
- ✓

- B

- Csingle product is obtained in both the reactions

- Dsingle product obtained in both the reactions








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$(1)$ $\begin{matrix}
C{{H}_{3}} \\
|\,\,\,\,\,\, \\
ZNH\,C\,HC{{O}_{2}}H \\
\end{matrix}$ $(2)$ $\begin{matrix}
C{{H}_{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
{{H}_{2}}N\,C\,HC{{O}_{2}}C{{H}_{2}}{{C}_{6}}{{H}_{5}} \\
\end{matrix}$
$(3)$ $\begin{matrix}
\,\,\,\,\,\,\,\,\,\,\,\,C{{H}_{3}}{{C}_{6}}{{H}_{5}} \\
|\,\,\,\,\, \\
ZNH\,C\,HC{{O}_{2}}H \\
\end{matrix}$ $(4)$ $\begin{matrix}
C{{H}_{2}}{{C}_{6}}{{H}_{5}}\,\,\,\,\, \\
|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
{{H}_{2}}N\,CHC{{O}_{2}}C{{H}_{2}}{{C}_{6}}{{H}_{5}} \\
\end{matrix}$
$ + \frac{1}{2}\frac{{d\left[ C \right]}}{{dt}} = - \frac{1}{3}\frac{{d\left[ D \right]}}{{dt}} = + \frac{1}{4}\frac{{d\left[ A \right]}}{{dt}} = \frac{{ - d\left[ B \right]}}{{dt}}$
the reaction is