na + X
(a)
(i) Test of nitrogen from Lassaigne's filtrate. The filtrate is tested for alkaline nature, if it is not
alkaline it is made so by adding a few drops of sodium hydroxide solution. Now 1-2 ml of freshly
prepared ferrous sulphate solution is added, the contents are boiled and then cooled. Dilute sulphuric
or hydrochloric acid is added to disslove ferrous hydroxide. Appearance of green or prussian blue
colour confirms the presence of nitrogen in the organic compound.
10
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Chemistry of the test: Sodium cyanide formed during the fusion of the compound with sodium reacts
with ferrous sulphate to form sodium ferrocyanide, Na4[Fe(CN)6]. The latter reacts with ferric
sulphate (produced by the oxidation of ferrous sulphate by air in presence of OH-) and give a
precipitate of prussian blue coloured compound, ferric ferrocyanide.
2NaCN + FeSO4 → Fe (CN)2↓ + Na2SO4
Ferrous cyanide
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Fe(CN)2 + 4NaCN → Na4[Fe(CN)6]
Sod. Ferrocyanide
3Na4[Fe(CN)6] + 2Fe2(SO4)3 → Fe4[Fe(CN)6]3 + 6Na2SO4
Ferric ferrocyanide (Prussian blue)
When compound contains both N and S then sodium sulphocyanide is formed which gives a blood red
colour on the addition of ferric chloride solution.
3NaCNS + FeCl3
Sod. Sulphocyanide
→
Fe(CNS)3 + 3 NaCl
Ferric sulphocyanide(blood red)
But the absence of blood red colouration does not necessarily mean that sulphur is absent. This is
due to the fact that in the presence of excess of sodium, sodium thiocyanate decomposes to form
sodium cyanide and sodium sulphide.
2Na + NaCNS → Na2S + NaCN.
21. Differentiate between the principle of estimation of nitrogen in an organic compound
by (i) Dumas method and (ii) Kjeldahl’s method.
Solution:
(i) Principal of Duma's method
A known mass of the given organic compound is heated strongly with excess of cupric oxide in an
atmosphere strongly with excess of cupric oxide in an atmosphere of CO2. Carbon and hydrogen are
oxidized to CO2 and H2O respectively while nitrogen present in the compound is set free. A small
amount of the oxides of nitrogen if formed are reduced copper gauze.
Heat
C + 2CuO
(Compound)
CO2 + Cu
in
Heat
2H + CuO
(Compound)
H2O + Cu
Heat
2N + CuO
N2 + small amount of oxides of -nitrogen.
Heat
Oxides of nitrogen + Cu
N2 + CuO
11
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(Reduced)
The gaseous mixture is passed through the rough concentrated KOH solution which absorb both
CO2 and H2O vapours. Nitrogen (N2) is not absorbed by KOH and gets collected over it. the volume of
nitrogen evolved is noted and from this the amount of nitrogen or percentage can be calculated.
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(ii) Principal of Kjeldahl's method
A known mass of the organic compound is digested (heated strongly) with conc. H2SO4 and a little
potassium sulphate (to increase the b.p. of H2SO4) and a little mmercury or copper sulphate (a
catalyst) in a long necked flask called Kjeldahl's flask. As a result of digestion, the nitrogen present in
the organic compound is converted into ammonium sulphate. The compound so obtained in the form
of a solution ammonia goes thus, liberated is absorbed in a known volume of a standard solution
(molarity known) of an acid such as H2SO3
Digertiony
N + H2SO4 → (NH4)2SO4
(from organic
compound)
(NH4)2SO4 + 2NaOH
2NH3 + H2SO4
Na2SO4 + 2H2O +2NH3
(NH4)2SO4
The volume of the acid left unused is determined by titrating it against a standard alkali solution such
as NaOH.
H2SO4 + 2NaOH → Na2SO4 + H2O.
22. Discuss the principle of estimation of halogens, sulphur and phosphorus present in
an organic compound.
Solution:
Principle of estimation of halogens
Carius method involves conversion of the halogen present in any organic compound into the
corresponding silver halide. From the mass of the organic compound taken and the mass of silver
halide formed, the percentage of halogen in the compound can be calculated.
in
Principle of estimation of sulphur
Sulphur present in any organic compound is oxidized to sulphuric acid by disgesting with fuming
nitric acid. Sulphuric acid so formed is then quantitatively precipitated as barium sulphate by the
addition of excess barium chloride. The precipitate of barium sulphate is filtered, washed, dried and
weighed.
