See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Concept: Geometrical (cis/trans) isomerism arises at each C=C double bond where both substituents on each sp2 carbon are different. For a molecule with multiple double bonds, the total number of geometrical isomers is 2^n, where n is the number of double bonds capable of showing geometrical isomerism, but we must check if any combination produces an identical (meso-like) structure that reduces the count. The compound is hexa-2,4-diene: CH3–CH=CH–CH=CH–CH3. Step 1: Identify the double bonds. - Double bond 1: C2=C3 (between CH3–CH= and =CH–) - Double bond 2: C4=C5 (between –CH= and =CH–CH3) Step 2: Check each double bond for geometrical isomerism. - At C2=C3: C2 has CH3 and H; C3 has –CH=CH–CH3 and H → both carbons have two different groups → can show cis/trans. - At C4=C5: C4 has CH3–CH=CH– and H; C5 has CH3 and H → both carbons have two different groups → can show cis/trans. Step 3: Enumerate possible combinations. With 2 double bonds, we get 2^2 = 4 combinations: (2Z,4Z), (2Z,4E), (2E,4Z), (2E,4E). Step 4: Check for internal symmetry (identical isomers). The molecule CH3–CH=CH–CH=CH–CH3 is symmetric (same groups on both ends: CH3 on C1 and C6). Because of this symmetry: - (2Z,4E) and (2E,4Z) are mirror images of the same compound (they are identical due to the molecular symmetry — rotating 180° interconverts them). So these two count as ONE isomer. - (2Z,4Z) and (2E,4E) are each unique. Step 5: Count distinct geometrical isomers. - (2E,4E): 1 isomer - (2Z,4Z): 1 isomer - (2E,4Z) and (2Z,4E): these are the same compound due to the C2 symmetry of the molecule → 1 isomer Total = 3 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 (2E,4Z) and (2Z,4E) identical. - (d) 6: Far too many; no basis for this count. Therefore, the correct answer is B.