Since H 2 SO 4 is an oxidizing agent, it oxidizes HI (produced in the reaction to I 2 ). 2HI + H 2 SO 4 → I 2 + SO 2 + H 2 O. As a result, the reaction between alcohol and HI to produce alkyl iodide cannot occur. Therefore, sulphuric acid is not used during the reaction of alcohols with KI.

The correct combination of names for isomeric alcohols with molecular formula C4H10O is/are (A) tert-butanol and 2-methylpropan-2-ol (B) tert-butanol and 1, 1-dimethylethan-1-ol (C) n-butanol and butan-1-ol (D) isobutyl alcohol and 2-methylpropan-1-ol jee advanced 1 Answer +1 vote answered Sep 16, 2022 by AnmolSingh (85.1k points)

There is no reason why you could not use phosphoric (V) acid in the bromide case instead of sulfuric acid if desired. When alcohols react with a hydrogen halide, a substitution takes place producing an alkyl halide and water:

This is the reason why sulphuric acid (${H_2}{SO_4}$) is not used during the reaction of alcohol (OH) with potassium iodide (KI). In the presence of a dilute acid, KI would produce HI. $ 2KI + {H_2}{SO_4}\to 2KHSO_4 + HI $ If the acid used is sulphuric acid, the HI gets used up to produce I2 gas. $ 2HI + {H_2}{SO_4}\to I_2 + SO_2 + H_2O $

In conclusion, sulphuric acid is not used during the reaction of alcohols with KI because it can interfere with the reaction by protonating the alcohol to form alkyl sulfates. Instead, alternative reagents such as sodium hydroxide, pyridine, and acetone can be used to facilitate the reaction of alcohols with KI. Community Answer