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If you are asked to make benzyl alcohol from aniline, the question is testing one idea. You need one extra carbon, and a diazonium salt is how you attach it. NCERT's Class 12 Amines chapter sets this exact conversion as Exercise 9.8 (ix), "Aniline to benzyl alcohol" (NCERT Chemistry Part II, Unit 9: Amines). Here is the route step by step, with the reasons an examiner wants to see.
Why can't aniline become benzyl alcohol in one step?
In aniline (C₆H₅NH₂) the nitrogen is bonded straight to a ring carbon. In benzyl alcohol (C₆H₅CH₂OH) the oxygen sits on a CH₂ carbon outside the ring. The molecule has to grow from six carbons to seven, and no single reagent does that.
The obvious shortcut gives the wrong product. NCERT notes that a diazonium salt solution hydrolyses to phenol if its temperature is allowed to rise to 283 K (NCERT Unit 9). Phenol has the OH on the ring, not on a CH₂, so it is not benzyl alcohol.
What are the steps from aniline to benzyl alcohol?
The reagents and conditions are NCERT's (NCERT Unit 9).
Diazotisation. Aniline with NaNO₂ and HCl at 273–278 K gives benzenediazonium chloride, C₆H₅N₂⁺Cl⁻. Keep it cold: it is stable in cold water but reacts with water when warmed.
Sandmeyer reaction. Treat the diazonium salt with CuCN/KCN. The diazo group leaves as nitrogen gas and CN takes its place, giving benzonitrile, C₆H₅CN. This step adds the seventh carbon.
Reduction. Reduce benzonitrile with LiAlH₄, with H₂/Ni, or with sodium amalgam in ethanol. The C≡N becomes CH₂NH₂, giving benzylamine, C₆H₅CH₂NH₂.
Nitrous acid. Treat benzylamine with NaNO₂ and HCl, which make nitrous acid in the flask. The aliphatic diazonium salt that forms is unstable, loses N₂ and gives benzyl alcohol, C₆H₅CH₂OH.
In one line: C₆H₅NH₂ → C₆H₅N₂⁺Cl⁻ → C₆H₅CN → C₆H₅CH₂NH₂ → C₆H₅CH₂OH.
Why go through a diazonium salt at all? NCERT points out that a cyano group cannot be put on a benzene ring by nucleophilic substitution of chlorine in chlorobenzene, but cyanobenzene is easily obtained from a diazonium salt. The reaction is named after Traugott Sandmeyer, who described replacing the amino group in aromatic compounds with chlorine, bromine and cyano in 1884 (Berichte der deutschen chemischen Gesellschaft, 1884).
Why does nitrous acid treat aniline and benzylamine so differently?
Both are primary amines, and both meet nitrous acid in this route (NCERT Unit 9). The difference is what the NH₂ is attached to.
On a ring carbon, as in aniline: at 273–278 K you get an aromatic diazonium salt stable enough to carry into the next reaction.
On an alkyl carbon, as in benzylamine's CH₂: the aliphatic diazonium salt is unstable, releases nitrogen gas and leaves an alcohol behind.
NCERT gives both results side by side, and notes that the quantitative release of nitrogen from primary aliphatic amines is used to estimate amino acids and proteins (NCERT Unit 9). Benzylamine behaves as an aliphatic amine here because its nitrogen sits on the CH₂, not on the ring.
To see what a stable aromatic diazonium salt does instead, read our explainer on the azo dye test for aromatic primary amines.
What mistakes cost marks in this conversion?
Run through this checklist before you finalise a CBSE board or JEE Main answer. The conditions are the ones in NCERT Unit 9.
Writing phenol as the product. That is what warming the diazonium salt with water gives, and it skips the extra carbon.
Leaving out the temperature in step 1. Write 273–278 K (0–5 °C).
Writing KCN alone in step 2. The Sandmeyer reaction needs the copper(I) salt, CuCN.
Hydrolysing the nitrile with H₂O/H⁺ instead of reducing it. You need CH₂NH₂, and only a reducing agent gives that.
Forgetting to show N₂ leaving in steps 2 and 4.
How should you practise conversions like this?
Work backwards from the product. What makes a CH₂OH here? A CH₂NH₂ with nitrous acid. What makes a CH₂NH₂? A C≡N with a reducing agent. What puts a CN on a ring? A diazonium salt with CuCN. What makes the diazonium salt? Aniline. Four questions give you four steps.
Then try the pattern on a cousin problem. NCERT Exercise 9.9 (iii) takes CH₃CH₂Br through KCN, LiAlH₄ and HNO₂ at 0 °C, the same add-a-carbon, reduce, nitrous-acid chain starting from an alkyl halide (NCERT Unit 9). Compare it with converting toluene to benzyl alcohol in two steps, where the extra carbon is already in place, and with the Balz-Schiemann route to fluorobenzene, another reaction that depends on a diazonium salt.
If you get stuck on a step, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer. If the gap is in the basics, the Fermi Concept Library gives each chemistry concept its own lesson.
