MCQMediumJEE 2026Aldehydes & Ketones

JEE Chemistry 2026 Question with Solution

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\alpha, \beta-unsaturated methylketone (CH3CH=CHC=OCH3CH_3-CH=CH-C=O-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

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Statements about the Haloform reaction are to be checked for truth.

Find: Which option contains the true statements.

The solution states that hypohalites are oxidizing agents and evaluates each statement one by one.

  • A. NaOCl\text{NaOCl} reaction with KI\text{KI} is taken in the solution as generating the active intermediate OI\text{OI}^- in educational context, so it is treated as True.
  • B. KOI\text{KOI} has iodine in the +1+1 oxidation state and acts as an oxidizing agent, not a reducing agent. So False.
  • C. The compound CH3CH=CHC=OCH3CH_3-CH=CH-C=O-CH_3 contains the required methyl ketone unit, so it gives the iodoform reaction. Hence True.
  • D. Isopropyl alcohol (CH3CH(OH)CH3)\left(\text{CH}_3\text{CH(OH)CH}_3\right) contains the methyl carbinol group and gives a positive iodoform test. So the statement is False.
  • E. Methanoic acid (HCOOH)\left(\text{HCOOH}\right) does not contain the required CH3CO\text{CH}_3\text{CO}- or CH3CH(OH)\text{CH}_3\text{CH(OH)}- unit. So False.

Therefore, the true statements are A and C.

The correct option is C.

Statement-wise Analysis

Given: A multiple-statement question on the Haloform reaction.

Find: Identify the true statements using the condition for a positive iodoform test.

A compound gives the iodoform test if it contains a methyl ketone group or can be oxidized to one, as in secondary alcohols of the form CH3CH(OH)R\text{CH}_3\text{CH(OH)R}.

For statement C, the compound CH3CH=CHC=OCH3CH_3-CH=CH-C=O-CH_3 is an α,β\alpha, \beta-unsaturated methyl ketone, but it still contains the essential CH3CO\text{CH}_3\text{CO}- unit. Hence it gives the test.

For statement D, isopropyl alcohol is of the form CH3CH(OH)CH3\text{CH}_3\text{CH(OH)CH}_3, so it is oxidized to acetone and therefore gives a positive iodoform test. Hence the statement is false.

For statement E, methanoic acid lacks the necessary structural requirement, so it does not respond positively.

The provided solution also marks A as true in the intended educational context and B as false because hypoiodite behaves as an oxidizing agent.

Thus, only A and C are true, so the correct option is C.

Common mistakes

  • Assuming every alcohol fails the iodoform test is incorrect. Secondary alcohols of the form CH3CH(OH)R\text{CH}_3\text{CH(OH)R}, such as isopropyl alcohol, give a positive test because they are oxidized to methyl ketones. Check the functional group before deciding.

  • Thinking an α,β\alpha, \beta-unsaturated methyl ketone will not respond because of the double bond is wrong. The key requirement is the presence of the CH3CO\text{CH}_3\text{CO}- unit, which is still present here. Focus on the methyl ketone moiety, not the extra unsaturation.

  • Confusing hypohalites with reducing agents is a conceptual error. Species such as hypoiodite act as oxidizing agents in haloform chemistry. Use oxidation state and reaction role to classify the reagent correctly.

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