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Offline Rutherford

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NMR spectra problem
« on: August 12, 2012, 01:04:05 PM »
Why is in the following spectrum http://arrhenius.rider.edu/nmr/nmr_tutor/selftests/h22/h22_fs_1.html the solution 2,4-dichloro acetophenone. Couldn't it be 3,4-dichloro acetophenone, too?
« Last Edit: August 12, 2012, 06:50:22 PM by Arkcon »

Offline sjb

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Re: MNR spectra problem
« Reply #1 on: August 12, 2012, 02:32:45 PM »
Why is in the following spectrum http://arrhenius.rider.edu/nmr/nmr_tutor/selftests/h22/h22_fs_1.html the solution 2,4-dichloro acetophenone. Couldn't it be 3,4-dichloro acetophenone, too?

Consider the splitting you'd get. 3,4-dichloro is more likely to have a singlet due to the proton at C-2 (not sure why the peak at δ7.34 is "labelled" as a singlet, tbh)

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Offline Rutherford

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Re: MNR spectra problem
« Reply #3 on: August 12, 2012, 03:06:28 PM »
Why is in the following spectrum http://arrhenius.rider.edu/nmr/nmr_tutor/selftests/h22/h22_fs_1.html the solution 2,4-dichloro acetophenone. Couldn't it be 3,4-dichloro acetophenone, too?

Consider the splitting you'd get. 3,4-dichloro is more likely to have a singlet due to the proton at C-2 (not sure why the peak at δ7.34 is "labelled" as a singlet, tbh)
Both compounds should have singlets.
http://www.sigmaaldrich.com/spectra/fnmr/FNMR001843.PDF
I can't see on this spectrum well, there is no zoom.

Offline AWK

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Re: MNR spectra problem
« Reply #4 on: August 12, 2012, 03:13:09 PM »
Quote
I can't see on this spectrum well, there is no zoom.
This is qood quality PDF. You can zoom it even 10 times.
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Offline Rutherford

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Re: MNR spectra problem
« Reply #5 on: August 12, 2012, 03:39:51 PM »
Okay I see two singlets and two doublets, same amount of both were on the link I give, they are only on a little different shifts, but how could I guess what was the correct compound then from such a small difference?

Offline orgopete

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Re: NMR spectra problem
« Reply #6 on: August 13, 2012, 12:26:53 AM »
If I had this spectrum and had to determine its structure from its molecular formula, I would have a difficult time to decide. If I looked up comparable spectra (e.g. authentic IR), I could decide.
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Offline Rutherford

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Re: NMR spectra problem
« Reply #7 on: August 13, 2012, 04:32:30 AM »
When I was trying to solve it I was left with the two possibilities I mentioned and I couldn't decide which would be the correct one. In the answer there was only 1 correct compound so I was confused.
If I had this spectrum and had to determine its structure from its molecular formula, I would have a difficult time to decide. If I looked up comparable spectra (e.g. authentic IR), I could decide.
Yes, with an IR specter it would be an easier choice.
Anyway, thanks for the help.

Offline AWK

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Re: NMR spectra problem
« Reply #8 on: August 13, 2012, 05:15:58 AM »
2',4'-dichloroacetophenone shows completely different proton NMR spectrum
Search:
http://riodb01.ibase.aist.go.jp/sdbs/cgi-bin/cre_index.cgi
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Offline Rutherford

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Re: NMR spectra problem
« Reply #9 on: August 13, 2012, 05:46:31 AM »
I am not sure, I didn't know how to use the application on the link you gave, so I tried in chemdraw and the results are attached. Multiplets can't be distinguished, however, they don't matter because they are same in both molecules. From the picture, I see that the shifts are slightly different.

Offline orgopete

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Re: NMR spectra problem
« Reply #10 on: August 13, 2012, 09:30:18 AM »
2',4'-dichloroacetophenone shows completely different proton NMR spectrum

Agreed. I didn't see that as the question. If you had these two spectra and had to assign one to the 2',4' and the other to the 3',4'-dichloroacetophenone, explain how you would do it? The question is not about looking them up, it is about how would you unequivocally assign them?
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Offline Rutherford

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Re: NMR spectra problem
« Reply #11 on: August 13, 2012, 10:09:09 AM »
No, I had one spectra and I gave it link in the first post, just had the molecular formula and to determine the structural formula of the compound. I still don't understand where is the big difference.

