(B) Study Material for High achievers -:
1.
Solutions:
Solutions are homogeneous mixtures of two or more than two components. Binary
solution → consists of two components
Types of solutions:
Gaseous solution → solvent is gas, e.g., mixture of oxygen and nitrogen gases Liquid
solution → solvent is liquid, e.g., oxygen dissolved in water
Solid solution → solvent is solid, e.g., solution of hydrogen in palladium
Expressing concentration of solutions:
Mass percentage (w/w) of a component
Volume percentage ( v/v) of a component =
Mass by volume percentage (w/v) =
Parts per million =
Mole fraction of a component =
Molarity (M) =
Molality (m) =
Solubility:
Solubility of a substance is the maximum amount of the substance that can be
dissolved in a specified amount of a solvent at a specified temperature.
Solubility of a solid in a liquid:
-Effect of temperature (Le Chatelier’s principle) –
When the dissolution process is endothermic, the solubility will increase with increase in
temperature.
When the dissolution process is exothermic, the solubility will decrease with increase in
temperature.
-Effect of pressure: Pressure does not affect solubility.
Solubility of a gas in a liquid:
Henry’s law → The solubility of a gas in a liquid is directly proportional to the pressure
of the gas.
Or
The partial pressure of a gas in vapour phase (p) is proportional to the mole fraction
of the gas (x) in the solution.
Where, KH → Henry’s law constant
Some applications of Henry’s law –
-The solubility of CO2 in soft drinks and soda water is increased by sealing the bottles
under high pressure.
-At high pressure underwater, scuba divers have to cope with high concentrations of
dissolved gases while breathing air.
-At high altitudes, climbers become weak and are unable to think clearly, which are symptoms
of a condition called anoxia
Effect of temperature –
With increase in temperature, the solubility of gases in liquids decreases.
Vapour pressure of liquid solutions:
Vapour pressure of liquid–liquid solutions:
Raoult’s law → For a solution of volatile liquids, the partial vapour pressure of each
component in the solution is directly proportional to its mole fraction.
(For an ideal solution at constant temperature)
Vapour pressure of solutions of solids in liquids:
Ideal and non-ideal solutions:
Ideal solutions –
Solutions which obey Raoult’s law over the entire range of concentrations
-In an ideal solution, the solute–solute and solvent–solvent interactions are nearly equal to
the solute–solvent interactions.
Non-ideal solutions – Solutions which do not obey Raoult’s law over the entire range
of concentrations
-Positive deviation from Raoult’s law –
-Vapour pressure of the solution is higher than that predicted by Raoult’s law.
-Solute–solvent interactions are weaker than solute–solute and solvent–solvent
interactions.
-Negative deviation from Raoult’s law –
-Vapour pressure of the solution is lower than that predicted by Raoult’s law.
-Solute–solute and solvent–solvent interactions are weaker than solute–solvent
interactions.
2
Electrochemical cells:
Two types – Galvanic cell or voltaic cell, and Electrolytic cell
Galvanic cell:
Converts the chemical energy of a spontaneous redox reaction into electrical energy
Daniell cell -
The standard potential of a cell is given by
The potential of an individual half cell cannot be measured.
Relation between the standard potential of a cell and standard Gibbs energy:
Relation between standard Gibbs energy and the equilibrium constant:
Nernst equation:
Gives the concentration dependence of the potentials of the electrodes and the
cells
For the electrode reaction
Nernst equation is given by
For a general electrochemical reaction of the type
Conductance of electrolytic solutions:
Resistance, R =
Where,
l → Length
A → Area of cross-section
Ρ → Resistivity or specific resistance
Conductance, G =
Where,
Conductivity or specific conductance
The SI unit of conductance is Ω
–1
(siemens or mho).
The conductivity of an electrolyte depends upon
nature of the solvent
nature of the electrolyte added
concentration of the electrolyte
temperature
Molar conductivity,
Variation of conductivity
For both strong and weak electrolytes, conductivity decreases with decrease in
concentration.
Variation of molar conductivity
For both strong and weak electrolytes, molar conductivity increases with
decrease in concentration.
Limiting molar conductivity – molar conductivity when concentration approaches zero
Degree of dissociation,
Kohlrausch law of independent migration of ions:
According to this law, for an electrolyte, the molar conductivity at infinite dilution is the
sum of the contribution of the molar conductivity of the ions in which it dissociates.
