Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: aldoxime_amine on April 21, 2009, 02:17:42 PM
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1.The 13C-NMR spectrum of a commercial sample of 2,4-pentanediol shows not three but five peaks, at 23.3, 23.9, 46.5, 64.8, and 68.1 ppm. Why?
I guess H-bonding is playing spoilsport here, but can't think beyond that.
2. Identify each of the following two isomers of molecular formula C20H18O from the 1H-NMR data given:
Isomer P: 2.23 (singlet, 1H), 3.92 (doublet, 1H, J = 7 Hz), 4.98 (doublet, 1H, J = 7 Hz), 6.81 (singlet, 10H), 6.99 (singlet, 5H).
Isomer Q: 2.14 (singlet, 1H), 3.55 (singlet, 2H), 7.25 (broad peak, 15H).
These two isomers could be distinguished by a single chemical test. What is it?
The 6.81 singlet in P and 7.25 in Q is throwing me off. Can anyone push me in the right direction?
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Re 1) Good question. Nuanced, but good question. You're right, H-bonds are going to be key. My initial thought was mixture of diastereomers... and that may still play a part, but I think H-bonding's a better answer. Maybe try drawing some structures with H-bonds and try describing them in words... see if anything jumps out at you.
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Re 2) unless there are two sets of doublets down there, I don't think I ever take splitting into account between 6.5-8. Do you know what protons usually resonate in that region - which might lend itself to having integrations in multiples of 5?
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1) I'd think the mix of diastereomers would be more likely, but then why would you have 5 and not 6 signals?
Here's Aldrich's spectrum of the mix: http://www.sigmaaldrich.com/spectra/fnmr/FNMR001680.PDF
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1) I'd think the mix of diastereomers would be more likely, but then why would you have 5 and not 6 signals?
Yeah, after doing a bit of research (http://dx.doi.org/10.1021/jo026652i, http://dx.doi.org/10.1016/0040-4020(96)00373-0) (don't worry, op, they're not spoilers), it seems the mix of diastereomers is the primary 'right' answer. but I think H-bonding is still important... and it answers why only 5 signals. So let's assume it is a mix of diastereomers. WHY are the carbon atoms still not magnetically equivalent. there's a hint in white text below. I don't know if it's giving away too much, so I'll hide it from op. highlight it to read.
(edit: on second thought, I retract my hint because in retrospect I think it's a bad hint. Let's just say I think the constitutional isomer given was chosen for a reason)
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I am not sure if I understood much of what was given in that link. But I have attached a diagram. Could it be the right reason?
I know it won't be an exact chair, just guessing around.
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Do you know what protons usually resonate in that region - which might lend itself to having integrations in multiples of 5?
Monosubstituted phenyl...
So, Q is probably 2,2,2-triphenyl ethanol and P 1,2,2 triphenyl ethanol.
In P, the alpha H (to OH) gives 4.98?
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yes, and yes. The papers were just showing that pure meso diol and pure "trans" diol give a single 13C spectra with only 3 lines, so the 5 lines is mainly due to the mix of diastereomers. I think the 6-membered H-bonded conformation is important, because it shows how the middle carbon might be the same in both. I think your chair form of the meso diol is wrong... not that it matters, but the di-equatorial is probably favored.
Those are the structures I drew for 2)
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Hydroxylic peak in proton nmr ranges from 1 to 6 ppm. I do think 2.23 in P and 2.14 in Q would be more likely to be hydroxylic peaks.
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I think your chair form of the meso diol is wrong... not that it matters, but the di-equatorial is probably favored.
Yeah, I drew it wrong. The -Me should be equatorial; sorry the diagram doesn't make sense. :P
Thanks for the article. :)
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I think Your case is similar to that in case of 2,4 pentadione( kindly see the attached files)
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Would H-bonding just lead to broadning of your peaks, instead of giving seperate peaks?
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Would H-bonding just lead to broadning of your peaks, instead of giving seperate peaks?
perhaps in the 1H spectrum, prolly not for the 13 spectrum.
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Would H-bonding just lead to broadning of your peaks, instead of giving seperate peaks?
perhaps in the 1H spectrum, prolly not for the 13 spectrum.
Yes Its nothing I've seen but you can't define a hydrogen bond as a certain state. It's always a summation of several states. At least when its not locked.
But if you talk about H-bonding do you mean intramolecular?