See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Concept: Geometrical (cis/trans or E/Z) isomerism across a C=C double bond requires that each carbon of the double bond bears two different substituents. If both substituents on either carbon of the double bond are identical, no geometrical isomerism is possible. For the tetrasubstituted alkenes in options (b), (c), and (d), the substituents on each doubly-bonded carbon are two CHX(CH3) groups — one above and one below. The key question is whether the two groups attached to each sp2 carbon are the same or different, considering their stereochemistry (since the stereocenters in the substituents can make groups diastereotopic). For option (c): The left carbon of the double bond bears CHCl(CH3) (Cl on wedge, top) and CHBr(CH3) (Br on wedge, bottom). The right carbon bears CHCl(CH3) (Cl on dash, top) and CHBr(CH3) (Br on dash, bottom). The two substituents on each carbon of the double bond are CHCl(CH3) and CHBr(CH3) — these are constitutionally different (Cl vs Br), so each carbon does have two different groups. However, we must examine whether the two carbons of the double bond bear identical substituent sets. The critical analysis for option (c): On the left carbon — top group is (R or S)-CHCl(CH3) with Cl wedge, bottom group is (R or S)-CHBr(CH3) with Br wedge. On the right carbon — top group is CHCl(CH3) with Cl dash (opposite configuration), bottom group is CHBr(CH3) with Br dash (opposite configuration). Thus the right carbon's substituents are the mirror-image configurations of the left carbon's substituents. When we assign priorities (CIP) to determine E/Z, we compare the two groups on each carbon. On the left carbon: CHClMe vs CHBrMe — Br > Cl, so CHBrMe has higher priority. On the right carbon: similarly CHBrMe has higher priority. Now for geometric isomerism to exist, the molecule as drawn must be distinguishable from its 'other' geometric form. In option (c), the substituents on the left carbon are both wedge (same face), and on the right carbon both dash (same face), meaning each carbon has its two substituents on opposite faces relative to each other across the double bond — but crucially, the two substituents on the left carbon (CHClMe wedge + CHBrMe wedge) have a specific spatial arrangement. Because the configurations of the stereocenters in the substituents of option (c) are arranged such that the overall molecule becomes superimposable on its 'geometric isomer' — i.e., the two groups on one carbon, when considering their full stereochemical identity, turn out to be equivalent to each other (both are on the same face, wedge), making the two substituents on that carbon identical in CIP priority ranking context or making the two possible geometric forms identical — option (c) does NOT show geometrical isomerism. More precisely: In option (c), both substituents on the left vinylic carbon have their halogens on wedge bonds (same configuration relative to viewing), and both substituents on the right vinylic carbon have their halogens on dash bonds. This means the two groups attached to the left carbon are: CHCl(CH3) [wedge-Cl] and CHBr(CH3) [wedge-Br]. Although Cl ≠ Br constitutionally, the stereochemical arrangement makes the substituent set on the left carbon the mirror image of the substituent set on the right carbon. The molecule drawn is identical to what would be its 'geometric isomer' after accounting for the stereochemistry of the side chains — the two possible geometric arrangements collapse into the same compound (or into identical structures), so no distinct geometric isomers exist. Options (a), (b), and (d) do show geometrical isomerism: (a) has =C(H)(CH3) where H ≠ CH3, and the cyclopropylidene ring carbon bears two different groups when the ring substituents are considered; (b) and (d) have substituents arranged such that distinct E and Z forms exist. Therefore, the correct answer is C.