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NCERT Solutions · Class 12 Chemistry · NCERT Chemistry Part I and II Part II · Chapter 9

Chapter 9: Amines (Organic Chemistry)

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

C₆H₅N⁺(CH₃)₃I⁻, N,N,N-trimethylanilinium iodide.

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

N-Phenylbenzamide (benzanilide).

Q: Convert (i) 3-Methylaniline into 3-nitrotoluene. (ii) Aniline into 1,3,5-tribromobenzene.

(i) Diazotise, convert to the fluoroborate and treat with NaNO₂/Cu. (ii) Tribrominate, diazotise, then remove N₂⁺ with H₃PO₂.

Q: 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.

(i) carbylamine; (ii) Hinsberg; (iii) azo dye test; (iv) HNO₂ (N₂ from benzylamine); (v) carbylamine.

Q: Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.

The Hinsberg test: 1° gives an alkali-soluble sulphonamide, 2° an alkali-insoluble one, 3° no reaction.

Q: Write short notes on the following: (i) Carbylamine reaction (ii) Diazotisation (iii) Hofmann's bromamide reaction (iv) Coupling reaction (v) Ammonolysis (vi) Acetylation (vii) Gabriel phthalimide synthesis.

As explained, with the equations above.

Q: Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?

Aryl halides do not undergo nucleophilic substitution with the phthalimide anion.

Q: Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.

(i) Arenediazonium salt. (ii) Alcohol and N₂ gas.

Answers to all 9 intext questions and 14 exercises of Chapter 9, Amines (NCERT Class 12 Chemistry Part II, 2026-27): classification and IUPAC names, isomers of C₄H₁₁N, preparation, basic strength order, Hinsberg, carbylamine and nitrous acid tests, diazotisation and diazonium salt reactions, Hoffmann bromamide and Gabriel synthesis, step-by-step conversions and identification of A, B and C. All 23 questions are answered, with the key answer highlighted.

NCERT textbook Chapter 9 (PDF)

Exam revision? Chapter 9 notes with key points, diagrams and practice questions: see the free sample · ₹199 for all 10 chapters

Free NCERT solutions by Notes Bazar · www.notesbazar.in/ncert-solutions/class-12-chemistry/chapter-9-amines

Basicity in water: (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃, and (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃ (inductive effect, solvation and steric effects). Aromatic amines are weaker bases than NH₃ (lone pair delocalised into the ring). Tests: carbylamine (CHCl₃ + alc. KOH, foul smell) for 1° amines; Hinsberg (C₆H₅SO₂Cl): 1° product dissolves in alkali, 2° product does not, 3° does not react; HNO₂: aliphatic 1° amines give N₂, aromatic 1° amines give diazonium salts at 273–278 K.

Intext questions

9.1

Classify the following amines as primary, secondary or tertiary: (i) naphthalen-1-amine (1-naphthylamine) (ii) N,N-dimethylnaphthalen-1-amine (iii)

(C X_{2} H X_{5}) X_{2} CHNH X_{2}

(iv)

(C X_{2} H X_{5}) X_{2} NH

Solution

(i) Primary (–NH₂)
(ii) Tertiary (–N(CH₃)₂)
(iii) Primary (–NH₂ on carbon)
(iv) Secondary (>NH)

(i) primary; (ii) tertiary; (iii) primary; (iv) secondary.

9.2

(i) Write structures of different isomeric amines corresponding to the molecular formula,

C X_{4} H X_{11} N

. (ii) Write IUPAC names of all the isomers. (iii) What type of isomerism is exhibited by different pairs of amines?

Solution

Structure	IUPAC name	Type
CH₃CH₂CH₂CH₂NH₂	Butan-1-amine	1°
CH₃CH₂CH(NH₂)CH₃	Butan-2-amine	1°
(CH₃)₂CHCH₂NH₂	2-Methylpropan-1-amine	1°
(CH₃)₃CNH₂	2-Methylpropan-2-amine	1°
CH₃NHCH₂CH₂CH₃	N-Methylpropan-1-amine	2°
CH₃NHCH(CH₃)₂	N-Methylpropan-2-amine	2°
CH₃CH₂NHCH₂CH₃	N-Ethylethanamine	2°
(CH₃)₂NCH₂CH₃	N,N-Dimethylethanamine	3°

(iii) Chain isomerism (e.g. butan-1-amine and 2-methylpropan-1-amine), position isomerism (butan-1-amine and butan-2-amine), metamerism (N-methylpropan-1-amine and N-ethylethanamine) and functional isomerism (primary, secondary and tertiary amines of the same formula).

