See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Step 1: Identify the structure. The compound is CH3-CH=C(Br)-C(Cl)=CH-CH3, a diene with two double bonds: one between C2-C3 and another between C4-C5 (using a 6-carbon chain numbering). Each double bond is a site of potential geometric (cis/trans or E/Z) isomerism. Step 2: Check each double bond for geometric isomerism. - Double bond 1: CH3-CH= C(Br)-. The two groups on C2 are H and CH3; the two groups on C3 are Br and the rest of the chain. Since C2 has two different substituents (H and CH3) and C3 has two different substituents (Br and the C4 portion), this double bond can exhibit E/Z isomerism. So 2 configurations here. - Double bond 2: -C(Cl)=CH-CH3. The two groups on C4 are Cl and the C3 portion; the two groups on C5 are H and CH3. Since C4 has two different substituents (Cl and C3 portion) and C5 has two different substituents (H and CH3), this double bond can also exhibit E/Z isomerism. So 2 configurations here. Step 3: Calculate total geometrical isomers. With two independent double bonds, each capable of E or Z configuration, the total number of geometrical isomers = 2 × 2 = 4. Step 4: Check for any meso-form reduction. In this compound, the two double bonds bear different substituents (Br vs Cl), so no meso cancellation occurs. All 4 combinations (E,E), (E,Z), (Z,E), (Z,Z) are distinct. Step 5: Why other options fail. - (a) 2: Undercounts by considering only one double bond. - (b) 3: Incorrect; no meso form eliminates one of the four isomers here. - (d) 6: Overcounts; there is no additional source of isomerism beyond the two double bonds. Therefore, the correct answer is C.