See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

To determine E/Z configuration, we apply CIP priority rules to the two substituents on each carbon of the indicated double bond. E isomer means the two higher-priority groups are on opposite sides of the double bond. Concept: E (entgegen) configuration means higher-priority substituents on each sp2 carbon are on opposite sides of the double bond plane. Structure (1): At the indicated double bond: - Left vinyl carbon: substituents are (i) the carbon bearing OCH3, CH3, and CH3 [higher priority due to O atom] and (ii) the rest of the chain. Right vinyl carbon: substituents are (i) CH2Br [higher priority due to Br] and (ii) CH(F)(CH3) [has F]. - On the left carbon: the group with oxygen (H3CO-C) has higher priority over the alkyl. - On the right carbon: CH2Br has higher priority than CH(F)(CH3) because Br > F by atomic number. - Looking at the drawn structure: CH2Br (high priority, right carbon) and the OCH3-bearing group (high priority, left carbon) are on opposite sides of the double bond → E configuration. Structure (2): At the indicated double bond: - One vinyl carbon has CH2Br and the peroxymethyl-methylene chain; other has CHBr(CH3) and CHO. - Higher priority on each carbon: CH2Br side has the peroxy/ether oxygen chain vs CH2Br — Br wins; CHO carbon: O of aldehyde vs CHBr(CH3) — Br (atomic number 35) vs O (atomic number 8), so CHBr(CH3) wins. - The two higher-priority groups (CH2Br and CHBr(CH3)) appear to be on the same side (Z) based on the drawn structure → Z configuration, NOT E. Structure (3): At the indicated double bond: - One vinyl carbon: Cl and CH2Cl; other carbon: two branched alkyl groups. - Higher priority on left carbon: Cl > CH2Cl (direct Cl attachment wins). - Higher priority on right carbon: the larger branched alkyl. - From the structure as drawn, Cl (higher priority) and the higher-priority alkyl are on opposite sides → E configuration. - Wait, re-examining: both Cl substituents are on the same carbon (one is directly Cl, one is CH2Cl), and the geometry of the drawn structure places Cl on top and the alkyl chains distributed — the higher-priority groups (Cl on one carbon, higher alkyl on other) appear to be on the same side → Z. Thus structure 3 is Z, not E. Structure (4): At the indicated double bond (the one with the arrow): - One vinyl carbon: D and H — D has higher priority than H (deuterium atomic mass > protium). - Other vinyl carbon: -CH=CHCH3 (vinyl/propenyl, higher priority due to extended conjugation counted by CIP) and -C(CH3)(H3C)- bearing isopropyl. - CIP for the propenyl vs isopropyl: propenyl =CH-CH=CH2 duplicated atoms give (C,C,C) at first shell repeated; isopropyl gives (C,C,H) — propenyl wins. - D (higher priority) is on the same side as the propenyl group (higher priority) based on structure as drawn? No — D and the higher-priority propenyl group are on opposite sides → E configuration. Thus structures 1 and 4 are E isomers, corresponding to answer choice (c). Therefore, the correct answer is C.