Question

Consider the following substances
$1.$ $CH_{3}C\equiv CCH_{3}$
$2.$ $CH_{3}CH_{2}C\equiv CH$
$3.$ $CH_{3}CH=CHCH_{3}$
$4.$ $CH_{3}CH_{2}CH=CH_{2}$
Which of the following reagent can be used to distinguish the compound $(2)$ from the rest of the compounds?

• A. Bromine/ $CCl_{4}$
• B. Bromine/ $CH_{3}COOH$
• C. Alk. $KMnO_{4}$
• D. Ammoniacal $AgNO_{3}$ or ammoniacal cuprous chloride

Answer 1

Answer: (D)

Compound ($2$) has acidic hydrogen (acetylinic hydrogen), which can be replaced.
$CH_{3}-CH_{2}-C\equiv CH+\left(\left[A g \left(N H_{3}\right)_{2}\right]\right)^{+}OH^{-} \rightarrow \underset{\left(\text{White ppt}\right)}{C H_{3} - C H_{2} C \equiv C^{-} A g^{+}}+H_{2}O+2NH_{3}$
$CH_{3}CH_{2}C\equiv CH+\left(\left[C u \left(N H_{3}\right)_{2}\right]\right)^{+}OH^{-} \rightarrow \underset{\left(\text{Red ppt.}\right)}{C H_{3} C H_{2} C \equiv C C u}+H_{2}O+2NH_{3}$
Compounds $1, 3$ and $4$ have no acidic hydrogen. Therefore, they do not react with reagent option $D$.