Question

Compound "A" reacts with \(NaNO_2\) and \(HCl\) to give a diazonium salt, which on reaction with N,N-dimethylaniline produces methyl orange. What is the compound "A"?

• A. Benzoic acid
• B. Salicylic acid
• C. Sulphanilic acid
• D. Picric acid

Hint:

Methyl orange is made by coupling a diazonium salt of an aromatic amine that also bears a sulphonic acid group with N,N-dimethylaniline.

Answer 1

The Correct Option is C

Step 1: Methyl orange is prepared by an azo-coupling reaction. Its structure has a sulphonate group on one benzene ring and a dimethylamino group on the other ring, connected through an azo (\(-N=N-\)) linkage.

Step 2: \(NaNO_2 + HCl\) at low temperature converts a primary aromatic amine, \(Ar-NH_2\), into its diazonium salt through diazotization.

Step 3: For "A" to give the diazonium salt required for methyl orange, "A" must already carry both a primary aromatic \(-NH_2\) group and a \(-SO_3H\) group on the same benzene ring.

Step 4: The compound with exactly this combination is sulphanilic acid, para-aminobenzenesulphonic acid, \(H_2N-C_6H_4-SO_3H\).

Step 5: Diazotizing sulphanilic acid and coupling with N,N-dimethylaniline forms methyl orange.

Benzoic acid, salicylic acid, and picric acid have no free primary aromatic amine group, ruling them out.

\[\boxed{\text{A = Sulphanilic acid}}\]