See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Solution ## Part 1: Benzenoid Isomers of $C_6H_3ClBrI$ For a trisubstituted benzene with three different substituents (Cl, Br, I), count positional isomers: **Possible positions:** - **1,2,3-** (ortho-ortho) - **1,2,4-** (ortho-meta) - **1,3,5-** (meta-meta) Since all three substituents are different, each arrangement gives one distinct isomer. $$\boxed{\text{3 benzenoid isomers}}$$ --- ## Part 2: Stability Comparisons **(a) Ketone (I) > Enol (II)** Ketone form is much more stable; aromatic rings resist forming enols. ✓ **(b) Phenol (I) > Phenoxide tautomer (II)** Phenol is stable; the tautomeric form disrupts aromaticity. ✓ **(c) Imidate (II) > Formamide (I)** The imidate form (resonance-stabilized $C=NH$) is more stable due to better charge distribution. ✓ **(d) Enol form (II) > Amide (I)** ✗ **Incorrect.** Amides (I) are highly stable due to resonance ($C=O$ ↔ $C^+-O^-$). Enols are unstable and tautomerize to ketones. Amide (I) is significantly **more stable**. **(e) Carboxylic acid (I) > Enediol (II)** Carboxylic acid is far more stable; enediols are highly unstable. ✓ **Answer: D** — The correct statement should reverse the stability order.