A straight-chain alcohol C5H12O is optically active and gives a yellow precipitate with I2/NaOH. It is oxidised by Jones reagent to a ketone Z. Z is then treated with I2/NaOH and the mixture is acidified. The carboxylic acid formed is

JEE Advanced levelSingle correct optionDifficulty 3 of 3ro072

A straight-chain alcohol $C X_{5} H X_{12} O$ is optically active and gives a yellow precipitate with $I X_{2}$ / $NaOH$ . It is oxidised by Jones reagent to a ketone Z. Z is then treated with $I X_{2}$ / $NaOH$ and the mixture is acidified. The carboxylic acid formed is

A straight-chain, chiral

C X_{5} H X_{12} O

giving iodoform must be pentan-2-ol, because pentan-3-ol is achiral and pentan-1-ol has no

CH X_{3} CH (OH)

unit. Jones reagent converts it to pentan-2-one,

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

. The haloform reaction removes

CHI X_{3}

and leaves

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

, which gives butanoic acid on acidification.