April 19, 2024, 04:22:04 PM
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Topic: Acid-Catalyzed Alcoholysis of δ-Valerolactone -- Need help with mechanism  (Read 10623 times)

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Offline Fozzie

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Hello, I'm working on a mechanism for the acid-catalyzed reaction of δ-Valerolactone with ethanol. I have the end result. The major product should be 5-hydroxypentanoic acid. I have the general mechanism for acid-catalyzed ester hydrolysis, and I know that alcoholysis reacts similarly. The cyclic ester is throwing me a curve though. This is the mechanism I've worked out this far:



The acid protonates the carbonyl oxygen. The nucleophile (ethanol) attacks the carbonyl carbon, forming a protonated tetrahedral intermediate. Here's the first place it gets a little confusing for me. In order for the ring to break where it's supposed to, the protonated ether would have to deprotonate, and the cyclic ether would have to protonate. This leads me to my first question. Why does the cyclic oxygen preferentially protonate rather than the already protonated ether? Is there something about the positive charge in the ring that makes it more stable? Does hyperconjugation from the CH2CH3 not stabilize the charged oxygen enough? Putting a pin in that, I moved forward with the mechanism. The carbon-oxygen ring bond breaks, donating electrons to the oxygen to stabilize the positive charge. This would leave a carbocation, though, so one of the other oxygens would have to form a double bond. Between the OH and OCH2CH3, does the OCH2CH3 form the double bond because it can more adequately support the resulting positive charge? And in the final step, would the resulting by-product be a diethyl ether (may be protonated to further propagate the reaction?)?

I hope my description is clear. It helps to look at the mechanism I drew when reading it. If you have any questions for me or need clarification, feel free to let me know. I'll be checking on this periodically throughout the night, as I'm studying for a test tomorrow.

Thanks!
Fozzie

Offline Dan

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The major product should be 5-hydroxypentanoic acid

It isn't, it ethanol it will be ethyl 5-hydroxypentanoate.

The mechanism is the same as the acid-catalysed ester hydrolysis (AAC2), but with ethanol as the nucleophile instead of water.

Your mechanism is OK up until the point where you use ethanol as the nucleophile in an SN2 reaction.
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Offline Fozzie

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Thank you for your reply. I spoke with my organic chemistry professor. She agrees. So I assume that when the ring breaks, the double bond would form with the hydroxyl group, allowing deprotonation by ethanol? That makes more sense the more I think about it. H+ would be much easier to remove than +CH2CH3. I think my textbook might have mistakenly given the hydrolysis product. Am I correct in assuming that using water in place of the alcohol, the lactone would be hydrolyzed into 5-hydroxypentanoic acid?

Offline orgopete

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Am I correct in assuming that using water in place of the alcohol, the lactone would be hydrolyzed into 5-hydroxypentanoic acid?

Without researching the cited chemistry, I would say the mechanism would be correct, however I don't think your textbook is correct. 5 and 6-membered lactones form very easily. The usual method to isolate the hydroxy acid is to use a base catalyzed opening of the lactone and then a careful neutralization. Excess acid will catalyze lactonization. I think that is the actual chemistry you may find in the literature, but for the purpose of writing the mechanism of an acid catalyzed hydrolysis, it would be fine.
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