For $\ce{C11H13NO}$, I have the following $\ce{^1H}$ NMR data:

shift (ppm)

  • 3.00 (6H, s)
  • 6.55 (1H, dd, $J=15.5, \pu{9 Hz}$)
  • 6.70 (2H, d, $J=\pu{8Hz}$)
  • 7.35 (1H, d, $J=\pu{15.5Hz}$)
  • 7.45 (2H, d, $J=\pu{8Hz}$)
  • 9.60 (1H, d, $J=\pu{9Hz}$)

I also know there are two absorptions from IR at $1600$ and $\pu{1660cm^{-1}}$.

From the IR it seems reasonable to suggest an aromatic amide ($\pu{1660cm^{-1}}$). The shift of $\pu{9.60ppm}$ is huge and I think this could really only come from a proton connected to the carbon of a carbonyl. The 6H peak could be two equivalent $\ce{CH3}$ groups, and as it is singlet, I thought they could hang off the nitrogen of the amide, then it wouldn't split. The two sets of 2H atoms I thought could be from the aromatic ring. The problem is piecing all of it together!

The closest structure I have is this:


but I know this is wrong as it lacks a $\pu{9.60ppm}$ shift. How can I arrive at the correct structure?

  • 2
    $\begingroup$ Well, those with a 15.5 Hz coupling are probably on an alkene trans to one another. And in the 6–8 ppm region there are only 4 other protons. So... <draw your own conclusion here>. The next step is then to determine the substitution pattern on the ring. Is it ortho, meta, or para - which substitution pattern is consistent with two doublets? $\endgroup$ Sep 26, 2017 at 17:58
  • 1
    $\begingroup$ ok so there is 4! and it would be para substituted $\endgroup$
    – gamma1
    Sep 26, 2017 at 18:06
  • 2
    $\begingroup$ Exactly. Now, for the 9.60 ppm proton, you've already correctly said that that's an aldehyde. So, that NMe2 group in your proposed structure should be somewhere else. $\endgroup$ Sep 26, 2017 at 18:34
  • 1
    $\begingroup$ Excellent point, I missed that. The 9 Hz is clearly a coupling to the aldehydic proton. $\endgroup$ Sep 26, 2017 at 18:59
  • 1
    $\begingroup$ Once you figure out it, you might consider posting your solution and your logic as an answer. This may help another person who is struggling with a similar problem. $\endgroup$
    – Zhe
    Sep 26, 2017 at 19:25

1 Answer 1


It always helps to start with the ‘blooming obvious’: the signal at $\pu{9.60ppm}$. This is obviously an aldehyde. It also obviously couples ($^3J=\pu{9Hz}$) with the signal at $\pu{6.55 ppm}$. This proton is either in the range of a dihalogenated carbon, an electron-rich aromatic system or an alkene. We can rule out all possibilities except for the alkene: it can’t be aromatic because there would be no $^3J$ coupling to a non-aromatic proton (there has to be at least one quarternary carbon in-between).

Moving on, we have a very high coupling constant of $J = \pu{15.5 Hz}$. While it would, in theory, be possible that this arises from a geminal $^2J$ coupling of two diastereotopic protons (although those are typically on the order of $\pu{12Hz}$, so slightly less), the more reasonable explanation is an (E)-configured alkene. If it were a $\ce{CH2}$ group, we not only would get into trouble explaining the extremely high chemical shift but also why the aldehyde only couples to one of those protons. Thankfully, an alkene is what we already guessed from the chemical shift. So we now have three protons out of the way and a $\ce{?-CH=CH-CH=O}$ substructure.

Now we might choose to take double bond equivalents into play. The sum formula $\ce{C11H13NO}$ evaluates to 6 double bond equivalents. We have used two for the alkene and the aldehyde. The four remaning double bond equivalents together with the region the chemical shifts are in scream substituted benzene. We may now take a look at the actual integrals and couplings of the protons we deem can still be aromatic (all except for the 6 singlet protons at $\pu{3.00ppm}$). We have two sets of two and each are an apparant doublet. Therefore, we must have a disubstituted with a para-substitution pattern.

The two fragments we have so far, a $\ce{C6H4}$ and a $\ce{C3H3O}$ one, add up to $\ce{C9H7O}$, meaning that $\ce{C2H6N}$ is still missing. We should also note that all these six hydrogens are chemically equivalent. You can play around with this, but the only way to have six chemically equivalent hydrogens and a nitrogen in a group that connects only once is in a dimethylamino group. Thus, the final structure is:

*para*-dimethylaminocinnamaldehyde with the proton shifts highlighted.

Finally, allow me a few words on why your structure is incorrect apart from the missing proton corresponding to $\pu{9.60ppm}$:

  1. The phenyl protons would show up as a set of three (so far so good). However, they would be a doublet and two triplets (one of the triplets corresponding to the single proton). Your spectrum does not contain any apparant triplets. Also, monosubstituted phenyl groups tend to have much more similar shifts to the point where they often only give a $\ce{5H}$ multiplet.

  2. The six protons on the $\ce{NMe}$ groups must be equivalent. Since they are obviously on different carbons, there must be some rotation that maps one set onto the other. In my structure, you can easily rotate the $\ce{Ph-N}$ bond to transform one group onto the other. In your structure, rotation around the $\ce{C(O)-N}$ bond is restricted due to the partial double bond character. Therefore, the two methyl groups would show up as two separate signals of three hydrogens each. Look up the spectrum of DMF (dimethylformamide) to confirm this.

Final note: I know nothing about IR and I do not consider it a helpful tool in structure determination. Thus, I cannot comment on the IR frequences and what they can, would or cannot correspond to.


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