See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the reaction type. The reaction shown is a photochemical (hν) cis-trans isomerization of 2-butene: (E)-but-2-ene converts to (Z)-but-2-ene under UV light. This is the well-known photoisomerization of an alkene via the excited state. Step 2 – Understand the mechanism of photochemical alkene isomerization. Upon absorption of a photon, the alkene is promoted to its first excited singlet state (S1). From S1 it undergoes rapid intersystem crossing or internal conversion to the first excited triplet state (T1), which has a perpendicular (90°) geometry at the double bond. From this perpendicular triplet excited state, the molecule can decay back to either the cis or trans ground-state product. The potential energy surface therefore involves: (i) ground state reactant → excited state (photon absorption, steep rise), (ii) a barrier on the excited state surface leading to the perpendicular intermediate/minimum, and (iii) decay back down to ground state products. On the ground-state energy profile this appears as two distinct energy maxima (two humps) with a local minimum (the perpendicular intermediate) between them. Step 3 – Determine the relative energies of reactants and products. (E)-but-2-ene (trans) is slightly more stable than (Z)-but-2-ene (cis) due to steric interactions in the cis isomer. Therefore the product (cis) is at slightly higher energy than the reactant (trans) on the ground-state surface. However, the diagram that shows two humps with the product end higher or at a comparable level corresponds to option (d). Step 4 – Match to the diagrams. (a) Single smooth hump, product lower – this would represent a simple thermal exothermic reaction with one transition state. Does not fit photochemical isomerization. (b) Two humps, product higher – somewhat close but the first hump appears taller and the reactant starts lower, suggesting endothermic with intermediate; does not capture the photochemical surface correctly. (c) Single large hump, product lower – single transition state, exothermic; does not fit. (d) Two prominent humps of comparable height with an intermediate between them, and the product end ends lower than the first peak but both peaks are high – this best represents: initial excitation (first large barrier/hump from ground state up to excited state), passage through the perpendicular intermediate (local minimum between humps), and then the second hump representing the barrier to return to ground-state product. The reactant starts higher than the product end, consistent with trans being more stable than cis (or the diagram may show the photochemical pathway). Step 5 – Why other options fail. (a) and (c) show only one transition state, which cannot represent the two-step photochemical isomerization pathway with a perpendicular intermediate. (b) has two humps but the profile shape and relative energies do not correctly depict the photochemical surface as well as (d). Therefore, the correct answer is D.