Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Organic Spectroscopy => Topic started by: Sub on September 27, 2019, 05:09:48 PM
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I'm trying to match up the peaks of (1S) Borneol. But I'm getting confused on the splitting for some of these peaks. Specifically the A proton. I've always known it to be that a proton's signal will be split if there are protons 3 atoms away. The only atoms within this distance would be the B/I protons which seem to be equivalent so I figured I wouldn't need to do a branching diagram. But apparently the A hydrogen matches up with the 4.00 ppm which has 8 peaks. Using N+1 rule why is the splitting not a triplet? If N+1 doesn't work how can I figure out which other protons are splitting hydrogen A?
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Why do think that the B and the I hydrogen atoms are equivalent?
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Why do think that the B and the I hydrogen atoms are equivalent?
After working through it a bit more I figured that they aren't actually equivalent because one will have OH cis to it while the other will be trans. Would it be right to say that none of the hydrogens are equivalent?
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I am a little rusty at this sort of analysis, but I am inclined to say that they are not.
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The only equivalent hydrogens are the 3 on each of the methyls.
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They are not equivalent due to the rigidity of the camphane ring.
Axial and equatorial protons are magnetically equivalent, due to the rapid flipping of the “chair ↔ boat ↔ chair” conformers, which is impossible for the bridged camphane ring.
PS: Besides, magnetic inequivalence of axial/equatorial protons of flexible rings like cyclohexane, is visible at very low temperatures (say -70oC) that the conformers flipping rate is significantly decreased.