Amines — NCERT Solutions

Concept: An amine is classified based on the number of hydrogen atoms of NH₃ replaced by alkyl/aryl groups.

• Primary (1°): one H replaced → R–NH₂
• Secondary (2°): two H replaced → R₂NH
• Tertiary (3°): three H replaced → R₃N

(i) The structure shown in the image is that of a cyclic secondary amine (cyclohexylamine type) or an N-substituted compound. Based on standard NCERT context, image (i) represents a compound where nitrogen bears two carbon substituents and one H → Secondary amine.

(ii) The structure shown in image (ii) represents a compound where nitrogen bears three carbon substituents and no H → Tertiary amine.

(iii) $(C_2H_5)_2CHNH_2$
The nitrogen atom is bonded to two H atoms and one carbon group (the $–CH(C_2H_5)_2$ group). Since only one H of NH₃ is replaced by an alkyl group, this is a Primary amine (1°).

(iv) $(C_2H_5)_2NH$
The nitrogen atom is bonded to two ethyl groups and one H atom. Two H atoms of NH₃ are replaced → Secondary amine (2°).

Given: Molecular formula $C_4H_{11}N$

Degree of unsaturation $= \frac{2(4)+2-11+1}{2} = 0$, so all isomers are saturated amines.

(i) Structures of all isomeric amines:

Primary amines (R–NH₂):

1. $CH_3CH_2CH_2CH_2NH_2$ — Butan-1-amine
2. $CH_3CH_2CH(NH_2)CH_3$ — Butan-2-amine
3. $(CH_3)_2CHCH_2NH_2$ — 2-Methylpropan-1-amine
4. $(CH_3)_3CNH_2$ — 2-Methylpropan-2-amine

Secondary amines (R₂NH):

5. $CH_3NHCH_2CH_2CH_3$ — N-Methylpropan-1-amine
6. $CH_3NHCH(CH_3)_2$ — N-Methylpropan-2-amine
7. $(C_2H_5)_2NH$ — N-Ethylethanamine (Diethylamine)

Tertiary amines (R₃N):

8. $(CH_3)_2NCH_2CH_3$ — N,N-Dimethylethanamine
9. $(CH_3)_3N$ — Wait, $(CH_3)_3N$ has formula $C_3H_9N$, not $C_4H_{11}N$.

Correct: $(CH_3)_2NC_2H_5$ — N,N-Dimethylethanamine ✓

(ii) IUPAC names:

(iii) Types of isomerism:

• Chain isomerism: Isomers 1 and 3 (different carbon skeletons, both primary amines). E.g., butan-1-amine and 2-methylpropan-1-amine.
• Position isomerism: Isomers 1 and 2 (same chain, $-NH_2$ at different positions). E.g., butan-1-amine and butan-2-amine.
• Metamerism: Isomers among secondary amines having different alkyl groups on either side of nitrogen. E.g., N-methylpropan-1-amine and N-ethylethanamine.
• Functional isomerism: Primary, secondary and tertiary amines with the same molecular formula are functional isomers of each other.

(i) Benzene → Aniline:

Step 1: Nitration of benzene to give nitrobenzene.
$$C_6H_6 \xrightarrow{\text{conc. HNO}_3/\text{conc. H}_2\text{SO}_4} C_6H_5NO_2$$

Step 2: Reduction of nitrobenzene to aniline.
$$C_6H_5NO_2 \xrightarrow{\text{Fe/HCl or Sn/HCl}} C_6H_5NH_2$$

(ii) Benzene → N,N-Dimethylaniline:

Step 1: Benzene → Nitrobenzene (as above)

Step 2: Nitrobenzene → Aniline (reduction as above)

Step 3: Aniline is treated with excess methyl iodide (CH₃I) in the presence of $Na_2CO_3$:
$$C_6H_5NH_2 \xrightarrow{2CH_3I / Na_2CO_3} C_6H_5N(CH_3)_2$$
(N,N-Dimethylaniline)

