A compound X, C6H14O2, is cleaved by HIO4 to give only propanone (2 mol per mol of X). Heating X with dilute H2SO4 gives Y, C6H12O, which gives a yellow precipitate with I2/NaOH. Y is

JEE Advanced levelSingle correct optionDifficulty 3 of 3ap198

A compound X, $C X_{6} H X_{14} O X_{2}$ , is cleaved by $HIO X_{4}$ to give only propanone (2 mol per mol of X). Heating X with dilute $H X_{2} SO X_{4}$ gives Y, $C X_{6} H X_{12} O$ , which gives a yellow precipitate with $I X_{2}$ / $NaOH$ . Y is

Cleavage to two propanone molecules shows X is the symmetrical vicinal diol pinacol,

(CH X_{3}) X_{2} C (OH) C (OH) (CH X_{3}) X_{2}

. Acid-catalysed rearrangement (methyl shift) gives

(CH X_{3}) X_{3} C - CO - CH X_{3}

, which has the

CH X_{3} CO X^{-}

group needed for the iodoform test. 4-Methylpentan-2-one is also a methyl ketone but cannot be derived from X.