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Topic: e1 mechanism  (Read 6364 times)

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myelver10

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e1 mechanism
« on: October 22, 2004, 10:26:11 AM »
do the conditions for an e1 mechanism favor a secondary and primary alcohol in a weak base or strong base  or tertiary alcohols in a weak or strong base????
     
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Offline Donaldson Tan

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Re:e1 mechanism
« Reply #1 on: October 23, 2004, 05:52:04 AM »
elimination is favoured for primary alcohol in the presence of a strong base.
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Re:e1 mechanism
« Reply #2 on: October 24, 2004, 01:57:08 PM »
I think that E1 is favored when you can form a more stable carbocation, like with a tertiary alcohol, right?

Offline Donaldson Tan

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Re:e1 mechanism
« Reply #3 on: October 24, 2004, 04:45:58 PM »
I think that E1 is favored when you can form a more stable carbocation, like with a tertiary alcohol, right?

isn't nucleophilic substitution (SN2) favoured if a carbocation is more likely to form, especially if this carbocation is trigonal planar?
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Re:e1 mechanism
« Reply #4 on: October 24, 2004, 08:26:59 PM »
E1 requires a full carbocation to form, right?  SN2 doesn't form a full carbocation, right?  You have direct inversion and displacement.  E1 begins by losing whatever leaving group to make a cation, then a base deprotonates next to the cation to make the alkene.

dexangeles

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Re:e1 mechanism
« Reply #5 on: November 11, 2004, 01:40:01 AM »
E1 dehydration E1 dehalogenation SN1 all form carbocations from the dissociation of the oxonium ion
the 1 means it's unimollecular in the slow step

E2 dehydration E2 dehalogenation SN2 don't form carbocations (bimolecular)
primary alcohols or alkyl halides undergo this reaction because there is only one beta carbon that tries to stbilize the (+) charge of the alpha Carbon, which in turn is not enough to create a stable carbocation

no carbocations are formed thru an E2 dehydrohalogenation in the presence of a strong base, the strong base (eg. any butoxide) forces a direct bimolecular reaction on the slow step (rate determining step)

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