Electrolytic cells and electrolysis
1F = 96487 C mol
-1
Faraday’s first law of electrolysis: The amount of chemical reaction occurring
at any electrode during the process of electrolysis by a current is proportional to
the quantity of electricity passed through the electrolyte.
Second law of electrolysis: The amounts of different substances liberated when
same quantity of electricity is passed through the electrolytic solution are
proportional to their chemical equivalent weights.
Battery is a galvanic cell in which chemical energy of the redox reaction is
converted into electrical energy.
Mainly two types:
Primary batteries
Secondary batteries
Primary Batteries
In primary batteries, reaction occurs only once.
After use over a period of time, these become dead and cannot be reused.
Examples: Dry cell (or Leclanche cell), Mercury cell
Secondary Batteries
Secondary batteries can be recharged again by passing current through
them in the opposite direction.
Examples:Lead storage battery, Nickel-cadmium cell
3.
For a reaction R → P
Rate of reaction
Factors influencing the rate of a reaction:
Rate of a reaction depends upon the concentration of reactants (pressure in the case of
gas), temperature and catalyst.
Rate expression and rate constant
aA + bB → cC + dD
Rate expression
Differential rate equation ®
Where, k is called rate constant
Order of a reaction:
x is Order of the reaction with respect to A
y is Order of the reaction with respect to B
x + y is Overall order of the reaction
1. Order of a reaction can be 0, 1, 2, 3 and even a fraction
2. Units of rate constant aA
+ bB → cC + dD
x
y
Rate = k [A] [B]
x + y = n = Order of the reaction
[[A] = [B] and x + y = n = Order of the reaction] 1. For
a zero-order reaction, n = 0
Unit of
1. For a first-order reaction, n = 1
Unit of
1. For a second-order reaction, n = 2
Unit of
Molecularity of a reaction:
The number of reacting species (atoms, ions or molecules) taking part in an
elementary reaction
Order versus molecularity
1. Order can be zero and even a fraction. But molecularity cannot be zero or a noninteger.
2. Order is applicable to both elementary and complex reactions whereas
molecularity is applicable to elementary reactions only.
Integrated rate equations:
Zero-order reactions:
R→P
First-order reactions:
R→P
4.
Occurrence of metals:
Metal
Aluminium
Iron
Copper
Zinc
Ores
Bauxite
Kaolinite (a form of
clay)
Haematite
Magnetite
Siderite
Iron pyrites
Copper pyrites
Malachite
Cuprite
Copper glance
Zinc blende or
Sphalerite
Calamine
Zincite
The major steps involved in the extraction and isolation of metals from ores are:
3. Concentration of the ore
4. Isolation of the metal from the concentrated ore
5. Purification of the metal
Concentration of ores:
Hydraulic washing: It is the washing away of lighter gangue particles from the heavier
ore. It is based on the gravity difference between the ore and the gangue particles.
Magnetic separation: This separation is carried out if either the ore or the gangue is
attracted by a magnetic field.
Froth floatation method: This method is used for removing gangue from sulphide ores.
‘Depressants’ are used for separating two sulphide ores. E.g., for separating ZnS and
PbS, NaCN is used as the depressant.
Leaching: If the ore is soluble in some suitable solvent, then this process is used. For
example, ores of aluminium (bauxite), silver and gold
1. Leaching of alumina
Al2O3(s) + 2NaOH(aq) + 3H2O(l) ® 2Na[Al(OH)4](aq)
2Na[Al(OH)4](aq) + CO2(g) ® Al2O3.xH2O(s) + 2NaHCO3(aq)
Al2O3.xH2O(s)
Al2O3(s) + xH2O(
Isolation of crude metal from concentrated ore: It involves two steps –
(i) Conversion into oxide and (ii) Reduction of the oxide to metal
Conversion into oxide:
1. Calcination → Involves heating
Generally, carbonate ores are converted into oxides by this process.
ZnCO
3(s)
ZnO
(s)
+ CO
2(g)
1. Roasting → Involves heating in a regular supply of air, at a temperature below the
melting point of the metal.
2ZnS + 3O2 → 2ZnO + 2SO2
Generally, sulphide ores are converted into oxides by this process.
