Amines

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.

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(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 ✓

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(ii) IUPAC names:

| S.No. | Structure | IUPAC Name |
|---|---|---|
| 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 |
| 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 |
| 8 | $(CH_3)_2NC_2H_5$ | N,N-Dimethylethanamine |

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(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.
$$C_6H_5NO_2 \xrightarrow{\text{Fe/HCl or Sn/HCl}} C_6H_5NH_2$$

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(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)

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(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).

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(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}

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(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}

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(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)

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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 ✓

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(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)

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(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).

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(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.

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(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.

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(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).

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(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.

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(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)

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(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.)

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(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.

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(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.

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(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.

(i) Decreasing order of pKb (i.e., increasing order of basicity reversed):

Higher $pK_b$ = weaker base.

- $C_6H_5NH_2$: lone pair delocalised into ring → weakest base → highest $pK_b$ ($pK_b = 9.38$)
- $C_6H_5NHCH_3$: N-methyl group slightly increases electron density but ring still withdraws → weak base ($pK_b \approx 9.15$)
- $C_2H_5NH_2$: alkyl group, primary amine ($pK_b = 3.29$)
- $(C_2H_5)_2NH$: two alkyl groups, secondary amine → strongest base → lowest $pK_b$ ($pK_b = 3.0$)

\boxed{C_6H_5NH_2 > C_6H_5NHCH_3 > C_2H_5NH_2 > (C_2H_5)_2NH}

(Decreasing order of $pK_b$)

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(ii) Increasing order of basic strength:

- $C_6H_5NH_2$: weakest (resonance withdrawal)
- $CH_3NH_2$: primary alkylamine
- $C_6H_5N(CH_3)_2$: N,N-dimethylaniline — two methyl groups increase electron density on N, but ring still withdraws; stronger than aniline but weaker than aliphatic amines
- $(C_2H_5)_2NH$: strongest (two ethyl groups + solvation)

\boxed{C_6H_5NH_2 < C_6H_5N(CH_3)_2 < CH_3NH_2 < (C_2H_5)_2NH}

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(iii)(a) Aniline, p-nitroaniline, p-toluidine — increasing basic strength:

- $p$-Nitroaniline: $-NO_2$ is strongly electron-withdrawing (−M and −I), further reduces lone pair availability → weakest base.
- Aniline: lone pair in resonance with ring.
- $p$-Toluidine: $-CH_3$ is electron-donating (+I), increases electron density on ring and on N → strongest base among the three.

\boxed{p\text{-Nitroaniline} < \text{Aniline} < p\text{-Toluidine}}

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(iii)(b) $C_6H_5NH_2,\ C_6H_5NHCH_3,\ C_6H_5CH_2NH_2$ — increasing basic strength:

- $C_6H_5NH_2$: lone pair delocalised into ring → weakest.
- $C_6H_5NHCH_3$: N-methyl group donates electrons to N, but ring still withdraws; slightly stronger than aniline.
- $C_6H_5CH_2NH_2$: $-NH_2$ is not directly on ring; benzyl group has mild −I effect but no resonance withdrawal of lone pair; behaves more like aliphatic amine → strongest.

\boxed{C_6H_5NH_2 < C_6H_5NHCH_3 < C_6H_5CH_2NH_2}

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(iv) Decreasing order of basic strength in gas phase:

In gas phase, there is no solvation effect. Basic strength depends purely on electron-donating ability of alkyl groups (inductive effect).

More alkyl groups → more electron donation → stronger base.

\boxed{(C_2H_5)_3N > (C_2H_5)_2NH > C_2H_5NH_2 > NH_3}

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(v) Increasing order of boiling point:

Factors: molecular mass, hydrogen bonding.
- $(CH_3)_2NH$: secondary amine, one N–H bond, weak H-bonding, MW = 45.
- $C_2H_5NH_2$: primary amine, two N–H bonds, stronger H-bonding than secondary, MW = 45.
- $C_2H_5OH$: alcohol, O–H bond, strongest H-bonding (O more electronegative than N), MW = 46.

