Question

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. $\alpha, \beta$-unsaturated methylketone ($\text{CH}_3-\text{CH}=\text{CH}-\text{C=O}-\text{CH}_3$) will give iodoform reaction.
D. Isopropyl alcohol will not give iodoform test.
E. Methanoic acid will give positive iodoform test.

• A. A, B & C Only
• B. B, D & E Only
• C. A & C Only
• D. A, C & E Only

Hint:

The Haloform reaction is an oxidation reaction followed by nucleophilic substitution and elimination. Hypohalites are oxidizing agents.

Answer 1

The Correct Option is C

A. Haloform reaction is typically carried out using $X_2/NaOH$ (or $NaOX$). $\text{NaOCl}$ reaction with KI yielding KOI is not standard, but $\text{I}_2 + \text{NaOH} \to \text{NaOI}$ (hypoiodite) is the active intermediate. Assuming A refers to the generation of the active intermediate $\text{OI}^-$, it is often considered true in educational contexts (True).
B. $\text{KOI}$ contains Iodine in the $+1$ oxidation state and acts as an oxidizing agent. False.
C. $\alpha, \beta$-unsaturated methylketone has the required $\text{CH}_3\text{C=O}$ group. True.
D. Isopropyl alcohol ($\text{CH}_3\text{CH(OH)CH}_3$) has the methyl carbinol group and gives a positive test. False.
E. Methanoic acid ($\text{HCOOH}$) lacks the required $\text{CH}_3\text{C=O}$ or $\text{CH}_3\text{CH(OH)}$ group. False.
The technically true statement is C. Given the options, the intent must be A and C.