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In the present challenge, bicyclic molecules are the topic. And please note that there is a prize to be won (a Springer book of your choice up to a value of €100 — given to one winner selected randomly). Please read on.

Meet the challenge

As the name implies, a bicyclic molecule contains two joined rings [1]. Bridged bicyclic structures occur widely in many biologically important natural molecules such as borneol and camphor. Indeed, such molecules are object of high interest in organic chemistry and drug discovery owing to their attractive pharmacokinetic properties.

A common challenge for NMR scientists concerning bridged bicyclic molecules is the determination of the relative orientation of substituents in such molecules, as shown in Fig. 1.

Fig. 1
figure 1

Example of a bicyclo[2.2.1]alkane having endo-oriented (left) and exo-oriented (right) substituent

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The standard and likely the most convenient approach to NMR structure elucidation of such substances is the NOESY experiment to observe NOE correlations of protons neighboring the substituent group R with the protons of the nearby CH2 groups belonging to either the short bridge (for endo isomer) or the long bridge (for exo isomer) of the bicycle.

But there is often a simpler way for the structure elucidation of such molecules, especially if the short bridge contains a heteroatom. In such cases, the spin-coupling picture of the protons located at the bridgehead positions becomes simpler.

The challenge

A colleague obtained two isomers of a bridged bicyclic compound (Fig. 2). But time is short! All we have before setting up next batch of overnight reactions are the 1H-NMR spectra of these isomers in chloroform-d3 (Fig. 3).

Fig. 2
figure 2

Structures of two bridged bicyclic isomers A (endo isomer) and B (exo isomer)

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Fig. 3
figure 3

600 MHz.1H-NMR spectra of substances A and B in CDCl3

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Using only these two 1H-NMR spectra without peak-picking and detailed multiplet analysis, try to determine which spectrum (top and bottom) corresponds to which isomer. Explain your reasoning. A hint: find the key difference in the spin-coupling patterns, not the chemical shifts!