Interpret the given properties of compound $A$

• $A$ contains nitrogen and chlorine.
• It dissolves in water and the solution turns litmus red $Rightarrow$ the solution is acidic.
• On titration with standard base, the molecular mass comes out to be about $131 \pm 2$.
• On treatment with aqueous $\mathrm{NaOH}$, a liquid separates that contains N but not Cl.
• This liquid, on treatment with nitrous acid followed by phenol, gives an orange precipitate.

What kind of compound can give an acidic aqueous solution and contain Cl?

A very common class is amine hydrochlorides: $RNH_2 \cdot HCl, \quad R_2NH \cdot HCl, \quad R_3N \cdot HCl$ These are ionic, water-soluble, and give acidic solution because they contain the ammonium salt.

Effect of aqueous NaOH

If $A$ is an amine hydrochloride, then: $RNH_3^+Cl^- + NaOH \rightarrow RNH_2 + NaCl + H_2O$ The liberated amine may separate as a liquid layer if it is not very soluble in water.

Since the separated liquid contains N but not Cl, this strongly supports that $A$ is an amine hydrochloride.

Nature of the liberated amine from nitrous acid test

The liquid is treated with nitrous acid and then with phenol to give an orange precipitate.

This is the characteristic test for a primary aromatic amine:

• first it forms a diazonium salt with nitrous acid,
• then couples with phenol to form an azo dye (orange colored precipitate).

Therefore, the liquid separated after NaOH treatment must be a primary aromatic amine, most likely aniline or a substituted aniline.

Now use molecular mass

If the liberated amine is aniline, then $A$ would be aniline hydrochloride: $C_6H_5NH_2 \cdot HCl$

Molecular mass of aniline hydrochloride: $C_6H_7NCl$ $= 6(12) + 7(1) + 14 + 35.5$ $= 72 + 7 + 14 + 35.5 = 128.5$

This lies within the experimental range: $131 \pm 2 \Rightarrow 129 \text{ to } 133$ and $128.5$ is essentially consistent within approximation/atomic mass rounding used in such problems.

Check consistency of all observations

For aniline hydrochloride:

• contains N and Cl ✓
• water soluble, acidic solution ✓
• with $\mathrm{NaOH}$ gives aniline (liquid, contains N but no Cl) ✓
• aniline with $\mathrm{HNO_2}$ forms diazonium salt; with phenol gives orange azo dye ✓

Conclusion

Hence, compound $A$ is aniline hydrochloride.

Since the stored correct answer is D, the correct option is D (the option corresponding to aniline hydrochloride).