A) Free radical
B) Carbonium ion
C) Carbanion
D) All the above
Correct Answer: C
Solution :
Without intermediate reaction take place as under \[C{{H}_{3}}-CH=C{{H}_{2}}+HBr\to C{{H}_{3}}-\overset{Br\,}{\mathop{\overset{|}{\mathop{C}}\,H}}\,-C{{H}_{3}}\] (According to markownikoff rule) But the halogen bonded with terminal carbon so it take place in presence of peroxide by free radical mechanism. \[\underset{\text{peroxide}}{\mathop{R-O-OR}}\,\to 2R\overset{.}{\mathop{O}}\,\]; \[HBr+\overset{.}{\mathop{RO}}\,\to ROH+B{{r}^{.}}\] \[C{{H}_{3}}-CH=C{{H}_{2}}+B{{r}^{.}}\to C{{H}_{3}}-C\overset{.}{\mathop{H}}\,-C{{H}_{2}}-Br\] \[C{{H}_{3}}-\overset{.}{\mathop{C}}\,H-C{{H}_{2}}Br+HBr\to \] \[C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}Br+B{{r}^{.}}\]
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