Solution to bridged bicyclic molecule NMR challenge

The Bridged bicyclic molecule NMR Challenge asked readers to identify isomers of bicyclo[2.2.1]alkane from the ¹H-NMR spectrum. The most deshielded protons with signals in the vicinity of 4.5 ppm are from protons located at the base of the bridge. The key difference between the two spectra is that one of them shows two triplets, while the other shows a triplet and a doublet (Fig. 1).

At this point, one has to remember the Karplus equation that describes the variation of the coupling constants between two hydrogen atoms as a function of the dihedral angle between them. The magnitude of this coupling is generally smallest when the torsion angle is close to 90° and largest at angles of 0 and 180° (Fig. 2).

Figure 2 illustrates the dihedral angles between the epoxy bridge hydrogen atoms and their neighboring hydrogen atoms in endo- and exo-isomers. Hence, according to the Karplus equation, in the case of the endo-isomer, the proton at the base of the bridge (closer to CH₂OH) position has more complex multiplicity than in the exo-isomer, as illustrated in Fig. 3.

Keeping this in mind, let us analyze the ¹H-NMR spectra of these isomers (Fig. 1). So, as can be seen from Fig. 1, in the endo-isomer of the bicyclo[2.2.1]alkane, both ¹H signals of hydrogen atoms at the base of the bridge (4.4…4.6 ppm) will appear as triplets (A, Fig. 1, left spectrum). At the same time, for the exo-isomer, they appear as one triplet and one doublet (B, Fig. 1, right spectrum).