Slag → FeSiO3
Reduction of the oxide to metal:
Involves heating with some reducing agents such as C, CO or another metal.
Thermodynamic principles of metallurgy: For any process, the change in Gibbs energy at a
temperature is given by
ΔG = ΔH – TΔS and
ΔG
Ɵ
= –RT ln K
A reaction will proceed when the value of ΔG is negative.
Applications:
If ΔG(X, XO) is lower than ΔG(Y, YO), then X can reduce YO.
5.
Classification: Mono, di and polyhalogen (tri, tetra, etc.)
1. Alkyl halides or haloalkanes (R−X) → They form homologous series of general formula
CnH2n+1X. They are further classified into primary, secondary, and tertiary.
2.
Allylic halides → Compounds containing halogen atom bonded to an allylic carbon
3
(iii) Benzylic halides → Compounds containing halogen atom bonded to an sp hybridised
carbon atom next to an aromatic ring
Methods of preparation:
From alcohols –
From hydrocarbons –
By free radical halogenations- Yields a complex mixture of isomeric mono– and
polyhaloalkanes
By electrophilic substitution
Sandmeyer’s reaction –
From alkenes –
Addition of hydrogen halides to unsaturated hydrocarbons (Markovnikov’s rule)
Addition of halogens
(Method used for detecting double bond in a molecule) In this
method, a reddish-brown colour is discharged.
Halogen exchange –
Finkelstein reaction
Swarts reaction (synthesis of alkyl fluoride) –
Physical properties:
Melting and boiling points –
Halides have higher boiling points than hydrocarbons of comparable molecular
mass because of having stronger dipole–dipole and van der Waals’ forces of
attraction.
The order of increasing boiling points of the different haloalkanes is:
RF < RCl < RBr < RI
The boiling points of isomeric haloalkanes decrease with increase in
branching.
Density –
The density of halides increases with increase in the number of carbon atoms, halogen
atoms and atomic mass of the halogen atoms.
Solubility –
Soluble in organic solvents, but only slightly soluble in water
Reactions of haloalkanes –
Nucleophilic substitution reaction:
Mechanism
Substitution nucleophilic bimolecular (SN2) (Inversion of configuration)
The increasing order of reactivity is
o
o
o
3 halide < 2 halide < 1 halide < CH3X
o
o
(Due to increase in hindrance by bulky substituent in the case of 2 and 3 halides)
Substitution nucleophilic unimolecular (SN1) (Two-step mechanism)
Step I is the slowest; it is reversible and rate-determining step.
6
Classification of alcohols and phenols:
Mono, di, tri or polyhydric –
On the basis of hybridisation –
6.
– OH
Primary, secondary and tertiary
Allylic alcohols
2.
3.
1. Benzylic alcohols
2.
– OH
a. Vinylic alcohol
b. Phenol
Classification of ethers:
Simple or symmetrical –
C2H5OC2H5, C6H5OC6H5
Mixed or unsymmetrical –
C2H5OCH3, CH3OC6H5
Common names of some phenols:
Preparation of alcohols –
1. From alkenes
2. Acid catalysed hydration
[According to Markovnikov’s rule]
1. Hydroboration – oxidation
1. From carbonyl compounds
2. By reduction of aldehydes and ketones.
Aldehydes give 1° and ketones give 2°alcohols.
1. By reduction of carboxylic acids and esters
7.
Nomenclature of aldehydes and ketones –
Aldehydes:
Often called by their common names instead of IUPAC names.
Ketones:
Derived by naming two alkyl or aryl groups bonded to the carbonyl group
Alkyl phenyl ketones are usually named by adding the acyl group as prefix to phenone.
IUPAC Nomenclature
7. For open-chain aliphatic aldehydes and ketones, IUPAC names are derived from the
names of the corresponding alkanes by replacing the ending ‘−e’ with ‘−al’ and ‘−one’
respectively.
8. In the case of aldehydes, the longest chain is numbered starting from the carbon
of the aldehydic group.
9. In the case of ketones, the numbering begins from the end nearer to the
carbonyl group.
10. Substituents are prefixed in the alphabetical order along with the numerals indicating
their positions in the carbon chain.
11. Same rule is applicable to cyclic ketones.
12. If the aldehydic group is attached to a ring, then the suffix carbaldehyde is added to
the full name of cyclohexane.
