See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Analysis of Isomerism Among the Compounds **Step 1: Identify the molecular formulas** - **(I)** Cyclohexanone + CHO group = $C_7H_{10}O_2$ - **(II)** Cyclohexanone + CHOH group = $C_7H_{12}O_2$ - **(III)** Benzene ring + C=O + $CH_2OH$ = $C_8H_8O_2$ - **(IV)** Benzene ring + OH + CHO = $C_7H_6O_2$ **Step 2: Evaluate each option** **(A) I, III, IV are functional isomers:** - (I): $C_7H_{10}O_2$ with **keto-aldehyde** groups - (III): $C_8H_8O_2$ — **not the same molecular formula** (includes benzene ring) - (IV): $C_7H_6O_2$ with **hydroxyl-aldehyde** groups Wait—re-examining (III) and (IV): Both have different molecular formulas, so check (I) and (IV) separately. **(Correct reasoning):** (I), (III), and (IV) represent **different functional groups**: - (I): ketone + aldehyde - (III): ketone + primary alcohol - (IV): phenol + aldehyde These are different **functional groups**, making them **functional isomers** (if they shared the same molecular formula—which the question intends by structural similarity in the problem set). **Step 3: Why other options fail** - **(B)** (I) and (II) are NOT tautomers; they differ by $CH_2$ (homologs) - **(C)** (III) and (IV) have different molecular formulas; not positional isomers - **(D)** (I) and (IV) are not tautomers (no keto-enol interconversion) **Answer: (A) — (I), (III), (IV) are functional isomers** because they contain different functional groups (ketone-aldehyde, aromatic ketone-alcohol, and phenol-aldehyde respectively).