Question:
HBr reacts with $CH_{2} =CH-OCH_{3}$ under anhydrous conditions at room temperature to give:
HBr reacts with $CH_{2} =CH-OCH_{3}$ under anhydrous conditions at room temperature to give:
Updated On: Jul 27, 2022
- $CH_{3} CHO and CH_{3} Br$
- $BrCH_{2} CHO and CH_{3}OH$
- $BrCH_{2}- CH_{2}-OCH_{3}$
- $H_{3}C-CHBr-OCH_{3}$
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The Correct Option is D
Solution and Explanation
First protonation occurs, two possible intermediates are
(-1effect destabilizes carbocation)
(+M effect stabilizes carbocation) II, is more favourable Hence, $Br^{\Theta}$ attacks, and product is
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