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Topic: epoxidation reaction  (Read 2216 times)

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Offline iScience

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epoxidation reaction
« on: January 22, 2014, 08:43:44 PM »


2 questions:

1.) the H indicated by the red arrow, where does that go? does it just get pulled away by an X-?

2.) there are two reaction pathways in this picture, can someone explain to me the mechanism of the top one? ie the one with the peroxy acid?

Offline AlphaScent

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Re: epoxidation reaction
« Reply #1 on: January 22, 2014, 09:45:01 PM »
Would the electrons of the olefin attack an oxygen or the oxygen attack the olefin?  Think about this also.  Where does the oxygen come from to form the epoxide?  R-O-O-H or R-O-O-H? 
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Offline orgopete

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Re: epoxidation reaction
« Reply #2 on: January 22, 2014, 11:15:37 PM »

1.) the H indicated by the red arrow, where does that go? does it just get pulled away by an X-?


The structure with the red arrow is the product of that reaction. In order to form the epoxide, a base must be used. For this reaction, NaOH is typically used.
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Offline PhDoc

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Re: epoxidation reaction
« Reply #3 on: February 08, 2014, 04:03:42 PM »
You're using a peroxy acid in your example, not a peroxide. The two behave very differently.

Draw resonance structures for the peroxy acid. Which oxygen is electrophilic? Why? The mechanism is concerted, and is not amenable to a nucleophile-electrophile approach, although the "butterfly mechanism" can make it appear like such an approach.

Look up the mechanism of epoxidation on Wikipedia. Although it's always better to use your textbook, this one is correct.
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