Q. A keto ester $(A)$ with molecular formula $C_{6}H_{10}O_{3}$ on treatment with $NaOH + I_{2}$ does not give iodoform but on boiling with dilute KOH gives a compound $(B)$ with molecular formula $C_{4}H_{5}O_{3}K$ which upon acidification followed by heating undergoes decarboxylation to give acetone. The keto ester $(A)$ is
A) $CH_{3}COCH_{2}CH_{2}COOCH_{3}$
B) $CH_{3}COCH_{2}COOC_{2}H_{5}$
C) $CH_{3}CH_{2}OCH_{2}COOCH_{3}$
D) $CH_{3} - COCH\left( CH_{3} \right)COOCH_{3}$
Explanation
(b)
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The keto-ester $(A)$ does not give haloform reaction inspite of the presence of $CH_{3}CO -$ group in it. The reason is the presence of active methylene group $(ie,\ - CH_{2} -$), which prevents the conversion of $CH_{3}CO - to\ CX_{3}CO -$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.