A) \[C{{H}_{3}}CHC{{H}_{2}}C{{H}_{2}}BrCOOH\]
B) \[C{{H}_{3}}CHBrC{{H}_{2}}C{{H}_{2}}COOH\]
C) \[C{{H}_{2}}BrC{{H}_{2}}{{(C{{H}_{2}})}_{2}}COOH\]
D) \[C{{H}_{3}}CHC{{H}_{2}}BrC{{H}_{2}}COOH\]
Correct Answer: B
Solution :
\[HBr\] gets added to \[C{{H}_{2}}=CH{{(C{{H}_{2}})}_{2}}COOH\]. It has no effect on \[-COOH\] group. \[C{{H}_{2}}=CHC{{H}_{2}}C{{H}_{2}}COOH+HBr\] \[\underset{\begin{smallmatrix} 4-bromopen\operatorname{ta}noic\,acid \\ [major\,product] \end{smallmatrix}}{\mathop{\xrightarrow{{}}{{H}_{3}}C-CH-\underset{\begin{smallmatrix} | \\ Br \end{smallmatrix}}{\mathop{C{{H}_{2}}C{{H}_{2}}COOH}}\,}}\,\] The above addition follows Markownikoffs rule during addition on unsymmetrical alkene. The negative part of the addendum is added to that C-atom which has lesser number of H atoms.
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