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Topic: Breaking Cyclic Ester Bond  (Read 6730 times)

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Offline glycogrl07

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Breaking Cyclic Ester Bond
« on: February 21, 2008, 05:06:26 PM »
I am trying to figure out the mechanism for breaking the cyclic ester bond with HCl and CH3OH to form C7H13O2Cl from 3-ethyl-2-oxacyclopentanone.  I thought at first the Cl would act as a Nu on the system, but I don't know then how the extra methyl group is added also. 

Thank you
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Offline azmanam

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Re: Breaking Cyclic Ester Bond
« Reply #1 on: February 21, 2008, 05:31:03 PM »
first, cyclic esters are named 'lactones.'

second, if those are your only two reagents, then you've got me stumped.  I've got C7H14O3 as my product.  Are you sure there isn't a second step with PCl3 or PPh3/CCl4 or something?
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Offline glycogrl07

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Re: Breaking Cyclic Ester Bond
« Reply #2 on: February 21, 2008, 05:39:50 PM »
sorry...breaking lactone bond

As for the reagents, that is what is given in the question... I actually want to do halogenation, but the alcohol does something in the reaction also, and the bond between the alpha and beta carbons is broken.   I found the reference for the reaction, but I am trying to understand it now.
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Offline azmanam

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Re: Breaking Cyclic Ester Bond
« Reply #3 on: February 21, 2008, 05:46:17 PM »
Wow.  Those are some pretty harsh conditions.

The first step is just an acid catalyzed transesterification.  Second step is probably chloride mediated SN2 hydronium ion displacement of the protonated alcohol from the first step.

Bond b/t alpha/beta carbons is not broken.
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Offline glycogrl07

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Re: Breaking Cyclic Ester Bond
« Reply #4 on: February 24, 2008, 10:28:45 PM »
Thank you for your help.  I think I was getting confused before. But you broke it down and now I wonder what I missed.  It's so simple, now.

Thanks again
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