Propanone with $Ba (OH) X_{2}$ gives P. P is warmed with a trace of iodine to give Q by loss of water. Q is treated with excess $NaOCl$ (haloform reaction) and then acidified. The organic acid obtained is

Answer: A

P is 4-hydroxy-4-methylpentan-2-one (diacetone alcohol) and Q is

(CH X_{3}) X_{2} C = CHCOCH X_{3}

, 4-methylpent-3-en-2-one. The haloform reaction converts the methyl ketone

COCH X_{3}

into

COO X^{-}

and gives

CHCl X_{3}

. Acidification gives

(CH X_{3}) X_{2} C = CHCOOH

, 3-methylbut-2-enoic acid.