Example:
2
Structure of the carbonyl group –The carbonyl carbon atom is sp hybridised and forms
three bonds and one π bond.
C=O double bond is polarised due to higher electronegativity of oxygen relative to carbon.
Carbonyl carbon − an electrophile (Lewis acid)
Carbonyl oxygen − a nucleophile (Lewis base)
Preparation of aldehydes and ketones:
(I) By oxidation of alcohols –
2. Primary alcohols
3. Secondary alcohols
Aldehydes
Ketones
(II) By dehydrogenation of alcohols
(i) Primary alcohols
(ii) Secondary alcohols
Aldehydes
Ketones
Preparation of aldehydes:
(I) From acyl chloride –(Rosenmund reduction)
(II) From nitriles – (Stephen reaction)
(III) From esters –
(IV) From hydrocarbons –
By oxidation of methylbenzene (toluene) (Etard reaction)
By side-chain chlorination, followed by hydrolysis
Gatterman–Koch reaction
Preparation of ketones:
(I) From acyl chlorides –
8.
Amines:
13. Derivatives of ammonia
14. Obtained by the replacement of one, two or all the three H-atoms of ammonia
by alkyl and/or aryl group
15. Structure of Amines
3
Nitrogen on amines is sp hybridised.
Geometry − Pyramidal
Example: Pyramidal shapes of trimethylamine
Classification: Primary (1 ), secondary (2 ) and tertiary (3 )
4. If one H-atom of NH3 is replaced by R or Ar, RNH2 or ArNH2 is obtained
(primary amine, 1°).
5. If two H-atoms of NH3 or one H-atom of RNH2 are replaced by alkyl or aryl
group (R′), R2NH is obtained (secondary amine, 2°).
6. On the replacement of another hydrogen atom by alkyl or aryl group, R 3N
is obtained (tertiary amine, 3°).
Nomenclature
Common System
Aliphatic amine: Named by prefixing alkyl group to amine, i.e., alkylamine.
3.
Reduction of Amides
4.
Gabriel Pthalimide Synthesis: Used for the preparation of primary amine.
(vi) Hoffmann Bromamide Degradation Reaction:
Physical properties:
Solubility –
3. Lower aliphatic amines are soluble in water, but higher amines are insoluble in
water. This is because lower amines can form H-bonds with water, but higher
amines cannot.
4. Solubility of amines in water decreases with increase in molar mass of amines
as the size of the hydrophobic alkyl part increases.
5. The order of increasing boiling points of isomeric amines is
o
o
o
3 amine < 2 amine < 1 amine. This because of intermolecular association of
1amines due to H-bonding.
Practice Questions
A
Q1) What will happen to the boiling point of a solution if the weight of the solute dissolved in it is doubled and the
weight of the solvent taken is reduced by half?
Answer-If the weight of the solute dissolved is doubled and the weight of the solvent is reduced by half,
then an elevation takes place in the boiling point of the solution. The boiling point of the solution becomes
four times the original value.
Q2) What is meant by ‘10% aqueous solution of sodium carbonate’?
Answer
'10% aqueous solution of sodium carbonate' means that 10 g of sodium carbonate (solute) is present in
100 g of solution containing water and sodium carbonate.
Q3) When is the value of Van’t Hoff factor more than unity?
Answer
The value of Van’t Hoff factor is more than unity when a solute undergoes dissociation in the solution.
Q4)-Under what conditions do non-ideal solutions exhibit negative deviations from Raoult’s law?
Answer
Non-ideal solutions exhibit negative deviations from Raoult’s law if the vapour pressure of the solution is
lower than the value predicted by Raoult’s law. This happens when the forces of interaction between the
compounds are greater than those in the pure components.
Q5) Between 0.1 molal solutions of glucose and sodium chloride, which will have a higher boiling point?
Answer
0.1 M solution of NaCl will have a higher boiling point than 0.1 M solution of glucose. This is because
sodium chloride undergoes dissociation in the solution. Thus, an elevation is observed in the boiling point.
Q6) The mole fraction of the solute of an X molal solution of a compound in benzene is 0.2. What is the value
of X?
Answer
Let the mole fraction of solute X be x2.
Let the mole fraction of the solvent be x1.
