Reduction of aromatic nitro compounds using $S n$ and $H C l$ gives:

aromatic primary amines

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aromatic secondary amines

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aromatic tertiary amines

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aromatic amides

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Solution

The correct option is A aromatic primary amines
(I) $2 C_{6} H_{5} {N O}_{2} + 3 S n + 12 H C l ⟶ 2 C_{6} H_{5} {N H}_{2} + 3 S n {C l}_{4} + 4 H_{2} O$
(II) $S n {C l}_{4} + 2 H C L = H_{2} [S n {C l}_{6}]$
Chlorostanic Acid
(III) $2 C_{6} H_{5} {N H}_{2} + H_{2} [S n {C l}_{6}] ⟶ (C_{6} H_{5} \oplus N H_{3})_{2} S n {C l}_{6}^{2 -}$
Aniline Stanicchloride
(IV) $(C_{6} H_{5} \oplus N H_{3})_{2} S n {C l}_{6}^{2 -} + 8 N a O H ⟶ 2 C_{6} H_{5} - N H_{2} {N a}_{2} {S n}^{+} O_{3} + 6 N a C l + 5 H_{2} O$

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