Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Organic Spectroscopy => Topic started by: aspiringphysician on June 19, 2012, 04:41:47 PM
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I am having a really difficult time understanding the following question:
"The following NMR spectrum contains a mixture of both stereoisomers (E and Z) of β-bromostyrene. Since the integration of a peak in a 1H NMR spectrum is proportional to the number of hydrogens in resonance, if one knows the number of protons represented by a particular integral value for more than one compound, one can use those values to determine the relative amounts of those compounds in the sample. For example, a 1H NMR spectrum is run on a sample containing two different methyl ester compounds. Ester A gives a peak for the methyl group at 3.9 ppm with an integration of 30, while ester B gives a peak for the methyl group at 3.8 ppm whose integral value is 10. Therefore, the ratio of A:B must be 30:10 (3:1) or 75% A and 25% B. Use the integration to determine the relative amount of the two isomers."
I have attached a picture of the NMR spectrum...where do I begin??? :-\
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I am having a really difficult time understanding the following question:
"The following NMR spectrum contains a mixture of both stereoisomers (E and Z) of β-bromostyrene. Since the integration of a peak in a 1H NMR spectrum is proportional to the number of hydrogens in resonance, if one knows the number of protons represented by a particular integral value for more than one compound, one can use those values to determine the relative amounts of those compounds in the sample. For example, a 1H NMR spectrum is run on a sample containing two different methyl ester compounds. Ester A gives a peak for the methyl group at 3.9 ppm with an integration of 30, while ester B gives a peak for the methyl group at 3.8 ppm whose integral value is 10. Therefore, the ratio of A:B must be 30:10 (3:1) or 75% A and 25% B. Use the integration to determine the relative amount of the two isomers."
I have attached a picture of the NMR spectrum...where do I begin??? :-\
Firstly identify the peaks beloning to the E and Z isomers in your spectrum. Look for the C=C proton signals.
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Firstly identify the peaks beloning to the E and Z isomers in your spectrum. Look for the C=C proton signals.
Okay, so does the peak at 7.1 correspond to the E isomer and the one at 6.4 correspond to the Z isomer? If so, since both of their integrations are 1, does that mean the sample is a 1:1 mixture (50% E and 50% Z)?
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Firstly identify the peaks beloning to the E and Z isomers in your spectrum. Look for the C=C proton signals.
Okay, so does the peak at 7.1 correspond to the E isomer and the one at 6.4 correspond to the Z isomer? If so, since both of their integrations are 1, does that mean the sample is a 1:1 mixture (50% E and 50% Z)?
The coupling constants should be different, E being larger than Z.
These two look the same to me. But it looks like a 1:1 mixture.
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Thank you very much! :)
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Okay, going back to this topic again... ::)
The NMR spectrum in my first post in this thread corresponds to the reaction pictured below. I know that in this reaction, an E1 mechanism produces both E and Z β-bromostyrene, while an E2 mechanism produces just the Z isomer of β-bromostyrene. Since the NMR spectrum has proton peaks corresponding to both isomers, it seems that the reaction proceeds through an E1 mechanism, but what really confuses me is that the reaction is occurring in a polar aprotic solvent (acetone) with a strong base (K2CO3 is a strong base, isn't it???), so that definitely favors an E2 mechanism, correct?
Because of this confusion, I am struggling to figure out how to answer the following question on my lab report: Based on the NMR spectrum, is the stereochemistry of this reaction consistent with an E1 or an E2 mechanism?
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Well, it turns out, according to my professor, that the NMR spectrum does NOT in fact contain both stereoisomers of β-bromostyrene, contrary to what the lab manual stated. ::) No wonder I was so confused! The two α-proton peaks correspond only to the Z-isomer, the major product of this reaction, which is formed through an E2 mechanism.