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Topic: dehydration of alcohols  (Read 5993 times)

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myelver10

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dehydration of alcohols
« on: October 21, 2004, 06:56:45 PM »
what is the first in the mechanism of the dehydration reaction of a tertiary alcohol with sulfuric acid to form an alkene?

The loss of OH to form a carbocation?
Protonation of a hydroxyl group?
« Last Edit: October 21, 2004, 07:12:32 PM by myelver10 »

Demotivator

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Re:dehydration of alcohols
« Reply #1 on: October 21, 2004, 07:17:54 PM »
Protonation of a hydroxyl group.
Then loss of H2O to form a carbocation.

wgmwwggmmwgm

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Re:dehydration of alcohols
« Reply #2 on: November 08, 2004, 09:48:43 AM »
it is no doubt that protonation of a hydroxyl group is the first step,but
 loss of H2O to form a carbocation is doubtable, maybe there is a intermediate
like the SN2 mechanism.  [Nu---Sub----H3O+].

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Re:dehydration of alcohols
« Reply #3 on: November 08, 2004, 11:40:57 AM »
SN2 chemistry is nearly impossible with a tertiary alcohol.  The backside of the substrate is too sterically hindered for the nucleophile to attack.  The SN1 mechanism requires the formation of a discrete carbocation which is then much less hindered towards attack by a nucleophile.

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Re:dehydration of alcohols
« Reply #4 on: November 09, 2004, 08:40:06 AM »
SN1 is favoured as the R groups in the alkyl group would stabilise the carbocations formed, in virtue of their 'electron-donating' effect. Pressence of H+ favours the abstraction of water.
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dexangeles

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Re:dehydration of alcohols
« Reply #5 on: November 10, 2004, 01:22:24 AM »
Protonation is always your first step on E1/E2/SN1/SN2

Although forming an alkene from an alcohol is not SN1 nor SN2
It is considered an Elimination Reaction E1/E2
dehydration means removal of _________?
i'll leave that up to you, it should be easy to answer

SN1/SN2 is a substitution nucleophilic meaning to say you substitute the OH grp w/ a halogen to form an alkyl halide

E1/E2 is an elimination meaning your OH and another H (thats should give you a hint about the above answer) is removed thus a double bond is formed to form an alkene

you need to see your reactants and products first b4 you decide what reaction mechanism it is.  There too many...good luck when you get to Zaitsev and Markovnikov ;)

remember 2ndary and tertiary alcohols go thru E1 because the formation of the carbocation by the dissociation of the alkyloxonium ion is stable enough, and yes the 3 beta alkyl groups do stabilize the carbocation thru hyperconjugation.

primary alcohols always go thru E2 because the carbocation is not going to be stable enough to be a intermediate in the reaction.

also remember, E1/E2 are all considered Beta elimination meaning the hydrogens of the beta carbons which are directly connected to the alpha carbon is removed to form a double bond.

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