MCQ
In the reaction sequence
$\begin{array}{*{20}{c}}
{OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\
{\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\
{{C_6}{H_5} - CH - C{H_2} - C - C{H_3}}
\end{array}\mathop {\xrightarrow{{(i)\,NaOBr}}}\limits_{(ii)\,{H_2}O/{H^ + }\,(iii)\,\Delta } $ product
product will be :
- A$\begin{array}{*{20}{c}}
{OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{{C_6}{H_5} - CH - C{H_2} - COOH}
\end{array}$ - B$C_6H_5-COOH, COOH-COOH$ and $CHBr_3$
- ✓$\begin{array}{*{20}{c}}
O \\
{||} \\
{{C_6}{H_5} - C - C{H_3}}
\end{array}$ and $CHBr_3$ - DOnly $CHBr_3$
