MCQ
The product obtained from the reaction is


- ✓

- B

- C

- D








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$\begin{array}{*{20}{c}}
\,\,\,{{C_2}{H_5}} \\
{|\,\,\,\,\,\,\,} \\
{C{H_3} - C - COOH} \\
{|\,\,\,\,\,\,} \\
\,\,\,{{C_3}{H_7}}
\end{array}\xrightarrow{{{N_3}H/Conc.\,{H_2}S{O_4}}}\begin{array}{*{20}{c}}
\,\,\,\,\,\,\,\,\,{{C_2}{H_5}} \\
| \\
{C{H_3} - C - N{H_2}} \\
| \\
\,\,\,\,\,\,\,\,\,{{C_3}{H_7}}
\end{array}$
is called :

$\left( C _6 H _5\right)_3 C - Cl \frac{ OH ^{-}}{\text {Pyridine }}\left( C _6 H _5\right)_3 C - OH$