Eight isomers (four 1°, three 2°, one 3°); they show chain, position, metamerism and functional isomerism.

9.3

How will you convert (i) Benzene into aniline (ii) Benzene into N,N-dimethylaniline (iii)

Cl - (CH X_{2}) X_{4} - Cl

into hexan-1,6-diamine?

Solution

(i) $C X_{6} H X_{6} conc \cdot HNO X_{3} / H X_{2} SO X_{4} C X_{6} H X_{5} NO X_{2} Sn / HCl C X_{6} H X_{5} NH X_{2}$
(ii) Make aniline as in (i), then $C X_{6} H X_{5} NH X_{2} + 2 CH X_{3} Cl C X_{6} H X_{5} N (CH X_{3}) X_{2} + 2 HCl$ (or with $CH X_{3} I$ )
(iii) $Cl (CH X_{2}) X_{4} Cl + 2 KCN NC (CH X_{2}) X_{4} CN LiAlH X_{4} (or H X_{2} / Ni) H X_{2} N (CH X_{2}) X_{6} NH X_{2}$

(i) Nitration then reduction. (ii) Aniline + 2 CH₃Cl. (iii) Replace Cl by CN, then reduce.

9.4

Arrange the following in increasing order of their basic strength: (i)

C X_{2} H X_{5} NH X_{2}

C X_{6} H X_{5} NH X_{2}

NH X_{3}

C X_{6} H X_{5} CH X_{2} NH X_{2}

and

(C X_{2} H X_{5}) X_{2} NH

(ii)

C X_{2} H X_{5} NH X_{2}

(C X_{2} H X_{5}) X_{2} NH

(C X_{2} H X_{5}) X_{3} N

C X_{6} H X_{5} NH X_{2}

(iii)

CH X_{3} NH X_{2}

(CH X_{3}) X_{2} NH

(CH X_{3}) X_{3} N

C X_{6} H X_{5} NH X_{2}

C X_{6} H X_{5} CH X_{2} NH X_{2}

Solution

(i) $C X_{6} H X_{5} NH X_{2} < NH X_{3} < C X_{6} H X_{5} CH X_{2} NH X_{2} < C X_{2} H X_{5} NH X_{2} < (C X_{2} H X_{5}) X_{2} NH$
(ii) $C X_{6} H X_{5} NH X_{2} < C X_{2} H X_{5} NH X_{2} < (C X_{2} H X_{5}) X_{3} N < (C X_{2} H X_{5}) X_{2} NH$
(iii) $C X_{6} H X_{5} NH X_{2} < C X_{6} H X_{5} CH X_{2} NH X_{2} < (CH X_{3}) X_{3} N < CH X_{3} NH X_{2} < (CH X_{3}) X_{2} NH$

Aniline is weakest (lone pair delocalised into the ring). Among alkylamines in water, the order depends on the +I effect, solvation of the ammonium ion and steric crowding.

As ordered above.

9.5

Complete the following acid-base reactions and name the products: (i)

CH X_{3} CH X_{2} CH X_{2} NH X_{2} + HCl

(ii)

(C X_{2} H X_{5}) X_{3} N + HCl

Solution

(i) $CH X_{3} CH X_{2} CH X_{2} NH X_{2} + HCl CH X_{3} CH X_{2} CH X_{2} NH X_{3} X^{+} Cl X^{-}$ : propylammonium chloride
(ii) $(C X_{2} H X_{5}) X_{3} N + HCl (C X_{2} H X_{5}) X_{3} NH X^{+} Cl X^{-}$ : triethylammonium chloride

(i) Propylammonium chloride. (ii) Triethylammonium chloride.

9.6

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

Solution

Aniline is methylated step by step until the quaternary salt forms; sodium carbonate neutralises the HI produced:

$C X_{6} H X_{5} NH X_{2} + 3 CH X_{3} I Na X_{2} CO X_{3} C X_{6} H X_{5} N X^{+} (CH X_{3}) X_{3} I X^{-}$

The final product is N,N,N-trimethylanilinium iodide (a quaternary ammonium salt).

C₆H₅N⁺(CH₃)₃I⁻, N,N,N-trimethylanilinium iodide.

