Chemistry · Amines · NEET
Both replace the -N2+ group of a diazonium salt with Cl or Br. The only difference is the copper reagent. In the Sandmeyer reaction you use cuprous halide (cuprous chloride Cu2Cl2 or cuprous bromide Cu2Br2) with the matching acid HCl or HBr. In the Gattermann reaction you use fine copper metal powder plus HCl or HBr. Gattermann uses cheaper copper powder but usually gives a lower yield. So: Sandmeyer = Cu(I) salt, Gattermann = Cu powder. Same product, different form of copper.
Learn this list, NEET loves it. For -Cl: Cu2Cl2/HCl (Sandmeyer) or Cu powder/HCl (Gattermann). For -Br: Cu2Br2/HBr (Sandmeyer) or Cu powder/HBr (Gattermann). For -CN: Cu2(CN)2/KCN (Sandmeyer) gives the nitrile Ar-CN. For -I (iodobenzene): just warm with KI, NO copper needed. For -F (fluorobenzene): use fluoroboric acid HBF4 to get the diazonium fluoroborate, then heat it (this is the Balz-Schiemann reaction). Notice iodine and fluorine do NOT use the Sandmeyer method.
Iodide ion (I-) from KI is a strong enough nucleophile / reducing agent to react with the diazonium ion on its own, so simply shaking the diazonium salt with KI gives iodobenzene and N2 gas. Chloride and bromide ions are not reactive enough by themselves, so a copper catalyst (Cu2Cl2, Cu2Br2 or Cu powder) is needed to help the -Cl or -Br replace the -N2+ group. This is why NCERT says replacement by iodine does not require any cuprous halide.
In azo coupling the diazonium ion Ar-N2+ acts as a weak electrophile (E+) and attacks a second aromatic ring to form a coloured -N=N- (azo) bridge. The diazonium ion is a weak electrophile, so it can only attack rings that are strongly activated. Phenol (-OH) and aniline (-NH2) are strong activating, electron-donating groups, so they make the ring reactive enough. The attack happens at the para position (or ortho if para is blocked). The product is a bright orange, red or yellow azo dye. This reaction retains nitrogen; no N2 is lost.
No. In Sandmeyer, Gattermann, iodobenzene (KI) and phenol formation (warm water), the -N2+ group leaves as N2 gas and is replaced by another group. But in azo coupling the nitrogen is KEPT: both nitrogen atoms stay in the product as the -N=N- azo link. This is the key point examiners test. If you see a coloured dye product like C12H10N2O, no N2 was released and it was a coupling reaction.
A given nitrogen-containing aromatic compound A reacts with Sn/HCl, followed by HNO2, to give an unstable compound B. B, on treatment with phenol, forms a beautiful coloured compound C with molecular formula C12H10N2O. The structure of compound A is
Identify the final product in the following reaction sequence: C6H5N2+Cl- --(i) Cu2Br2/HBr--(ii) Mg/dry ether--(iii) H2O--> Product
Statement I: Benzenediazonium salt is prepared by the reaction of aniline with acid at 273-278 K. It decomposes easily in the dry state. Statement II: Insertion of iodine into the benzene ring is difficult, and hence iodobenzene is prepared through the reaction of the benzenediazonium salt with KI. In the light of the above statements, choose the most appropriate answer.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
No, do not mix them up. The Gattermann reaction here uses copper powder + HX to turn a diazonium salt into an aryl halide. The Gattermann-Koch reaction is a totally different reaction that adds a -CHO group to benzene using CO, HCl and AlCl3. In the Amines chapter, Gattermann means the diazonium-to-halide reaction only.
Azo dyes are usually bright orange, red or yellow. The colour comes from the extended conjugation across the -N=N- azo bridge joining two aromatic rings. This large conjugated system absorbs visible light, so the compound appears coloured. This is why the products in exam questions are called beautiful coloured compounds.
Simply warm the benzenediazonium chloride solution with water. The -N2+ group is replaced by -OH, giving phenol and N2 gas. C6H5N2+Cl- + H2O gives C6H5OH + N2 + HCl. This is different from coupling: here N2 is released, no dye forms.
Azo coupling normally occurs at the para position of the activated ring, because -OH and -NH2 are ortho/para directors and the bulky diazonium electrophile prefers the less crowded para spot. If the para position is already blocked, coupling shifts to the ortho position.
Because they let you make many benzene compounds you cannot make directly: chlorobenzene, bromobenzene, iodobenzene, fluorobenzene, phenol, benzene itself and azo dyes all start from one diazonium salt. NEET often gives a multi-step sequence and asks for the final product, so knowing which reagent gives which group is high-scoring.
