Amines

1. Classification and Structure

Amines are derivatives of ammonia: 1° ($RNH_2$), 2° ($R_2NH$), 3° ($R_3N$). The nitrogen is $sp^3$ hybridised with a lone pair, giving a pyramidal shape. That lone pair makes amines basic and nucleophilic.

2. Preparation

• Reduction of nitro compounds: $Fe$/HCl or $Sn$/HCl, or catalytic hydrogenation. Fe/HCl is preferred industrially because the $FeCl_2$ formed is hydrolysed, regenerating HCl.
• Reduction of nitriles and amides with $LiAlH_4$.
• Gabriel phthalimide synthesis: gives pure 1° aliphatic amines. It cannot make aromatic amines, because aryl halides do not undergo nucleophilic substitution with the phthalimide anion.
• Hoffmann bromamide degradation: $RCONH_2 + Br_2 + 4NaOH \to RNH_2 + Na_2CO_3 + 2NaBr + 2H_2O$. The amine has one carbon fewer than the amide.

3. Basicity

In the gas phase basicity follows the +I effect: $3^\circ > 2^\circ > 1^\circ > NH_3$. In water, solvation (H-bonding of the protonated amine) and steric hindrance also count, so for methyl amines the order is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$, and for ethyl amines $(C_2H_5)_2NH > (C_2H_5)_3N > C_2H_5NH_2 > NH_3$.

Aniline is much weaker than aliphatic amines: its lone pair is delocalised into the benzene ring. Electron-donating groups (−$CH_3$, −$OCH_3$) on the ring increase basicity; electron-withdrawing groups (−$NO_2$) decrease it.

4. Key Reactions

• Carbylamine test (isocyanide test): 1° amines (aliphatic and aromatic) + $CHCl_3$ + alcoholic KOH give a foul-smelling isocyanide.
• Hinsberg's test with benzenesulphonyl chloride: 1° amines give a sulphonamide soluble in alkali; 2° amines give an insoluble one; 3° amines do not react.
• Nitrous acid: 1° aliphatic amines give alcohols and $N_2$ gas; 1° aromatic amines at 273–278 K give diazonium salts.
• Acylation: converting $-NH_2$ to $-NHCOCH_3$ moderates its activating effect, which is how p-bromoaniline is made selectively.

5. Electrophilic Substitution in Aniline

$-NH_2$ is a strong ortho/para director. Bromine water gives a white precipitate of 2,4,6-tribromoaniline at once. Direct nitration gives a significant amount of the meta product too, because in strong acid aniline is protonated to the anilinium ion ($-NH_3^+$), which is meta-directing.

6. Diazonium Salts

Benzenediazonium chloride, $C_6H_5N_2^+Cl^-$, is stable only at low temperature but is extremely useful because $N_2$ is an excellent leaving group:

• Sandmeyer: $CuCl/HCl$, $CuBr/HBr$ or $CuCN/KCN$ replace $-N_2^+$ with Cl, Br or CN. (Gattermann uses Cu powder with HX.)
• KI gives iodobenzene; $HBF_4$ then heat gives fluorobenzene (Balz–Schiemann).
• Warm water gives phenol; $H_3PO_2$ or ethanol replaces it with H.
• Coupling: with phenol (in alkali) or aniline (in mild acid) gives brightly coloured azo dyes, e.g. p-hydroxyazobenzene (orange).

Practice questions

1. Which amine gives a positive carbylamine test?

   1. (CH₃)₂NH
   2. (CH₃)₃N
   3. C₆H₅NH₂
   4. C₆H₅NHCH₃
   Answer

   C. Only primary amines give the carbylamine test; aniline is a primary aromatic amine.

2. The most basic compound in aqueous solution is:

   1. NH₃
   2. CH₃NH₂
   3. (CH₃)₂NH
   4. (CH₃)₃N
   Answer

   C. For methyl amines in water, the balance of +I effect, solvation and steric hindrance makes the 2° amine strongest.

3. Hoffmann bromamide degradation of propanamide gives:

   1. Propanamine
   2. Ethanamine
   3. Butanamine
   4. Propanoic acid
   Answer

   B. The amine has one carbon fewer than the amide: C₂H₅CONH₂ → C₂H₅NH₂.

4. Aniline is a weaker base than ethylamine because:

   1. Aniline has a higher molar mass
   2. The lone pair on N in aniline is delocalised into the ring
   3. Aniline forms H-bonds
   4. Ethylamine is aromatic
   Answer

   B. Resonance with the benzene ring makes the lone pair less available for protonation.

5. Benzenediazonium chloride reacts with CuCN/KCN to give:

   1. Benzonitrile
   2. Phenyl isocyanide
   3. Benzamide
   4. Aniline
   Answer

   A. This is a Sandmeyer reaction replacing -N₂⁺ with -CN.

6. Gabriel phthalimide synthesis is used to prepare:

   1. Primary aromatic amines
   2. Primary aliphatic amines
   3. Secondary amines
   4. Tertiary amines
   Answer

   B. It gives pure 1° aliphatic amines. Aryl halides do not react with the phthalimide anion.