9.7

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

Solution

$C X_{6} H X_{5} NH X_{2} + C X_{6} H X_{5} COCl pyridine C X_{6} H X_{5} NHCOC X_{6} H X_{5} + HCl$

The product is N-phenylbenzamide (benzanilide). (This is benzoylation.)

N-Phenylbenzamide (benzanilide).

9.8

Write structures of different isomers corresponding to the molecular formula,

C X_{3} H X_{9} N

. Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid.

Solution

$CH X_{3} CH X_{2} CH X_{2} NH X_{2}$ : propan-1-amine (1°)
$CH X_{3} CH (NH X_{2}) CH X_{3}$ : propan-2-amine (1°)
$CH X_{3} NHCH X_{2} CH X_{3}$ : N-methylethanamine (2°)
$(CH X_{3}) X_{3} N$ : N,N-dimethylmethanamine (3°)

Primary aliphatic amines liberate nitrogen with nitrous acid, so propan-1-amine and propan-2-amine give $N X_{2}$ .

Propan-1-amine and propan-2-amine liberate N₂.

9.9

Convert (i) 3-Methylaniline into 3-nitrotoluene. (ii) Aniline into 1,3,5-tribromobenzene.

Solution

(i) $3 - CH X_{3} C X_{6} H X_{4} NH X_{2} NaNO X_{2} / HCl 273 - 278 K 3 - CH X_{3} C X_{6} H X_{4} N X_{2} X^{+} Cl X^{-} HBF X_{4} 3 - CH X_{3} C X_{6} H X_{4} N X_{2} X^{+} BF X_{4} X^{-} NaNO X_{2} / Cu heat 3 - CH X_{3} C X_{6} H X_{4} NO X_{2}$
(ii) $C X_{6} H X_{5} NH X_{2} Br X_{2} / H X_{2} O 2, 4, 6 - Br X_{3} C X_{6} H X_{2} NH X_{2} NaNO X_{2} / HCl 273 - 278 K 2, 4, 6 - Br X_{3} C X_{6} H X_{2} N X_{2} X^{+} Cl X^{-} H X_{3} PO X_{2} / H X_{2} O 1, 3, 5 - Br X_{3} C X_{6} H X_{3}$

(i) Diazotise, convert to the fluoroborate and treat with NaNO₂/Cu. (ii) Tribrominate, diazotise, then remove N₂⁺ with H₃PO₂.

Exercises

9.1

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

(CH X_{3}) X_{2} CHNH X_{2}

(ii)

CH X_{3} (CH X_{2}) X_{2} NH X_{2}

(iii)

CH X_{3} NHCH (CH X_{3}) X_{2}

(iv)

(CH X_{3}) X_{3} CNH X_{2}

(v)

C X_{6} H X_{5} NHCH X_{3}

(vi)

(CH X_{3} CH X_{2}) X_{2} NCH X_{3}

(vii) m-

BrC X_{6} H X_{4} NH X_{2}

Solution

(i) Propan-2-amine: primary
(ii) Propan-1-amine: primary
(iii) N-Methylpropan-2-amine: secondary
(iv) 2-Methylpropan-2-amine: primary
(v) N-Methylbenzenamine (N-methylaniline): secondary
(vi) N-Ethyl-N-methylethanamine: tertiary
(vii) 3-Bromobenzenamine (3-bromoaniline): primary

As listed above.

9.2

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.

Solution

(i) Carbylamine test (warm with $CHCl X_{3}$ and alcoholic KOH): methylamine (1°) gives a foul-smelling isocyanide; dimethylamine does not.
(ii) Hinsberg test (benzenesulphonyl chloride): a secondary amine gives a sulphonamide that is insoluble in alkali; a tertiary amine does not react.
(iii) Azo dye test: aniline, diazotised at 273–278 K and added to alkaline β-naphthol, gives an orange-red dye; ethylamine gives $N X_{2}$ bubbles and no dye.
(iv) Nitrous acid at low temperature: benzylamine (aliphatic type) gives brisk effervescence of $N X_{2}$ ; aniline forms a diazonium salt (gives an azo dye with β-naphthol).
(v) Carbylamine test: aniline (1°) gives the foul smell of phenyl isocyanide; N-methylaniline (2°) does not.

(i) carbylamine; (ii) Hinsberg; (iii) azo dye test; (iv) HNO₂ (N₂ from benzylamine); (v) carbylamine.

9.3

Account for the following: (i)

p K_{b}

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.

