See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Concept: Geometrical (cis/trans or E/Z) isomerism arises at each C=C double bond where both carbons of that double bond bear two different substituents. Step 1: Identify the double bonds in CH2=CH-CH=CH-CH=CH2 (hexa-1,3,5-triene). - Double bond 1: C1=C2 (positions 1,2) - Double bond 2: C3=C4 (positions 3,4) - Double bond 3: C5=C6 (positions 5,6) Step 2: Check each double bond for geometrical isomerism. - C1=C2: C1 bears two H atoms (CH2=), so both substituents on C1 are identical. No geometrical isomerism possible at this bond. - C5=C6: C6 bears two H atoms (=CH2), so both substituents on C6 are identical. No geometrical isomerism possible at this bond. - C3=C4: C3 is connected to C2 (which is part of the vinyl group CH=CH2) and H; C4 is connected to C5 (which is part of the vinyl group CH=CH2) and H. Both carbons of this double bond bear two different groups, so geometrical isomerism IS possible here. Step 3: Count the geometrical isomers. Only the central double bond (C3=C4) shows geometrical isomerism, giving 2 isomers: cis (Z) and trans (E). Step 4: Why other options fail. - Option (b) 3, (c) 4, (d) 8 would require additional stereocenters or additional double bonds capable of E/Z isomerism, but the terminal =CH2 groups eliminate isomerism at bonds 1 and 3. Only 1 double bond (central) contributes, giving 2^1 = 2 isomers. Therefore, the correct answer is A.