Boiling points: $(CH_3)_2NH$ (7°C) < $C_2H_5NH_2$ (17°C) < $C_2H_5OH$ (78°C)

\boxed{(CH_3)_2NH < C_2H_5NH_2 < C_2H_5OH}

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(vi) Increasing order of solubility in water:

- $C_6H_5NH_2$ (aniline): large hydrophobic benzene ring → least soluble.
- $(C_2H_5)_2NH$: two ethyl groups, larger hydrophobic part than $C_2H_5NH_2$, only one N–H for H-bonding → less soluble.
- $C_2H_5NH_2$: smaller molecule, two N–H bonds for H-bonding → most soluble.

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

(i) Ethanoic acid → Methanamine ($CH_3NH_2$):

Ethanoic acid ($CH_3COOH$) → Ethanamide ($CH_3CONH_2$) → Methanamine ($CH_3NH_2$) via Hofmann bromamide reaction.

$$CH_3COOH \xrightarrow{NH_3/\Delta} CH_3CONH_2 \xrightarrow{Br_2/KOH} CH_3NH_2$$

Hofmann bromamide reaction reduces the carbon chain by one.

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(ii) Hexanenitrile → 1-Aminopentane ($CH_3(CH_2)_4NH_2$):

Hexanenitrile: $CH_3(CH_2)_4CN$ (6 carbons including CN)

Partial hydrolysis of hexanenitrile gives pentanamide, then Hofmann reaction gives 1-aminopentane.

OR — Direct reduction:
$$CH_3(CH_2)_4CN \xrightarrow{H_2/Ni \text{ or } LiAlH_4} CH_3(CH_2)_4CH_2NH_2$$
This gives 1-aminohexane (6 carbons), not 1-aminopentane.

Correct route: Hexanenitrile has 6 carbons. 1-Aminopentane has 5 carbons. So:
$$CH_3(CH_2)_4CN \xrightarrow{H_2O/H^+,\text{ partial}} CH_3(CH_2)_4CONH_2 \xrightarrow{Br_2/KOH} CH_3(CH_2)_4NH_2$$
(Hexanamide → 1-Aminopentane via Hofmann degradation)

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(iii) Methanol → Ethanoic acid:

$$CH_3OH \xrightarrow{[O]} HCHO \xrightarrow{[O]} HCOOH$$

Better route:
$$CH_3OH \xrightarrow{HI} CH_3I \xrightarrow{KCN} CH_3CN \xrightarrow{H_2O/H^+} CH_3COOH$$

Step 1: $CH_3OH + HI \rightarrow CH_3I + H_2O$
Step 2: $CH_3I + KCN \rightarrow CH_3CN + KI$
Step 3: $CH_3CN + H_2O + H^+ \rightarrow CH_3COOH$

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(iv) Ethanamine → Methanamine:

Ethanamine: $C_2H_5NH_2$ (2 carbons) → Methanamine: $CH_3NH_2$ (1 carbon)

$$C_2H_5NH_2 \xrightarrow{(CH_3CO)_2O} C_2H_5NHCOCH_3 \xrightarrow{Br_2/KOH} ?$$

Actually, the Hofmann reaction on an amide derived from ethanamine doesn't directly give methanamine.

Correct route:
$$C_2H_5NH_2 \xrightarrow{[O]} CH_3COOH \xrightarrow{NH_3} CH_3CONH_2 \xrightarrow{Br_2/KOH} CH_3NH_2$$

Step 1: Oxidise ethanamine to ethanoic acid: $C_2H_5NH_2 \xrightarrow{[O]} CH_3COOH$
Step 2: $CH_3COOH + NH_3 \rightarrow CH_3CONH_2$
Step 3: $CH_3CONH_2 + Br_2 + 4KOH \rightarrow CH_3NH_2 + K_2CO_3 + 2KBr + 2H_2O$

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(v) Ethanoic acid → Propanoic acid:

Ethanoic acid ($CH_3COOH$, 2C) → Propanoic acid ($CH_3CH_2COOH$, 3C)

$$CH_3COOH \xrightarrow{PCl_5} CH_3COCl \xrightarrow{KCN} CH_3COCN$$

Better route:
$$CH_3COOH \xrightarrow{NH_3/\Delta} CH_3CONH_2 \xrightarrow{Br_2/KOH} CH_3NH_2 \xrightarrow{HNO_2} CH_3OH \xrightarrow{HBr} CH_3Br \xrightarrow{KCN} CH_3CN \xrightarrow{H_2O/H^+} CH_3COOH$$

