- A(image) $+ CO + HCl =$ Sandmayer reaction

- B(image) $+ CuO_3 =$ Schotten Baumann reaction

- ✓(image) $\xrightarrow{{B{r_2} + KOH}} = $ Hoffmann Bromamide reaction

- DAll of these




Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - CH - C{H_3}(excess) + B{r_2}\xrightarrow{{hv}}}
\end{array}$ $\mathop {\begin{array}{*{20}{c}}
{\,\,\,\,\,C{H_3}} \\
| \\
{C{H_3} - C - C{H_3}} \\
| \\
{\,\,Br}
\end{array}}\limits_{(A)} $ + $\mathop {\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,\,\,} \\
{\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - CH - C{H_2} - Br}
\end{array}}\limits_{(B)} $
the percentage yields of the products $(A)$ and $(B)$ are expected to be