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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?

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  • $\begingroup$ 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
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    $\begingroup$ @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
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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".

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  • $\begingroup$ 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
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    $\begingroup$ NOESY experiment would clarify this splitting as well. $\endgroup$ Apr 2 '20 at 18:00
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    $\begingroup$ 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

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