Explanation

(d)
$‘X’$ is a three carbon compound with two halogen atom, so its molecular formula is $C_{3}H_{6}Cl_{2}$. Only terminal alkynes give red ppt. with ammoniacal $Cu_{2}{Cl}_{2}$, so the hydrocarbon produced by the reaction of $‘X’$ with alc. $KOH$, must be a terminal alkyne $(i.e.,\ CH_{3}C \equiv CH)$.
$C_{3}H_{6}Cl_{2}\overset{Alc.KOH}{\rightarrow}CH_{3}C \equiv CH\overset{Amm.Cu_{2}Cl_{2}}{\rightarrow}CH_{3}C \equiv CCu \downarrow$
red ppt.
Compound $(X)$ gives an aldehyde when reacts with aqueous $KOH$. This suggests that both the halogens are present on same terminal carbon atom. Thus, the formula of compound $(X)$ is
[image:13_image267.emf]and the reactions are as follows
[image:13_image268.emf]
$CH_{3}C \equiv CH\underset{Cu_{2}Cl_{2}}{\overset{Ammoniacal}{\rightarrow}}CH_{3}C \equiv CCu \downarrow$
red ppt.
[image:13_image269.emf]
[image:13_image270.emf]