question_answer

\[C{{H}_{3}}-\overset{C{{H}_{3}}}{\mathop{\overset{|}{\mathop{CH}}\,}}\,C\equiv CH~\xrightarrow{excess\,\,HBr}\] The product of the above reaction is:

A) \[C{{H}_{3}}-\overset{C{{H}_{3}}}{\mathop{\overset{|}{\mathop{CH}}\,}}\,-\overset{Br}{\mathop{\overset{|}{\mathop{C}}\,}}\,={{\overset{Br}{\mathop{\overset{|}{\mathop{CH}}\,}}\,}_{2}}\]

B) \[C{{H}_{3}}-\overset{C{{H}_{3}}}{\mathop{\overset{|}{\mathop{CH}}\,}}\,-\overset{Br}{\mathop{\overset{|}{\mathop{C}}\,}}\,=C{{H}_{2}}\]

C) \[C{{H}_{3}}-\overset{C{{H}_{3}}}{\mathop{\overset{|}{\mathop{CH}}\,}}\,-\underset{Br}{\mathop{\underset{|}{\mathop{\overset{Br}{\mathop{\overset{|}{\mathop{C}}\,}}\,}}\,}}\,=C{{H}_{3}}\]

D) \[C{{H}_{3}}-\overset{C{{H}_{3}}}{\mathop{\overset{|}{\mathop{CH}}\,}}\,-C{{H}_{2}}-\underset{Br}{\mathop{\underset{|}{\mathop{\overset{Br}{\mathop{\overset{|}{\mathop{CH}}\,}}\,}}\,}}\,\]

Correct Answer: C

Solution :

[c]