See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Analysis of Structures E, F, and G **Step 1: Identify the molecular formulas** - Structure E: $C_6H_{10}O$ (ketone with one degree of unsaturation) - Structure F: $C_6H_{10}O$ (enol form with one degree of unsaturation) - Structure G: $C_6H_{10}O$ (enol form with one degree of unsaturation) All three have identical molecular formulas. **Step 2: Examine structural relationships** **E and F:** Structure E is a ketone (keto form) while F is an enol. They differ only in the position of a hydrogen atom and a double bond: $$E \text{ (ketone)} \rightleftharpoons F \text{ (enol)}$$ This is **keto-enol tautomerism** — they are tautomers that can interconvert via hydrogen shift. **F and G:** Both are enols with the same connectivity but different spatial arrangements of substituents around the $C=C$ double bond (one has $OH$ and $CH_3$ in different geometric positions). However, since both F and G are highly unstable enol forms, they would rapidly tautomerize to the stable keto form (E). **Step 3: Evaluate options** - **(A)** False — they're not resonance structures (no electron rearrangement, involves H migration) - **(C)** False — geometric isomers have different spatial arrangements but are usually stable; these enols are unstable - **(D)** False — diastereomers apply to compounds with multiple stereocenters - **(B)** **True** — E, F are tautomers (keto-enol pair); E and G are also tautomers as G converts to E **Answer: (B) E, F and E, G are tautomers**