Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Organic Spectroscopy => Topic started by: Navidad on May 13, 2020, 12:05:22 PM
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I was given the molecular formula: C9H10O3
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The compound has both a benzene ring and 2 substituents.
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Where have you got stuck? Can you make a decision on the substitution pattern on the ring, for instance?
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I don't really understand how to analyze an H NMR spectrum.
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Where have you got stuck? Can you make a decision on the substitution pattern on the ring, for instance?
I don't really understand how to analyze an H NMR spectrum. I get why the peak at ~3.9 is a singlet, but I don't get the n+1 rule.
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One thing that is a good starting point before even looking at the IR or NMR is to calculate the number of degrees of unsaturation, a.k.a. the hydrogen deficiency index. This will suggest things to look for in each spectrum.
The n+1 rule is a way of saying that (given certain conditions) the number of lines within a signal is n+1, and n is the number of equivalent hydrogen atoms that are coupling with the signal in question. Most coupling encountered in introductory organic chemistry happens because one group of hydrogens is three bonds away from another group of hydrogens: Hx-C-C-Hy. The three bonds are the hydrogen-carbon, the carbon-carbon, and the carbon-hydrogen.
However, hydrogen atoms on aromatic rings often fail to follow perfectly the first-order rules that are taught in introductory organic courses. Therefore, I suggest that you look at other pieces of information available to you in the H-1 NMR, such as the number of peaks and the integrals. You also have a C-13 spectrum that is helpful.
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\Therefore, I suggest that you look at other pieces of information available to you in the H-1 NMR, such as the number of peaks and the integrals. You also have a C-13 spectrum that is helpful.
I know that the integral of the singlet is 5 and that there are 6 hydrogens associated with the neighbors, but where do I go after that?
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I know that the integral of the singlet is 5 and that there are 6 hydrogens associated with the neighbors, but where do I go after that?
What makes you believe that the singlet has an integral of 5?
Have you calculated the degrees of unsaturation?
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I know that the integral of the singlet is 5 and that there are 6 hydrogens associated with the neighbors, but where do I go after that?
Have you calculated the degrees of unsaturation?
My bad, I meant that the degree of unsaturation is 5 isn't it?
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Yes. How many degrees of unsaturation are accounted for by the aromatic ring?
In general integrals are relative. I am tempted to assume that the smallest integral corresponds to 1 hydrogen, and then I would scale the others accordingly.
A good place to start with respect to the C-13 spectrum is to draw an imaginary line near 100 ppm and count up the number of carbons which have a particular hybridization.
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An aromatic ring accounts for 4 degrees of unsaturation.
Also, sorry to be a bother, but how do I use the C-13 spectrum in the way you mentioned? I am unfamiliar with how to use it.
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As a rough guide, everything above 100 ppm is sp2-hybridized, and everything below 100 ppm is sp2-hybridized. I imagine that there are some odd cases at the margins.
So if you have 5 degrees of unsaturation, what can you conclude?
With respect to the H-1 NMR, what do you think is the geometry of the substitution?
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So if you have 5 degrees of unsaturation, what can you conclude?
With respect to the H-1 NMR, what do you think is the geometry of the substitution?
With 5 degrees of unsaturation, I think I have 4 double bonds and 1 ring.
My teacher told me that I have a benzene ring with two substituents, but where those substituents go, I have no idea.
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https://orgchemboulder.com/Spectroscopy/specttutor/arom.shtml
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https://orgchemboulder.com/Spectroscopy/specttutor/arom.shtml
I'm a senior in high school so I'm not really acquainted with aromatic splitting and shifts
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As a rough guide, everything above 100 ppm is sp2-hybridized, and everything below 100 ppm is sp2-hybridized. I imagine that there are some odd cases at the margins.
Which one is sp2- hybridized?
You said the same thing twice.
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https://orgchemboulder.com/Spectroscopy/specttutor/arom.shtml
Did you read this link. Patterns for HNMR of disubstituted benzenes are shown there.
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https://orgchemboulder.com/Spectroscopy/specttutor/arom.shtml
Patterns for HNMR of disubstituted benzenes are shown there.
Sorry for the stupid question, but in the example of the disubstituted benzene, how does that peak show 4 protons? I am confused.
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Sometimes peaks can overlap. On proton spectra, integration helps to notice peak overlapping.
In your proton spectrum, you have twice two superimposed signals with very similar chemical shifts. And you don't have to consider any coupling between protons for this. In my opinion, on the C-13 spectrum, you also have two signals superimposed and additionally covered with a 120 ppm line. The benzene ring substitution pattern is also seen in the IR spectrum. Collecting all this information, it is possible to propose a meaningful structure of the studied compound.
But you have done it.
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Which one is sp2- hybridized?
You said the same thing twice.
Those above 100 ppm. Those below 100 ppm are sp3-hybridized. Carbon-13 spectra tend to spread things over a greater ppm range, which is helpful in this instance.
With respect to the degrees of unsaturation, there is one degree that is not yet accounted for. This suggests that there is one more ring or one more double bond. The IR spectrum and the C-13 NMR are quite helpful in identifying it.
One other point which can be helpful in interpreting C-13 NMR spectra. Carbons with no directly attached hydrogen atoms often show lower intensity than those that have attached hydrogen atoms.