MCQ
Product of this Hoffmann bromamide reaction is


- A$\begin{matrix}
O\,\, \\
||\,\, \\
Ph-C-C{{H}_{3}} \\
\end{matrix}$ - ✓$Ph-CHO$
- C

- D$Ph-CH_2 -NH_2$


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$A\xrightarrow[{(ii)\,\,Conc.\,{H_2}S{O_4}/\Delta }]{{{\text{(i)}}\,{\text{C}}{{\text{H}}_3}MgBr/{H_2}O}}$
$B\xrightarrow[{(ii)\,Zn/{H_2}O}]{{(i)\,{O_3}}}C + D$
$D\xrightarrow[\Delta ]{{Ba\left( {OH} \right)}}\begin{array}{*{20}{c}} {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} \\ {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\ {{H_3}C - C} \end{array}$$\begin{array}{*{20}{c}} {\,\,\,\,\,\,\,O} \\ {\,\,\,\,\,||} \\ { = CH - C - C{H_3}} \end{array}$
$X$ is