An unknown alkyl nitrile 'A' undergoes complete hydrolysis to form a carboxylic acid 'B'. Compound 'B' is then treated with bromine and red phosphorus followed by water workup, yielding 2 -bromo- 3 -methylbutanoic acid as the major product.

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An unknown alkyl nitrile 'A' undergoes complete hydrolysis to form a carboxylic acid 'B'. Compound 'B' is then treated with bromine and red phosphorus followed by water workup, yielding 2 -bromo- 3 -methylbutanoic acid as the major product. The IUPAC name of the starting alkyl nitrile 'A' is:

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Correct answer: A. 3 -Methylbutanenitrile

Step 1: The final product is 2 -bromo- 3 -methylbutanoic acid, which is formed by the Hell-Volhard-Zelinsky (HVZ) reaction of carboxylic acid 'B'. The HVZ reaction selectively introduces a bromine atom at the -carbon. Working backwards, replacing the -bromine with a hydrogen atom gives the structure of 'B' as 3 -methylbutanoic acid ( CH₃-CH(CH₃)-CH₂-COOH ). Step 2: Carboxylic acid 'B' is formed by the complete hydrolysis of alkyl nitrile 'A'. Replacing the carboxyl group ( -COOH ) with a nitrile group ( -C N ) give

An unknown alkyl nitrile 'A' undergoes complete hydrolysis to form a carboxylic acid 'B'. Compound 'B' is then treated with bromine and red phosphorus followed by water workup, yielding 2 -bromo- 3 -methylbutanoic acid as the major product. The IUPAC name of the starting alkyl nitrile 'A' is:

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