What will be the major product formed in the following reaction?

The correct answer is Option (4) → Cyclopentanecarboxylic acid (C₅H₉–COOH)

Bromocyclopentane is a secondary alkyl halide. With sodium cyanide the cyanide ion replaces bromide by nucleophilic substitution, giving the nitrile X (cyanocyclopentane). Acid hydrolysis of a nitrile on heating converts –C≡N into –COOH (through the amide), so A is cyclopentanecarboxylic acid:

    \( \ce{C5H9Br + NaCN -> C5H9CN + NaBr} \)

    \( \ce{C5H9CN} \xrightarrow[\Delta]{\ce{H3O+}} \ce{C5H9CONH2} \xrightarrow[\Delta]{\ce{H3O+}} \ce{C5H9COOH + NH4+} \)

Why the other options are incorrect

• Option (1) C₅H₉CN — Incorrect. This is the intermediate X; the second step, hydrolysis, changes it further.

• Option (2) C₅H₉OH — Incorrect. Cyclopentanol would come from substitution by \( \ce{OH^{-}} \) (aqueous KOH), not from \( \ce{CN^{-}} \) and hydrolysis; it also has one carbon less than the product.

• Option (3) C₅H₉CH₂OH — Incorrect. This would need reduction of the acid (for example with \( \ce{LiAlH4} \)); hydrolysis does not reduce.

NCERT (2026-27 edition):
From Nitriles and Amides