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. α,β-unsaturated methylketone (CH₃-CH=CH-C(∥(O))-CH₃) 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 :

Solution

Chemistry ma’am · 3 steps

Hypoiodite is an oxidiser, and iodoform needs only a CH₃CO⁻ or CH₃CH(OH)⁻ group.

1. The reagent. Sodium hypochlorite passes its oxygen to iodide: NaOCl + KI → KOI + NaCl. The hypoiodite OI⁻ then oxidises a CH₃CH(OH)⁻ group to CH₃CO⁻ and iodinates the methyl group, so it is not a reducing agent.
2. The structural test. A compound gives iodoform (CHI₃, a yellow precipitate) if it has a methyl ketone group CH₃CO⁻. It also gives it if it has a CH₃CH(OH)⁻ group that can be oxidised to one. Check each statement against this.

   Statement	Check	Verdict
   A	NaOCl + KI → KOI + NaCl	True
   B	KOI oxidises the substrate	False
   C	CH₃-CH=CH-CO-CH₃ has a CH₃CO⁻ end	True
   D	CH₃CH(OH)CH₃ has CH₃CH(OH)⁻, so it does give the test	False
   E	HCOOH has no CH₃ group at all	False

3. So. Only A and C are true, which is option (d).

The idea to keep. For any iodoform question, look for CH₃CO⁻ or CH₃CH(OH)⁻. The rest of the molecule, even a C=C double bond, does not matter. The hypohalite always acts as the oxidiser.

Answer from the NTA final answer key. Solution by PrepWiser.