Alkyl Halides - Result Question 10
####10. Which hydrogen in compound $(E)$ is easily replaceable during bromination reaction in presence of light?
(2019 Main, 10 Jan I)
$$ \begin{array}{llll} \delta & \gamma & \beta _{(E)} & \alpha \end{array} $$
(a) $\beta$-hydrogen
(b) $\delta$-hydrogen
(c) $\gamma$-hydrogen
(d) $\alpha$-hydrogen
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Solution:
- The compound $(E)$ has two allyl-hydrogen atoms $(\gamma)$. When $E$ reacts with $Br _2 / h v$, it readily undergoes allylic free radical substitution and forms 3, 3-dibromobut-1-ene
$$ \begin{aligned} & \stackrel{\delta}{C} H _3-\underset{\text { But-1-ene }}{\stackrel{\gamma}{C} H _2-\stackrel{\beta}{C}} H=\stackrel{\alpha}{C} H _2 \xrightarrow[-HBr]{\stackrel{Br _2}{ } / \frac{h v}{\longrightarrow}} \ & Br \ & \underset{\delta}{C _3} H _3 \underset{\substack{\gamma \ \text { 3-bromo-but-1-ene }}}{\stackrel{\mid}{C} H}-\underset{\beta}{C} H=\underset{-HBr}{C} H _2 \xrightarrow{Br _2 / h \nu} \ & \underset{\delta}{CH _3}-\underset{Br^{C}}{\stackrel{Br}{\mid}}-\underset{\beta}{Cr} H=\underset{\alpha}{C} H _2 \end{aligned} $$
$$ \text { 3,3-dibromobut-1-ene } $$
