Question

Write short notes on the following:
(i) Gabriel phthalimide reaction
(ii) Hoffmann bromamide reaction
OR
Write the following chemical reactions:
(i) Reaction of ethanolic NH₃ with C₂H₅Cl.
(ii) Reaction of ammonia with benzyl chloride, followed by reaction with two moles of CH₃Cl.

Hint:

Gabriel synthesis makes pure aliphatic primary amines (not aromatic ones) via potassium phthalimide; Hoffmann bromamide degrades an amide with Br₂/NaOH to an amine with one carbon less. In the alternative, ammonolysis gives ethylamine, and benzyl chloride + NH₃ then 2 CH₃Cl gives N,N-dimethylbenzylamine.

Answer 1

Option 1

(i) Gabriel synthesis, re-explained.
Step 1: Think of phthalimide as a protected form of the \(-NH_2\) group. Its acidic imide hydrogen is knocked off by alcoholic KOH to make potassium phthalimide.
Step 2: This nitrogen nucleophile then does an \(S_N2\) attack on an alkyl halide \(R\text{-}X\), planting the alkyl group on nitrogen (N-alkylphthalimide).
Step 3: Finally the protecting phthalimide is cut off by acid/alkaline hydrolysis (or hydrazinolysis), unmasking a clean primary amine \(R\text{-}NH_2\) with no over-alkylation.
Step 4: Because it needs an \(S_N2\) step, only \(1^\circ\) (and some \(2^\circ\)) alkyl halides work; aryl halides fail, so aniline-type amines cannot be prepared.

(ii) Hoffmann bromamide, re-explained.
Step 1: Start from a \(1^\circ\) amide \(R\text{-}CONH_2\) and shake it with \(Br_2\) in excess \(NaOH\).
Step 2: The reaction shortens the chain by one carbon; the old carbonyl carbon departs as \(Na_2CO_3\).
\[R\text{-}CONH_2 + Br_2 + 4NaOH \rightarrow R\text{-}NH_2 + Na_2CO_3 + 2NaBr + 2H_2O\]
Step 3: Memory aid: propanamide (\(C_3\)) \(\rightarrow\) ethylamine (\(C_2\)); benzamide \(\rightarrow\) aniline. It is a standard step-down (descent of series) tool.

Option 2

(i) Ammonia acts as a nucleophile on ethyl chloride, replacing \(Cl\) to give \(C_2H_5NH_2\) (ethylamine); with more halide present the amine keeps reacting to give \(2^\circ\), \(3^\circ\) amines and the quaternary salt \((C_2H_5)_4N^+Cl^-\).
\[C_2H_5Cl + NH_3 \rightarrow C_2H_5NH_2 + HCl\]
(ii) Benzyl chloride first gives benzylamine with ammonia; the two remaining N–H bonds are then both methylated by two moles of \(CH_3Cl\) to furnish N,N-dimethylbenzylamine.
\[C_6H_5CH_2Cl \xrightarrow{NH_3} C_6H_5CH_2NH_2 \xrightarrow{2\,CH_3Cl} C_6H_5CH_2N(CH_3)_2\]
\[\boxed{\text{N,N-dimethylbenzylamine}}\]