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Topic: Why do Cl-,Br- give Sn2 with 2' alkyl halide while Sn1 with 2' alcohol ??!  (Read 5523 times)

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Offline Babcock_Hall

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Re: Why do Cl-,Br- give Sn2 with 2' alkyl halide while Sn1 with 2' alcohol ??!
« Reply #15 on: January 29, 2018, 08:13:11 AM »
I will assume that the secondary alcohol is its own solvent, and the nucleophile is HX.  Then I would expect the SN1 mechanism to predominate.
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Offline Babcock_Hall

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Re: Sn1/Sn2 , E1/E2 exercise correction
« Reply #16 on: January 29, 2018, 08:16:28 AM »
So if you perform a ring-flip on the compound in 5, what do you see?
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Offline xshadow

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Re: Sn1/Sn2 , E1/E2 exercise correction
« Reply #17 on: January 29, 2018, 02:25:43 PM »
Quote from: Babcock_Hall on January 29, 2018, 08:16:28 AM
So if you perform a ring-flip on the compound in 5, what do you see?

With a ring flip I should have the  E2 mechanism...because the leaving group needs to stay in an axial position in order to give E2 ,for a cyclo....

So  I have Sn2 when the leaving group is equatorial(the  more stable conformer)while an elimination E2 in the other conforme...???
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Offline Babcock_Hall

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Re: Why do Cl-,Br- give Sn2 with 2' alkyl halide while Sn1 with 2' alcohol ??!
« Reply #18 on: January 29, 2018, 02:40:46 PM »
Is the SN2 reaction fast or slow for tertiary alkyl halides?  How fast are ring flips?
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