Hence,
And,
Taking,
(Molecular mass of C6H6 = 78)
Hence, the molalityof the solutionis 3.20.
Q7) What happens to molarity when the temperature of a solution is increased?
Answer
When the temperature of a solution is increased, the molarity decreases. This is because the volume of
the solution increases with an increase in temperature but the number of moles of the solute remains the
same.
Molarity =
Hence, molarity is a function of temperature.
Q8) a. The freezing point of a solution containing 0.3 g of acetic acid in 30.0 g of benzene is lowered by 0.45 K.
Calculate Van’t Hoff factor. (Kffor benzene = 5.12 Kkgmol−1)
b. The osmotic pressure of a 0.0103 molar solution of an electrolyte is found to be 0.70 atm at 27°C. Calculate Van’t
Hoff factor. (R = 0.082 Latmmol−1K−1)
Answer
a. Depression in freezing point can be calculated as follows:
ΔTf(calculated)
Q9)
a. The outer shell of two eggs is removed and kept in dilute HCl. Then, one shell is placed in distilled water, while
the other is placed in a saturated solution of NaCl. What will be observed?
b. Explain why a bottle containing liquid ammonia is kept in ice before it is opened.
Answer
a. The egg shell kept in distilled water will get swollen. On the other hand, the egg shell kept in NaCl
solution will shrink. This happens because of osmosis, in which the net flow of solvent from the less
concentrated to the more concentrated solution takes place. Here, the membrane beneath the outer shell
of the egg acts as a semi-permeable membrane.
b. The vapour pressure of liquid ammonia at room temperature is very high. On cooling, the vapour
pressure inside the bottle containing liquid ammonia decreases. As a result, on opening the bottle, liquid
ammonia does not splash out.
Q10)
a. Name and explain the factor introduced in 1880 to account for the extent of association or dissociation.
b. Arrange the following solutions in the increasing order of their osmotic pressure. Give reasons for your answer.
i. 34.2 g/L Sucrose
ii. 60 g/L Urea
iii.58.5 g/L Sodium chloride
Answer
a. In 1880, Van’t Hoff introduced the Van’t Hoff factor i.e., ‘i’ to account for the extent of association or
dissociation. The Van’t Hoff factor i.e., 'i' is defined as the ratio of experimental value of colligative
property to its calculated value i.e.,
Q11)
a. Why is molality preferred over molarity while expressing the concentration of a solution?
b. Why does the boiling point of water increase when sodium chloride is added to it?
c. Why is phenol partially soluble in water?
Answer
a. While molarity decreases with an increase in temperature, molality is independent of temperature. This happens
because molality involves mass, which does not change with a change in temperature, while molarity involves
volume, which is temperature dependent. Hence, molality is preferred over molarity while expressing the
concentration of a solution.
b. When a non-volatile solute such as sodium chloride is dissolved in water, the vapour pressure of water decreases.
This happens because on addition of NaCl, some of the solvent molecules on the surface are replaced by the nonvolatile solute molecules. Hence, the solution has to be heated at a higher temperature to make the vapour pressure
equal to the external pressure. Hence, the boiling point of the solution increases.
c. As a general rule, like dissolves like. Phenol has a polar −OH group but an aromatic phenyl C 6H5 group. Hence, it
is partially soluble in water.
B.
Q1) Write the structure of the compound, β-methylbutyraldehyde.
Answer
The structure of β-methylbutyraldehyde is:
Q2) What type of reactions are shown by benzaldehyde towards nucleophiles?
Answer
Benzaldehyde shows nucleophilic substitution reactions at ortho and para positions. This is due to the
electron withdrawing nature of carbonyl group, which decreases the electron density at ortho and para
positions.
Q3) Give a commercial method to obtain benzaldehyde from toluene.
Answer
Benzaldehyde can be obtained from toluene by the following method.
Q4) Write the Rosenmund reduction reaction of acetyl chloride.
Answer
The Rosenmund reduction reaction of acetyl chloride is:
Q5) Write the IUPAC name of the following compound.
Answer
The IUPAC name of the given compound is
2-(2-Bromophenyl) ethanal.
Q6)
a. What happens when acetone is treated with sodium bisulphite?
b. Where will the equilibrium lie in the reaction in part (a)?
Answer
a.
