Q. $HBr$ reacts with $CH_{2} = CH - OCH_{3}$ under anhydrous conditions at room temperature to give
A) $CH_{3}CHO\ and\ CH_{3}Br$
B) ${BrCH}_{2}CHO\ and\ {CH}_{3}OH$
C) ${BrCH}_{2} - {CH}_{2} - {OCH}_{3}$
D) $H_{3}C - CHBr - {OCH}_{3}$
Explanation
(d)
$CH_{2} = CH - O - CH_{3}$
$\overset{HBr}{\rightarrow}CH_{3} - CH - O - CH_{3}$
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$Br$
First protonation occurs, two possible intermediates are
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II, is more favourable. Hence, $Br^{-}$ attacks, and product is
$CH_{3} - CH - O - CH_{3}$
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$Br$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.