March 28, 2024, 08:23:46 AM
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Topic: Have you seen such axial bond in a substituted cyclohexane chair conformation?  (Read 7821 times)

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Offline BillJames

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Hey all, this diagram has bedazzled me:

Accordingly, the axial bond (i.e. the Br) is pointing "towards the left" when in such conformation, but from what I have found / googled, such axial bond points "to the right". What's going on?

Bill

Offline azmanam

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Both the Br and O- are equatorial bonds, not axial.  'to the left' and 'to the right' are totally relative based on the arbitrary orientation of your chair.  In this arbitrary orientation, both the O- and Br are pointed 'to the right.'  But rotate, oh, 10o about the vertical axis and the Br might just point 'to the left.'

Build a model, do a chair flip such that the O- and Br are actually axial (in this orientation, O- would be 'up', and Br would be 'down'), and everything will be right with the world again.  That O- as drawn will not participate in that SN2 attack; it can't reach the C-Br σ* orbital.  But it can after a chair flip to the axial positions.
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Offline BillJames

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Oopsy, I did mean equatorial rather than axial (blond moment...)

Thanks!

Offline BillJames

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Would someone care to tell me how would the axial bonds stick up / down on the Carbons where Bromine and Oxide are attached respectively? I am trying to figure out how Br bond and Oxide bond are not trans / anti.

Bill

Offline Dan

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URL=http://img526.imageshack.us/my.php?image=weirdconformationnb1.jpg][/URL]

I am trying to figure out how Br bond and Oxide bond are not trans / anti.

They are trans. In that picture, the O- is "up" and the Br is "down".
My research: Google Scholar and Researchgate

Offline BillJames

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Hmmm... Then why can't the molecule posted self-react to form an epoxide? The lecturer said that it is due to the non-anti position.

Unless I have missed out something from the original slide?

Offline azmanam

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see above

Quote
Build a model, do a chair flip such that the O- and Br are actually axial (in this orientation, O- would be 'up', and Br would be 'down'), and everything will be right with the world again.  That O- as drawn will not participate in that SN2 attack; it can't reach the C-Br σ* orbital.  But it can after a chair flip to the axial positions.

What I was describing is depicted in the green box in your image.  The posted conformation can interconvert with the molecule in the green box.  The molecule in the green box is now antiperiplanar and can close to the epoxide.

To contrast, the cis diastereomer cannot form the epoxide.  The cis diastereomer also has two chair forms (the two on the right), but the O- cannot add into the C-Br σ* orbital, and cannot form the epoxide.
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Offline BillJames

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Thanks for the prompt reply. Could you please further elaborate on this point that I have missed earlier?

"That O- as drawn will not participate in that SN2 attack; it can't reach the C-Br σ* orbital"

Bill

Offline azmanam

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SN2 reactions require the nucleophile (O-)to approach the carbon atom bearing the leaving group (the electrophile, Br) in a 'backside attack.'  the trajectory of the nucleophile needs to be approximately 180o from the C-LG bond.  If we number the compounds in the middle row of your slide 1 through 4, in compound 1, the O- approaches the C-LG bond at close to 90o.  compound 2 is the only one where the O- approaches the C-LG bond at or near 180o.  Compound 3 is closer to 45o, and compound 4 is back to approximately 90o

This is because the electrons from the nucleophile need to add into the C-LG σ* orbital.  The σ* orbital is co-linear with the C-LG bond, but points in the opposite direction of C→LG bond.  The only way to add the electrons from the nucleophile into that C-LG σ* orbital is if the nucleophile approaches at a trajectory nearly co-linear with the C→LG bond, but from the backside.
Knowing why you got a question wrong is better than knowing that you got a question right.

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