Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: JoeyBob on March 14, 2021, 07:41:30 PM
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So Im trying to figure out the mechanism where ethyl 3-hydroxy-2-methyl-3-(4-methylphenyl)propanoate undergoes a dehydration reaction under basic conditions yet forms 4-methylacetophenone.
My issue is that I dont understand how the dehydration product is cleaved or really converts into 4-methylacetophenone at all. Attached I have a mechanism of how I would expect a basic dehydration reaction to go, but I really dont see how it would lead to decomposition.
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I would say the dehydration product is not cleaved, its the starting material directly. Think about how this is formed and then do the steps in opposite direction
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I thought alcohol dehydration reactions formed alkenes, but if I understand you correctly then that’s not always the case.
I do know how the reverse of a Reformatsky reaction will form this “dehydration” product, but that would require zinc and wouldn’t be a dehydration reaction. If the reverse reformatsky reaction had a name I could maybe find literature that’d give me insight into the solution. But that would require zinc still and it would really just be the backwards mechanism, so probably wouldn’t be helpful...
Initially i wasn’t sure if the question itself was wrong, but I guess the dehydration proceeds in manner where no alkene intermediate is formed?
Basic conditions aren’t even favourable for dehydration reactions though... this question is so frustrating.
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If you think of not using zinc, its more a reverse aldol reaction. I wonder how often this happens?
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I did some searching, and it seems to be something called a retro-aldol reaction?
Essentially, as I understand it, the base cleaves the hydrogen on the -OH group, leaving the negatively charged oxygen atom, which is pushed to form a carbonyl. This splits the C-C bond. I tried to draw it, and while it's not 100% right, I think it's the correct mechanism.
(https://i.imgur.com/B2nNY6k.png)
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How is that a dehydration reaction? Like I guess you form water, but water isn't what is being released from the compound...
I would have never thought of the carbon bonds being broken in such a manner... arent they horrible leaving groups? like Id imagine that the ester is more reactive under basic conditions, but I guess if your using the right base (NaOEt) the reaction about the ester wont make anything new.
Ill read up on the retro aldol, would've of never guessed that a the leaving group would be at the carbon. I guess the negative charge is stabilized by the ester.
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How is that a dehydration reaction? Like I guess you form water, but water isn't what is being released from the compound...
I would have never thought of the carbon bonds being broken in such a manner... arent they horrible leaving groups? like Id imagine that the ester is more reactive under basic conditions, but I guess if your using the right base (NaOEt) the reaction about the ester wont make anything new.
Ill read up on the retro aldol, would've of never guessed that a the leaving group would be at the carbon. I guess the negative charge is stabilized by the ester.
It was what I could find looking around. My textbook doesn't seem to mention anything about it, so it might be a bit more advanced than what I'm used to dealing with. Although, since your product has fewer Cs than your reactant, something has to leave, and it seems like the ester is the obvious leaving group then.
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Thanks. On a sidenote, how do you get such perfect curves on chemdraw (if thats what youre using)? If you look at mine you can see how rough the notation is.
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Thanks. On a sidenote, how do you get such perfect curves on chemdraw (if thats what youre using)? If you look at mine you can see how rough the notation is.
I use the arrows in the toolbar. There's a few to choose between.
Also, google "retro-aldol addition". It's what I used as a reference. I even found your exact problem on there! I'm very unsure why you're expected to solve the problem using a dehydration reaction.
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And I was using the pen tools the whole time...
At this point Im hoping its just a very poorly worded question. Ive wasted so much time looking at all the different dehydration reactions that might apply to no avail. I guess a condensation reaction implies dehydration (aldol condensation), but this is the reverse.
I suppose the enolate could become dehydrated.
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I really wonder if this reaction is practically usefull? You cant drive the equilibrium? I have done a lot of nitro-aldol reaction and never se any retro-reaction? I use nitromethane and different benzaldehydes.
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At this point Im hoping its just a very poorly worded question. Ive wasted so much time looking at all the different dehydration reactions that might apply to no avail. I guess a condensation reaction implies dehydration (aldol condensation), but this is the reverse.
I suppose the enolate could become dehydrated.
It really seems like a poorly worded question. Its either dehydration by the mechanism you posted which leads to αβ-unsaturated ketone or decomposition via retro-aldol to get the acetophenone. It just all depends on conditions and the retro reaction can be forced such as in this paper
https://pubs.rsc.org/en/content/articlelanding/2016/ob/c6ob01198e#!divAbstract
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But its not a pure retro reaction, its a transfer reaction?