Question

Identify $Z$ in the following reaction sequence

• 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)

$\mathrm{CH}\mathrm{CH}\mathrm{CH}\mathrm{OH}\underset{160{}^{\circ }-180{}^{\circ }\mathrm{C}}{\overset{\mathrm{Conc}\mathrm{H}{}_2\mathrm{SO}{}_4}{\to }}\mathrm{CH}{}_3\mathrm{CH}=\mathrm{CH}{}_2$
$\stackrel{\mathrm{Br}{}_2}{\to }\stackrel{\mathrm{alc}\cdot \mathrm{KOH}}{\to }\underset{\text{ (Y) }}{\mathrm{CH}{}_3\mathrm{CHBr}}-\mathrm{CH}{}_2\mathrm{Br}$
$[\underset{\text{ (A) (B) }}{\mathrm{CH}{}_3\mathrm{C}(\mathrm{Br})=\mathrm{CH}{}_2}+\mathrm{CH}{}_3\mathrm{CH}=\mathrm{CHBr}]$
$\underset{-\mathrm{HBr}}{\overset{\mathrm{NaNH}{}_2}{\to }}\mathrm{CH}{}_3\mathrm{C}\equiv \mathrm{CH}$
Alcoholic $KOH$ brings about dehydrobromination of $Y$ and give a mixture of vinyl bromide ($A$ and $B$) while $NaN{H}_2$ being a strong base than $alc.KOH$ readily brings about dehydrobromination of less reactive vinyl bromide to give propyne $\mathrm{CH}{}_3\mathrm{C}\equiv \mathrm{CH}$ i.e. $(Z)$.