$H B r$ reacts with $C H_{2} = C H - O C H_{3}$ under anhydrous conditions at room temperature to give:

C H_{3} C H O

and

C H_{3} B r

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B r C H_{2} C H O

and

C H_{3} O H

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B r C H_{2} C H_{2} O C H_{3}

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C H_{3} - C H (B r) - O C H_{3}

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Solution

The correct option is D $C H_{3} - C H (B r) - O C H_{3}$
$C H_{3} O - C H = C H_{2} + H^{+} B r^{-} \to C H_{3} O - C H B r - C H_{3}$

$H B r$ protonates the double bonded $C$ of alkene which make a resonance-stabilized cation

$C H_{3} - C H^{+} - O C H_{3} ⇌ C H_{3} C H = O^{+} C H_{3}$ . This can add $B r^{-}$ at the $C H$ carbon.

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