See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Analysis of Isomerism between Compounds II, I, III & IV **Molecular Formula Check:** All four compounds have the formula $C_{15}H_{14}N_2O$ (verify by counting atoms), confirming they are isomers. **II vs I: Functional Isomerism** - **II:** Contains $C=N-OH$ (oxime functional group) - **I:** Contains $C(=O)-NH$ (amide functional group) Different functional groups → **functional isomers** **II vs III: Tautomerism** - **II:** $C=N-OH$ (oxime form) - **III:** $C(H)-N=O$ (nitrosomethine form) The oxime and nitroso forms are tautomers related by proton shift and electron rearrangement: $$C=N-OH \rightleftharpoons C(H)-N=O$$ This is **tautomerism** **II vs IV: Identical Compounds** - **II:** Phenyl group attached to $C=N$, methylphenyl attached to $N-OH$ - **IV:** Methylphenyl attached to $C=N$, phenyl attached to $N-OH$ Both have identical connectivity and functional groups when drawn correctly → **identical compounds** (just redrawn perspectives) **Answer: Option (C) — Functional, tautomer, identical** - II & I: functional isomers - II & III: tautomers - II & IV: identical compounds