Offline AWK

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Re: NMR spectra problem
« Reply #12 on: August 13, 2012, 01:58:17 PM »
Just click "I agree the disclaimer..."
then put correct name: 3',4'-dichloroacetophenone or 2',4'-dichloroacetophenone
and click : search (down left corner
Next page gives you informations what typr of spectra are accessible

SDBS No   Molecular Formula   Molecular Weight   MS   CNMR   HNMR   IR   Raman   ESR       Compound Name
3065           C8H6CL2O   189.0       Y   Y   Y   Y   Y   N       2',4'-dichloroacetophenone

or
SDBS No   Molecular Formula   Molecular Weight   MS   CNMR   HNMR   IR   Raman   ESR       Compound Name
6087           C8H6CL2O   189.0       N   N   Y   Y   N   N       3',4'-dichloroacetophenone

and click on "Y" under HNMR

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Offline Rutherford

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Re: NMR spectra problem
« Reply #13 on: August 14, 2012, 05:50:28 AM »
I tried there, it's better in quality, but there is no information about the singlets, doublets...
Still, I don't understand how could I just by looking at the spectra in my first post determine the structural formula of the compound. I don't see any big difference.

Offline fledarmus

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Re: NMR spectra problem
« Reply #14 on: August 14, 2012, 09:42:23 AM »
It really depends a lot on how much calculating your teacher expects you to do. There are a few ways of looking at substituent effects on the NMR of protons attached to a benzene ring, depending on how deep you get.

The first is exactly what you did, and it is an excellent start and usually as far as you need to look. Recognize splitting patterns among the protons, and use that and the integration to determine which carbons of the benzene ring are substituted. It is very easy to recognize a 1,4; 1,3; or 1;2 substituted benzene ring based on splitting patterns, and in this case you recognized a 1,2,4 substituted benzene ring. You just can't tell what groups are where. You do know, however, that the semi-singlet will be the proton by itself in between two substituents (4 position), that the semi-doublet will be para to the semi-singlet (1 position), and that there will be a slightly more complicated peak, sort of a semi-doublet of doublets that is next to it (2 position)

Now you have to look at the groups themselves, and what effect they will have on protons at various positions around the ring. These are, fortunately, basically additive. If you pick a proton on a benzene ring, there is a Z value for substituents at each point on the ring that will give you a fairly good prediction of where the proton will be. If there are no substituents at all, the proton will be at 7.36 ppm, which is the chemical shift for benzene.

Electron withdrawing substituents, like carbonyls, will pull electrons away from the benzene ring, reducing the effect of the ring current. Due to resonance effects, this is stronger in the ortho and para positions. Protons on a benzene ring which contains a carbonyl group will have a higher chemical shift than 7.36, and the ortho and para protons will be higher than the meta protons. Electron donating groups, like hydroxyls and amines, have the opposite effect, strengthening the ring current and reducing the chemical shift of protons on the ring, but it is still most pronounced in the ortho and para positions. The halogens fit somewhere between the two extremes, having little effect on chemical shift, and due to back bonding, the effects on ortho, para, and meta positions aren't always as clear cut. For chlorine, for example, the effect is weak in the para position, weaker in the meta position, and even weaker in the ortho position.

So let's look at the two possibilities. For the 3,4-dichloroacetophenone, you would expect the 2 proton (semi-singlet) to be raised significantly by the carbonyl in the ortho position, lowered slightly by the 4 chloro in the meta position, and lowered just a trace by the 3 chloro in the ortho position. The 5 proton (semi-doublet) would be raised slightly by the meta ketone, and lowered slightly by an ortho chloro and a meta chloro. The 6 proton (more complicated semi-dd) would be raised significantly by the ortho ketone, and lowered by the para and meta chloro. So you would expect the 6 proton (dd) to be highest, then the 2 proton (s), then the 5 proton (d).

For the 2,4-dichloroacetophenone, you would expect the 3 proton (semi s) to be raised slightly by the meta ketone, and lowered a tiny bit by the two ortho chlorines. The 5 proton (semi-dd) is raised slightly by the meta ketone, lowered a tiny bit by the ortho chloring, and lowered more by the para chloring. The 6 proton (semi-d) is raised quite a bit by the meta ketone, and lowered somewhat by the two meta chlorines. That puts the 6 proton highest (d), then the 3 proton (s), then the 5 proton (dd).

In your spectrum, you show the semi-d at around 7.5 ppm, the semi-s  around 7.4, and the semi-dd around 7.3. This is the order found in the second example (6-3-5), so my guess would be that this is the 2,4- rather than the 3,4-dichloroacetophenone.

Of course, this whole analysis is easier if you can actually attach some numbers to it. I like this site - http://www.chem.wisc.edu/areas/reich/handouts/nmr-h/hdata.htm - scroll down on the left side to "benzene - substituent effects". Using this table, you can calculate (for 2,4-dichloroacetophenone) the 6 proton to be about 7.82, the 3 proton to be about 7.50, and the 5 proton to be 7.31.


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