Sandmeyer vs. Gattermann Transformations

The given sequence starts from aniline and involves the standard reactions of aromatic amines containing a -$$NH_2$$ group.

Step 1 : Diazotisation
$$C_6H_5NH_2 \xrightarrow[\;0-5^{\circ}\text{C}\;]{NaNO_2/HCl} C_6H_5N_2^{+}Cl^{-}$$

The diazonium salt $$C_6H_5N_2^{+}Cl^{-}$$ is the first isolated product and is labelled P.
• P contains only one diazonium group and no $$NO_2$$ groups, hence it is not a dinitro compound. Therefore Option C is wrong.

Step 2 : Sandmeyer cyanation
$$C_6H_5N_2^{+}Cl^{-} \xrightarrow{CuCN/HCl} C_6H_5C\equiv N$$

The product of this Sandmeyer reaction is benzonitrile, designated Q.

Step 3 : Stephen reduction (SnCl2/HCl, ethanol)
In ethanol, stannous chloride and concentrated HCl reduce a nitrile to an iminium chloride (Stephen reaction):
$$C_6H_5C\equiv N \xrightarrow[\text{EtOH}]{SnCl_2/HCl} C_6H_5CH{=}NH\cdot HCl$$

The iminium chloride so obtained is compound S.
• S possesses no phenolic $$-OH$$ group, so it will not give the phthalein-dye (phenol) test. Hence Option B is wrong.

Step 4 : Hydrolysis of the iminium salt
$$C_6H_5CH{=}NH\cdot HCl \xrightarrow{H_2O,\; \Delta} C_6H_5CHO + NH_4Cl$$

The final product is benzaldehyde and is denoted T.
• Benzaldehyde is a colourless liquid, so T is not coloured; Option D is wrong.

Verification of the statements
A. Q (benzonitrile) on treatment with ethanol in the presence of $$SnCl_2/HCl$$ indeed furnishes benzaldehyde, an aromatic aldehyde. Correct.
B. False, as discussed.
C. False, P is a diazonium salt, not a dinitro compound.
D. False, benzaldehyde is colourless.

Hence the only correct choice is:
Option A which is: $$\mathbf{Q}$$ on treatment with ethanol generates an aromatic aldehyde.

Four classic name reactions of nitrogen-containing compounds are involved: Sandmeyer cyanation, Hoffmann bromamide degradation, Gabriel phthalimide synthesis and the carbylamine test. By recalling the characteristic product of each reaction, we can match every entry of List-I (P, Q, R, S) with one structure from List-II (1 - 5).

An aryl diazonium salt reacts with $$CuCN$$ to give the corresponding aryl nitrile.

$$C_6H_5N_2^+Cl^- \;\xrightarrow[\,]{\,CuCN\,}\; C_6H_5CN + N_2\uparrow$$

The product is benzonitrile, structure (1). Hence $$P \to (1)$$.

When an amide is treated with $$Br_2$$ and alcoholic $$KOH$$, the carbonyl carbon is lost as $$CO_3^{2-}$$ and a primary amine containing one carbon fewer is produced.

$$CH_3CONH_2 \;\xrightarrow[\,4\,KOH\,]{\,Br_2\,}\; CH_3NH_2 + K_2CO_3 + 2\,KBr + 2\,H_2O$$

The product is methylamine, structure (2). Hence $$Q \to (2)$$.

Step 1 - phthalimide is converted to its potassium salt, which undergoes $$S_N2$$ substitution with an alkyl halide (say $$C_2H_5Br$$).
Step 2 - hydrazinolysis (or acid/base hydrolysis) liberates the primary amine.

Overall: $$C_6H_4(CO)_2NH \;\xrightarrow[\,KOH\,]{\,C_2H_5Br\,}\; C_6H_4(CO)_2N-C_2H_5 \;\xrightarrow[\,NH_2NH_2\,]{}\; C_2H_5NH_2$$

The product is ethylamine, structure (4). Hence $$R \to (4)$$.

Primary amines, chloroform and alcoholic $$KOH$$ give the offensive-smelling isocyanide (carbylamine).

$$C_6H_5NH_2 + CHCl_3 + 3\,KOH \rightarrow C_6H_5NC + 3\,KCl + 3\,H_2O$$

The product is phenyl isocyanide, structure (5). Hence $$S \to (5)$$.

Collecting the four matches: $$P\to(1),\; Q\to(2),\; R\to(4),\; S\to(5).$$

Therefore the correct option is
Option B which is: (P) → (1), (Q) → (2), (R) → (4), (S) → (5).