Search any question & find its solution
Question:
Answered & Verified by Expert
In the series of reaction
$$
\begin{array}{r}
\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} \frac{\mathrm{NaNO}_{2} / \mathrm{HCl}}{0-5^{\circ} \mathrm{C}} \mathrm{X} \\
\qquad \frac{\mathrm{HNO}_{2}}{\mathrm{H}_{2} \mathrm{O}} \longrightarrow \mathrm{Y}+\mathrm{N}_{2}+\mathrm{HCl}, \mathrm{X} \text { and }
\end{array}
$$
$Y$ are respectively
Options:
$$
\begin{array}{r}
\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} \frac{\mathrm{NaNO}_{2} / \mathrm{HCl}}{0-5^{\circ} \mathrm{C}} \mathrm{X} \\
\qquad \frac{\mathrm{HNO}_{2}}{\mathrm{H}_{2} \mathrm{O}} \longrightarrow \mathrm{Y}+\mathrm{N}_{2}+\mathrm{HCl}, \mathrm{X} \text { and }
\end{array}
$$
$Y$ are respectively
Solution:
2954 Upvotes
Verified Answer
The correct answer is:
$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}_{2}^{\oplus} \mathrm{Cl}^{-}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.