Organic Chemistry - Mechanism of Diol Cyclization
Uploader Comments (wolfram184)
Top Comments
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that dude sounds creeping.. still a useful vid..
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If the mechanism were SN2, a chiral substrate would yield a chiral product no matter which hydroxyl were protonated. Conversely, if the mechaniusm were SN1, a chiral substrate in this case would yield racemic product.
I would like to see the result of a study with a labelled, chiral substrate. The mechanism in the video may be the best, but it doesn't rule out alternative pathways.
All Comments (21)
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i don't get it... why is the video 5 seconds long?? are we supposed to pause it? lolol
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dude, why you talking like a fucking horny queer.
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can i ask what will be the reaction if a cyclic ether opens to form a diol?
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Chemistry, can't live with it, can't live without it. I'm just in a bad mood, don't mind me. Study on.
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there is a pathway to a diprotonated intermediate, but as it would be at a higher energy, it would also have a higher energy barrier than the cyclization step.
molecules, like students, follow the path of least resistance.
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after one OH gets protonated, the second (in principle any subsequent) protonation will be more difficult, as you're trying to add a H+ to a positively charged species. It's all down to the position of the equilibrium favouring a single protonated species.
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good question. im waiting for a reply
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im sorry, just completely ignore what I just said.
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I would imagine it has something to do with the inductive effect of the oxygen.
you can find out which way it goes by considering the secondary alcohol as chiral centre. If the primary alcohol is protonated there will be retention of the chiral centre. If the secundary alcohol is protonated, the cyclic product will became racemic because of Sn1.
ginogigo 4 years ago
Nicely put! The stereochemistry, while not well defined in the schematics, is an important consideration as it would allow for assessment of Sn1 versus Sn2 as you have pointed out. Thanks for your input!
wolfram184 4 years ago
The protonation and substitution of the secondary alcohol should be favoured because secondary carbocations are more stable than primary carbocations, right?
andreasryden 4 years ago
Yes that is correct that secondary carbocations are more favorable than primary. However since the mechanism does not involve formation of a carbocation intermediate, this will not be a consideration. You make a good point, as carbocation intermediates may be observed under appropriate conditions (secondary & tertiary carbocations, polar aprotic solvents). Thanks for your feedback!
wolfram184 4 years ago
Cool Thank you
cruxis7856 4 years ago
you're very welcome, did you find the video useful? Thanks!
wolfram184 4 years ago