Q. A compound $X(C_{5}H_{8})$ reacts with ammoniacal$AgNO_{3}$ to give a white precipitate, and on oxidation with hot alkaline $KMnO_{4}$ gives the acid,$\left( CH_{3} \right)_{2}CHCOOH$. Therefore, $X$ is
A) $CH_{2} = CHCH = CHCH_{3}$
B) $CH_{3}\left( CH_{2} \right)_{2}C \equiv CH$
C) $\left( CH_{3} \right)_{2}CH - C \equiv CH$
D) $\left( CH_{3} \right)_{2}C = C = CH_{2}$
Explanation
(c)
Terminal alkyne reacts with ammoniacal$AgNO_{3}$ solution give a whilteppt, therefore $x$ must be a terminal alkyne. Thus, $x$ must be$\left( CH_{3} \right)_{2}CHC \equiv CH$
$\left( CH_{3} \right)_{2}CHC \equiv CH\overset{{KMnO}_{4}}{\rightarrow}\left( {CH}_{3} \right)_{2}CHCOOH + \lbrack HCOOH\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.