Question 1

(i) C6H5NH2 < NH3 < C6H5CH2NH2 < C2H5NH2 < (C2H5)2NH
(ii) C6H5NH2 < C2H5NH2 < (C2H5)3N < (C2H5)2NH
(iii) C6H5NH2 < C6H5CH2NH2 < (CH3)3N < CH3NH2 < (CH3)2NH

Accepted Answer

These are comparative orders of basicity of different amines. The order is based on the electron donating or withdrawing effects of substituents and steric factors affecting the availability of the lone pair on nitrogen for protonation.

(i) C6H5NH2 < NH3 < C6H5CH2NH2 < C2H5NH2 < (C2H5)2NH
Explanation:
- Aniline (C6H5NH2) is least basic due to resonance delocalization of lone pair into the benzene ring.
- NH3 is more basic than aniline as lone pair is localized.
- Benzylamine (C6H5CH2NH2) is more basic than NH3 because the benzyl group is electron donating via inductive effect.
- Ethylamine (C2H5NH2) is more basic than benzylamine due to stronger +I effect.
- Diethylamine ((C2H5)2NH) is more basic than ethylamine due to more alkyl groups donating electron density.

(ii) C6H5NH2 < C2H5NH2 < (C2H5)3N < (C2H5)2NH
Explanation:
- Aniline is least basic due to resonance.
- Ethylamine is more basic than aniline.
- Triethylamine ((C2H5)3N) is less basic than diethylamine ((C2H5)2NH) due to steric hindrance which reduces availability of lone pair.

(iii) C6H5NH2 < C6H5CH2NH2 < (CH3)3N < CH3NH2 < (CH3)2NH
Explanation:
- Aniline is least basic.
- Benzylamine is more basic than aniline.
- Trimethylamine ((CH3)3N) is less basic than methylamine (CH3NH2) due to steric hindrance.
- Methylamine is less basic than dimethylamine ((CH3)2NH) because two alkyl groups donate more electron density.

Thus, the orders reflect the combined effects of resonance, inductive effects, and steric hindrance on basicity. — Basicity depends on availability of lone pair on nitrogen. Resonance with aromatic ring reduces basicity (aniline). Alkyl groups (+I effect) increase electron density and basicity. Steric hindrance in tertiary amines reduces basicity despite more alkyl groups. Hence, the given orders follow these principles.

Question 2

1 Classify the following amines as primary, secondary or tertiary:

(i) ![img-3.jpeg](img-3.jpeg)

(ii) ![img-4.jpeg](img-4.jpeg)

(iii)  $(\mathrm{C}_2\mathrm{H}_5)_2\mathrm{CHNH}_2$

(iv)  $(\mathrm{C}_2\mathrm{H}_5)_2\mathrm{NH}$

Accepted Answer

Solution:
(i) Structure not visible here but classify based on number of alkyl groups attached to nitrogen.
(ii) Same as above.
(iii) $(\mathrm{C}_2\mathrm{H}_5)_2\mathrm{CHNH}_2$ is a primary amine because the nitrogen is attached to one carbon (the CH group) and two ethyl groups are on the carbon, not on nitrogen.
(iv) $(\mathrm{C}_2\mathrm{H}_5)_2\mathrm{NH}$ is a secondary amine because nitrogen is attached to two ethyl groups and one hydrogen.

Explanation:
Primary amine: Nitrogen attached to one alkyl or aryl group.
Secondary amine: Nitrogen attached to two alkyl or aryl groups.
Tertiary amine: Nitrogen attached to three alkyl or aryl groups. — Classify amines by counting number of alkyl groups attached to nitrogen atom:
- Primary amine: 1 alkyl group
- Secondary amine: 2 alkyl groups
- Tertiary amine: 3 alkyl groups
Apply this to each given structure.

Question 3

2 (i) Write structures of different isomeric amines corresponding to the molecular formula,  $\mathrm{C}_4\mathrm{H}_{11}\mathrm{N}$ .

(ii) Write IUPAC names of all the isomers.

(iii) What type of isomerism is exhibited by different pairs of amines?