Solution

(i) In aniline, the nitrogen lone pair is delocalised into the benzene ring, so it is less available to accept a proton; aniline is a weaker base (higher $p K_{b}$ ). In methylamine, the +I effect of $CH X_{3}$ increases the electron density on N.

(ii) Ethylamine forms hydrogen bonds with water. Aniline also can, but its large hydrophobic phenyl group outweighs this, so it is hardly soluble.

(iii) Methylamine is basic and produces $OH X^{-}$ in water: $CH X_{3} NH X_{2} + H X_{2} O CH X_{3} NH X_{3} X^{+} + OH X^{-}$ . The hydroxide ions precipitate iron: $FeCl X_{3} + 3 OH X^{-} Fe (OH) X_{3} ↓ + 3 Cl X^{-}$ (hydrated ferric oxide).

(iv) Nitration is done in strongly acidic medium, where much of the aniline is protonated to the anilinium ion, $C X_{6} H X_{5} NH X_{3} X^{+}$ . The $- NH X_{3} X^{+}$ group is electron-withdrawing and meta-directing, so a large amount (about 47%) of m-nitroaniline forms.

(v) Aniline is a Lewis base and reacts with the catalyst $AlCl X_{3}$ (a Lewis acid) to form a salt. The nitrogen then carries a positive charge, which strongly deactivates the ring, so no Friedel–Crafts reaction occurs.

(vi) In arenediazonium ions, the positive charge is delocalised over the benzene ring by resonance, which stabilises them (at 273–278 K). Alkyldiazonium ions have no such stabilisation and decompose at once.

(vii) It gives pure primary amines, without the mixture of secondary, tertiary amines and quaternary salts formed in ammonolysis of alkyl halides.

(i) Lone pair delocalisation in aniline; (ii) the large hydrophobic phenyl group; (iii) OH⁻ from methylamine precipitates Fe(OH)₃; (iv) anilinium ion is meta-directing; (v) aniline forms a salt with AlCl₃; (vi) resonance stabilisation; (vii) only primary amine forms.

9.4

Arrange the following: (i) In decreasing order of the

p K_{b}

values:

C X_{2} H X_{5} NH X_{2}

C X_{6} H X_{5} NHCH X_{3}

(C X_{2} H X_{5}) X_{2} NH

and

C X_{6} H X_{5} NH X_{2}

(ii) In increasing order of basic strength:

C X_{6} H X_{5} NH X_{2}

C X_{6} H X_{5} N (CH X_{3}) X_{2}

(C X_{2} H X_{5}) X_{2} NH

and

CH X_{3} NH X_{2}

(iii) In increasing order of basic strength: (a) Aniline, p-nitroaniline and p-toluidine (b)

C X_{6} H X_{5} NH X_{2}

C X_{6} H X_{5} NHCH X_{3}

C X_{6} H X_{5} CH X_{2} NH X_{2}

(iv) In decreasing order of basic strength in gas phase:

C X_{2} H X_{5} NH X_{2}

(C X_{2} H X_{5}) X_{2} NH

(C X_{2} H X_{5}) X_{3} N

and

NH X_{3}

(v) In increasing order of boiling point:

C X_{2} H X_{5} OH

(CH X_{3}) X_{2} NH

C X_{2} H X_{5} NH X_{2}

(vi) In increasing order of solubility in water:

C X_{6} H X_{5} NH X_{2}

(C X_{2} H X_{5}) X_{2} NH

C X_{2} H X_{5} NH X_{2}

Solution

(i) $C X_{6} H X_{5} NH X_{2} > C X_{6} H X_{5} NHCH X_{3} > C X_{2} H X_{5} NH X_{2} > (C X_{2} H X_{5}) X_{2} NH$ (a higher $p K_{b}$ means a weaker base)
(ii) $C X_{6} H X_{5} NH X_{2} < C X_{6} H X_{5} N (CH X_{3}) X_{2} < CH X_{3} NH X_{2} < (C X_{2} H X_{5}) X_{2} NH$
(iii)(a) p-Nitroaniline < aniline < p-toluidine
(iii)(b) $C X_{6} H X_{5} NH X_{2} < C X_{6} H X_{5} NHCH X_{3} < C X_{6} H X_{5} CH X_{2} NH X_{2}$
(iv) $(C X_{2} H X_{5}) X_{3} N > (C X_{2} H X_{5}) X_{2} NH > C X_{2} H X_{5} NH X_{2} > NH X_{3}$ (in the gas phase, only the inductive effect matters)
(v) $(CH X_{3}) X_{2} NH < C X_{2} H X_{5} NH X_{2} < C X_{2} H X_{5} OH$
(vi) $C X_{6} H X_{5} NH X_{2} < (C X_{2} H X_{5}) X_{2} NH < C X_{2} H X_{5} NH X_{2}$

As ordered above.

