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Is my drawing correct? Thanks in advance!

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  • $\begingroup$ When you say "peaks", you count a doublet as one, right? You should call it "signal" then. In that case, you are right, there are five different chemical environments for protons in the ortho and meta isomers, and three in para. $\endgroup$ – Karl May 17 '16 at 10:47
  • $\begingroup$ Ok! lemme edit my question @Karl $\endgroup$ – mystreet123 May 17 '16 at 10:57
  • $\begingroup$ @Karl is it also true that the indane-1 3-dione derivative has 5 signals instead of 3? as there is no vertical line of symmetry $\endgroup$ – mystreet123 May 17 '16 at 11:07
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    $\begingroup$ Correct, the left and right side are not chemically equivalent. But the signals will again be complex and overlapping. I don't think anyone can predict any of the spectra by hand. Those are tough cases. $\endgroup$ – Karl May 17 '16 at 11:29
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In a word, no, your drawings are not correct. And incorrect enough for me to have a rant about NMR teaching at the introductory level.

Some background knowledge before tackling your molecules.


'Environments' is a vague term, often misused and misunderstood, unless you specify what type of environment: chemical or magnetic. As an argument about chemical environments alone, you need only look at the symmetry of the molecule to make a judgement. This can be understood and determined by someone who has never heard of NMR. It might also be an appropriate assessment if you do not intend to make any interpretation of the splittings of peaks from coupling. If you want to look at it from an NMR perspective (and you ARE asking as an NMR question), then you have to consider magnetic environments also. In all four molecules you have drawn, every single proton is in a separate environment.

Equivalence of the proton environments can be assessed on whether a particular nucleus is identical to another site, based on a substitution test. If you substitute, in turn, one of the 1H nuclei in the molecule below, you end up with identical molecules, therefore the protons are chemically equivalent. They will have the same chemical shift.

enter image description here

An additional element of equivalence needs to be considered also, and this relates to magnetic equivalence. Nuclei that are chemically equivalent may not couple to other nuclei in an identical manner, and these nuclei are therefore magnetically non-equivalent. Although they hgave the same chemical shift, their coupling patterns are very complicated, and usually involve coupling to each other.

enter image description here

In the case above, the two protons shown in red are in different environments; the same chemical environment, but different magnetic environments. Without understanding this key concept, you would not be able to rationalise the appearance of the spectrum, which actually looks like:

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Some standard couplings that you should commit to memory for aromatic rings:

  • 3JH-H ortho couplings are typically 7-8 Hz
  • 4JH-H meta couplings are typically 2-3Hz
  • 5JH-H para couplings are typically <1Hz, and are not commonly observed.

Splittings in a spectrum can be reported as either

  • observed splitting pattern so a doublet of doublets with similar couplings is likely to be reported as a triplet
  • expected splittings based on molecular structure. So a reported doublet of doublets with similar couplings may look like a triplet.

Although both methods are used in the literature, the strong preference is for the latter, as it provides structural information, and a means for justifying the appearance of the spectrum. It is much more difficult to start from the appearance of a spectrum and work backwards to determine/justify structure. A triplet coupling implies coupling to 2 identical nuclei


...And you're back in the room.... So, coming to your molecules......

  • the 1,2 disubstituted and 1,3 disubstituted case

You can work through predicting what you would expect based on known typical couplings. Do this for each proton. A couple of workings are shown, and the theoretical expectation of these peaks is shown.

enter image description here

Now, unfortunately, this particular group of substituents are both strongly electron withdrawing, and what you would probably get is compounded complications though strong coupling as the chemical shift differences are not sufficient enough to allow simple first order analysis ie a dog's breakfast, as shown below.

enter image description here

Having spent time determining the number of chemical and magnetic environments is pointless because the spectrum looks so complicated that it looks nothing like you would expect, although this is hardly presumed knowledge for NMR101.

A more suitable choice of substituents, -NH2 and -COOH or -CH3 and -COOH, would have provided a much better example.

  • 1,4 disubstituted and the indanedione thingy

There are two chemical environments directly on the ring, but they are magnetically non-equivalent. So again, time spent trying to predict what the splitting might look like leads to frustration because we get a second order AA'BB' system that, although some might try to convince themselves looks like a doublet, is in fact a much more complicated spectrum (like in the case for furan above). Both give textbook splitting patterns that indicate a 1,4-disubstituted or symmetrical 1,2-disubstituted rings, but nothing like what you might expect if you trying to determine expected splittings for the first time.

And so the correct answer for how many proton 'environments' in these compounds is:

enter image description here

.End Rant.

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You'll see no triplets, because there are no two chemically equivalent protons in ortho and meta.

I guess you might be right in the case of the para-isomer, because the 5J coupling through the ring might be too weak to see (it is not zero, 3/5 are not magnetically equivalent, neither are 2/6).

In the meta isomer you will see some 4J coupling between position 2 and 4/6, and 5 will definitely show a dd signal.

ortho: 3+6 will each be dd (3J+4J), and 4+5 will each be ddd (2x3J+4J). Plus you'll see a lot of second order coupling, because the CS difference is so small. That means the dd will not look like a regular dd.

("dd" means doublet of doublets, i.e. when a proton has two different neighbours. dd can look like a triplet, when the two coupling constants are (nearly) identical.)

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  • $\begingroup$ How about low resolution nmr? Sorry I'm still a high school student I don't quite understand about 4J coupling and dd signal... $\endgroup$ – mystreet123 May 17 '16 at 10:06
  • $\begingroup$ At 60 MHz or so, second order coupling will likely melt all peaks into one multiplet. The normal rules for doublets, triplets etc. only work if the signals of each coupling group are well separated in the spectrum. 0.5 ppm are only 30 Hz at 60 MHz. That is not a lot more than the regular 3J coupling in the aromatic ring. $\endgroup$ – Karl May 17 '16 at 10:19
  • $\begingroup$ Can you tell me how many proton environments there are for each isomer please? @Karl $\endgroup$ – mystreet123 May 17 '16 at 10:21
  • $\begingroup$ That you can work out yourself, im sure? Hint: Only the para isomer is symmetric. $\endgroup$ – Karl May 17 '16 at 10:25
  • $\begingroup$ So is it 5 for ortho meta isomers and 3 for para isomer?:) @Karl $\endgroup$ – mystreet123 May 17 '16 at 10:26

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