JEE Main 2025 · 23 January · morning · Single correct · +4 / −1

The major product of the following reaction is :
CH₃ CH₂ CH=O →(excess HCHO alkali) ?

  1. (a) CH₃-C-CH=O
  2. (b) CH₃-CH₂-CH₂-OH
  3. (c) CH₃-C(|)(CH₂-OH)-CH₂-OH
  4. (d) CH₃, -, C, -, CH=O; |; CH₂-OH
Answer and solution

Answer (c) CH₃-C(|)(CH₂-OH)-CH₂-OH

Solution

Chemistry ma’am · 4 steps

Use up every alpha-hydrogen with HCHO first, then let HCHO run a crossed Cannizzaro on what is left.

  1. Crossed aldol first. Propanal, CH₃CH₂CHO, has two α-hydrogens on its CH₂. In alkali this carbon forms an enolate that attacks HCHO, the most reactive carbonyl present. Each attack adds one CH₂OH group, and excess HCHO lets this happen twice.

    CH₃CH₂CHO →(HCHO, OH⁻) CH₃CH(CH₂OH)CHO →(HCHO, OH⁻) CH₃C(CH₂OH)₂CHO

  2. No alpha-H left. The α-carbon now carries CH₃, two CH₂OH groups and CHO, and no hydrogen. It cannot do another aldol and cannot dehydrate. It now behaves like an aldehyde with no α-H.
  3. Crossed Cannizzaro. With excess HCHO and alkali, OH⁻ adds to HCHO, which then transfers a hydride to the other aldehyde. HCHO is oxidised to formate, and the CHO group is reduced to CH₂OH.

    CH₃C(CH₂OH)₂CHO + HCHO + OH⁻ → CH₃C(CH₂OH)₃ + HCOO⁻

  4. So. The major product is CH₃C(CH₂OH)₃, 2-(hydroxymethyl)-2-methylpropane-1,3-diol. That is option (c).

The idea to keep. Aldehyde with α-H plus excess HCHO in alkali: put one CH₂OH on the α-carbon for every α-H. Then the CHO becomes CH₂OH by crossed Cannizzaro, and HCHO goes to formate.

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

Study Aldehydes & Ketones

Asked about 130 times in JEE Main since 2020. This question is from 2025.