How to differentiate diastereotopic protons by NMR in flexible groups?

In (1⁠R,2⁠S)-2-(hydroxymethyl)cyclohexan-1-ol shown below, Ha and Hb are diastereotopic (they will have different chemical shifts), but because of fast rotation, NOESY should show that both couple with Hc, right? How to differentiate diastereotopic protons by NMR in flexible groups?

• Hmm... I am wondering if the rotation really will be that fast. I guess there will be some preferred position of the hydroxyl group. If you would then perform low-temperature NMR you could maybe be able to freeze the molecule in that preferred conformation. The only way I could imagine you would be able to differentiate between $\mathsf{H_a}$ and $\mathsf{H_b}$ though would be to perform ab initio calculations and compare these to the experimental chemical shifts. Why are you interested in that anyway? – logical x 2 Dec 1 '16 at 18:23

I don’t think you can distinguish which is which in a freely rotatable group like the one you drew. They will be different, both signals will split up as dd (potentially ddd, if coupling to the hydroxy proton is observed). There will be significant roofing. Careful analysis of the spectrum should reveal two different shifts coupling to each other but very close.

Rotation around the $\ce{C-C}$ axis should be rapid on the NMR timescale resulting in the couplings $^3J_{\ce{HaHc}}$ and $^3J_{\ce{HbHc}}$ averaging out to one common value and the signal of $\ce{Hc}$ having a dddt-type pattern (three doublets to the neighbouring ring protons, a triplet to the side-chain protons).

All NMR methods I know to determine relative stereochemistry (e.g. NOESY, Murata, …) rely on rotation being slow on the NMR timescale.

Potentially cooling your sample down to very low temperatures ($-50~\mathrm{^\circ C}$ or less) could inhibit the rotation and allow for intermittent resolution of the protons. However, that should also produce a more distinct chemical shift difference, so it may actually not help.

• In the frozen-in state, shouldn't the coupling constants to Hc give a good idea which is which (Karplus equation)? – Karl Dec 1 '16 at 22:10
• @Karl Yes, imho. However, I’m not sure if the chemical shifts in frozen state can be mapped back to those of the non-frozen state. (I never did low-temp NMR, so … ^^') – Jan Dec 1 '16 at 22:25
• Right. The average chemical shifts in the freely rotating hydroxymethyl group might easily reverse the peak order in the spectrum. – Karl Dec 1 '16 at 22:40

There are a number of methods that have proven successful in a number of cases, but no universally applicable method exists. Some thoughts to throw into the mix, to add to Jan's answer.

The greater the rate of rotation, or conformational exchange, the more difficult to implement a method. That is, if no preferred conformation exists, it becomes very difficult. The methods developed by Murata (J. Org. Chem., 1999, 64 (3), pp 866–876 DOI: 10.1021/jo981810k) make use of the 2 and 3 bond couplings to carbon, as well as the 3J H-H coupling. A very good reference that covers much of the recent work is by Bifulco, G. et al. (Chem. Rev. 2007, 107, 3744-3779 DOI: 10.1021/cr030733c).

For complicated natural product systems, often the difficulty is accurately determining the H-C coupling constants, and a lot of work is being done in this area of NMR methodology. The review by Parella et al (Prog in Nucl Magn Res Spec 73 (2013) 17–55 DOI:10.1016/j.pnmrs.2013.07.001) provides some good insights here, and the very recent review by Tormena (Prog in Nucl Magn Res Spec 96 (2016) 73-88 DOI:10.1016/j.pnmrs.2016.04.001) discusses some complementary methods of analysis using modelling and theoretical calculations.

Freely rotating chains of course exist as a staggered average of rotational conformers. Sometimes, even in acyclic systems, specific conformers are preferred, and the coupling constants reflect this. However, in fast rotating systems, the couplings average across the lifetimes of all conformers - if you are seeing 3JH-H coupling of about 7 Hz, you have a very fast averaging system, with no strongly preferred conformation. If you have couplings that are more toward 3-4 (gauche) or 10-12 (anti), then it is likely that a preferred conformation exists. As Jan suggests, cooling can help, but even simply changing solvent can have benefits. Going to a non-polar solvent will promote intramolecular H-bonding, which can help stabilise a preferred conformer. Even going to a higher magnetic field can help favour the one population.