See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Concept: Geometrical (cis-trans) isomerism in cycloalkanes arises when two or more ring carbons each bear two different substituents, creating restricted rotation and the possibility of substituents being on the same side (cis) or opposite sides (trans) of the ring plane. Step 1: Identify the compound. The structure shown is 1-methyl-4-bromocyclohexane. Carbon C1 bears a CH3 group (and H), and carbon C4 bears a Br atom (and H). Both substituted ring carbons have two different groups attached (one substituent + one H), satisfying the requirement for geometric isomerism. Step 2: Determine possible geometric isomers. For a 1,4-disubstituted cyclohexane with two different substituents (CH3 and Br), two arrangements are possible: - cis-1-methyl-4-bromocyclohexane: CH3 and Br on the same face of the ring. - trans-1-methyl-4-bromocyclohexane: CH3 and Br on opposite faces of the ring. Step 3: Check for symmetry/identity. In 1,4-disubstituted cyclohexane with two DIFFERENT groups (CH3 ≠ Br), the cis and trans isomers are distinct, non-superimposable structures. Neither is identical to the other, and neither is a meso compound (since the two substituents differ). So both are valid, distinct geometric isomers. Step 4: Count total geometric isomers. The compound itself as drawn represents one form (either cis or trans). The total number of geometric isomers possible for this compound is 2 (one cis and one trans). Why other options fail: - (a) 0: Incorrect; geometric isomerism is possible because both C1 and C4 carry two different groups. - (c) 3 and (d) 4: Incorrect; with only two stereocenters in a simple 1,4-disubstituted system bearing different groups, only cis and trans forms exist — giving exactly 2 geometric isomers. Therefore, the correct answer is B.