Which of the following methods of preparation of amines will not give same number of carbon atoms in the chain of amines as in the reactant?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Treatment of an amide with bromine in an aqueous solution of sodium hydroxide
This is the Hoffmann bromamide degradation. The alkyl group moves from the carbonyl carbon to the nitrogen, and the carbonyl carbon is lost as carbonate, so the primary amine has one carbon atom less than the amide:
\( \displaystyle \ce{R-CONH2 + Br2 + 4NaOH -> R-NH2 + Na2CO3 + 2NaBr + 2H2O} \)
For example, ethanamide (\( \ce{CH3CONH2} \), two carbons) gives methanamine (\( \ce{CH3NH2} \), one carbon). In the other three methods the carbon chain is kept whole.
Option (1) Reaction of an alkyl nitrile with LiAlH4 — Gives the same number of carbon atoms. The \( \ce{C#N} \) carbon is reduced to \( \ce{CH2} \) and stays in the chain:
\( \displaystyle \ce{R-C#N} \xrightarrow{\ce{LiAlH4}} \ce{R-CH2-NH2} \)
The amine has the same carbons as the nitrile, which is why this reaction is used for the ascent of the amine series.
Option (2) Reaction of an amide with LiAlH4 followed by treatment with water — Gives the same number of carbon atoms. The C=O group is reduced to CH2; no carbon is removed:
\( \displaystyle \ce{R-CONH2} \xrightarrow[\text{(ii) } \ce{H2O}]{\text{(i) } \ce{LiAlH4}} \ce{R-CH2-NH2} \)
Option (3) Heating an alkyl halide with the potassium salt of phthalimide followed by hydrolysis — Gives the same number of carbon atoms. In the Gabriel phthalimide synthesis the alkyl group R of the halide is joined to nitrogen as N-alkylphthalimide, and hydrolysis releases it as \( \ce{R-NH2} \); the phthalimide carbons leave as phthalate:
\( \displaystyle \ce{R-X} \xrightarrow[\text{(ii) alkaline hydrolysis}]{\text{(i) potassium phthalimide}} \ce{R-NH2} \)
Option (4) Treatment of an amide with bromine in an aqueous solution of sodium hydroxide — Does not give the same number of carbon atoms. Hoffmann bromamide degradation removes the carbonyl carbon, so the amine has one carbon less than the amide.
NCERT (2026-27 edition):
6. Hoffmann bromamide degradation reaction – “Hoffmann developed a method for preparation of primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide. In this degradation reaction, migration of an alkyl or aryl group takes place from carbonyl carbon of the amide to the nitrogen atom. The amine so formed contains one carbon less than that present in the amide.”