See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Enol Content Analysis **Understanding Enol Form:** The enol tautomer forms when a carbonyl compound undergoes keto-enol tautomerism: $R_2C=O \rightleftharpoons R_2C(OH)$ The stability and abundance of the enol form depends on: 1. Resonance stabilization of the conjugate base 2. Conjugation with $\pi$ bonds 3. Absence of competing stabilization of the keto form **Analyzing Each Option:** **(A) $CH_3CHO$ (Acetaldehyde)** - Simplest aldehyde with no electron-withdrawing groups nearby - Enol form: $CH_2=CH-OH$ is relatively stable - **No competing resonance stabilization** of keto form - Enol content: ~0.0001-0.001% but **highest among these** **(B) $CH_3COCH_3$ (Acetone)** - Ketone with electron-donating methyl groups - Keto form strongly stabilized by inductive effects - Enol form is destabilized - Enol content: very low **(C) $CH_3-C(=O)-CH_2CHO$ (Methylglyoxal)** - Two electron-withdrawing carbonyl groups - Both stabilize the keto form through resonance and induction - Enol form is significantly destabilized - Enol content: negligible **(D) $CH_3-CO-CH_2-CO_2CH_3$ (Methyl acetoacetate)** - Both carbonyls present (ketone + ester) - Ester group is electron-withdrawing - Keto form strongly stabilized - Enol content: very low **Answer: (A)** is correct because acetaldehyde lacks electron-withdrawing substituents and competing stabilization of the keto form, making the enol tautomer relatively more stable and abundant compared to the other compounds.