enter image description here

What does the 1H NMR spectroscopy coupling pattern look like for the hydrogen attached to the carbon that the blue arrow is pointing to? There is a $\ce{CH2}$ attached to ether side of the carbon, but each of those carbons is in a different chemical environment from the other. Does that mean that it is a doublet of doublets? Or just a doublet?


1 Answer 1


The two $\ce{CH2}$ groups are, as you mentioned, in chemically different environments. Thus, each one will couple separately. But within each of the two groups, there are two protons. So you will not get a doublet of doublets, but a triplet of triplets.

Note that for each $\ce{CH2}$ group the hydrogens are chemically equivalent but not magnetically. Therefore, if you include a chiral environment, the pattern will further complexify itself into a higher-order signal that cannot easily be analysed. But since you’re probably measuring in chloroform you need not worry — there is no chiral information present and the couplings will be degenerate.

If I had to label a spin system here, I would probably choose $\ce{A3BB'CC'MM'X3}$.

By the way, it can be assumed that both coupling constants to the two different $\ce{CH2}$ groups be reasonably similar that the overall pattern looks like a quintet without actually being one.

  • $\begingroup$ If I replaced the oxygen and the methyl group to its left with, lets say a single methyl group, would the two CH2 groups be in chemically identical environments, thus giving me a lone triple? $\endgroup$
    – Lahav
    Dec 19, 2015 at 3:07
  • 1
    $\begingroup$ @Lahav If you replace the $\ce{OMe}$ with $\ce{CH3}$, then not only are you creating two chemically equivalent but also two magnetically equivalent environments (okay, it is slightly more complicated when dealing with the magnetic equivalence). The resulting pattern should be a near-perfect quintet (4 nuclei to couple to). However, the actual NMR spectrum of pentane is not easy to analyse because the signals all overlap. Pentane’s spin system should be $\ce{A6B4C2}$. $\endgroup$
    – Jan
    Dec 19, 2015 at 14:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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