Principle of estimation of phosphorus
The phosphorous present in an organic compound is oxidized to phosphoric acid by heating with
fuming nitric acid. The phosphoric acid so obtained is precipiitated as MgNH4PO4, which on ignition is
convered into Mg2P2O7. Then, from the mass of Mg2P2O7, the mass of phosphorous present in the
sample of the organic compound can be calculated.
23. Explain the principle of paper chromatography.
12
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Solution:
Paper chromatography is a special type of partition chromatography. It is based on the differential
(continuous) partitioning of components of a mixture between stationary phase and mobile phase.
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In paper chromatography, a strip of special paper spotted at the base with the solution of the
mixture is suspended in a suitable solvent or a mixture of solvents. This solvent acts as the mobile
phase. The solvent rises up the paper by capillary action and flows over the spot. The paper
selectively retains different components according to their differing partition in the two phases. The
paper strip so developed is known as chromatogram. The spots of the separate coloured compounds
are visible at different heights from the position of initial spot on the chromatogram. The spot of the
separate colourless compounds may be observed either under ultraviolet light or by the use of an
appropriate spray reagent.
24. Why is nitric acid added to sodium extract before adding silver nitrate for testing
halogens?
Solution:
When we test halogens from sodium extract, a small amount of dil. HNO3 is added. When we test
from Lassaigne’s extract, it contains halogens. But if nitrogen and sulphur are also present along
with halogens in the organic compound, the Lassaigne’s extract contains sodium sulphide and sodium
cyanide (Na2S and NaCN) along with sodium halide. Nitric acid decomposes sodium cyanide and
sodium sulphide which otherwise form precipitate with silver nitrate and thus interfere with the test.
NaCN + HNO3 → NaNO3 + HCI
Na2S + 2HNO3 → 2NaNO3 + H2S
Therefore, dil nitric acid is added before testing halogens to expel all the gases if evolved.
25. Explain the reason for the fusion of an organic compound with metallic sodium for
testing nitrogen, sulphur and halogens.
Solution:
The elements, nitrogen, sulphur and halogens are covalently bonded to carbon atoms in the organic
compounds. To make their detection easy, these are to be first converted into the ionic forms.
This is done by fusing the organic compound with sodium metal. The ionic compounds formed during
the fusion are extracted in aqueous solution, and can be detected by simple chemical tests.
C+N+ Na
From organic
compound
extracted
+Na+(aq) + CN-(aq)
NaCN
in
Fusion
sodium in water cyanide
Fusion
X (Cl, Br, I) + Na
Fusion
NaX (X= Cl, BrI)
Extracted in water
Na+(aq) + X-(aq)
extracted
13
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S + 2Na
From organic
compound
Na2S
in Water
sodium sulphide
2Na+(aq) + S2-(aq)
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If nitrogen and sulphur both are present in any organic compound, sodium thiocyanate (NaSCN) is
formed during fusion which is the presence of excess sodium, forms sodium cyanide and sodium
sulphide.
Fusion extracted
Na + C + N + S → NaCNS → Na+(aq) + CNS-(aq)
In the organic
Compound
Sodium in water
thiocyanate
NaCNS + 2Na
3Na+(aq) + CN-(aq) + S2- (aq)
extracted in water
NaCN + Na2S
12.27 Name a suitable technique of separation of the components from a mixture of calcium sulphate
and camphor.
Calcium sulphate and camphor present in a mixture can be separated by sublimitation. Camphor
sublimes in heating and can be recovered in crystalline form as sublimate.
26. Explain, why an organic liquid vaporises at a temperature below its boiling point in
steam distillation?
Solution:
In steam distillation, the liquid starts boiling when sum of the vapour pressures of the organic liquid
(p2) and that due to water vapour (p1) becomes equal to the atmosphere (p) i.e.,
P= p1+ p2
Since a liquid boils as soon as its vapour pressure becomes equal to the atmospheric pressure, a
mixture of two immiscible liquids will boil at a temperature lower than the normal boiling points of
both the liquids. The mixture will continue to boil at the same temperature until one of the liquids is
completely distilled out.
in
27. Will CCl4 give white precipitate of AgCl on heating it with silver nitrate? Give a reason
for your answer.
Solution:
CCl4 will not give white precipitate of AgCl due to the absence of electron releasing group on the
carbon atom bonded to the Cl atoms (CCl4 is a non-polar solvent.)
28. Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved
during the estimation of carbon present in an organic compound?
Solution:
During the estimation of percentage of carbon in the organic compound, carbon is oxidised by
heating the substance in excess of oxygen.