9.5

How will you convert: (i) Ethanoic acid into methanamine (ii) Hexanenitrile into 1-aminopentane (iii) Methanol to ethanoic acid (iv) Ethanamine into methanamine (v) Ethanoic acid into propanoic acid (vi) Methanamine into ethanamine (vii) Nitromethane into dimethylamine (viii) Propanoic acid into ethanoic acid?

Solution

(i) $CH X_{3} COOH NH X_{3} heat CH X_{3} CONH X_{2} Br X_{2} / KOH CH X_{3} NH X_{2}$ (Hoffmann bromamide)
(ii) $CH X_{3} (CH X_{2}) X_{4} CN H X_{2} O partial hydrolysis CH X_{3} (CH X_{2}) X_{4} CONH X_{2} Br X_{2} / KOH CH X_{3} (CH X_{2}) X_{4} NH X_{2}$
(iii) $CH X_{3} OH PCl X_{5} CH X_{3} Cl KCN CH X_{3} CN H X_{3} O X^{+} CH X_{3} COOH$
(iv) $C X_{2} H X_{5} NH X_{2} HNO X_{2} C X_{2} H X_{5} OH KMnO X_{4} CH X_{3} COOH NH X_{3} heat CH X_{3} CONH X_{2} Br X_{2} / KOH CH X_{3} NH X_{2}$
(v) $CH X_{3} COOH LiAlH X_{4} CH X_{3} CH X_{2} OH SOCl X_{2} C X_{2} H X_{5} Cl KCN C X_{2} H X_{5} CN H X_{3} O X^{+} C X_{2} H X_{5} COOH$
(vi) $CH X_{3} NH X_{2} HNO X_{2} CH X_{3} OH PCl X_{5} CH X_{3} Cl KCN CH X_{3} CN LiAlH X_{4} CH X_{3} CH X_{2} NH X_{2}$
(vii) $CH X_{3} NO X_{2} Sn / HCl CH X_{3} NH X_{2} CHCl X_{3} / alc \cdot KOH CH X_{3} NC LiAlH X_{4} CH X_{3} NHCH X_{3}$
(viii) $C X_{2} H X_{5} COOH NH X_{3} heat C X_{2} H X_{5} CONH X_{2} Br X_{2} / KOH C X_{2} H X_{5} NH X_{2} HNO X_{2} C X_{2} H X_{5} OH KMnO X_{4} CH X_{3} COOH$

As in the sequences above.

9.6

Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.

Solution

Hinsberg test with benzenesulphonyl chloride ( $C X_{6} H X_{5} SO X_{2} Cl$ ):

Primary amine: forms an N-alkylbenzenesulphonamide, which still has an acidic H on N and dissolves in alkali: $C X_{6} H X_{5} SO X_{2} Cl + C X_{2} H X_{5} NH X_{2} C X_{6} H X_{5} SO X_{2} NHC X_{2} H X_{5} + HCl$
Secondary amine: forms an N,N-dialkylbenzenesulphonamide with no H on N, insoluble in alkali: $C X_{6} H X_{5} SO X_{2} Cl + (C X_{2} H X_{5}) X_{2} NH C X_{6} H X_{5} SO X_{2} N (C X_{2} H X_{5}) X_{2} + HCl$
Tertiary amine: has no H on N and does not react.

The Hinsberg test: 1° gives an alkali-soluble sulphonamide, 2° an alkali-insoluble one, 3° no reaction.

9.7

Write short notes on the following: (i) Carbylamine reaction (ii) Diazotisation (iii) Hofmann's bromamide reaction (iv) Coupling reaction (v) Ammonolysis (vi) Acetylation (vii) Gabriel phthalimide synthesis.