Accepted Answer

Solution:
(i) Isomeric amines with formula C4H11N include:
- Butan-1-amine (CH3CH2CH2CH2NH2)
- Butan-2-amine (CH3CH2CH(NH2)CH3)
- 2-Methylpropan-1-amine ((CH3)2CHCH2NH2)
- 2-Methylpropan-2-amine ((CH3)3CNH2) - Note: This is tertiary amine, so not primary amine.
- N-Methylpropan-1-amine (CH3CH2CH2NHCH3) (secondary amine)
- N,N-Dimethylmethanamine (CH3NHCH3) (tertiary amine)

(ii) IUPAC names:
- Butan-1-amine
- Butan-2-amine
- 2-Methylpropan-1-amine
- N-Methylpropan-1-amine
- N,N-Dimethylmethanamine

(iii) The isomerism exhibited is chain isomerism (different carbon chain arrangement) and position isomerism (position of amino group). Also, functional group isomerism between primary, secondary and tertiary amines.

Explanation:
Isomers differ in the arrangement of carbon skeleton or position of amino group. Different types of isomerism include chain, position and functional group isomerism. — List all possible amines with formula C4H11N, name them according to IUPAC, and identify isomerism types based on structure differences.

Question 4

Write chemical equations for the following reactions:

(i) Reaction of ethanolic $\mathrm{NH_3}$ with $\mathrm{C_2H_5Cl}$.

(ii) Ammonolysis of benzyl chloride and reaction of amine so formed with two moles of $\mathrm{CH_3Cl}$.

Accepted Answer

Solution:
(i) Reaction of ethanolic NH3 with C2H5Cl:
$$\mathrm{C_2H_5Cl} + \mathrm{NH_3} \xrightarrow{ethanol, 373K} \mathrm{C_2H_5NH_2} + \mathrm{HCl}$$
Further reaction with alkyl halide:
$$\mathrm{C_2H_5NH_2} + \mathrm{C_2H_5Cl} \rightarrow \mathrm{C_2H_5NH C_2H_5} + \mathrm{HCl}$$
$$\mathrm{C_2H_5NH C_2H_5} + \mathrm{C_2H_5Cl} \rightarrow \mathrm{C_2H_5N(C_2H_5)_2} + \mathrm{HCl}$$
$$\mathrm{C_2H_5N(C_2H_5)_2} + \mathrm{C_2H_5Cl} \rightarrow \mathrm{[C_2H_5]_4N^+Cl^-}$$

(ii) Ammonolysis of benzyl chloride and reaction with two moles of CH3Cl:
$$\mathrm{C_6H_5CH_2Cl} + \mathrm{NH_3} \rightarrow \mathrm{C_6H_5CH_2NH_2} + \mathrm{HCl}$$
$$\mathrm{C_6H_5CH_2NH_2} + 2\mathrm{CH_3Cl} \rightarrow \mathrm{C_6H_5CH_2N(CH_3)_2} + 2\mathrm{HCl}$$ — Ammonolysis involves nucleophilic substitution of halogen by NH3 forming primary amine, which can further react with alkyl halides to form secondary, tertiary amines and quaternary ammonium salts.

Question 5

Write chemical equations for the following conversions:

(i) $\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{Cl}$ into $\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2}$

(ii) $\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}_{2}-\mathrm{Cl}$ into $\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2}$

Accepted Answer

Solution:
(i) Conversion of CH3-CH2-Cl to CH3-CH2-CH2-NH2:
Step 1: Reaction with ethanolic NaCN:
$$\mathrm{CH_3CH_2Cl} + \mathrm{NaCN} \rightarrow \mathrm{CH_3CH_2CN} + \mathrm{NaCl}$$
Step 2: Reduction of nitrile to amine:
$$\mathrm{CH_3CH_2CN} + 4[H] \xrightarrow{LiAlH_4} \mathrm{CH_3CH_2CH_2NH_2}$$

(ii) Conversion of C6H5-CH2-Cl to C6H5-CH2-CH2-NH2:
Step 1: Reaction with ethanolic NaCN:
$$\mathrm{C_6H_5CH_2Cl} + \mathrm{NaCN} \rightarrow \mathrm{C_6H_5CH_2CN} + \mathrm{NaCl}$$
Step 2: Reduction of nitrile to amine:
$$\mathrm{C_6H_5CH_2CN} + 4[H] \xrightarrow{LiAlH_4} \mathrm{C_6H_5CH_2CH_2NH_2}$$ — First convert alkyl halide to nitrile by nucleophilic substitution with cyanide ion. Then reduce nitrile to primary amine using LiAlH4 or catalytic hydrogenation.

