= Solution
\b[C is the best geometrical match for a strongly bound 1:1 complex with E.] Both have an adjacent <amide> donor–acceptor edge. Arrange them head-to-tail so that the N–H of E donates to the <carbonyl> <oxygen> of C, and the N–H of C donates to the <carbonyl> <oxygen> of E. This closes the familiar two-<hydrogen>-bond cyclic <amide> motif, illustrated below.
C combines a strong acceptor with a useful donor in the correct spacing. B has a stronger individual donor, but its <pyridine> N and OH are separated around the ring and do not reproduce the compact adjacent <amide> edge of E. A lacks a matching donor, and D's N–H is weak and has the wrong geometry for the same <amide>–<amide> motif. \b[Site strength and spatial complementarity must be considered together]; multiplying two isolated probe constants is not a reliable prediction of the complete association equilibrium. Self-association of C or E can compete and must be included in a quantitative binding analysis.
\Image[/past-exam-of-the-chemistry-b-course-of-the-university-of-cambridge/2024/paper-1-recognition.png]
{title=Complementary amide hydrogen bonds, Watson–Crick pairs and guanine recognition}
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