I have a molecule

enter image description here

and i need to assign the hnmr. I have a printed spectrum

enter image description here

So far my answer is:

chemical shift_______splitting pattern_______coupling constant

 2.32                quartet            6.6 Hz & 13.2 Hz

 2.66                singlet                   -

 3.66                triplet                6.5 Hz

 5.11         triplet of doublets         1.5 Hz & 16.8 Hz

 5.81   doublet of doublet of triplets   6.9 Hz & 10.2 Hz & 13.8 Hz & 17.2 Hz

I am having trouble assigning them to the hydrogens. I think the H on the OH is a singlet and the H on the right hand side of the double bond is a ddt as it has cis (13.8 Hz) and trans (17.2 Hz). Then on the left of the double bond, the cis H would be the quartet as it has cis coupling (13.2) and the trans H would be the td as it has trans coupling (16.8 Hz).

But if these are correct then does that mean the four hydrogens on carbons 2 and three are the triplet group?


1 Answer 1


You haven't quite rationailsed the spectrum correctly; If you look at your molecule you will see that you have 6 distinct 1H species, but you have only described 5. Take particular note of the signals that you have described as a td at 5.11ppm. This should be symmetrical about the centre peak if it was a true td, and this signal clearly is not. So, it actually comes from two overlapping signals; one with a large coupling (~17Hz) and the other a medium coupling (~10Hz). Also, a quartet will only have one coupling constant - this will be the separation of any of two adjacent peaks in the quartet, and so is 6.6Hz, as you have described. So perhaps this is not a true quartet.

When assigning your peaks, you should first look at your chemical shifts;

  • Vinylic protons typically come in the range 3.5-8.5, but commonly 5-7ppm
  • alkyl protons alpha to a heteroatom 3-4.5ppm
  • alkyl protons alpha to vinyl group 1.5-2.5ppm
  • hydroxyl protons can come just about anywhere, but true alkyl-OH 1H will typically come 1-5 ppm

So from this information, you should be able to broadly assign the 6 protons in your molecule.

You then need to look at coupling constants to further refine your assignments. For vinylic groups, you can rely on the following general coupling constants:

enter image description here

  • trans coupling Jbc 14-19Hz
  • cis coupling Jac 4-12 Hz
  • geminal coupling Jab -4-4 Hz
  • vicinal coupling Jcd 4-10 Hz

This should allow you to complete the assignment of this spectrum. See how you go from here.

A useful approach to these type of assignment problems, especially if you know the structure, is to look at the molecule and try to predict the spectrum you'd expect, rather than looking at the spectrum and working out what's what.

enter image description here

Looking at your molecule, I'd expect:

  • Ha to come in the vinylic region, and appear as a dd (cis and geminal coupling to Hb and Hc)
  • Hb to come in the vinylic region, and appear as a dd (trans and geminal coupling to Ha and Hc)
  • Hc to come in the vinylic region, and appear as a ddt (cis, trans and vicinal coupling to Ha, Hb and Hd)
  • Hd to come in the alkyl region alpha to a double bond, and appear as a dt (coupling to Hc and He)
  • He to come in the alkyl region alpha to a heteroatom, and appear as a t (coupling to Hd)
  • Hf to come somewhere and appear most likely as an exchange broadened singlet


Edits to answer questions raised in comment to the original answer.

Below are two splitting diagrams to help you rationalise the splitting for the peaks centred around 2.3 ppm and 5.10 ppm.

enter image description here

Let's first look at your upfield peaks (2.10ppm), which is the splitting diagram on the left. This comes from signal Hd, and should show coupling to Hc (which will give rise to a doublet) and also He (giving a triplet). So this signal is actually a doublet of triplets. Normally if the doublet coupling and triplet coupling are quite different, then you observe two clear sets of triplet peaks. However, in this case, Jdc is approximately the same as Jde (within the natural linewisth of the peaks), and so you get overlap of some of peaks, creating what looks like a quartet. It is still a dt, with 6 lines, but you can't resolve them all. So, you still have two coupling constants, but both couplings are about 6.6Hz.

The second diagram (on the right) is for your sets of signals around 5.10 ppm. This is actually the overlap of two independent signals. Ha will occur as a dd, with cis coupling to Hc (~14Hz), and geminal coupling to Hb (~2Hz, so very small, and realy only broadens the peak in this case.) This is the signal on the right. Hb will also occur as a dd, with trans coupling to Hc (~17Hz), and again, geminal coupling to Ha. This is the signal on the left. It just so happens (life is like that sometimes), that one set of lines for each signal overlap, to create what looks like an aysmmetrical triplet. So, here are your two signals, Ha and Hb. Integrating the area under the peaks serves as a valuable tool to tell you how many signals each peak represents, and in this case you would have clearly seen that the signals at 5.10ppm come from 2 protons.

As a note, geminal couplings in vinylic systems is usually quite small, and often is of similar magnitude to allylic coupling (4J couplings, such as Jad, and this is why the peaks for protons a and b actually look a little more complicated than just doublets of doublets. The main take home message about double bond couplings at an introductory level really comes from the cis and trans couplings.

  • $\begingroup$ I dont understand how you tell which hydrogen doesn't have a peak. So you are saying that the td at 5.11ppm is two peaks, how do you know what it really is and how would you write this down as I have only been given 5 peaks with 6 distinct hydrogen environments. Also, for the quartet I got a coupling constant of 6.6 and 13.2Hz so do i take an average or just ignore the 13.2Hz one. $\endgroup$
    – user5181
    Commented Oct 18, 2014 at 10:56

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