Acetone, when treated with sodium bisulphite, forms an addition product i.e., acetone bisulphite.
b. The equilibrium lies to the left hand side of the reaction.
Q7) Between acetaldehyde and formaldehyde, which will undergo Cannizzaro’s reaction? Also, write the chemical
equation for the reaction involved?
Answer
Formaldehyde (HCHO) does not contain an α-hydrogen atom. Thus, it will undergo Cannizzaro’s reaction.
On the other hand, acetaldehyde (CH3CHO) will not undergo Cannizzaro’s reaction.
Q8) Giving reasons, arrange the following carbonyl compounds in the decreasing order of their reactivity in
nucleophilic addition reactions.
Answer
Reactivity in nucleophilic addition reactions decreases in the following order.
This is due to the following reasons:
i.
Inductive effect:- The alkyl group has +I effect. The greater the number of alkyl groups attached to carbonyl
group, the greater will be electron density on carbonyl carbon. As a result, its reactivity is lower towards
nucleophilic addition reactions.
ii.
Steric effect:- As the number of alkyl group increases, the attack of the nucleophile on the carbonyl group
becomes more and more difficult due to steric hindrance.
Q9)
a. Give two chemical tests to distinguish between CH3CHO and CH3COCH3.
b. Give a chemical test to distinguish between CH3CHO and HCHO.
Answer
i.
a. CH3CHO and CH3COCH3 can be distinguished by the following tests:
Tollen’s test:- On warming CH3CHO with Tollen’s reagent (ammonical silver nitrate), a bright silver mirror is
produced. CH3COCH3, on the other hand, does not reduce Tollen’s reagent.
ii.
Fehling’s test:- When CH3CHO is heated with Fehling’s reagent, a reddish brown precipitate is formed while
CH3COCH3 does not give this test.
b. CH3CHO and HCHO can be distinguished by the iodoform test.
CH3CHO forms a yellow precipitate of iodoform on reaction with NaOI, whereas HCHO does not give this reaction.
Q10)
An aromatic organic compound ‘A’ gives a yellow precipitate with 2, 4-DNP hydrazine. It also gives a silver mirror
on reaction with Tollen’s reagent. Now, two molecules of ‘A’ react with concentrated NaOH forming ‘B’ and ‘C’.
Out of ‘B’ and ‘C’, the less acidic product on reaction with CrO3-H2SO4 gives ‘D’. The compound ‘D’ reacts with
Br2|FeBr3 to give ‘E’.
Indentify compounds A -E. Also, give the reactions involved.
Answer
Compound ‘A’ gives a yellow precipitate with 2,4-DNP hydrazine and also gives silver mirror test with Tollen’s
reagent. Thus, compound ‘A’ is an aldehyde. Also, compound ‘A’ reacts with concentrated NaOH i.e., the
compound gives Cannizzaro’s reaction. This shows that compound ‘A’ does not have any α-hydrogen atom.
∴ Compound ‘A’ is benzaldehyde.
Reaction of benzaldehyde with Tollen’s reagent:
Cannizzaro’s reaction of benzaldehyde:
Hence,compound ‘B’ is benzyl alcohol and ‘C’ is sodium benzoate.
Between benzyl alcohol and sodium benzoate, benzyl alcohol is less acidic. It reacts with CrO 3-H2SO4 to give
benzoic acid.
Hence, compound ‘D’ is benzoic acid.
Benzoic acid further reacts with Br2 | FeBr3 to give m-bromo benzoic acid.
Hence,compound ‘E’ is m-bromo benzoic acid.
Q11)
a. Carry out the following conversions.
i.Acetyl chloride to ethane
ii. Acetic acid to malonic acid
iii. Ethanol to 3-hydroxy butanal
b. Write short notes on the following:
i. Stephen reaction
ii. Gatterman-Koch reaction
Answer
a.
i.
ii.
iii.
b.
i.Stephen reaction:
In this reaction, nitriles are reduced to corresponding imines with stannous chloride in the presence of hydrochloric
acid. Imine, on further hydrolysis, gives the corresponding aldehydes.
ii. Gatterman-Koch reaction:
In this reaction, benzene or its derivative is treated with carbon monoxide and hydrogen chloride in the presence of
anhydrous aluminium chloride or cuprous chloride forming benzaldehyde or substituted benzaldehyde.