Question 6

4 Arrange the following in increasing order of their basic strength:

(i)  C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2 and (C2H5)2NH
(ii)  C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2
(iii)  CH3NH2, (CH3)2NH, (CH3)3N, C6H5NH2, C6H5CH2NH2

Accepted Answer

Solution: Basic strength of amines depends on availability of lone pair on nitrogen for protonation. (i) C2H5NH2 (ethylamine) > C6H5CH2NH2 (benzylamine) > NH3 (ammonia) > (C2H5)2NH (diethylamine) > C6H5NH2 (aniline) Explanation: - Ethylamine is more basic than benzylamine due to +I effect. - Benzylamine is more basic than ammonia. - Diethylamine is more basic than ethylamine generally, but in this order, aniline is least due to resonance. - Aniline is least basic because lone pair is delocalized into aromatic ring. (ii) C2H5NH2 < (C2H5)2NH < (C2H5)3N < C6H5NH2 Explanation: - Basicity increases from primary to tertiary alkyl amines due to +I effect. - Aniline is least basic due to resonance. (iii) (CH3)2NH > CH3NH2 > C6H5CH2NH2 > (CH3)3N > C6H5NH2 Explanation: - Secondary amine is more basic than primary. - Tertiary amine is less basic than secondary due to steric hindrance. - Aniline is least basic. Hence, the increasing order of basic strength is as above. — Basic strength depends on availability of lone pair on nitrogen. Alkyl groups (+I effect) increase electron density, increasing basicity. Aromatic ring delocalizes lone pair, decreasing basicity. Steric hindrance in tertiary amines can reduce basicity. Thus, orders are derived accordingly.

Question 7

5 Complete the following acid-base reactions and name the products:

(i)  CH3CH2CH2NH2 + HCl →
(ii)  (C2H5)3N + HCl →

Accepted Answer

Solution:
(i) CH3CH2CH2NH2 + HCl → CH3CH2CH2NH3+ Cl−
Product: Propylammonium chloride (a salt formed by protonation of propylamine)

(ii) (C2H5)3N + HCl → (C2H5)3NH+ Cl−
Product: Triethylammonium chloride (a salt formed by protonation of triethylamine)

Explanation:
Both primary and tertiary amines act as bases and accept a proton from HCl to form their respective ammonium salts. — Amines react with acids to form ammonium salts.
Primary amine + HCl → primary ammonium salt.
Tertiary amine + HCl → tertiary ammonium salt.
This is an acid-base neutralization reaction.

Question 8

6 Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.

Accepted Answer

Solution:
Aniline (C6H5NH2) reacts with excess methyl iodide (CH3I) in the presence of sodium carbonate (Na2CO3) to give quaternary ammonium salt.

Stepwise reaction:
C6H5NH2 + CH3I → C6H5NHCH3+ I− (N-methylaniline iodide)
C6H5NHCH3 + CH3I → C6H5N(CH3)2+ I− (N,N-dimethylaniline iodide)
C6H5N(CH3)2 + CH3I → C6H5N+(CH3)3 I− (N,N,N-trimethylanilinium iodide)

Since excess methyl iodide is used, the final product is the quaternary ammonium salt: N,N,N-trimethylanilinium iodide.

Sodium carbonate is used to neutralize HI formed during the reaction. — Aniline undergoes stepwise alkylation with methyl iodide.
Each step adds one methyl group to nitrogen.
Excess methyl iodide leads to quaternary ammonium salt formation.
Sodium carbonate neutralizes acidic HI formed.

Question 9

7 Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.

Accepted Answer

Solution:
Aniline reacts with benzoyl chloride (C6H5COCl) in the presence of a base (like pyridine) to form benzamide derivative called acetanilide.

Reaction:
C6H5NH2 + C6H5COCl → C6H5NHCOC6H5 + HCl

Product: Benzanilide (or N-phenylbenzamide)

Explanation:
The amino group of aniline acts as a nucleophile and attacks the carbonyl carbon of benzoyl chloride, replacing Cl and forming an amide bond. — Aniline's -NH2 group reacts with acid chloride to form amide.
This is an acylation reaction.
The product is benzamide derivative named benzanilide.

