Which hydrogen in compound (E) is easily replaceable during bromination reaction in presence of light ?
\(\mathrm{CH_3-CH_2-CH=CH_2}\) (E)
Labels: \(\mathrm{CH_3}\) = \(\delta\), \(\mathrm{CH_2}\) = \(\gamma\), \(\mathrm{CH}\) = \(\beta\), terminal \(\mathrm{=CH_2}\) = \(\alpha\).
\(\mathrm{CH_3-CH_2-CH=CH_2}\) (E)
Labels: \(\mathrm{CH_3}\) = \(\delta\), \(\mathrm{CH_2}\) = \(\gamma\), \(\mathrm{CH}\) = \(\beta\), terminal \(\mathrm{=CH_2}\) = \(\alpha\).
Show Hint
- \(\beta\) - hydrogen
- \(\alpha\) - hydrogen
- \(\gamma\) - hydrogen
- \(\delta\) - hydrogen
The Correct Option is C
Solution and Explanation
With Br2/light a free-radical substitution occurs at the allylic carbon, the sp3 carbon next to the C=C. This is the CH2 labelled \(\gamma\).
Removing an H from it gives the allylic radical \(\mathrm{CH_3-\dot{C}H-CH=CH_2}\leftrightarrow\mathrm{CH_3-CH=CH-\dot{C}H_2}\), stabilised by resonance, whereas vinylic (\(\alpha\), \(\beta\)) radicals are unstable and the \(\delta\) (primary alkyl) radical has no resonance.
Answer: (3) \(\gamma\) - hydrogen
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