(iii) Cl–(CH₂)₄–Cl → Hexan-1,6-diamine:

1,4-Dichlorobutane is treated with excess ethanolic ammonia in a sealed tube:
$$Cl(CH_2)_4Cl \xrightarrow{\text{excess alc. NH}_3, \Delta} H_2N(CH_2)_4NH_2$$

However, to get hexan-1,6-diamine (6 carbons) from 1,4-dichlorobutane (4 carbons), the correct route is via nitrile:

Step 1: React with NaCN:
$$Cl(CH_2)_4Cl + 2NaCN \rightarrow NC(CH_2)_4CN + 2NaCl$$
(Hexanedinitrile)

Step 2: Reduce with $H_2$/Ni or $LiAlH_4$:
$$NC(CH_2)_4CN \xrightarrow{4H_2/Ni \text{ or } LiAlH_4} H_2N(CH_2)_6NH_2$$
(Hexan-1,6-diamine)

Final Answer: $H_2N–(CH_2)_6–NH_2$ (Hexan-1,6-diamine)

Concept: Basic strength depends on the availability of the lone pair on nitrogen.

• Alkyl groups (electron-donating) increase basic strength.
• Aryl groups (electron-withdrawing by resonance) decrease basic strength.
• In aqueous solution, secondary alkylamines are stronger bases than primary, which are stronger than tertiary (due to solvation effects).

(i) $C_2H_5NH_2,\ C_6H_5NH_2,\ NH_3,\ C_6H_5CH_2NH_2,\ (C_2H_5)_2NH$

• $C_6H_5NH_2$: lone pair delocalised into ring → weakest base.
• $NH_3$: no alkyl group.
• $C_6H_5CH_2NH_2$: benzyl group is electron-withdrawing by induction but $-NH_2$ not directly on ring, so stronger than $NH_3$.
• $C_2H_5NH_2$: alkyl group donates electrons → stronger than $NH_3$.
• $(C_2H_5)_2NH$: two alkyl groups → strongest.

$$\boxed{C_6H_5NH_2 < NH_3 < C_6H_5CH_2NH_2 < C_2H_5NH_2 < (C_2H_5)_2NH}$$

(ii) $C_2H_5NH_2,\ (C_2H_5)_2NH,\ (C_2H_5)_3N,\ C_6H_5NH_2$

In aqueous solution:

• $C_6H_5NH_2$: weakest (resonance withdrawal).
• $(C_2H_5)_3N$: tertiary, less solvated → weaker than secondary.
• $C_2H_5NH_2$: primary.
• $(C_2H_5)_2NH$: secondary, best combination of induction and solvation → strongest.

$$\boxed{C_6H_5NH_2 < C_2H_5NH_2 < (C_2H_5)_3N < (C_2H_5)_2NH}$$

(iii) $CH_3NH_2,\ (CH_3)_2NH,\ (CH_3)_3N,\ C_6H_5NH_2,\ C_6H_5CH_2NH_2$

• $C_6H_5NH_2$: weakest (lone pair in resonance with ring).
• $C_6H_5CH_2NH_2$: $-NH_2$ not on ring, slightly stronger than aniline but weaker than alkylamines.
• $(CH_3)_3N$: tertiary, less solvated.
• $CH_3NH_2$: primary.
• $(CH_3)_2NH$: secondary → strongest.

$$\boxed{C_6H_5NH_2 < C_6H_5CH_2NH_2 < (CH_3)_3N < CH_3NH_2 < (CH_3)_2NH}$$

Concept: Amines are Lewis bases. They react with acids to form salts (ammonium salts).

