Chemistry · Amines · NEET
Iodide ion (I-) is a good reducing agent and a strong nucleophile, so it can replace the diazonium group by itself. Just warming the diazonium salt with potassium iodide (KI) gives iodobenzene: C6H5N2+ + KI to C6H5I + N2. Chlorine and bromine need cuprous salts (Cu2Cl2 or Cu2Br2, the Sandmeyer reaction), but iodine does not. This is a common NEET point of difference.
Direct iodination of benzene with I2 is very slow and reversible. The HI formed is a reducing agent and pushes the reaction backward, removing the iodine again. You would need an oxidant to remove HI. The diazonium route with KI avoids all this and gives iodobenzene cleanly, so NCERT prefers it.
First convert the diazonium chloride to benzenediazonium fluoroborate by adding fluoroboric acid (HBF4). This salt is stable and insoluble. Then heat it with aqueous sodium nitrite (NaNO2) in the presence of copper powder. The diazonium group is replaced by the nitro group, giving nitrobenzene plus N2.
The plain diazonium chloride is unstable and decomposes easily in solution, giving side products like phenol. The fluoroborate salt is stable and can be isolated as a solid. Starting from this stable salt makes the nitro replacement clean and controlled.
No. Nitration with a mixture of HNO3 and H2SO4 adds NO2 wherever the ring directs it, and on aniline it gives mostly meta product because of protonation. The diazonium route places the nitro group exactly where the amino group was, so you control the position. This positional control is the whole reason diazonium salts are useful in synthesis.
Given below are two statements. 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. Iodide replacement uses only warm potassium iodide (KI). Copper (as Cu2Cl2 or Cu2Br2) is needed only for chloride and bromide replacement, which is the Sandmeyer reaction.
First make benzenediazonium fluoroborate using fluoroboric acid (HBF4), then heat it with aqueous NaNO2 and copper powder to get the nitro compound plus N2.
Nitrogen gas (N2) is always released when the diazonium group is replaced. This loss of stable N2 is the driving force for all diazonium substitution reactions.
They let you place halogens, OH, NO2, CN, and H at an exact position on the ring. Multi-step conversion questions in NEET often need this positional control, so these reactions appear regularly.
