See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

The compound is CH3-CH=C(Br)-C(Br)=CH-CH3, which is 3,4-dibromo-2,4-hexadiene (a conjugated diene with two C=C double bonds, each bearing a bromine substituent). Step 1: Identify the double bonds and check for geometric isomerism at each. - Double bond 1: C2=C3, i.e., CH3-CH= C(Br)-. For geometric isomerism, each carbon of the double bond must have two different groups. - C2 has: CH3 and H (two different groups) ✓ - C3 has: Br and C4 chain (two different groups) ✓ - So double bond 1 can show E/Z isomerism. - Double bond 2: C4=C5, i.e., -C(Br)=CH-CH3. For geometric isomerism: - C4 has: Br and C3 chain (two different groups) ✓ - C5 has: H and CH3 (two different groups) ✓ - So double bond 2 can also show E/Z isomerism. Step 2: Count possible combinations. With two independent double bonds each showing E/Z isomerism, theoretically there are 2 × 2 = 4 combinations: (E,E), (E,Z), (Z,E), (Z,Z). Step 3: Check for any equivalences due to molecular symmetry. The molecule CH3-CH=C(Br)-C(Br)=CH-CH3 is symmetric about the C3-C4 bond (both ends are CH3-CH= and =CH-CH3 with Br on each inner carbon). Due to this symmetry, the (E,Z) and (Z,E) combinations are actually identical (one is the mirror image/superimposable form of the other through the symmetry axis). Therefore, these two count as only one distinct isomer. Step 4: Final count of distinct geometrical isomers. - (E,E): 1 isomer - (Z,Z): 1 isomer - (E,Z) = (Z,E): 1 isomer (these are the same compound due to molecular symmetry) Total = 3 distinct geometrical isomers. Why other options fail: - (a) 2: Undercounts; ignores the mixed E/Z isomer. - (c) 4: Overcounts; does not account for the molecular symmetry that makes (E,Z) and (Z,E) identical. - (d) 6: Greatly overcounts; no basis for 6 isomers here. Therefore, the correct answer is B.