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Topic: Alcohol reactions by oxonium ion and active esters  (Read 2988 times)

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Offline elerock21227

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Alcohol reactions by oxonium ion and active esters
« on: December 08, 2010, 01:00:29 AM »
Oxonium part:

Say there is a t-butyl alcohol reacting with  a non-nucleophilic acid like H2SO4.

The alcohol is protinated, forming a good leaving group and leaving. A tertiary carbocation is formed and H2O grabs the Hydrogen on the β-carbon and elimintion occurs . The products are an alkene and H3O+

I'm wondering why elimination product is the major yield of this rxn? What is stopping the carbocation on the substrate from going SN1 and reverting back to a t-butyl alcohol ?

Active ester part:

If there is a primary alcohol reacting with Phosphorous Oxychloride & pyridine the reaction yields E2 products. However, if that primary alcohol reacts with Thionyl chloride or Phosphorous trihalide, SN2 products are yielded.

Why there are 2 different mechanisms possible?

Offline Doc Oc

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Re: Alcohol reactions by oxonium ion and active esters
« Reply #1 on: December 08, 2010, 09:04:39 AM »
For the first question, there is a lot of other factors that come into play, the most significant being the strength of the base and the polarity of the solvent.  Check out this table, I think it gives a decent summary on considerations of SN1 vs E1

http://www.wfu.edu/~wrightmw/chm122/handouts/sn1sn2e1e2%20summary.pdf

The 2nd question is somewhat related, there's a strong base in the 1st reaction condition, but in the 2nd there's an absence of strong base and you're left with something that is a better nucleophile than base.

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