Question

Which of the following is $\gamma$- elimination?

• A. $C{H}_3-C{H}_2-\stackrel{H}{\stackrel{|}{CH}}-\stackrel{Br}{\stackrel{|}{C{H}_2}}\underset{}{\overset{KOH\left(alc\right)}{\to }}C{H}_{3}C{H}_{2}CH=C{H}_2+HBr$
• B. $R_2-\stackrel{H}{\stackrel{|}{C}}-Cl\to R_{2}C:$
• C. $\underset{Br}{\underset{|}{C{H}_2}}-C{H}_2-\underset{Br}{\underset{|}{C{H}_2}}\underset{}{\overset{Zndust}{\to }}△+B{r}_2+Zn$
• D. None of these

Answer 1

The correct option is C

$\underset{Br}{\underset{|}{C{H}_2}}-C{H}_2-\underset{Br}{\underset{|}{C{H}_2}}\underset{}{\overset{Zndust}{\to }}△+B{r}_2+Zn$

We know, all three reactions are $\gamma$ - elimination. Now, $\gamma$ - elimination is the $\to$ Loss of 2 atoms or group of atoms from 2 bonds away (next to adjacent)$\alpha$positopn
$\to$ It leads to formation of 3 - membered ring in this case.
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