I was supposed to identify this compound by knowing its molecular formula and the following spectra:

$1\mathrm{H}\ (\mathrm{s},\ 9.67\,\mathrm{ppm})\\ 1\mathrm{H}\ (\mathrm{quartet},\ 7.73\,\mathrm{ppm})\\ 1\mathrm{H}\ (\mathrm{quartet},\ 7.30\,\mathrm{ppm})\\ 1\mathrm{H}\ (\mathrm{quartet},\ 6.63\,\mathrm{ppm})$

My question is: how come we get quartets for the protons on the furan ring? I can see it getting doublets and triplets, but it seems impossible for them to couple with the aldehyde proton (which, in itself, doesn't couple with any of them).

  • 7
    $\begingroup$ Who decided that these peaks are quartets? Was this provided to you, or was it your interpretation? They will not be (and are not) quartets, but doublets of doublets. On a good day of shimming you will see some coupling to the aldehyde. $\endgroup$
    – long
    Commented May 5, 2016 at 6:58
  • $\begingroup$ This was provided to me by a problem. Indeed, they should've specified it's a doublet doublet which looks like a quartet. therefore, my confusion arised from the fact that the coupling constants would be the same. $\endgroup$
    – L3ul
    Commented May 5, 2016 at 7:10
  • 1
    $\begingroup$ The coupling constants are no the same, because then you would indeed see a triplett. If the four peaks are indeed equidistant, then one coupling constant would be twice as large as the other. $\endgroup$
    – Karl
    Commented May 6, 2016 at 11:54
  • 1
    $\begingroup$ It sounds like the problem is really one of terminology: you would expect to see a doublet of doublets for each of the three aromatic protons, and a doublet of doublets (typically) has 4 peaks, but it is not correct to call such a multiplet a quartet. The integral ratios would be 1:1:1:1 instead of 1:3:3:1 (maybe 1:2:1 if the two inner peaks coalesce). Sometimes TAs (and even faculty) get sloppy/lazy with their notation. $\endgroup$
    – S. Burt
    Commented May 9, 2016 at 4:48

1 Answer 1


They are not coupling with the aldehyde proton. Because the three protons around the ring are locked in such close proximity to one another, they will all couple each other. Let us consider these protons:


As by the $(n+1)$ rule of multiplicity, a proton will couple with each nearby, chemically unique proton and be split into $(n+1)$ peaks. These split signals can then also be split into $(n+1)$ more peaks by other nearby, chemically unique proton. Since all three protons are in the same region of space around the ring, they all couple each other, splitting each proton's signal into $(1+1)(1+1)=4$ peaks, making each signal a doublet of doublets (a quartet).

If you'd like a detailed analysis of the NMR of furfanal, you can read about it here.

  • 5
    $\begingroup$ A doublet of doublets is not a quartett. Even if one coupling constant is twice that of the other, the four peaks still have the same integral each, not 1:3:3:1. $\endgroup$
    – Karl
    Commented May 6, 2016 at 11:51

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