Solution

(i) Carbylamine reaction: primary amines (aliphatic or aromatic) heated with chloroform and alcoholic KOH give foul-smelling isocyanides; a test for 1° amines: $RNH X_{2} + CHCl X_{3} + 3 KOH RNC + 3 KCl + 3 H X_{2} O$
(ii) Diazotisation: conversion of a primary aromatic amine into a diazonium salt with nitrous acid ( $NaNO X_{2} + HCl$ ) at 273–278 K: $C X_{6} H X_{5} NH X_{2} + NaNO X_{2} + 2 HCl C X_{6} H X_{5} N X_{2} X^{+} Cl X^{-} + NaCl + 2 H X_{2} O$
(iii) Hofmann bromamide reaction: an amide with bromine and NaOH gives a primary amine with one carbon less: $RCONH X_{2} + Br X_{2} + 4 NaOH RNH X_{2} + Na X_{2} CO X_{3} + 2 NaBr + 2 H X_{2} O$
(iv) Coupling reaction: a diazonium salt reacts with phenol (in alkali) or aniline (mildly acidic) at the para position to give coloured azo compounds: $C X_{6} H X_{5} N X_{2} X^{+} Cl X^{-} + C X_{6} H X_{5} OH OH X^{-} C X_{6} H X_{5} N = NC X_{6} H X_{4} OH$ (p-hydroxyazobenzene, orange)
(v) Ammonolysis: an alkyl halide heated with ethanolic ammonia in a sealed tube (373 K) gives a mixture of primary, secondary, tertiary amines and the quaternary ammonium salt: $RX + NH X_{3} RNH X_{2} + HX$ , and further alkylation
(vi) Acetylation: replacement of an H of –NH₂ by an acetyl group using acetyl chloride or acetic anhydride (with pyridine): $C X_{6} H X_{5} NH X_{2} + (CH X_{3} CO) X_{2} O C X_{6} H X_{5} NHCOCH X_{3} + CH X_{3} COOH$
(vii) Gabriel phthalimide synthesis: phthalimide is converted to potassium phthalimide (KOH), alkylated with an alkyl halide, and the N-alkylphthalimide is hydrolysed (NaOH) to give a pure primary amine

As explained, with the equations above.

9.8

Accomplish the following conversions: (i) Nitrobenzene to benzoic acid (ii) Benzene to m-bromophenol (iii) Benzoic acid to aniline (iv) Aniline to 2,4,6-tribromofluorobenzene (v) Benzyl chloride to 2-phenylethanamine (vi) Chlorobenzene to p-chloroaniline (vii) Aniline to p-bromoaniline (viii) Benzamide to toluene (ix) Aniline to benzyl alcohol.

Solution

(i) $C X_{6} H X_{5} NO X_{2} Sn / HCl C X_{6} H X_{5} NH X_{2} NaNO X_{2} / HCl 273 K C X_{6} H X_{5} N X_{2} Cl CuCN C X_{6} H X_{5} CN H X_{3} O X^{+} C X_{6} H X_{5} COOH$
(ii) $C X_{6} H X_{6} HNO X_{3} / H X_{2} SO X_{4} C X_{6} H X_{5} NO X_{2} Br X_{2} / FeBr X_{3} m - BrC X_{6} H X_{4} NO X_{2} Sn / HCl m - BrC X_{6} H X_{4} NH X_{2} NaNO X_{2} / HCl 273 K m - BrC X_{6} H X_{4} N X_{2} Cl H X_{2} O warm m - BrC X_{6} H X_{4} OH$
(iii) $C X_{6} H X_{5} COOH SOCl X_{2} C X_{6} H X_{5} COCl NH X_{3} C X_{6} H X_{5} CONH X_{2} Br X_{2} / KOH C X_{6} H X_{5} NH X_{2}$
(iv) $C X_{6} H X_{5} NH X_{2} Br X_{2} / H X_{2} O 2, 4, 6 - Br X_{3} C X_{6} H X_{2} NH X_{2} NaNO X_{2} / HCl 273 K ArN X_{2} X^{+} Cl X^{-} HBF X_{4} ArN X_{2} X^{+} BF X_{4} X^{-} heat 2, 4, 6 - Br X_{3} C X_{6} H X_{2} F$
(v) $C X_{6} H X_{5} CH X_{2} Cl KCN C X_{6} H X_{5} CH X_{2} CN LiAlH X_{4} C X_{6} H X_{5} CH X_{2} CH X_{2} NH X_{2}$
(vi) $C X_{6} H X_{5} Cl HNO X_{3} / H X_{2} SO X_{4} p - ClC X_{6} H X_{4} NO X_{2} Sn / HCl p - ClC X_{6} H X_{4} NH X_{2}$
(vii) $C X_{6} H X_{5} NH X_{2} (CH X_{3} CO) X_{2} O C X_{6} H X_{5} NHCOCH X_{3} Br X_{2} / CH X_{3} COOH p - BrC X_{6} H X_{4} NHCOCH X_{3} H X_{3} O X^{+} or OH X^{-} p - BrC X_{6} H X_{4} NH X_{2}$ (acetylation controls the activation so only one Br enters)
(viii) $C X_{6} H X_{5} CONH X_{2} Br X_{2} / KOH C X_{6} H X_{5} NH X_{2} NaNO X_{2} / HCl 273 K C X_{6} H X_{5} N X_{2} Cl H X_{3} PO X_{2} C X_{6} H X_{6} CH X_{3} Cl anhyd \cdot AlCl X_{3} C X_{6} H X_{5} CH X_{3}$
(ix) $C X_{6} H X_{5} NH X_{2} NaNO X_{2} / HCl 273 K C X_{6} H X_{5} N X_{2} Cl CuCN C X_{6} H X_{5} CN H X_{3} O X^{+} C X_{6} H X_{5} COOH LiAlH X_{4} C X_{6} H X_{5} CH X_{2} OH$

