

MCQOPTIONS
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1. |
\[C{{H}_{3}}-\overset{C{{H}_{3}}}{\mathop{\overset{|}{\mathop{CH}}\,}}\,C\equiv CH~\xrightarrow{excess\,\,HBr}\] 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. | \[C{{H}_{3}}-\overset{C{{H}_{3}}}{\mathop{\overset{|}{\mathop{CH}}\,}}\,-C{{H}_{2}}-\underset{Br}{\mathop{\underset{|}{\mathop{\overset{Br}{\mathop{\overset{|}{\mathop{CH}}\,}}\,}}\,}}\,\] |
Answer» D. \[C{{H}_{3}}-\overset{C{{H}_{3}}}{\mathop{\overset{|}{\mathop{CH}}\,}}\,-C{{H}_{2}}-\underset{Br}{\mathop{\underset{|}{\mathop{\overset{Br}{\mathop{\overset{|}{\mathop{CH}}\,}}\,}}\,}}\,\] | |