14
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C + O2
CO2
CO2, being acidic oxide is absorbed by KOH. So CO2 formed in the estimation of carbon, is absorbed
in a known weight of KOH. From the increase in the weight of KOH, percentage of carbon in the
organic substance is estimated.
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29. Why is it necessary to use acetic acid and not sulphuric acid for acidification of
sodium extract for testing sulphur by lead acetate test?
Solution:
For testing sulphur from sodium extract, it is acidified with acetic acid but not sulphuric acid .
Because SO42-ions from H2SO4 reacts with Pb++ from Lead Acetate, giving a white ppt of PbSO4,
though sulphur is not present in the organic compound.
H2SO4
2H+ + SO42-
Pb2 +SO42PbSO4
(from Pb acetate)
30. An organic compound contains 69% carbon and 4.8% hydrogen and the remaining
percentage of oxygen. Calculate the masses of carbon dioxide and water produced when
0.20 g of this substance is subjected to complete combustion.
Solution:
Percentage of carbon = 69%
Percentage of hydrogen = 4.8%
Mass of substance = 0.2g
Mass of CO2, formed
C + O2
12g
CO2
44g
Amount of carbon present in 0.2g of the substance =
12g of carbon on complete combustion gives = 44 g of CO2
= 0.138g
in
0.138g of carbon gives =
x 0.138
= 0.506 g of CO2
Mass of water formed
Amount of hydrogen, present in 0.2 g of the substance =
2H2 + O2
4g
= 0.0096g
2H2O
36g
4g of H2 on complete combustion gives = 36g of water
0.0096g of H2 …………..
= 0.0864 g of water.
31. A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s
15
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method. The ammonia evolved was absorbed in 50 ml of 0.5 M H2SO4. The residual acid
required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage
composition of nitrogen in the compound.
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Solution:
Mass of organic compound = 0.50 g
Initial volume of acid = 50 mL
Molarity of acid = 0.5 M
Volume of NaOH of 0.5 M molarity required for neutralization of residual acid = 60 mL
= 30 mL of 0.5 M H2SO4
Volume of H2SO4(0.5 M) consumed = (50 - 30) = 20
20 ml of 0.5 M H2SO4 = 40 mL of 0.5 M NH3 or 20 mL of 1 M NH3.
1000 mL of 1 M NH3 contains 14 g Nitrogen (1 mol of N2)
20 mL of 1 M NH3 contains
g Nitrogen
0.5g of organic compound contains
gN
100 g of organic compound contains =
g= 56% of N2.
32. 0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius
estimation. Calculate the percentage of chlorine present in the compound.
Solution:
Mass of silver chloride = 0.5740 g
Molecular mass of silver chloride = 108 + 35.5 = 143.5
143.5g of silver chloride contains = 35.5 g
∴ .5740 g of AgCl will contain =
Percentage of chlorine =
= .142 g
x 100 = 37.57 %
33. In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur
compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the
given compound.
Molecular mass of barium sulphate = 233 g
in
Solution:
Mass of barium sulphate = 0.668 g
233 g of barium sulphate contains sulphur = 32g
0.668 g of barium sulphate contains =
16
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=
= 19.66%
34. In the organic compound CH2 = CH - CH2 - CH2 - C = CH, the pair of hybridised
orbitals involved in the formation of: C2 - C3 bond is:
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(a) sp - sp2
(b) sp - sp3
(c) sp2 - sp3
(d) sp3 - sp3
Solution:
sp3 – sp3.
35. In the Lassaigne’s test for nitrogen in an organic compound the Prussian blue colour
is obtained due to the formation of:
Na4 [Fe(CN)6]
Fe4[Fe(CN)6]3
Fe2[Fe(CN)6]
Fe3[Fe(CN)6]4
Solution:
Fe4[Fe(CN)6]3.
36. Which of the following carbocation is most stable ?
a) (CH3)3C.+CH2
b) (CH3)3 C+
c) CH3CH2+CH2
d) CH3+CHCH2CH3
in
Solution:
b) (CH3)3 C+ .
37. The best and latest technique for isolation, purification and separation of organic
compounds is:
(a) Crystallisation
(b) Distillation
(c) Sublimation
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(d) Chromatography
Solution:
(d) Chromatography.
38. The reaction
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is classified as
(a) electrophilic substitution reaction
(b) nucleophilic substitution reaction
(c) elimination reaction
(d) addition reaction
Solution:
(b) nucleophilic substitution reaction.
in
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