HBr reacts with $\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{OCH}_{3}$ under anhydrous conditions at room temperature to give

(D) $\mathrm{H}_{3} \mathrm{C}-\mathrm{CHBr}-\mathrm{OCH}_{3}$

(A) $\mathrm{CH}_{3} \mathrm{CHO}$ and $\mathrm{CH}_{3} \mathrm{Br}$

(B) $\mathrm{BrCH}_{2} \mathrm{CHO}$ and $\mathrm{CH}_{3} \mathrm{OH}$

(C) $\mathrm{BrCH}_{2}-\mathrm{CH}_{2}-\mathrm{OCH}_{3}$

JEE Main · 2006 · 30 Apr 2006 · Offline · Hydrocarbons

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) $H_{3} C - CHBr - OCH_{3}$ ✓ Correct

Answer: (D)

Solution

The substrate

C H_{2} = CH - OC H_{3}

is a vinyl ether, and HBr adds across its double bond by electrophilic addition. The proton attacks first, and its direction of addition is decided by which carbocation is more stable: when

H^{+}

adds to the terminal

C H_{2}

, the positive charge falls on the carbon bearing the

OC H_{3}

group, where an oxygen lone pair stabilises it strongly as an oxocarbenium ion. Bromide then captures that carbon, so bromine and the methoxy group end up on the same carbon, giving

H_{3} C - CHBr - OC H_{3}

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option

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JEE Main · 2006 · 30 Apr 2006 · Offline · Hydrocarbons

Solution

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