Simplest route:
$$CH_3COOH \xrightarrow{PCl_5} CH_3COCl \xrightarrow{\text{diazomethane}} CH_3COCHN_2 \xrightarrow{H_2O} CH_3CH_2COOH$$

Standard NCERT route:
$$CH_3COOH \xrightarrow{LiAlH_4} CH_3CH_2OH \xrightarrow{HBr} CH_3CH_2Br \xrightarrow{KCN} CH_3CH_2CN \xrightarrow{H_2O/H^+} CH_3CH_2COOH$$

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(vi) Methanamine → Ethanamine:

Methanamine ($CH_3NH_2$, 1C) → Ethanamine ($C_2H_5NH_2$, 2C)

$$CH_3NH_2 \xrightarrow{HNO_2} CH_3OH \xrightarrow{HBr} CH_3Br \xrightarrow{KCN} CH_3CN \xrightarrow{LiAlH_4} C_2H_5NH_2$$

Step 1: $CH_3NH_2 + HNO_2 \rightarrow CH_3OH + N_2 + H_2O$
Step 2: $CH_3OH + HBr \rightarrow CH_3Br + H_2O$
Step 3: $CH_3Br + KCN \rightarrow CH_3CN + KBr$
Step 4: $CH_3CN \xrightarrow{LiAlH_4} C_2H_5NH_2$

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(vii) Nitromethane → Dimethylamine:

Nitromethane ($CH_3NO_2$) → Methanamine ($CH_3NH_2$) → Dimethylamine ($(CH_3)_2NH$)

Step 1: Reduce nitromethane:
$$CH_3NO_2 \xrightarrow{Zn/HCl \text{ or } LiAlH_4} CH_3NH_2$$

Step 2: React methanamine with excess $CH_3I$ (controlled) or via reductive amination:
$$CH_3NH_2 + CH_3I \xrightarrow{Na_2CO_3} (CH_3)_2NH + HI$$

(Control the amount of $CH_3I$ to get secondary amine)

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(viii) Propanoic acid → Ethanoic acid:

Propanoic acid ($CH_3CH_2COOH$, 3C) → Ethanoic acid ($CH_3COOH$, 2C)

$$CH_3CH_2COOH \xrightarrow{NH_3/\Delta} CH_3CH_2CONH_2 \xrightarrow{Br_2/KOH} CH_3CH_2NH_2$$
$$CH_3CH_2NH_2 \xrightarrow{HNO_2} CH_3CH_2OH \xrightarrow{[O]} CH_3COOH$$

Step 1: Propanoic acid → Propanamide ($NH_3/\Delta$)
Step 2: Propanamide → Ethanamine (Hofmann bromamide reaction)
Step 3: Ethanamine → Ethanol ($HNO_2$)
Step 4: Ethanol → Ethanoic acid (oxidation with $K_2Cr_2O_7/H_2SO_4$)

Method: Hinsberg's Test

Hinsberg's reagent is benzenesulphonyl chloride ($C_6H_5SO_2Cl$).

The amine is treated with Hinsberg's reagent in the presence of aqueous KOH.

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Primary amine ($RNH_2$):

Reacts with $C_6H_5SO_2Cl$ to form a sulphonamide which is soluble in KOH (due to acidic N–H).

$$RNH_2 + C_6H_5SO_2Cl \xrightarrow{KOH} C_6H_5SO_2NHR \xrightarrow{KOH} C_6H_5SO_2NR^-K^+ + H_2O$$

The product dissolves in KOH → Primary amine confirmed.

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Secondary amine ($R_2NH$):

Reacts with $C_6H_5SO_2Cl$ to form a sulphonamide which is insoluble in KOH (no acidic N–H).

$$R_2NH + C_6H_5SO_2Cl \xrightarrow{KOH} C_6H_5SO_2NR_2 + HCl$$

The product is insoluble in KOH → Secondary amine confirmed.

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Tertiary amine ($R_3N$):

Does not react with $C_6H_5SO_2Cl$ (no N–H bond). The amine layer separates out.

$$R_3N + C_6H_5SO_2Cl \rightarrow \text{No reaction (or forms a soluble salt that separates on acidification)}$$

No sulphonamide formed → Tertiary amine confirmed.

