Question

Identify $Z$ in the following reaction sequence
$ CH_3CH_2CH_2OH \xrightarrow[160 - 180^{\circ}C]{Conc. H_2SO_4} X \rightarrow Y \xrightarrow[(ii) NaNH_2]{(i) Alc. KOH} $

• A. $CH _{3}- CH \left( NH _{2}\right)- CN _{2} NH _{2}$
• B. $CH _{3}- CHOH - CH _{2} OH$
• C. $CH _{3}- C ( OH )= CH _{2}$
• D. $CH _{3}- C \equiv CH$

Answer 1

Answer: (D)

$CH _{3} CH _{2} CH _{2} OH \xrightarrow[160- 180^{\circ}C]{\text { conc. } H _{2} SO _{4}}CH _{3} CH = CH _{2} \xrightarrow{ Br _{2}}\xrightarrow{ alc . KOH } \underset{Y}{CH_3Br}- CH_2Br$
$[\underset{A}{CH _{3} C ( Br )} = CH _{2}+\underset{B}{CH _{3} CH} = CHBr ] \xrightarrow[- HBr ]{ NaNH _{2}} \underset{Z}{CH _{3} C} \equiv CH$
Alc. $KOH$ smoothly brings about dehydrobromination of y to give a mixture of vinyl bromide (A and B) while $NaNH _{2}$ being a strong base than alcohol $KOH$ readily brings about dehydrobromination of less reactive vinyl bromide to give propyne (Z).