MCQ
Consider the following reaction
(image) $\frac{{{H_2}}}{{pd - BaS{O_4}}}\,'A'$
The product $'A'$ is
- A$C_6H_5COCH_3$
- B$C_6H_5Cl$
- ✓$C_6H_5CHO$
- D$C_6H_5OH$
(image) $\frac{{{H_2}}}{{pd - BaS{O_4}}}\,'A'$
The product $'A'$ is

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$\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - CH - C - C{H_2} - C{H_2}OH}
\end{array}$