A) \[{{H}_{3}}C-\underset{D}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-\overset{\bullet }{\mathop{C}}\,{{H}_{2}}\]
B) \[{{H}_{3}}C-\underset{D}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-\overset{\bullet }{\mathop{\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,}}\,-\overset{{}}{\mathop{C}}\,{{H}_{2}}\]
C) \[{{H}_{3}}C-\overset{\bullet }{\mathop{\underset{D}{\mathop{\underset{|}{\mathop{C}}\,}}\,}}\,-\overset{{}}{\mathop{\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{CH}}\,}}\,}}\,-\overset{{}}{\mathop{C}}\,{{H}_{3}}\]
D) \[{{H}_{3}}C-\overset{\bullet }{\mathop{CH}}\,-\overset{{}}{\mathop{\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{CH}}\,}}\,}}\,-\overset{{}}{\mathop{C}}\,{{H}_{3}}\]
Correct Answer: B
Solution :
[b] \[B{{r}^{\bullet }}\] is less reactive and more selective and so the most stable free radical \[\left( 3{}^\circ \right)\] will be the major product.
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