Chemistry · Amines · NEET
Both do the same job: they replace the diazonium group (-N2+) of a benzene diazonium salt with -Cl or -Br to give chlorobenzene or bromobenzene. The difference is only in the copper reagent. Sandmeyer uses a cuprous halide (Cu2Cl2 / CuCl or Cu2Br2 / CuBr). Gattermann uses copper metal powder together with the halogen acid (HCl or HBr). The Gattermann version is a cheaper, simpler variant, but its yield is usually lower than Sandmeyer's.
In the Gattermann method, finely divided copper powder plus HX generates the active cuprous halide in situ (during the reaction). Because the reagent is made on the spot and the copper surface area is limited, the reaction is slower and gives a lower yield than the Sandmeyer method, where you add a ready-made, freshly prepared cuprous halide. For NEET, just remember: same product, Gattermann = lower yield.
A cuprous (copper-I) halide. For chlorobenzene you use cuprous chloride (Cu2Cl2 / CuCl in HCl); for bromobenzene you use cuprous bromide (Cu2Br2 / CuBr in HBr). Note it is cuprous (Cu+), not cupric (Cu2+). The freshly prepared diazonium solution is mixed with this cuprous halide, and the -N2+ group is replaced by -Cl or -Br with loss of N2 gas.
No. For iodine, no copper reagent is needed at all: you simply shake the diazonium salt with potassium iodide (KI) and the group is replaced by -I. For fluorine you use the Balz-Schiemann reaction (heat the diazonium fluoroborate, ArN2+BF4-). Sandmeyer and Gattermann are only for -Cl and -Br. This exact split is a favourite NEET trap.
No, they are different. The Gattermann reaction (this page) converts a diazonium salt to a haloarene using Cu powder + HX. The Gattermann-Koch reaction is a totally separate reaction that makes benzaldehyde from benzene using CO + HCl with anhydrous AlCl3/CuCl. Same chemist name, different reactions. NEET can list both to confuse you, so read the reagents, not the name.
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. Choose the most appropriate answer.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes. For the same halogen, both give the same haloarene (chlorobenzene or bromobenzene). Only the reagent and yield differ: Sandmeyer uses cuprous halide and gives higher yield; Gattermann uses Cu powder + HX and gives lower yield.
Cuprous, i.e. copper(I). You use Cu2Cl2 (CuCl) for -Cl and Cu2Br2 (CuBr) for -Br. Writing cupric (Cu2+/CuCl2) is wrong and is a common NEET trap.
You add plain copper metal powder along with HCl or HBr. The Cu powder plus the acid generates the active cuprous halide in situ during the reaction, so no separately prepared cuprous salt is needed.
Cu-mediated substitution does not work well for fluorine. Fluoroarenes are made by the Balz-Schiemann reaction: the diazonium fluoroborate ArN2+BF4- is heated and decomposes to give the aryl fluoride, N2, and BF3.
Yes. NCERT (Haloalkanes and Haloarenes) teaches Sandmeyer's reaction with cuprous halides and notes iodobenzene via KI; the Gattermann variant (Cu powder + HX) is the standard companion. NEET regularly tests these in diazonium reaction sequences and match-the-reagent questions.
