Which of the following statements are TRUE about Haloform reaction?: A. Sodium hypochlorite reacts with KI to give KOI. B. KOI is a reducing agent. C. , -unsaturated methylketone ( CH _ 3 - CH = CH - O O - CH _ 3 ) will give iodoform reaction. D. Isopropyl alcohol will not give iodoform test. E. Methanoic acid will give positive iodoform test. Choose the correct answer from the options given below:
C. A & C Only
In the haloform reaction: (A) Sodium hypochlorite reacts with KI in acidic medium to generate the active halogenating species, forming KOI (potassium iodite/hypoidite). TRUE. (B) KOI serves as an oxidizing agent (not reducing) in the reaction, converting methyl ketones to carboxylic acids. FALSE. (C) α,β-unsaturated methylketones possess the essential methyl ketone group (CH_3CO-) required for haloform reaction, allowing iodination and subsequent cleavage. TRUE. (D) Isopropyl alcohol (CH_3)_2CHOH lacks a methyl ketone group, so it cannot undergo the haloform reaction. TRUE. (E) Formic acid has no methyl ketone functionality, preventing haloform test. FALSE. Therefore, statements A and C are true.
Related: Chemistry — Aldehydes and Ketones · All PYQ Banks