I have been using bromoacetaldehyde diethyl acetal recently and noticed something interesting about the 1H NMR. The methylene protons of the ethyl acetal are magnetically non equivalent. You see two quartets integrating for two protons with different shifts.

You can see this in the spectra here.

I have found a short description, however it is a little unclear:

Diethyl acetals such as (1 6) and triglycerides such as triacetin (1 7) have a molecular symmetry plane and consequently are optically inactive. However, this plane does not bisect the connecting line of the geminal protons, which are magnetically non equivalent. This is manifested by the relative complexity of the AA'BB'X type proton spectrum of the glyceryl moiety of triacetin (Figure). Similar magnetic non-equivalence can be induced across an ester bond; e.g., the two methyl groups of the isopropyl ester (18) area non-equivalent.

I don't really understand as there is no restriction of rotation. And does this mean that this would disappear using high temperature NMR?

  • $\begingroup$ This issue has been addressed. The methylene hydrogens on the acetal ethyl groups are diastereotopic, i.e., different. chemistry.stackexchange.com/questions/86205/… $\endgroup$
    – user55119
    Commented Apr 11, 2018 at 17:10
  • $\begingroup$ Sorry! Didn't see this when I searched $\endgroup$
    – loltim
    Commented Apr 12, 2018 at 10:22
  • $\begingroup$ See ADDENDUM in my answer. $\endgroup$
    – user55119
    Commented Apr 12, 2018 at 16:15

1 Answer 1


This link on SE explains why the methylene hydrogens of the ethyl groups are different or diastereotopic. For your specific case, if a deuterium is substituted for Ha on the left side of the structure, it leads to an R-configuration while substitution of deuterium at Hb leads to an S-configuration. Both of these hydrogen sites "see" the same handedness at the methine carbon. Thus, the two hydrogens, Ha and Hb, are in different environments, they have different chemical shifts and they are coupled to each other and the vicinal methyl group. The same argument applies to the ethyl group on the right side of the structure. This difference in chemical shift is more apparent at high field (e.g., 300 MHz spectrum) rather than at low field (60 MHz).

There is no restricted rotation and temperature will not make a difference. Don't feel bad, these arguments were made for a monoalkyl derivative of diethyl malonate by a chemistry Nobelist!

ADDENDUM: Here is an additional link to this answer. Here is a related spectrum: Intermediate Level #12 at this link.

enter image description here


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