MCQ
Product $(C)$ is
- A
- ✓
- C
- D$Ph - C \equiv C-Ph$
✓
Answer
Correct option: B.
b
$Ph - CH = CH - Ph\xrightarrow[{CC{l_4}}]{{B{r_2}}}\mathop {\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,Br} \\
{\,\,\,\,\,\,\,\,\,|} \\
{Ph - CH - CH - Ph} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{Br\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}}\limits_{(ha\log enation)} $ $\xrightarrow{{2NaN{H_2}\,({E_2}\,reaction)}}Ph - C \equiv C - Ph\xrightarrow[{Pd/CaC{O_3}}]{{{H_2}}}$ (figure)
$Ph - CH = CH - Ph\xrightarrow[{CC{l_4}}]{{B{r_2}}}\mathop {\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,Br} \\
{\,\,\,\,\,\,\,\,\,|} \\
{Ph - CH - CH - Ph} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{Br\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}}\limits_{(ha\log enation)} $ $\xrightarrow{{2NaN{H_2}\,({E_2}\,reaction)}}Ph - C \equiv C - Ph\xrightarrow[{Pd/CaC{O_3}}]{{{H_2}}}$ (figure)
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