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The compound 'A' when treated with $\mathrm{HNO}_3$ (in presence of $\mathrm{H}_2 \mathrm{SO}_4$ ) gives compound ' $\mathrm{B}$ ' which is then reduced with $\mathrm{Sn}$ and $\mathrm{HCl}$ to aniline. The compound ' $\mathrm{A}$ ' is
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Benzene
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