A) \[HBr\]
B) \[SOB{{r}_{2}}\]
C) \[B{{r}_{2}}\]
D) \[C{{H}_{3}}MgBr\]
Correct Answer: B
Solution :
[b] \[C{{H}_{3}}-\overset{C{{H}_{3}}}{\mathop{\overset{|}{\mathop{C}}\,}}\,H-OH\xrightarrow{SOB{{r}_{2}}}\] \[C{{H}_{3}}-\overset{C{{H}_{3}}}{\mathop{\overset{|}{\mathop{C}}\,}}\,H-Br+S{{O}_{2}}+HBr\] Side products are gases. In case of \[HBr\] elimination also favours to produce alkene.
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