Question 10

8 Write structures of different isomers corresponding to the molecular formula, C3H9N. Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid.

Accepted Answer

Solution:
Isomers of C3H9N are:
1. Propylamine (CH3CH2CH2NH2) - 1-aminopropane
2. Isopropylamine ((CH3)2CHNH2) - 2-aminopropane
3. Trimethylamine ((CH3)3N)

Among these, primary amines liberate nitrogen gas on treatment with nitrous acid.
Hence, 1-aminopropane and 2-aminopropane will liberate nitrogen gas.
Trimethylamine (tertiary amine) does not react with nitrous acid to liberate nitrogen gas.

Structures:
1. CH3-CH2-CH2-NH2
2. (CH3)2CH-NH2
3. (CH3)3N — Primary amines react with nitrous acid to form diazonium salts which decompose liberating nitrogen gas.
Secondary and tertiary amines do not liberate nitrogen gas.
Thus, only primary amines among isomers of C3H9N liberate nitrogen gas.

Question 11

Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

(i) (CH3)2CHNH2
(ii) CH3(CH2)2NH2
(iii) CH3NHCH(CH3)2
(iv) (CH3)3CNH2
(v) C6H5NHCH3
(vi) (CH3CH2)2NCH3
(vii) m-BrC6H4NH2

Accepted Answer

Solutions:
(i) (CH3)2CHNH2: IUPAC name is 2-Aminopropane. It is a primary amine because the amino group is attached to one alkyl group.

(ii) CH3(CH2)2NH2: IUPAC name is 1-Aminopropane. It is a primary amine.

(iii) CH3NHCH(CH3)2: IUPAC name is N-Methyl-2-aminopropane or N-Methylpropan-2-amine. It is a secondary amine because the nitrogen is attached to two alkyl groups.

(iv) (CH3)3CNH2: IUPAC name is 2-Amino-2-methylpropane (tert-butylamine). It is a primary amine.

(v) C6H5NHCH3: IUPAC name is N-Methylaniline. It is a secondary amine.

(vi) (CH3CH2)2NCH3: IUPAC name is N-Methyldiethylamine. It is a tertiary amine because nitrogen is attached to three alkyl groups.

(vii) m-BrC6H4NH2: IUPAC name is m-Bromoaniline. It is a primary amine. — The classification depends on the number of alkyl or aryl groups attached to the nitrogen atom:
- Primary amine: one alkyl/aryl group attached to N.
- Secondary amine: two alkyl/aryl groups attached to N.
- Tertiary amine: three alkyl/aryl groups attached to N.

IUPAC names are given by identifying the longest carbon chain attached to the amino group and naming substituents accordingly.

Question 12

Give one chemical test to distinguish between the following pairs of compounds.

(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylamine and aniline
(iv) Aniline and benzylamine
(v) Aniline and N-methylaniline.

Accepted Answer

Solutions:
(i) Methylamine and dimethylamine:
- Use Hinsberg test: Methylamine (primary amine) reacts with Hinsberg reagent (benzenesulfonyl chloride) to form a sulfonamide soluble in alkali, whereas dimethylamine (secondary amine) forms an insoluble sulfonamide.

(ii) Secondary and tertiary amines:
- Use Hinsberg test: Secondary amines form insoluble sulfonamides, tertiary amines do not react.

(iii) Ethylamine and aniline:
- Use bromine water test: Aniline reacts with bromine water to give a white precipitate of 2,4,6-tribromoaniline; ethylamine does not.

(iv) Aniline and benzylamine:
- Use nitrous acid test: Aniline forms diazonium salt at 0-5°C, benzylamine forms alcohol and nitrogen gas.

(v) Aniline and N-methylaniline:
- Use nitrous acid: Aniline forms diazonium salt, N-methylaniline does not form diazonium salt but gives a yellow oily layer. — Chemical tests exploit differences in reactivity:
- Hinsberg test differentiates primary, secondary and tertiary amines.
- Bromine water reacts with activated aromatic amines.
- Nitrous acid forms diazonium salts only with primary aromatic amines.
- Secondary and tertiary amines show different solubility and reactivity patterns.

Question 13

Account for the following:

(i) pKb of aniline is more than that of methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although amino group is o- and p- directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
(v) Aniline does not undergo Friedel-Crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.

