Spin spin coupling in a proton NMR of an ester?

Question

I am learning about proton NMR and spin-spin coupling, and am confused about whether splitting occurs over an ester bond. Specifically, in the case of ethyl methanoate, HCOOCH2CH3, if I were to number the proton environments going from left to right as 1,2 and 3, would 1 and 2 cause splitting of each other? If they would not cause splitting, then I would expect a singlet, a quartet and a triplet respectively. If they do cause splitting, then would you get a triplet, a quartet of doublets, and a triplet for 1,2 and 3 respectively? I know that with -OH and -NH the protons do not couple (although I am not sure why), so I was wondering whether the same principle may perhaps apply here...

Answer 1

In the most common case you can only observe ³J-couplings, any spins further away than three bonds do not cause a visible splitting in your spectra. The exception are couplings via double-bonds or aromatic systems, there you can often see small couplings over four bonds.

In this case the spins are too far away from each other to cause a visible splitting, you are correct that it would result in a singlet a quartet and a triplet.

In the case of OH and NH protons, chemical exchange can play a role. This is a bit more complicated, but the simple version is that protons that can exchange quickly with the solvent are often not observable in NMR (or just as a very broad signal) and due to that also don't couple. Though in the most common cases the OH proton is usually too far away from the other protons in the molecule to show a coupling anyway.