(i)
$$CH_3CH_2CH_2NH_2 + HCl \rightarrow CH_3CH_2CH_2\overset{+}{N}H_3\ Cl^-$$

Product: Propan-1-aminium chloride (or propylammonium chloride)

(ii)
$$(C_2H_5)_2NH + HCl \rightarrow (C_2H_5)_2\overset{+}{N}H_2\ Cl^-$$

Product: N-Ethylethanaminium chloride (or diethylammonium chloride)

Step 1: Aniline reacts with excess $CH_3I$ in the presence of $Na_2CO_3$:

$$C_6H_5NH_2 \xrightarrow{CH_3I} C_6H_5NHCH_3 \xrightarrow{CH_3I} C_6H_5N(CH_3)_2 \xrightarrow{CH_3I} [C_6H_5N(CH_3)_3]^+I^-$$

The final alkylation product is trimethylphenylammonium iodide: $[C_6H_5N(CH_3)_3]^+I^-$

Reaction of the final product:

Trimethylphenylammonium iodide is a quaternary ammonium salt. It does not react further with alkyl halides. However, it can react with $AgOH$ to give trimethylphenylammonium hydroxide:

$$[C_6H_5N(CH_3)_3]^+I^- + AgOH \rightarrow [C_6H_5N(CH_3)_3]^+OH^- + AgI\downarrow$$

The quaternary ammonium hydroxide is a strong base and can undergo Hofmann elimination on heating:
$$[C_6H_5N(CH_3)_3]^+OH^- \xrightarrow{\Delta} C_6H_5N(CH_3)_2 + CH_3OH$$

Note: The key point is that the final product $[C_6H_5\overset{+}{N}(CH_3)_3]I^-$ (trimethylphenylammonium iodide) is a quaternary ammonium salt with no further N-alkylation possible.

Concept: Aniline (a primary amine) undergoes acylation with benzoyl chloride (an acid chloride) to form an amide. This is the Schotten-Baumann reaction.

$$C_6H_5NH_2 + C_6H_5COCl \rightarrow C_6H_5NHCOC_6H_5 + HCl$$

Product: $N$-Phenylbenzamide (benzanilide)

The HCl produced is neutralised if the reaction is carried out in the presence of a base like pyridine or NaOH.

Molecular formula: $C_3H_9N$

Degree of unsaturation $= 0$, so all are saturated amines.

All isomers:

Primary amines:

1. $CH_3CH_2CH_2NH_2$ — Propan-1-amine
2. $(CH_3)_2CHNH_2$ — Propan-2-amine

Secondary amine:

3. $CH_3NHCH_2CH_3$ — N-Methylethanamine

Tertiary amine:

4. $(CH_3)_3N$ — N,N-Dimethylmethanamine (Trimethylamine)

Reaction with nitrous acid ($HNO_2 = NaNO_2 + HCl$):

• Primary aliphatic amines react with $HNO_2$ to liberate $N_2$ gas:

$$R-NH_2 + HNO_2 \rightarrow R-OH + N_2\uparrow + H_2O$$

• Secondary amines form N-nitrosamines (yellow oily liquid, no $N_2$ gas).
• Tertiary amines form ammonium salts (no $N_2$ gas).

Isomers that liberate $N_2$ gas (primary amines):

1. Propan-1-amine — $CH_3CH_2CH_2NH_2$
2. Propan-2-amine — $(CH_3)_2CHNH_2$

(i) 3-Methylaniline → 3-Nitrotoluene:

3-Methylaniline is $m$-toluidine: $3\text{-}CH_3C_6H_4NH_2$

Step 1: Protect the $-NH_2$ group by acetylation (to prevent oxidation during nitration):
$$3\text{-}CH_3C_6H_4NH_2 \xrightarrow{(CH_3CO)_2O} 3\text{-}CH_3C_6H_4NHCOCH_3$$

Step 2: Diazotisation — convert $-NH_2$ to diazonium salt:
$$3\text{-}CH_3C_6H_4NH_2 \xrightarrow{NaNO_2/HCl,\ 273\text{-}278K} 3\text{-}CH_3C_6H_4N_2^+Cl^-$$

Step 3: Replace $-N_2^+$ with $-NO_2$ using Sandmeyer-type reaction (treat with $HNO_2$ / $NaNO_2$, $Cu$ catalyst) — but the standard route is:

Actually, the correct approach:

Step 1: Diazotise 3-methylaniline:
$$3\text{-}CH_3C_6H_4NH_2 \xrightarrow{NaNO_2 + HCl,\ 273K} 3\text{-}CH_3C_6H_4N_2^+Cl^-$$

Step 2: Treat with $HBF_4$ then heat (Balz-Schiemann) — not for nitro.

