question_answer

\[HBr\]reacts with\[C{{H}_{2}}=CHOC{{H}_{3}}\]under anhydrous   conditions   at   room temperature to give     AIEEE  Solved  Paper-2006

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) \[{{H}_{3}}CCHBrOC{{H}_{3}}\]

Correct Answer: D

Solution :

Electrophilic addition reaction is more favorable \[C{{H}_{2}}=CH-O-C{{H}_{3}}\xrightarrow[{}]{HBr}\] \[C{{H}_{3}}-\underset{\begin{smallmatrix}  | \\  Br \end{smallmatrix}}{\mathop{CH}}\,-O-C{{H}_{3}}\] First protonation occurs, two possible intermediates are \[\underset{+}{\mathop{C}}\,{{H}_{2}}->\underset{\begin{smallmatrix}  | \\  H \end{smallmatrix}}{\mathop{CH}}\,->OC{{H}_{3}}\](I) (\[-I\]effect destabilises carbocation) and   (+M effect stabilises carbocation) II, is more favourable Hence,\[B{{r}^{\text{O-}}}\]attacks, and product is \[C{{H}_{3}}-\underset{\begin{smallmatrix}  | \\  Br \end{smallmatrix}}{\mathop{CH}}\,-O-C{{H}_{3}}\]