The following is a tutorial question I need to do:

NMR and IR spectra of unknown

I initially thought the answer was isobutyl formate (CAS:542-55-2) but what confuses me is the dd peak at $\pu{3.9-4.0 ppm}$. To my knowledge, isobutyl formate should give just a doublet at that region. What am I missing out?

  • $\begingroup$ The peak at 4 ppm is due to the 2 H atoms from CH2 which is bond to CH(CH3)2 on one side and to O-CHO on the other side. The molecule is HCOOCH2CH(CH3)2 $\endgroup$
    – Maurice
    Apr 2 '20 at 16:05
  • $\begingroup$ @Maurice Yes but shouldn't that CH2 give a doublet? Why dd? $\endgroup$
    – chemrese
    Apr 2 '20 at 16:06
  • $\begingroup$ @Maurice It can't be anything else but isobutyl formate, it's just I want to know the reasoning behind the dd peak. Can the CH2 couple with the aldehyde proton as well? How? $\endgroup$
    – chemrese
    Apr 2 '20 at 16:11
  • $\begingroup$ Your IR and 13C-NMR matches exactly with the known spectra of isobutyl formate, so your guess of the compound is most probably correct. However, I could not find its 1H NMR. Is it possible that your NMR was recorded with a chiral reagent? The two CH2 protons are enantiotopic and can couple to each other in presence of a chiral reagent. That's the only thing I can think of right now. $\endgroup$
    – S R Maiti
    Apr 2 '20 at 16:26
  • 1
    $\begingroup$ @ShoubhikRMaiti We should not forget that the compound in question is distributed commercially, e.g. by SigmaAldrich (e.g., sigmaaldrich.com/catalog/substance/…) and that this company has a tradition in publishing spectra catalogues (incl. IR and 1H/13C-NMR), too. Showing only a doublet around 4 ppm what could be a 300 MHz recording, see: sigmaaldrich.com/spectra/fnmr/FNMR001772.PDF $\endgroup$
    – Buttonwood
    Apr 2 '20 at 16:56

Your question is how can the CH2 be split into a dd when the -OCHO shows no triplett, right?

Well, the second splitting at the $\ce{CH2}$ is very small, and the $\ce{CHO}$ looks like it is very nearly split. That´s not just a broad peak, it has a rather clear shape imo. If you do just a bit of zero filling and apodisation, the triplett might well (should) appear.

But generally, you are right. Every splitting must have its counterpart.

Zero filling: Adding blanks to the end of your FID. Artificially increases the resolution, i.e. point density of your spectrum.

Apodisation: Changing the gain of your FID towards the end, by multiplying it with a decreasing or increasing function. The former gives you a less noisy spectrum, the latter a better resolution (sharper peaks actually), each at the cost of the other. The mathematical term for that is "convolution".

  • $\begingroup$ yes, that's what I meant. Maybe I should have worded my question better. The singlet at 8.07 ppm does indeed look nearly split, after reading your reply. Thanks. $\endgroup$
    – chemrese
    Apr 2 '20 at 17:33
  • 2
    $\begingroup$ NOESY experiment would clarify this splitting as well. $\endgroup$ Apr 2 '20 at 18:00
  • 1
    $\begingroup$ Drying the sample might make the splitting a lot clearer already. Or measureing the pure sample, with external lock. $\endgroup$
    – Karl
    Apr 2 '20 at 18:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.