Correct standard method:

Step 1: Acetylate $-NH_2$:
$$3\text{-}CH_3C_6H_4NH_2 \xrightarrow{Ac_2O} 3\text{-}CH_3C_6H_4NHCOCH_3$$

Step 2: Nitrate (the acetamido group directs ortho/para; since position 3 has $CH_3$, nitration occurs at position 4 relative to $NHCOCH_3$, i.e., position 4 of the ring).

Simpler NCERT approach:

Step 1: Diazotise:
$$3\text{-}CH_3C_6H_4NH_2 \xrightarrow{NaNO_2/HCl} 3\text{-}CH_3C_6H_4\overset{+}{N}_2Cl^-$$

Step 2: Replace $-N_2^+Cl^-$ with $-NO_2$ using $NaNO_2/Cu$ (Sandmeyer reaction with $NO_2^-$):
$$3\text{-}CH_3C_6H_4N_2^+Cl^- \xrightarrow{NaNO_2/Cu} 3\text{-}CH_3C_6H_4NO_2$$

Product: 3-Nitrotoluene ✓

(ii) Aniline → 1,3,5-Tribromobenzene:

Direct bromination of aniline gives 2,4,6-tribromoaniline. To get 1,3,5-tribromobenzene, the $-NH_2$ group must be removed after bromination.

Step 1: Bromination of aniline with excess $Br_2(aq)$:
$$C_6H_5NH_2 + 3Br_2 \rightarrow 2,4,6\text{-}Br_3C_6H_2NH_2 + 3HBr$$
(2,4,6-Tribromoaniline)

Step 2: Diazotisation of 2,4,6-tribromoaniline:
$$2,4,6\text{-}Br_3C_6H_2NH_2 \xrightarrow{NaNO_2/HCl,\ 273K} 2,4,6\text{-}Br_3C_6H_2N_2^+Cl^-$$

Step 3: Reductive removal of diazonium group using $H_3PO_2/H_2O$:
$$2,4,6\text{-}Br_3C_6H_2N_2^+Cl^- \xrightarrow{H_3PO_2/H_2O} 1,3,5\text{-}Br_3C_6H_3$$

Product: 1,3,5-Tribromobenzene ✓

Concept: IUPAC name of amine = name of parent alkane + suffix '-amine'. For secondary/tertiary amines, use N- prefix for substituents on nitrogen.

(i) $(CH_3)_2CHNH_2$

• Parent chain: propane (3 carbons including the CH)
• $-NH_2$ on C-2
• IUPAC Name: Propan-2-amine
• Classification: Primary amine (1°) — $-NH_2$ attached to one carbon

(ii) $CH_3(CH_2)_2NH_2 = CH_3CH_2CH_2NH_2$

• Parent chain: propane, $-NH_2$ on C-1
• IUPAC Name: Propan-1-amine
• Classification: Primary amine (1°)

(iii) $CH_3NHCH(CH_3)_2$

• Nitrogen has two different groups: methyl and isopropyl
• Larger group: propan-2-yl (isopropyl) → parent: propan-2-amine
• Smaller group on N: methyl → N-methyl
• IUPAC Name: N-Methylpropan-2-amine
• Classification: Secondary amine (2°)

(iv) $(CH_3)_3CNH_2$

• Parent chain: 2-methylpropane; $-NH_2$ on C-2
• IUPAC Name: 2-Methylpropan-2-amine
• Classification: Primary amine (1°)

(v) $C_6H_5NHCH_3$

• Parent: benzenamine (aniline); N-methyl substituent
• IUPAC Name: N-Methylaniline (or N-Methylbenzenamine)
• Classification: Secondary amine (2°)

