Compound A, $C X_{5} H X_{10} O$ , gives a yellow precipitate with $I X_{2}$ / $NaOH$ and does not reduce Tollens reagent. Clemmensen reduction of A gives straight-chain pentane. A is treated with excess $NaOI$ and then acidified to give acid B. B is

Answer: D

Iodoform positive and Tollens negative means a methyl ketone. Clemmensen reduction to unbranched pentane fixes A as pentan-2-one,

CH X_{3} COCH X_{2} CH X_{2} CH X_{3}

. The haloform reaction gives

CH X_{3} CH X_{2} CH X_{2} COO X^{-}

, so B is butanoic acid.