See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - E/Z nomenclature requires that each carbon of the double bond bears two DIFFERENT substituents. If one carbon has two identical substituents, E/Z designation is impossible and we assign 'none' (N). Structure 1 Analysis: The double bond indicated has one carbon bearing: (a) a phenyl group and (b) a benzyl (CH2Ph) group. The other carbon bears two hydrogen atoms (=CH2). Because the terminal carbon has two identical substituents (H and H), E/Z designation cannot be assigned. Wait - re-examining the structure: the arrow points to the internal double bond between the carbon bearing Ph and CH2Ph on one side, and the carbon bearing Ph on the other side. Looking more carefully: one sp2 carbon has a phenyl group and a benzyl group; the other sp2 carbon has a phenyl group and... Re-reading: the structure shows Ph-CH=C(CH2Ph) where both carbons are disubstituted with different groups. Carbon 1: phenyl (Ph) and H; Carbon 2: phenyl (Ph) and benzyl (CH2Ph). Priority on C1: Ph > H. Priority on C2: Ph > CH2Ph (Ph has higher priority than CH2Ph since direct attachment gives C,C,C vs C,H,H at first shell... actually Ph attached directly gives aromatic carbons). Assigning CIP priorities: C1 substituents: Ph (higher) vs H (lower). C2 substituents: Ph (higher, direct ring) vs CH2Ph (lower, CH2 then ring). Higher priority groups: Ph on C1 (top) and Ph on C2. Looking at the drawn structure, both Ph groups appear on the same side (both upper region) - this would be Z. Answer: Z. Structure 2 Analysis: The double bond has one carbon bearing two cyclopentadienyl groups (one above, one below-left). Because this carbon bears TWO cyclopentadienyl substituents that are identical (or even if slightly different connectivity, they are both cyclopenta-2,4-dien-1-yl groups attached the same way), E/Z cannot be assigned. The carbon bears two identical cyclopentadienyl rings, making E/Z designation impossible. Answer: N (none). Structure 3 Analysis: The double bond has one carbon bearing: phenyl group and H. The other carbon bearing: methyl (CH3) and OC(=O)CH3 (acetate). CIP priorities: C1: Ph > H. C2: OC(=O)CH3 > CH3 (oxygen beats carbon). The higher priority group on C1 is Ph; on C2 is OAc. In the drawn structure, Ph is on the left and OAc is on the right, with Ph and OAc on opposite sides of the double bond (the structure shows a trans-like arrangement where Ph and OAc are on opposite ends). Examining the geometry: Ph is on the left carbon pointing left, and OAc is on the right carbon pointing right - they are on opposite sides relative to the double bond axis. This means the two higher-priority groups (Ph and OAc) are on opposite sides = E configuration. Answer: E. Therefore, the correct answer is 1-Z; 2-N; 3-E.