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


Discussion

No Comment Found