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Summary Table:

| Amine | Reaction with $C_6H_5SO_2Cl$/KOH | Observation |
|---|---|---|
| Primary | Forms sulphonamide soluble in KOH | Clear solution |
| Secondary | Forms sulphonamide insoluble in KOH | Precipitate |
| Tertiary | No reaction | Amine layer separates |

(i) Carbylamine Reaction:

Primary amines (aliphatic or aromatic) react with chloroform ($CHCl_3$) and alcoholic KOH to form isocyanides (carbylamines), which have a very foul smell. This reaction is used as a test for primary amines.

$$RNH_2 + CHCl_3 + 3KOH(alc.) \xrightarrow{\Delta} RNC + 3KCl + 3H_2O$$

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

Secondary and tertiary amines do not give this test.

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(ii) Diazotisation:

The conversion of a primary aromatic amine into a diazonium salt by treatment with nitrous acid ($NaNO_2 + HCl$) at 273–278 K is called diazotisation.

$$C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273-278K} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$$

The diazonium salt is unstable and must be used immediately. It is stabilised by resonance in aromatic systems.

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(iii) Hofmann's Bromamide Reaction (Hofmann Degradation):

When a primary amide is treated with bromine and aqueous/ethanolic KOH, it gives a primary amine with one carbon less than the amide. This reaction involves migration of an alkyl or aryl group from carbonyl carbon to nitrogen.

$$RCONH_2 + Br_2 + 4KOH \rightarrow RNH_2 + K_2CO_3 + 2KBr + 2H_2O$$

Example:
$$CH_3CONH_2 + Br_2 + 4KOH \rightarrow CH_3NH_2 + K_2CO_3 + 2KBr + 2H_2O$$

This is useful for preparing primary amines with one carbon less than the starting amide.

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(iv) Coupling Reaction:

Arenediazonium salts react with activated aromatic compounds (phenols or aromatic amines) in weakly alkaline or acidic medium to form brightly coloured azo compounds ($-N=N-$). This reaction is called coupling reaction.

$$C_6H_5N_2^+Cl^- + C_6H_5OH \xrightarrow{NaOH} C_6H_5-N=N-C_6H_4OH + HCl$$
(p-Hydroxyazobenzene — orange dye)

$$C_6H_5N_2^+Cl^- + C_6H_5NH_2 \xrightarrow{\text{dil. HCl}} C_6H_5-N=N-C_6H_4NH_2 + HCl$$

Coupling occurs at the para position of phenol/amine. Azo dyes are used in the textile industry.

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(v) Ammonolysis:

The cleavage of the C–X bond of an alkyl halide by ammonia is called ammonolysis. The alkyl halide reacts with excess ethanolic ammonia in a sealed tube at 373 K.

$$RX + NH_3 \rightarrow RNH_2 + HX$$

The primary amine formed can further react:
$$RNH_2 + RX \rightarrow R_2NH \xrightarrow{RX} R_3N \xrightarrow{RX} R_4N^+X^-$$

A mixture of primary, secondary, tertiary amines and quaternary ammonium salt is obtained. Excess of ammonia favours primary amine formation.

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(vi) Acetylation:

Reaction of amines with acyl chlorides or acid anhydrides to form amides is called acetylation (acylation).

$$RNH_2 + CH_3COCl \rightarrow RNHCOCH_3 + HCl$$
$$RNH_2 + (CH_3CO)_2O \rightarrow RNHCOCH_3 + CH_3COOH$$

Example with aniline:
$$C_6H_5NH_2 + (CH_3CO)_2O \rightarrow C_6H_5NHCOCH_3 + CH_3COOH$$
(Acetanilide)

Acetylation is used to protect the $-NH_2$ group during reactions and to reduce the activating effect of $-NH_2$ in electrophilic substitution.

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(vii) Gabriel Phthalimide Synthesis:

This is a method for the preparation of pure primary amines. Phthalimide is treated with KOH to form potassium phthalimide, which then reacts with an alkyl halide. The N-alkyl phthalimide formed is hydrolysed with aqueous NaOH or $H_2SO_4$ to give the primary amine.

$$\text{Phthalimide} \xrightarrow{KOH} \text{Potassium phthalimide} \xrightarrow{RX} N\text{-alkylphthalimide} \xrightarrow{H_2O/H^+} RNH_2 + \text{phthalic acid}$$

This method gives exclusively primary amines and avoids secondary/tertiary amine formation. It cannot be used for aromatic primary amines because aryl halides do not undergo nucleophilic substitution easily.