A compound (X) $\left(\mathrm{C}_{5} \mathrm{H}_{8}ight)$ reacts with ammonical $\mathrm{AgNO}_{3}$ to give…

A compound (X) $\left(\mathrm{C}_{5} \mathrm{H}_{8}ight)$ reacts with ammonical $\mathrm{AgNO}_{3}$ to give a white precipitate, and on oxidation with hot alkaline $\mathrm{KMnO}_{4}$ gives the acid, $\left(\mathrm{CH}_{3}ight)_{2} \mathrm{CHCOOH}$, therefore $\mathrm{X}$ is $-$
  1. $\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3}$
  2. $\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{CH}_{3}$
  3. $\left(\mathrm{CH}_{3}ight)_{2} \mathrm{CH}-\mathrm{C} \equiv \mathrm{CH}$
  4. $\left(\mathrm{CH}_{3}ight)_{2} \mathrm{C}=\mathrm{C}=\mathrm{CH}_{2}$

Solution

Ammonical silver nitrate (Tollens reagent) reacts with 1-alkyne only and give white ppt so reactant should be terminal alkyne.
Reaction with hot $\mathrm{KMnO}_{4}$ is:
$$
\left(\mathrm{CH}_{3}ight)_{2} \mathrm{CH}-\mathrm{C} \equiv \mathrm{CH} ightarrow\left(\mathrm{CH}_{3}ight)_{2} \mathrm{CHCOOH}+\mathrm{CO}_{2}
$$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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