See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the two stereocenters/double bonds requiring E/Z designation. The molecule has two sites of geometric isomerism: (i) The exocyclic C=C double bond between the cyclohexane ring carbon and the sp2 carbon bearing –C≡CH and –CO2H. (ii) The C≡C (triple bond / cumulated or simply the allenic-type axis is not present here; rather the triple bond itself is linear and has no E/Z isomerism). Wait – re-examine: the triple bond (C≡CH) is terminal and linear, so it has no E/Z isomerism by itself. The two stereocenters are therefore: (i) The exocyclic alkene (C=C between ring and the carbon bearing ethynyl and CO2H). (ii) A second axis: the ring itself is a 1,4-disubstituted cyclohexane, where C1 bears the exocyclic double bond and C4 bears Cl. The ring substitution (1,4) creates a second geometric relationship. Step 2 – Assign priorities at the exocyclic double bond. The sp2 carbon outside the ring is bonded to: –C≡CH (ethynyl) and –CO2H, and =C(ring). The ring carbon (C1) is bonded to: two ring carbons (C2 and C6) and =C(exo). On the exo carbon: –CO2H > –C≡CH (oxygen atoms of CO2H give higher priority than carbons of C≡CH). On the ring carbon C1: the two ring chains lead to C4(Cl) on one path; by CIP, the path containing Cl has higher priority. Step 3 – Determine E or Z for the exocyclic double bond. Looking at the drawing: CO2H is on the lower right of the exo carbon, and the ring is on the left. The higher priority group on the exo carbon is CO2H (priority 1) vs. C≡CH (priority 2). The higher priority group on the ring carbon C1 is the ring chain leading to Cl (priority 1) vs. the other ring chain (priority 2). In the structure as drawn, CO2H and the Cl-bearing ring chain are on opposite sides of the double bond → E configuration for the exocyclic double bond. Step 4 – Assign E/Z for the ring (1,4-disubstituted cyclohexylidene / the ring double bond context). For a 1-chloro-4-(substituted methylidene)cyclohexane, the ring itself can be considered: C4-Cl relative to C1=C(exo). The Cl at C4 can be on the same side as CO2H (Z) or opposite side (E) relative to the ring plane and double bond geometry. In the structure shown, Cl is drawn to the left on C4, and the exocyclic double bond extends to the right side with CO2H below. Considering the 1,4-relationship in the ring for the second designation: Cl is on one face and the exo-alkylidene substituents determine the second E/Z. Analysis of the drawn structure places Cl and the C≡CH group on the same side → Z for the ring/double bond relationship, giving Z as the first designation. Step 5 – Compile designations. First designation (ring double bond, C4-Cl vs ring substituents relative to exo double bond): Z. Second designation (exocyclic double bond, CO2H vs C≡CH on exo carbon relative to ring path): E. This gives Z, E → answer (b). Why other options fail: (a) E, E – incorrect assignment of the first center. (c) E, Z – reverses both assignments. (d) Geometrical isomers are possible since both the exocyclic double bond and the 1,4-disubstituted ring create restricted rotation/geometric relationships. Therefore, the correct answer is B.