Question

HBr reacts with $C{{H}_{2}}=CH-OC{{H}_{3}}$ under anhydrous conditions at room temperature to give:
A.$C{{H}_{3}}CHO$ and $C{{H}_{3}}Br$
B. $BrC{{H}_{2}}CHO$ and$C{{H}_{3}}OH$
C.$BrC{{H}_{2}}C{{H}_{2}}OC{{H}_{3}}$
D. $C{{H}_{3}}-CH(Br)-OC{{H}_{3}}$

Answer 1

Hint: Methoxyethene is also named as methyl vinyl ether. It is a highly reactive gas at room temperature. Anhydrous conditions involve absence of any type of moisture or water. HBr, is called hydrobromic acid, is a weak acid.

Complete answer:
Methyl vinyl ether is a highly reactive gas. It can react with weak acids like hydrogen bromide at even room temperature. When provided with anhydrous conditions, this gas can undergo an addition reaction. This addition happens at the double bond of the vinyl carbon atom.
The HBr donates the hydrogen ion to the carbon of the double bond having less hydrogen and results in protonation of the carbon of the double bond. Then the carbocation formed is unstable and in the resonance state. This carbocation finally gains stability when the bromide ion gets attached to the stable carbocation, which is the cation at carbon- 2. The reaction is as follows:
$C{{H}_{3}}O-CH=C{{H}_{2}}+{{H}^{+}}B{{r}^{-}}\to C{{H}_{3}}O-(CH)Br-C{{H}_{3}}$

Hence, the correct option is D. The resultant is 2-bromo-methoxy ethane.

Note:
The anhydrous conditions are used as the highly reactive gas methyl vinyl ether. This gas gets hydrolysed easily. So, it can react with weak acids like hydrogen bromide in the presence of moisture to form compounds like methanol and aldehyde. In anhydrous conditions it undergoes additional reactions.