(vi) $(CH_3CH_2)_2NCH_3$

• Nitrogen has two ethyl groups and one methyl group
• Largest group: ethanamine as parent
• IUPAC Name: N-Ethyl-N-methylethanamine
• Classification: Tertiary amine (3°)

(vii) $m\text{-}BrC_6H_4NH_2$

• Bromine at meta position of aniline
• IUPAC Name: 3-Bromoaniline (or 3-Bromobenzenamine)
• Classification: Primary amine (1°)

(i) Methylamine ($CH_3NH_2$) and Dimethylamine ($(CH_3)_2NH$):

Test — Hinsberg's test (using benzenesulphonyl chloride, $C_6H_5SO_2Cl$):

• $CH_3NH_2$ (primary amine) reacts with $C_6H_5SO_2Cl$ to give a sulphonamide soluble in alkali (NaOH).
• $(CH_3)_2NH$ (secondary amine) reacts to give a sulphonamide insoluble in alkali.

Alternatively — Carbylamine test:

• $CH_3NH_2$ (primary amine) gives isocyanide (foul smell) with $CHCl_3$/alc. KOH.
• $(CH_3)_2NH$ (secondary amine) does not give carbylamine test.

$$CH_3NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} CH_3NC\uparrow + 3KCl + 3H_2O$$
(Foul-smelling isocyanide)

(ii) Secondary and Tertiary amines:

Test — Hinsberg's test:

• Secondary amine reacts with $C_6H_5SO_2Cl$ to give a sulphonamide insoluble in NaOH.
• Tertiary amine does not react with $C_6H_5SO_2Cl$ (no reaction or forms a soluble salt).

Alternatively — Nitrous acid test:

• Secondary amine + $HNO_2$ → N-nitrosamine (yellow oily liquid).
• Tertiary amine + $HNO_2$ → forms a salt (no yellow precipitate).

(iii) Ethylamine ($C_2H_5NH_2$) and Aniline ($C_6H_5NH_2$):

Test — Azo dye test (diazonium coupling):

• Aniline (aromatic primary amine) undergoes diazotisation with $NaNO_2/HCl$ at 273–278 K to form a diazonium salt, which couples with $\beta$-naphthol to give an orange-red azo dye.
• Ethylamine (aliphatic primary amine) forms an unstable diazonium salt that immediately decomposes to give $N_2$ gas and alcohol; no azo dye is formed.

$$C_6H_5NH_2 \xrightarrow{NaNO_2/HCl, 273K} C_6H_5N_2^+Cl^- \xrightarrow{\beta\text{-naphthol}} \text{Orange-red azo dye}$$

Alternatively: Aniline does not give carbylamine test (wait — aniline does give carbylamine test as it is a primary amine). Better test: Aniline gives orange precipitate with $Br_2$ water (2,4,6-tribromoaniline), while ethylamine does not give a precipitate.

(iv) Aniline ($C_6H_5NH_2$) and Benzylamine ($C_6H_5CH_2NH_2$):

Test — Azo dye test:

• Aniline (aromatic primary amine) forms a stable diazonium salt at 273–278 K, which couples with $\beta$-naphthol to give an orange-red azo dye.
• Benzylamine (aliphatic primary amine) forms an unstable diazonium salt that decomposes immediately; no azo dye formed.

Alternatively — Reaction with $FeCl_3$:

• Aniline gives a characteristic colour with $FeCl_3$.
• Benzylamine does not.

(v) Aniline ($C_6H_5NH_2$) and N-Methylaniline ($C_6H_5NHCH_3$):

Test — Carbylamine test:

• Aniline (primary amine) reacts with $CHCl_3$ and alc. KOH to give phenyl isocyanide (foul smell).

$$C_6H_5NH_2 + CHCl_3 + 3KOH \rightarrow C_6H_5NC + 3KCl + 3H_2O$$

• N-Methylaniline (secondary amine) does not give carbylamine test.