As in the sequences above.

9.9

Give the structures of A, B and C in the following reactions: (i)

CH X_{3} CH X_{2} I NaCN A OH X^{-} partial hydrolysis B NaOH + Br X_{2} C

(ii)

C X_{6} H X_{5} N X_{2} Cl CuCN A H X_{2} O / H X^{+} B NH X_{3} heat C

(iii)

CH X_{3} CH X_{2} Br KCN A LiAlH X_{4} B HNO X_{2} 0° C C

(iv)

C X_{6} H X_{5} NO X_{2} Fe / HCl A NaNO X_{2} + HCl 273 K B H X_{2} O / H X^{+} heat C

(v)

CH X_{3} COOH NH X_{3} heat A NaOBr B NaNO X_{2} / HCl C

(vi)

C X_{6} H X_{5} NO X_{2} Fe / HCl A HNO X_{2} 273 K B C X_{6} H X_{5} OH C

Solution

(i) A = $CH X_{3} CH X_{2} CN$ , B = $CH X_{3} CH X_{2} CONH X_{2}$ , C = $CH X_{3} CH X_{2} NH X_{2}$
(ii) A = $C X_{6} H X_{5} CN$ , B = $C X_{6} H X_{5} COOH$ , C = $C X_{6} H X_{5} CONH X_{2}$
(iii) A = $CH X_{3} CH X_{2} CN$ , B = $CH X_{3} CH X_{2} CH X_{2} NH X_{2}$ , C = $CH X_{3} CH X_{2} CH X_{2} OH$
(iv) A = $C X_{6} H X_{5} NH X_{2}$ , B = $C X_{6} H X_{5} N X_{2} X^{+} Cl X^{-}$ , C = $C X_{6} H X_{5} OH$
(v) A = $CH X_{3} CONH X_{2}$ , B = $CH X_{3} NH X_{2}$ , C = $CH X_{3} OH$
(vi) A = $C X_{6} H X_{5} NH X_{2}$ , B = $C X_{6} H X_{5} N X_{2} X^{+} Cl X^{-}$ , C = $C X_{6} H X_{5} N = N - C X_{6} H X_{4} OH$ (p-hydroxyazobenzene)

As listed above.

9.10

An aromatic compound 'A' on treatment with aqueous ammonia and heating forms compound 'B' which on heating with

Br X_{2}

and KOH forms a compound 'C' of molecular formula

C X_{6} H X_{7} N

. Write the structures and IUPAC names of compounds A, B and C.

Solution

C ( $C X_{6} H X_{7} N$ ) is aniline. It is formed by the Hoffmann bromamide reaction, so B is benzamide, and B comes from heating A with ammonia, so A is benzoic acid.

A = $C X_{6} H X_{5} COOH$ : benzoic acid (benzenecarboxylic acid)
B = $C X_{6} H X_{5} CONH X_{2}$ : benzamide (benzenecarboxamide)
C = $C X_{6} H X_{5} NH X_{2}$ : aniline (benzenamine)

A benzoic acid, B benzamide, C aniline.

