A) \[{{\left( C{{H}_{3}} \right)}_{3}}CBr\]
B) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Br\]
C) \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ Br \end{smallmatrix}}{\mathop{C}}\,H-C{{H}_{2}}C{{H}_{3}}\]
D) \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,H-C{{H}_{2}}Br\]
Correct Answer: C
Solution :
By reactively-selectively principle \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}\xrightarrow[{{130}^{o}}C]{B{{r}_{2}}}\] \[\underset{\text{(Major)}}{\mathop{C{{H}_{3}}-\overset{\begin{smallmatrix} Br \\ | \end{smallmatrix}}{\mathop{C}}\,H-C{{H}_{2}}C{{H}_{3}}}}\,+\underset{\text{(Minor)}}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Br}}\,\]
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