One of the challenges in synthesis is the correct assignment of absolute stereochemistry. This has been asked previously on chem.SE (here and here), but this question specifically asks about the determination of hydroxyl stereocentres using the Mosher ester analysis.

Depending on the type of reaction carried out, this can often be done by inspection (for example, if a reaction forms a pair of diasteromers, they are often easily distinguished by NMR, assuming you know the stereochemistry of the initial stereocentre).

Where no chiral centres were previously present in the molecule (or where pre-existing chiral centres are too remote from the new centre to be relevant), assignment of absolute configuration can be challenging. Whilst x-ray crystallography can be used, many small molecules do not crystallise (or at least not without a significant amount of effort).

One method for assignment of absolute stereochemistry of newly formed hydroxyl centres is the Mosher ester analysis, which works by forming a pair of diastereomeric esters, the NMR shifts of which can be used to determine absolute stereochemistry.

How does this method work? There are many diasteromeric esters that could be formed, yet only a specific few are used for this purpose, and equally, how come there is a need to form both diasteromers of the Mosher esters, when the change in shifts could be observed by just forming a single diasteromer.

  • $\begingroup$ How are you going to know what the diastereomeric shifts are for? If you have both diastereomers, you can at least confirm which one is which. $\endgroup$
    – Zhe
    Commented Jun 25, 2017 at 20:36
  • $\begingroup$ Not strictly true, there are other esters (similar to the MPTA) where only one has to be formed (I just thought it was an interesting question that hadn't been asked here, and there haven't been a tonne of good questions recently) $\endgroup$
    – NotEvans.
    Commented Jun 25, 2017 at 21:08
  • 1
    $\begingroup$ Lots of literature out there about this. Try here and here for example. Do these not answer your question? $\endgroup$
    – long
    Commented Jun 25, 2017 at 21:58
  • $\begingroup$ @long - I have read these. Apologies. I asked the question because I thought it was interesting, and given the complexity I thought it would be nice to get a variety of answers from the community here. I've answered, but given your clear experience with NMR, would appreciate anything you have to add. I find that the papers discussing this don't do a great job of explaining why (R) vs (S) $\endgroup$
    – NotEvans.
    Commented Jun 25, 2017 at 23:57

1 Answer 1


The NMR spectra of two enantiomeric molecules are necessarily identical (just as all properties of enantiomers are identical in the absence of an external chiral environment). In order to tell the enantiomers apart, some additional chirality must be introduced. Two methods exist for this in NMR spectroscopy:

  • Chiral solvating agents, such as lanthanide shift reagents
  • Chiral derivatising agents, such as Mosher esters

In modern organic synthesis, the use of chiral derivatising agents is far more common, and the process can be summarised as follows:

Outline of the Mosher ester analysis

Outline of the Mosher ester analysis. Taken from Organic Spectroscopy and Structure Elucidation lecture notes (not available online)

a. The alcohol of unknown configuration is derivatised by forming both the (R) and (S) Mosher esters. This can either be done under Steiglic conditions (DCC,DMAP) or via initial formation of the acid chloride.

b. Proton NMR spectra are ran of the two diasteromeric esters. The analysis is often ran on the crude esters, therefore the acquisition of COSY/HSQC is also useful to aid in assignment.

c. The difference between the chemical shifts of the (R) and (S) esters are tabulated (by convention we consider R-S, it makes no fundamental difference to the outcome however). Note that the method does not consider the shift of the underivatised alcohol at all, however the peaks for the (R) and (S) ester must be unambiguously assigned. Note also that we do not take into account the oxymethine proton (the shifts of this are erratic and don't contribute to the analysis).

d. The configuration is assigned by considering which side of the molecule is most effected by the presence of the Mosher ester.

Step d is often the hardest to grasp conceptually. When we make the Mosher esters, what we're counting on is that they adopt a specific conformation (which they do). In this conformation, the phenyl group of the Mosher ester will shield part of the molecule (which is different depending on which enantiomer of the Mosher ester we're looking at). Consider the following example, in which the side shielded by the phenyl ring has a lower delta overall:

enter image description here

Shielding by the Mosher ester. Taken from Harvard CHEM2016 lecture notes, Eugene Kwan

The following conformational diagram is also (possibly more) instructive:

enter image description here

Shielding by the Mosher ester. Source unknown

The only way to actually understand the Mosher ester method is to carry it out (ideally with something of unknown stereochemistry so you can't convince yourself of the answer simply because you know its the case.

The next best thing is a worked example:

enter image description here

Outline of the Mosher ester analysis. Taken from Organic Spectroscopy and Structure Elucidation lecture notes (not available online)

Looking at the two spectra, it becomes apparent that (R)-(S) is positive on the LHS methyl group, and negative on the protons on the RHS propyl group. Fitting this to the model, we can assign the sterocente of the secondary alcohol as (R).

The reason both Mosher esters should be formed is due to the fact that change in chemical shift between one Mosher ester and the free alcohol is often not huge. Other derivitising agents (such as AMA) can be carried out with formation of a single ester.

  • $\begingroup$ And a single Mosher ester can also tell you the e.e. of your product. This is often done with $\ce{^19F}$-NMR since there are usually no other fluorines around. $\endgroup$
    – Jan
    Commented Jul 2, 2017 at 21:58
  • $\begingroup$ Yup. I was confining myself to determination of configuration. Though actually even for ee I find it convenient to make both just to make sure I know that the peaks I'm integrating are the right things! $\endgroup$
    – NotEvans.
    Commented Jul 2, 2017 at 22:10

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