9.11

Complete the following reactions: (i)

C X_{6} H X_{5} NH X_{2} + CHCl X_{3} + alc \cdot KOH

(ii)

C X_{6} H X_{5} N X_{2} Cl + H X_{3} PO X_{2} + H X_{2} O

(iii)

C X_{6} H X_{5} NH X_{2} + H X_{2} SO X_{4} (conc \cdot)

(iv)

C X_{6} H X_{5} N X_{2} Cl + C X_{2} H X_{5} OH

(v)

C X_{6} H X_{5} NH X_{2} + Br X_{2} (aq)

(vi)

C X_{6} H X_{5} NH X_{2} + (CH X_{3} CO) X_{2} O

(vii)

C X_{6} H X_{5} N X_{2} Cl (i) HBF X_{4} (ii) NaNO X_{2} / Cu, heat

Solution

(i) $C X_{6} H X_{5} NH X_{2} + CHCl X_{3} + 3 KOH C X_{6} H X_{5} NC + 3 KCl + 3 H X_{2} O$ (phenyl isocyanide)
(ii) $C X_{6} H X_{5} N X_{2} Cl + H X_{3} PO X_{2} + H X_{2} O C X_{6} H X_{6} + N X_{2} + H X_{3} PO X_{3} + HCl$
(iii) $C X_{6} H X_{5} NH X_{2} + H X_{2} SO X_{4} C X_{6} H X_{5} NH X_{3} X^{+} HSO X_{4} X^{-}$ (anilinium hydrogensulphate; on heating at 453–473 K it gives sulphanilic acid)
(iv) $C X_{6} H X_{5} N X_{2} Cl + C X_{2} H X_{5} OH C X_{6} H X_{6} + N X_{2} + CH X_{3} CHO + HCl$
(v) $C X_{6} H X_{5} NH X_{2} + 3 Br X_{2} 2, 4, 6 - Br X_{3} C X_{6} H X_{2} NH X_{2} ↓ + 3 HBr$ (white precipitate)
(vi) $C X_{6} H X_{5} NH X_{2} + (CH X_{3} CO) X_{2} O C X_{6} H X_{5} NHCOCH X_{3} + CH X_{3} COOH$ (acetanilide)
(vii) $C X_{6} H X_{5} N X_{2} Cl HBF X_{4} C X_{6} H X_{5} N X_{2} X^{+} BF X_{4} X^{-} NaNO X_{2} / Cu heat C X_{6} H X_{5} NO X_{2}$ (nitrobenzene)

(i) phenyl isocyanide; (ii) benzene; (iii) anilinium hydrogensulphate; (iv) benzene; (v) 2,4,6-tribromoaniline; (vi) acetanilide; (vii) nitrobenzene.

9.12

Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?

Solution

The method needs the phthalimide anion to displace a halide from an alkyl halide by nucleophilic substitution. Aryl halides do not undergo nucleophilic substitution with the phthalimide anion (the C–X bond has partial double-bond character), so aromatic amines cannot be made this way.

Aryl halides do not undergo nucleophilic substitution with the phthalimide anion.

9.13

Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.

Solution

(i) Aromatic: diazonium salt at 273–278 K: $C X_{6} H X_{5} NH X_{2} + NaNO X_{2} + 2 HCl C X_{6} H X_{5} N X_{2} X^{+} Cl X^{-} + NaCl + 2 H X_{2} O$
(ii) Aliphatic: unstable diazonium salt decomposes, giving an alcohol and nitrogen gas: $RNH X_{2} + HNO X_{2} NaNO X_{2} + HCl ROH + N X_{2} ↑ + H X_{2} O$

(i) Arenediazonium salt. (ii) Alcohol and N₂ gas.

9.14

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?

Solution

(i) Nitrogen is less electronegative than oxygen, so the N–H bond is less polar than O–H and the amide ion ( $RNH X^{-}$ ) is less stable than the alkoxide ion ( $RO X^{-}$ ). Amines therefore lose a proton less easily.

(ii) Primary amines have two H atoms on N and form intermolecular hydrogen bonds. Tertiary amines have no H on N and cannot, so they boil lower.

(iii) In aromatic amines the nitrogen lone pair is delocalised into the benzene ring and is less available for a proton. In aliphatic amines, alkyl groups (+I effect) increase the electron density on nitrogen.

(i) N is less electronegative, so RNH⁻ is less stable. (ii) Primary amines hydrogen-bond; tertiary do not. (iii) Delocalisation in aromatic amines versus +I effect in aliphatic amines.

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