Q. The hydrocarbon which can react with sodium in liquid ammonia is
A) $CH_{3}CH_{2}CH_{2}C \equiv CCH_{2}CH_{2}CH_{3}$
B) $CH_{3}CH_{2}C \equiv CH$
C) $CH_{3}CH = CHCH_{3}$
D) $CH_{3}CH_{2}C \equiv CCH_{2}CH_{3}$
Explanation
(b)
$CH_{3}CH_{2}C \equiv CH\underset{\mathrm{\Delta}}{\overset{\frac{Na}{Liq.{NH}_{3}}}{\rightarrow}}CH_{3}CH_{2}C \equiv CNa^{+}$
Considering the options given it appears correct. $\frac{Na}{liq.{NH}_{3}}$ is known for metal dissolved reduction. Actually it is truth that $\frac{Na}{liq.{NH}_{3}}$ reduces internal triple bond and terminal double bond and do not reduce the terminal alkyne due to such alkylide formation.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.