A) \[C{{H}_{3}}C{{H}_{2}}CH=C{{H}_{2}}+HCl\]
B) \[C{{H}_{3}}C{{H}_{2}}CH=C{{H}_{2}}+HBr\]
C) \[C{{H}_{3}}CH=CH.C{{H}_{3}}+HBr\]
D) \[C{{H}_{3}}C{{H}_{2}}CH=C{{H}_{2}}+HI\]
Correct Answer: B
Solution :
When HBr reacts with unsymmetrical alkene, peroxide effect (.ie, addition of HBr opposite to the Markownikoffs rule) can operate, since it is only HBr for that both the propagation steps (the steps involved during the reaction) are exothermic. \[C{{H}_{3}}C{{H}_{2}}CH=C{{H}_{2}}+\overset{\delta \,+\,\delta }{\mathop{H\,Br}}\,\] \[\xrightarrow{\text{Peroxide}}C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Br\]
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