Question

Identify $Z$ in the following reaction sequence
$C{H}_{3}C{H}_{2}C{H}_{2}OH\underset{160-180^{\circ }C}{\overset{Conc.H_{2}S{O}_4}{\to }}X\to Y\underset{(ii)NaN{H}_2}{\overset{(i)Alc.KOH}{\to }}$

• A. $C{H}_3-CH\left(N{H}_2\right)-C{N}_{2}N{H}_2$
• B. $C{H}_3-CHOH-C{H}_{2}OH$
• C. $C{H}_3-C(OH)=C{H}_2$
• D. $C{H}_3-C\equiv CH$

Answer 1

Answer: (D)

$C{H}_{3}C{H}_{2}C{H}_{2}OH\underset{160-180^{\circ }C}{\overset{\text{ conc. }{H}_{2}S{O}_4}{\to }}C{H}_{3}CH=C{H}_2\stackrel{B{r}_2}{\to }\stackrel{alc.KOH}{\to }\underset{Y}{C{H}_{3}Br}-C{H}_{2}Br$
$[\underset{A}{C{H}_{3}C(Br)}=C{H}_2+\underset{B}{C{H}_{3}CH}=CHBr]\underset{-HBr}{\overset{NaN{H}_2}{\to }}\underset{Z}{C{H}_{3}C}\equiv CH$
Alc. $KOH$ smoothly brings about dehydrobromination of y to give a mixture of vinyl bromide (A and B) while $NaN{H}_2$ being a strong base than alcohol $KOH$ readily brings about dehydrobromination of less reactive vinyl bromide to give propyne (Z).