Alternatively — Hinsberg's test:

• Aniline gives sulphonamide soluble in NaOH.
• N-Methylaniline gives sulphonamide insoluble in NaOH.

(i) pKb of aniline > pKb of methylamine (aniline is a weaker base):

In aniline, the lone pair of electrons on the nitrogen atom is in conjugation with the $\pi$-electron system of the benzene ring. This delocalisation reduces the availability of the lone pair for protonation, making aniline a weaker base.

In methylamine, the methyl group is electron-donating (+I effect), which increases the electron density on nitrogen, making the lone pair more available for protonation. Hence methylamine is a stronger base.

Higher $pK_b$ means weaker base, so $pK_b$(aniline) $>$ $pK_b$(methylamine).

(ii) Ethylamine is soluble in water but aniline is not:

Ethylamine can form hydrogen bonds with water molecules due to the presence of $-NH_2$ group. The ethyl group is small, so the hydrophobic part does not significantly hinder dissolution.

Aniline also has $-NH_2$ group but the large hydrophobic benzene ring makes it predominantly non-polar. The hydrophobic interaction of the benzene ring with water outweighs the hydrogen bonding of $-NH_2$ with water, making aniline sparingly soluble in water.

(iii) Methylamine precipitates hydrated ferric oxide from FeCl₃ solution:

Methylamine is a stronger base than water. In aqueous solution, it produces $OH^-$ ions:
$$CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^-$$

The $OH^-$ ions react with $FeCl_3$ to precipitate hydrated ferric oxide (reddish-brown precipitate):
$$FeCl_3 + 3OH^- \rightarrow Fe(OH)_3\downarrow + 3Cl^-$$
(or $Fe_2O_3 \cdot xH_2O$, hydrated ferric oxide)

(iv) Aniline gives substantial m-nitroaniline on nitration:

Nitration is carried out with conc. $HNO_3$/conc. $H_2SO_4$ (a strongly acidic medium). In this medium, aniline gets protonated to form anilinium ion ($C_6H_5\overset{+}{N}H_3$).

The anilinium ion has $-\overset{+}{N}H_3$ group which is an electron-withdrawing group (−I and −M effect) and is a meta-director. Hence, nitration of the protonated aniline gives a substantial amount of m-nitroaniline.

(Some unprotonated aniline also undergoes nitration at o/p positions, giving o- and p-nitroaniline.)

(v) Aniline does not undergo Friedel-Crafts reaction:

Friedel-Crafts reaction requires a Lewis acid catalyst like $AlCl_3$. Aniline is a Lewis base; it donates its lone pair to $AlCl_3$ to form a complex:
$$C_6H_5NH_2 + AlCl_3 \rightarrow C_6H_5NH_2 \cdot AlCl_3$$

This complex formation deactivates the catalyst. Also, the $-NH_2$ group becomes $-\overset{+}{N}H_2AlCl_3^-$, which is an electron-withdrawing group and deactivates the ring. Hence, Friedel-Crafts reaction does not occur with aniline.

(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines:

In arenediazonium salts ($ArN_2^+$), the positive charge on nitrogen is stabilised by resonance with the $\pi$-electron system of the benzene ring. The $-N_2^+$ group is in conjugation with the ring, distributing the positive charge.

In alkyldiazonium salts ($RN_2^+$), no such resonance stabilisation is possible. Hence, they are highly unstable and decompose immediately to give $N_2$ gas and carbocation.

(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines:

Gabriel phthalimide synthesis involves the reaction of phthalimide (potassium salt) with an alkyl halide, followed by hydrolysis. This method gives exclusively primary amines because:

• The nitrogen in phthalimide has only one replaceable H (actually none — it is the K salt).
• Only one alkyl group can be introduced on nitrogen.
• Hydrolysis of the N-alkyl phthalimide gives the primary amine.

This avoids the formation of secondary and tertiary amines and quaternary ammonium salts, which are common side products in ammonolysis of alkyl halides. Hence, Gabriel synthesis gives pure primary amines.