See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Solution: Identifying the Compound with Four Diastereomers **Step 1: Understand what diastereomers are** Diastereomers are stereoisomers that are NOT mirror images. For alkenes, E/Z isomerism at each double bond creates diastereomers. **Step 2: Analyze Option (B)** This compound has **two C=C double bonds** in conjugation: $$H_3C-C(H)=C(H)-C(CH_3)(CH_2CH_3)$$ **Step 3: Count stereoisomers at each double bond** - **First C=C**: $H_3C$ vs $H$ on one end; $H$ vs the second double bond on the other → **E/Z isomerism** (2 possibilities) - **Second C=C**: $H$ vs second C=C on one end; $CH_3$ vs $CH_2CH_3$ on the other → **E/Z isomerism** (2 possibilities) **Step 4: Calculate total diastereomers** $$\text{Total stereoisomers} = 2 \times 2 = \boxed{4 \text{ diastereomers}}$$ **Why others fail:** - **(A)** One C=C only → 2 isomers (E/Z) - **(C)** Triple bond + C=C → No diastereomerism (triple bond lacks two substituents) - **(D)** Cumulated allenes ($C=C=C$) → No E/Z isomerism possible **Answer: (B)** ✓