Verified Solution • 2025–26 Session moderate

QJEEM24Chccd2: JEE Main 2024 Chemistry - Class 12 JEEM24Chccd2 Solution: m-chlorobenzaldehyde on treatment with 50% KOH solution yields :

m-chlorobenzaldehyde on treatment with 50% KOH solution yields :

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/a6d30cbb9ea9480ddc4613e5809edf9520ac01d86a515829b44cbb33adbba0d3.webp
  • B https://ik.imagekit.io/novatra/photos/f9a12e49a367366dd167f3669e650699123c143cab006c499bc29c36fe03ae5f.webp
  • C https://ik.imagekit.io/novatra/photos/944f0f327f60835fa55543b0d92b9d5e3a089fc8675ac660fac73683b0127afa.webp
  • D https://ik.imagekit.io/novatra/photos/7b025b98e90dc5d2026c1ca31b50cd5d14d88434d25c79b28d26a4f06780c032.webp

Step-by-Step Verified Solution

Meta-chlorobenzaldehyde will undergo

Cannizzaro reaction with 50% KOH50 \% \mathrm{~KOH} to give m−\mathrm{m}- chlorobenzoate ion and m-chlorobenzyl alcohol.

JEE Main 2024 (Online) 30th January Evening Shift Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 33 English Explanation

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