[aldoxime_amine]

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?

[azmanam]

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.

[azmanam]

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?

[Squirmy]

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

[azmanam]

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)

[aldoxime_amine]

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.

[aldoxime_amine]

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?

[azmanam]

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)

[Dolphinsiu]

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.

[aldoxime_amine]

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.

Thanks for the article.

[ziyadsara]

I think Your case is similar to that in case of 2,4 pentadione( kindly see the attached files)

[sanderol]

Would H-bonding just lead to broadning of your peaks, instead of giving seperate peaks?

[azmanam]

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.

[sanderol]

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?