A) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Br\]
B) \[C{{H}_{3}}C{{H}_{2}}\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,HBr\]
C) \[C{{H}_{3}}-C{{H}_{2}}\underset{C{{H}_{2}}Br}{\mathop{\underset{|}{\mathop{C}}\,}}\,HBr\]
D) \[C{{H}_{3}}C{{H}_{2}}\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,Br\]
Correct Answer: B
Solution :
[b] The reaction proceeds via free radical mechanism. As \[2{}^\circ \] free radical is more stable than\[1{}^\circ \], so \[C{{H}_{3}}C{{H}_{2}}CH(Br)C{{H}_{3}}\] would be formed.
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