MCQ
Rosenmund reduction cannot be used for the preparation of
- A$C_6H_5CHO$
- B$CH_3CHO$
- C$\begin{array}{*{20}{c}}
{C{H_3} - CH - CHO} \\
{|\,\,\,\,\,\,\,} \\
{C{H_3}}
\end{array}$ - ✓$HCHO$
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Reason $R:$ The increasing nuclear charge outweighs the shielding across the period.
In the light of the above statements, choose the most appropriate from the options given below:
$C{H_3} - CH = CH - COOH\xrightarrow{{B{r_2}}}$
