Question
Aniline is treated with at , and the resulting diazonium salt is then reacted with phenol in alkaline medium. Identify the major product and explain why the reaction must be carried out at low temperature. JEE Advanced 2023 pattern.
Solution — Step by Step
At , aniline + + HCl gives benzenediazonium chloride:
Diazonium salts are unstable above . They decompose to give phenol + gas. So we keep the temperature near 0°C to preserve the salt for the next step.
In dilute NaOH, phenol is deprotonated to phenoxide (), which is highly nucleophilic at the para position. Electrophilic attack by the diazonium cation gives a coupling product.
The major product is p-hydroxyazobenzene (an orange-yellow azo dye):
This is the basis of textile azo dyes — the same chemistry produces the colours in your kurta.
Why This Works
The diazonium cation is a weak electrophile, so it can attack only highly activated aromatic rings (phenols, anilines). Coupling occurs at the para position (or ortho if para is blocked) because the activating group strongly directs there.
Alkaline medium is essential: it converts phenol to phenoxide, dramatically increasing nucleophilicity. In neutral or acidic conditions, the coupling is too slow.
Alternative Method
Aniline (instead of phenol) couples in slightly acidic medium with diazonium salts to give p-aminoazobenzene. The choice of phenol-vs-aniline + alkaline-vs-acidic conditions is a classic NEET memorisation point.
Students often forget the temperature constraint. JEE has asked “why 0–5°C?” as a 2-mark explanation — the answer is thermal instability of the diazonium salt, not just “low temperature is needed for selectivity”.
For JEE Advanced, memorise the suite of reactions of diazonium salts: with (gives benzene), with (phenol), with (iodobenzene, Sandmeyer), with (benzonitrile), and the coupling reactions (azo dyes). They cover 80% of diazonium-related questions.