Question

Which of the following is a $\beta$ -elimination?

• A. $C{H}_3-C{H}_2-\stackrel{H}{\stackrel{|}{CH}}-\stackrel{Br}{\stackrel{|}{C{H}_2}}\underset{}{\overset{alc.KOH}{\to }}C{H}_{3}C{H}_{2}CH=C{H}_2+HBr$
• B. $\underset{H}{\underset{|}{C{H}_2}}-C{H}_2-\underset{Br}{\underset{|}{C{H}_2}}\underset{}{\overset{ZnDust}{\to }}\underset{H}{\underset{|}{C{H}_2}}-C{H}_2-\underset{H}{\underset{|}{C{H}_2}}$
• C. $C{H}_3-C{H}_2-C{H}_2-C{H}_3\underset{}{\overset{AlC{l}_3}{\to }}C{H}_3-\underset{C{H}_3}{\underset{|}{CH}}-C{H}_3$
• D. None of these

Answer 1

The correct option is A

$C{H}_3-C{H}_2-\stackrel{H}{\stackrel{|}{CH}}-\stackrel{Br}{\stackrel{|}{C{H}_2}}\underset{}{\overset{alc.KOH}{\to }}C{H}_{3}C{H}_{2}CH=C{H}_2+HBr$

We know about the elimination reaction. Option (C) is not an elimination reaction because no elimination of atoms or group of atoms occurs.

Now, what is $\beta$-elimination ?

$\to$ Elimination reaction in which loss of 2 atoms or group of atoms occurs from the adjacent $(\alpha and\beta$) position . Therefore,

$C{H}_3-C{H}_2-\underset{\beta }{\stackrel{H}{\stackrel{|}{CH}}}-\underset{\alpha }{\stackrel{Br}{\stackrel{|}{C{H}_2}}}\underset{}{\overset{alc.KOH}{\to }}C{H}_{3}C{H}_{2}CH=C{H}_2+HBr$ is the correct one.