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Here are is the NMR spectrum and the compound (both images are the same spectrum, just close up). I know the deshielded signals are the aromatic protons and the ones ~6ppm are the alkene protons but the rest is a bit muddy. I'm having trouble correctly assigning the signals to make them agree with the integrations (below the signals). I don't even see 5 equivalent protons anywhere despite the integration suggesting there are. I'm also confused about the multiplicity: the signal at ~1.35ppm looks like a quintet of doublet, but I don't see any protons that would make that. Can anyone help me?

NMR spectrum NMR spectrum (zoomed) Molecule

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  • $\begingroup$ It'd be helpful if you can calculate the coupling constants in each peak. Expand each region and get peak-picking in Hz. $\endgroup$ Oct 8 at 20:40
  • $\begingroup$ Next to the aniline-like aromatic H you have the three Me (2 Me with split of the isopropyl group, 1 Me isolated, singlet) for sure, too. If possible, collect complementary 1D data (DEPTQ, if not available an APT) or, with somewhat less resolution but higher signal sensitivity {1H,13C}-HSQC/HMBC to assign the other (more trickier) 6 protons signals in the multiplets (1.2 to 1.6 ppm) d), e), i), j); and somewhat shifted g) of iPr. Don't forget to integrate HSQC. $\endgroup$
    – Buttonwood
    Oct 8 at 20:54
  • $\begingroup$ The hydrogens labelled d and e should be designated as four unique hydrogens with different chemical shifts. Certainly the two proximate to the double bond are different from the distal pair. As Mathew has suggested, coupling constants would be useful. $\endgroup$
    – user55119
    Oct 9 at 1:48
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When I approach this type of question, the first thing I ask myself is whether there is a molecule-wide symmetry element present which would transform distant protons into chemically or magnetically equivalent protons. In your case, the molecule is chiral and does not have an axis of rotation so there are no helpful symmetry elements.

Next, I try to identify bonds around which free rotation is possible and which would transform hydrogens onto another.

  1. The most obvious candidate is the $\ce{{H^{(c)}}3C-C}$ bond. Free rotation around it transforms the three methyl hydrogens onto each other. We thus expect that these three hydrogens are magnetically equivalent and give only one signal.

    The same is true for the bonds to methyl groups (f) and (h), meaning that at the very least each of these should give a single signal.

  2. Less obvious but I hear you already identified it is the $\ce{N-Ph}$ bond. While rotation is slightly less free than in the previous cases as it is necessary to break the π system for rotation, this is nevertheless a very non-hindered rotation. Thus, the two (k) protons and the two (l) protons are expected to be the same.

    In general, this applies to all phenyl groups and almost all substituted benzenes.

  3. Finally, we have the $\ce{C_\mathrm{q}-CMe2H^{(g)}}$ candidate connecting the isopropyl group to the polycycle bridgehead. Free rotation around this bond might allow the (f) and (h) protons to become equivalent or near-equivalent. Often, isopropyl groups show just one methyl signal corresponding to $\mathrm{6H}$.

    However, a closer inspection of this group reveals that the methyl groups in this (and many other cases) are actually diastereotopic due to the compound already being chiral. Diastereotopic groups generally do not give the same signal although they may be similar enough to overlap at low resolutions.

As a sanity check at this point, I would go through protons connected to the same carbon atom and check whether I missed any equivalent ones. In your case, on the $\ce{C{H^{(d)}}2-C{H^{(e)}}2}$ remain. However, if you look at a three-dimensional structure (i.e. draw them with proper bond angles), you should see that they are not equivalent: One of the (d) protons points to the vinyl group, the other points to the succinimide. The same is true for the (e) protons. Thus, we expect four individual signals from these four protons (remember that the left and right hand side of the molecule are not chemically equivalent).

Now that we have covered all the protons in the structure, we can try to identify them in the spectrum. Of course, the phenyl and vinyl groups are easy as you have already stated. The two doublets at $\pu{3.1ppm}$ and $\pu{2.72ppm}$ with integral 1 will correspond to (i) and (j). The signal at $\pu{2.66ppm}$ that appears to be a quintet is probably a septet and corresponds to (g). The methyl group (c) must be a single peak (it does not couple with anything) so it must be in the $1.5{-}\pu{1.6ppm}$ multiplet.

We are still lacking the signals for the methyl groups (f) and (h). These must have an integral of 3 each unless they do not resolve in which case they would have a total integral of 6. Each of these signals must be a doublet (coupling to proton (g)). This narrows it down to the signals at $\pu{1.0ppm}$ and $\pu{1.12ppm}$. Given that these two are distinct, proton (g)’s signal is likely a quartet of quartets rather than a septet, but the coupling constants are likely indistinguishable which will lead to the apparent septet pattern.

Two protons (d) and two protons (e) remain to be identified. Thankfully, the area $1.2{-}\pu{1.6ppm}$ still contains an un-accounted-for integral of 4 protons. The four missing protons will be in this region. However, as they are all chemically (and magnetically) non-equivalent, they should each give a rather complicated ddd coupling pattern. With sufficient time and resources, it would probably be possible to identify the individual signals and coupling constants. However, it is probably not worth your time and I would report that area simply as a multiplet of integral 7 (methyl and bridge protons).

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    $\begingroup$ Why do you give methyl groups f and h the option of being equivalent? They are diastereotopic groups, as you state, inherently different. Rotation is not going to make them equivalent. At low field they may fortuitously appear as a 6H singlet but they are not equivalent. Perhaps we are dealing with semantics here. $\endgroup$
    – user55119
    Oct 11 at 15:22
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    $\begingroup$ @user55119 Mostly because I myself was surprised to realise that they are, indeed diastereotopic. I'm too used to isopropyl groups having just one methyl signal. $\endgroup$
    – Jan
    Oct 11 at 15:27
  • $\begingroup$ One upvote. Take a look here; chemistry.stackexchange.com/questions/48124/… $\endgroup$
    – user55119
    Oct 11 at 16:18

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