Hydrocarbons Question 247
Question: Consider the following reaction $ H_3C-\underset{D}{\mathop{\underset{|}{\mathop{CH}}}}-\underset{CH_3}{\mathop{\underset{|}{\mathop{CH}}}}-CH_3+\overset{\bullet }{\mathop{B}}r\xrightarrow{{}}‘X’+HBr $ Identify the structure of the major product ‘X’
Options:
A) $ H_3C-\underset{D}{\mathop{\underset{|}{\mathop{C}}}}H-\underset{CH_3}{\mathop{\underset{|}{\mathop{C}}}}H-\overset{\bullet }{\mathop{C}}H_2 $
B) $ H_3C-\underset{D}{\mathop{\underset{|}{\mathop{C}}}}H-\overset{\bullet }{\mathop{\underset{CH_3}{\mathop{\underset{|}{\mathop{C}}}}}}-\overset{{}}{\mathop{C}}H_2 $
C) $ H_3C-\overset{\bullet }{\mathop{\underset{D}{\mathop{\underset{|}{\mathop{C}}}}}}-\overset{{}}{\mathop{\underset{CH_3}{\mathop{\underset{|}{\mathop{CH}}}}}}-\overset{{}}{\mathop{C}}H_3 $
D) $ H_3C-\overset{\bullet }{\mathop{CH}}-\overset{{}}{\mathop{\underset{CH_3}{\mathop{\underset{|}{\mathop{CH}}}}}}-\overset{{}}{\mathop{C}}H_3 $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ B{r^{\bullet }} $ is less reactive and more selective and so the most stable free radical $ ( 3{}^\circ ) $ will be the major product.
BETA