Accepted Answer

Solutions:
(i) pKb of aniline is more than methylamine because the lone pair on nitrogen in aniline is delocalized into the benzene ring, reducing its availability to accept protons, thus making aniline less basic.

(ii) Ethylamine is soluble in water due to its ability to form hydrogen bonds with water molecules. Aniline is less soluble because the large hydrophobic benzene ring reduces its interaction with water.

(iii) Methylamine reacts with ferric chloride to form ferric hydroxide precipitate due to the formation of complex between methylamine and Fe3+ ions.

(iv) Although amino group directs electrophilic substitution to ortho and para positions, the protonated form of aniline under nitration conditions reduces electron density at ortho and para positions, leading to meta substitution.

(v) Aniline does not undergo Friedel-Crafts reaction because the amino group forms a complex with Lewis acid catalysts, deactivating the catalyst.

(vi) Diazonium salts of aromatic amines are more stable due to resonance stabilization of the diazonium group with the aromatic ring.

(vii) Gabriel phthalimide synthesis is preferred because it gives pure primary amines without contamination from secondary or tertiary amines. — The explanations involve electronic effects, solubility factors, reaction mechanisms, and stability considerations:
- Resonance delocalization affects basicity.
- Hydrophobic groups reduce solubility.
- Complex formation affects reactivity.
- Reaction conditions influence substitution patterns.
- Catalyst deactivation prevents Friedel-Crafts.
- Resonance stabilizes diazonium salts.
- Gabriel synthesis is selective for primary amines.

Question 14

Give plausible explanation for each of the following:

(i) Why are amines less acidic than alcohols of comparable molecular masses?
(ii) Why do primary amines have higher boiling point than tertiary amines?
(iii) Why are aliphatic amines stronger bases than aromatic amines?

Accepted Answer

Solutions:
(i) Amines are less acidic than alcohols because the conjugate base of alcohol (alkoxide ion) is more stabilized by electronegative oxygen, whereas the conjugate base of amines (amide ion) is less stabilized due to less electronegativity of nitrogen.

(ii) Primary amines have higher boiling points than tertiary amines because primary amines can form more extensive hydrogen bonding (both as donor and acceptor) compared to tertiary amines which have no N-H bonds and thus less hydrogen bonding.

(iii) Aliphatic amines are stronger bases than aromatic amines because in aromatic amines (like aniline), the lone pair on nitrogen is delocalized into the aromatic ring, reducing its availability to accept protons, whereas in aliphatic amines the lone pair is localized and more available. — Acidity depends on stability of conjugate base; oxygen stabilizes better than nitrogen.
Boiling points depend on hydrogen bonding capability.
Basicity depends on availability of lone pair on nitrogen for protonation.

Question 15

An aromatic compound 'A' on treatment with aqueous ammonia and heating forms compound 'B' which on heating with Br2 and KOH forms a compound 'C' of molecular formula C6H7N. Write the structures and IUPAC names of compounds A, B and C.

Accepted Answer

Solution:
Compound A is chlorobenzene (C6H5Cl).
Treatment with aqueous ammonia replaces Cl with NH2 to give compound B: aniline (C6H5NH2).
Heating B with Br2 and KOH leads to Hofmann rearrangement producing compound C: benzene with an amino group and one bromine substituted, but molecular formula C6H7N indicates aniline itself or its derivative.

More precisely, the reaction is the Hofmann reaction where aniline reacts with Br2 and KOH to give 2-aminophenol or 2-bromoaniline depending on conditions. However, given molecular formula C6H7N, compound C is likely to be aniline itself or a brominated derivative.

IUPAC names:
A: Chlorobenzene
B: Aniline (benzenamine)
C: 2-Bromoaniline (2-bromoaniline) — The reaction sequence:
- Aromatic halide (A) reacts with NH3 to give aromatic amine (B).
- Treatment of aromatic amine with Br2/KOH leads to substitution or rearrangement giving compound C.
- Molecular formula C6H7N corresponds to aniline or its mono-substituted derivative.

Hence, the structures and names are as above.

Amines

Amines — Study Notes

Objectives

I. AMINES

9.2 Classification

Practice Questions — Amines

All 5 Chapters in Chemistry-II