See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

## Solution: Geometrical Isomers of $CH_3-CH=CH-CH=CH-C_2H_5$ **Step 1: Identify the structure** The compound has two C=C double bonds: - Position 2-3: $CH=CH$ - Position 4-5: $CH=CH$ Both double bonds can exhibit cis-trans (or E-Z) isomerism. **Step 2: Analyze the first double bond ($C_2=C_3$)** Substituents: $-CH_3$ and $-CH=CH-C_2H_5$ This bond can be: - **cis** (both groups on same side) - **trans** (groups on opposite sides) **Step 3: Analyze the second double bond ($C_4=C_5$)** Substituents: $-CH=CH-CH_3$ and $-C_2H_5$ This bond can also be: - **cis** (both groups on same side) - **trans** (groups on opposite sides) **Step 4: Count total combinations** Since both double bonds are independent: $$\text{Total isomers} = 2 \times 2 = 4$$ The four isomers are: 1. (2Z, 4Z)-configuration 2. (2Z, 4E)-configuration 3. (2E, 4Z)-configuration 4. (2E, 4E)-configuration **Answer: (C) 4**