Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: workingundergrad1896 on January 18, 2019, 06:22:44 PM
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Hi,
I recently performed a DIBAL reduction of a lactone to lactol (5 membered). Both compounds contain a methyl group substituted onto the ring. In the lactone the methyl group appears as a singlet at 1.40 ppm. After performing the DIBAL reduction, a new chiral centre is introduced onto my molecule. In the spectrum for my lactol this methyl singlet is now split into two peaks at 1.40 and 1.18, where the integral of each peak is 1.5. My isolated lactol is pure according to TLC and IR as well as NMR.
Is this because my compound is a racemic mixture of the two possible enantiomers? (surely it can't be enantiomers though as the methyl groups in the two enantiomers couldn't have such a large difference in shift could they?? or could conformation differences in the ring actually cause such a significant change in shift due to different environments) Or is it due to some other effect, such as long range coupling to diastereotopic protons on my lactol ring?. The peaks are pretty much the same height asides from a roofing effect leaning upward toward the peak at 1.18.
EDIT: after further reading I've just realised something. Due to a fault on the 300 MHz spectrometer I ran my spectrum on the 600 MHz spectrometer. I think this stronger magnetic field may have resolved some longer range coupling? Hence the roofing effect??
Thanks for any help
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Posting an image of the spectrum would be REALLY helpful. NMR is hard enough without having to try to build the spectrum in my head.
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I am not sure that I understand your question. Are the two methyl groups enantiotopic?
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As far as could be understood without knowledge of the exact structure of your lactone:
1). A singlet methyl shift means hanging on a secondary lactone group, which also holds another substituent (aromatic?) and probably, located in α-carbonyl position.
2) Due to the carbonyl reduction, you have a mixture of diastereoisomers methyl-lactols.
3). Apart the α-neighboring groups (Shoorley’s rule), 1H-NMR proton shift also depends on the β-neighboring, functional groups and notably the β-hydroxyl groups, which are axial and equatorial and thus, a methyl shift difference of 0.22 ppm seems reasonable due to the influence of molecular symmetry in the 1H-NMR proton shift.
4). Given that your lactol is pure, the methyl shift = 1.40 ppm of both methyl-lactone and cis-methyl-lactol is rather a coincidence.
5). An alternative explanation could be based on the long distance influence of the hydroxyl group in the methyl proton shift, which is different for the cis-, trans- isomers of the above 3-methyl-(γ-)butyrolactol, as well as of 5-methyl-(γ-)butyrolactol diastereoisomers.
6). But all above are just hypotheses without knowledge of the exact structure of your lactone.
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... the spectrum for my lactol this methyl singlet is now split into two peaks at 1.40 and 1.18, where the integral of each peak is 1.5 ... Is this because my compound is a racemic mixture of the two possible enantiomers?
Absolutely not. Unless you specifically added a chiral shift reagent, it is impossible to distinguish enantiomers by NMR - they appear identical.
As pgk said, it sounds like (although I don't know your full structure) a mixture of diastereomers - ie if the reduction is non-stereoselective then you will obtain a mixture of syn (where the OH and the Me are on the same side of the ring) and anti (opposite sides) products. It isnt unusual for diastereomers to overlap on TLC, so the fact that the TLC appears "pure" doesn't necessarily rule this out.
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1). “A singlet methyl shift means hanging on a secondary lactone group, which also holds another substituent…..”
If this sounds incomprehensive, just replace it by:
“A singlet methyl shift means a methyl group that is hanged on a carbon atom of the lactone ring and which also holds another substituent….”
2). Yes indeed, syn-/anti- nomenclature is more correct but cis-/trans- nomenclature is still used for the description of same/opposite sided disubstitution of cycloaliphatic and heterocycloaliphatic compounds.
I fully apologize for any possible confusions.
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The new OH-group will flip up and down very fast, its a equilibrium. This will not be visible on a TLC-plate.
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Structural particularities that change molecular energy (e.g. intramolecular hydrogen bonds, substituents orientation-depended variations of dipole moment, etc.) may freeze the up/down flipping (conformers interconversion). But even so, TLC separation is dubious due to the short length (short development distance) of the TLC plate.