\(\mathrm{C}_6 \mathrm{H}_6 \underset{\mathrm{H}_2…

\(\mathrm{C}_6 \mathrm{H}_6 \underset{\mathrm{H}_2 \mathrm{SO}_4}{\stackrel{\mathrm{HNO}_3}{\longrightarrow}} X \underset{\mathrm{FeCl}_3}{\stackrel{\mathrm{Cl}_2}{\longrightarrow}} Y\). In the above sequence \(Y\) can be
  1. 3-nitrochlorobenzene
  2. 1-nitrochlorobenzene
  3. 4-nitrochlorobenzene
  4. none of these.

Solution

$\left(\mathrm{HNO}_3+\mathrm{H}_2 \mathrm{SO}_4\right)$ will generate $\mathrm{NO}_2^{+}$ (nitronium ion) which will give nitrobenzene, as $-\mathrm{NO}_2$ is deactivating and meta directing so, product formed will be 3-nitrochlorobenzene.

Asked in: NEET 2009 (Mains)

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