Nitro Compounds

Electrolytic reduction of nitrobenzene in strongly acidic medium (conc. $H_2SO_4$) gives phenylhydroxylamine, which rearranges (Bamberger rearrangement) to p-aminophenol:

$C_6H_5NO_2 + 4[H] \longrightarrow C_6H_5NHOH + H_2O$ $C_6H_5NHOH \xrightarrow{H^+} p\text{-}HO-C_6H_4-NH_2$

The $-NO_2$ group is strongly electron-withdrawing, through both the $-I$ and the $-R$ (resonance) effect. It makes the ring electron-deficient and deactivates it towards electrophiles.

In the resonance structures of nitrobenzene, a positive charge appears at the ortho and para positions:

   +        +        +
  ortho    para     ortho   (positive charge on C2, C4, C6)

The ortho and para positions are therefore more electron-deficient. The meta position (C3, C5) carries no positive charge and has relatively higher electron density. The electrophile ($E^+$) attacks the meta position, so the product is the m-substituted nitrobenzene (for example m-dinitrobenzene on nitration). The arenium ion for meta attack is also more stable than the ortho or para intermediates, which place + next to the electron-withdrawing $N^+$.

Plan: nitrobenzene to aniline, to benzenediazonium chloride, to benzonitrile, to benzaldehyde, then Perkin reaction.

1. Reduction: $C_6H_5NO_2 + 6[H] \xrightarrow{Sn/HCl} C_6H_5NH_2 + 2H_2O$
2. Diazotisation: $C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{0-5^\circ C} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$
3. Sandmeyer: $C_6H_5N_2^+Cl^- + CuCN \longrightarrow C_6H_5CN + N_2 + CuCl$
4. Stephen reduction: $C_6H_5CN \xrightarrow{SnCl_2/HCl,\ \text{then } H_3O^+} C_6H_5CHO$
5. Perkin reaction: $C_6H_5CHO + (CH_3CO)_2O \xrightarrow[453\text{ K}]{CH_3COONa} C_6H_5CH=CHCOOH + CH_3COOH$

The product is cinnamic acid.

(a) A base

In the nitrating mixture, $H_2SO_4$ is the stronger acid and protonates $HNO_3$, which acts as a base:

$HNO_3 + 2H_2SO_4 \longrightarrow NO_2^+ + H_3O^+ + 2HSO_4^-$

The nitronium ion $NO_2^+$ is the electrophile that attacks benzene.

(a) p-Aminophenol

In acidic medium, nitrobenzene is reduced electrolytically to phenylhydroxylamine, which undergoes Bamberger rearrangement to p-aminophenol. Azoxybenzene, azobenzene and hydrazobenzene form in neutral or alkaline media.

(a) $Na_2S$

Sodium sulphide (or ammonium sulphide, $(NH_4)_2S$) selectively reduces only one $-NO_2$ group of m-dinitrobenzene to give m-nitroaniline. Sn/HCl and $LiAlH_4$ would reduce both nitro groups, or give other products.

$m\text{-}C_6H_4(NO_2)_2 \xrightarrow{Na_2S \text{ or } (NH_4)_2S} m\text{-}O_2N-C_6H_4-NH_2$

Meta-directing nature. The $-NO_2$ group withdraws electrons by both the $-I$ and $-R$ effects. This deactivates the ring and leaves a positive charge at the ortho and para positions in the resonance structures. The meta positions have the highest electron density of the ring, so the electrophile attacks there. For example, nitration of nitrobenzene gives m-dinitrobenzene.

i) With Zn/$NH_4Cl$ (neutral medium). Nitrobenzene is reduced to N-phenylhydroxylamine:

$C_6H_5NO_2 + 4[H] \xrightarrow{Zn/NH_4Cl} C_6H_5NHOH + H_2O$

ii) With $LiAlH_4$. In ether, aromatic nitro compounds are reduced to azobenzene:

$2C_6H_5NO_2 + 8[H] \xrightarrow{LiAlH_4} C_6H_5N=NC_6H_5 + 4H_2O$