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Topic: Acetal Formation in Base (and lack thereof)  (Read 10254 times)

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Offline gingi85

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Acetal Formation in Base (and lack thereof)
« on: August 01, 2008, 03:59:51 AM »
To produce an acetal from a hemiacetal in base, we would need to replace the hydroxy group with an alkoxy group in an Sn2 type reaction. The book I'm reading says this is a most unfavorable reaction (which is why the equilibrium never goes past the hemiacetal stage in base). Why?

It would seem that an hydroxide ion is more stable than an alkoxide ion, b/c there are no methyl groups destabilizing the negative charge on the oxygen. Thus, the reaction should be favorable, no?

Offline azmanam

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Re: Acetal Formation in Base (and lack thereof)
« Reply #1 on: August 01, 2008, 07:50:46 AM »
Starting with a hemiacetal and an alcohol, what is the first step of the base catalyzed mechanism?  Is there anything else that can happen in this step?  Talk us through the mechanism and think about what else can happen in base.
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Offline gingi85

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Re: Acetal Formation in Base (and lack thereof)
« Reply #2 on: August 01, 2008, 08:14:05 AM »
Yes, it the base can deprotonate the hydroxy group ultimately bringing us back to the original carbonyl.

However, the same choice exists at the hemiacetal stage in acid. We can either protonate the OH group, leading to loss of water and ultimately acetal. Alternatively, we can protonate the OR group, lose the alchohol and return to the original carbonyl. In the end, both happen and we end up with some sort of equilibrium between carbonyl, hemiactal, and acetal.

Now, back to the hemiactal in base: Unless the activation energy for the the Sn2 displacement of the hydroxy group by the alkoxy group is especially high, we would expect to settle out at some sort of equilibrium between the three just like we do in acid.

Offline azmanam

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Re: Acetal Formation in Base (and lack thereof)
« Reply #3 on: August 01, 2008, 08:24:54 AM »
Quote
Yes, it the base can deprotonate the hydroxy group

This is the key here.  Once the hemiacetal alcohol is deprotonated, the forward reaction shuts down.  SN2 reaction will not displace O(2-) as a leaving group, and the electrons from the ether part of the hemiacetal will not collapse down to form an oxocarbenium ion (C-O+=C) because you won't see positive charges in base catalyzed mechanisms.

Now, as to why the book says the reaction will never proceed past the hemiacetal stage, I cannot say.  I'd hesitate to use the term never.  I'm sure there are special instances where an acetal can be formed under basic conditions.  But as far as the scope of the course is concerned, this equilibrium is sufficiently disfavored that acid catalyzed acetal formation is the mechanism we should concern ourselves with.
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Offline gingi85

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Re: Acetal Formation in Base (and lack thereof)
« Reply #4 on: August 01, 2008, 08:33:54 AM »
Thank you.

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