A) \[\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C{{H}_{3}}CBrC{{H}_{3}}}}\,\]
B) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Br\]
C) \[\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C{{H}_{3}}CHC{{H}_{2}}Br}}\,\]
D) \[\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C{{H}_{3}}C{{H}_{2}}CHC{{H}_{3}}}}\,\]
Correct Answer: C
Solution :
Key Idea: The reaction with\[HBr\]follows anti- Markownikoff rule. The rule states that negative part of adding reagent goes to carbon atom having more number of hydrogen atoms. \[C{{H}_{3}}-\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}\,=C{{H}_{2}}+HBr\xrightarrow[{}]{Peroxide}\]
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