Ethanoic acid is heated with methanol labelled as $CH X_{3} X_{18} X_{2}^{218} OH$ in the presence of conc. $H X_{2} SO X_{4}$ to give ester X. X is then completely hydrolysed with excess aqueous $NaOH$ under reflux. Assuming acyl-oxygen cleavage in both steps, the $X_{18} X_{2}^{218} O$ is finally found in

Answer: A

Esterification keeps the alcohol oxygen in the ester, giving

CH X_{3} CO X_{18} X_{2}^{218} OCH X_{3}

. In alkaline hydrolysis,

OH X^{-}

attacks the carbonyl carbon and the alkoxide (

CH X_{3} X_{18} X_{2}^{218} O X^{-}

) leaves, which gets protonated. So the label ends in methanol, not in the ethanoate.