Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: cuevas on June 07, 2011, 11:18:01 PM
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When Benzoic acid is treated with ethanol at pH=3 why does the ethanol
add to the carboxylic end and make an ester? I thought that COOH is a
moderately electron withdrawing group and would cause the EtOH add
to the "meta" position of the benzene part of the molecule. It's a doubt
that is killing me and I can't find the answer please help me!! Thank you
very VERY much!
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When Benzoic acid is treated with ethanol at pH=3 why does the ethanol
add to the carboxylic end and make an ester? I thought that COOH is a
moderately electron withdrawing group and would cause the EtOH add
to the "meta" position of the benzene part of the molecule. It's a doubt
that is killing me and I can't find the answer please help me!! Thank you
very VERY much!
Look up Fischer estericication
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Logic > Formulas & Rules
The only reason you gotta worry about that in the first place is to figure out which proton is gonna be the most reactive in the first place, right?
Are you adding a group that will effect electron availability to make an aromatic proton readily available? Maybe! You should do some guestimates as to how reactive it will then be...
Are you adding a group with a really reactive proton on it already? Definitely.
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Logic > Formulas & Rules
The only reason you gotta worry about that in the first place is to figure out which proton is gonna be the most reactive in the first place, right?
Are you adding a group that will effect electron availability to make an aromatic proton readily available? Maybe! You should do some guestimates as to how reactive it will then be...
Are you adding a group with a really reactive proton on it already? Definitely.
Somewhat confusing these comments!
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You just have to know to what functional group will ethanol react with. There are only few reactions in benzene chemistry.
You might want to know the hierarchy of carboxylic acids in terms of reactivity.
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Look up the mechanism of electrophilic aromatic substitution and you'll see why ethanol can't do it.
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Logic > Formulas & Rules
The only reason you gotta worry about that in the first place is to figure out which proton is gonna be the most reactive in the first place, right?
Are you adding a group that will effect electron availability to make an aromatic proton readily available? Maybe! You should do some guestimates as to how reactive it will then be...
Are you adding a group with a really reactive proton on it already? Definitely.
Somewhat confusing these comments!
I'm just saying that the proton on the COOH is the one that matters most unless something is done to tame it.
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Thanks for you help but I finally figured it out.....EtOH is not an electrophile so it will never add to the benzene ring that is why it undergoes the esterification reaction! 8)
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Thanks for you help but I finally figured it out.....EtOH is not an electrophile so it will never add to the benzene ring that is why it undergoes the esterification reaction! 8)
Never say never.
Selective formation of ethylbenzene using benzene and ethanol over HZSM-5 has been studied in the temperature range of 548–648 K at atmospheric pressure. The activity and selectivity of the catalyst showed maximum conversion at a particular Si/Al ratio of the HZSM-5 catalyst. The experimental data were analyzed based on the product pattern using the Langmuir-Hinshelwood-Hougen-Watson approach.
http://www.sciencedirect.com/science/article/pii/S0167299106811390
Thats why chemistry is still something you do rather than just read about (like math unless the person is some almost inhuman exception to the rule), that's also why its still fun.
Granted, there's not going to be a whole lot of vapors present at 1atm at those temperatures, but if it happened, it counts.
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I actually thought this was a very good question. Ethanol is more basic than a COOH group. Undoubtedly, the mineral acid protonates ethanol first. Why shouldn't it effect a Friedel Crafts alkylation reaction?
This is what I am presuming the original poster was thinking. The solution to knowing the answer is about the kinetics of the reactions. Generally, acid-base reaction are thermodynamic processes. There is virtually no barrier to a proton transfer. Therefore, although ethanol is protonated, the loss of water from protonated ethanol is slow. (If it were fast, we would find ethylene and ethyl ether products. Aromatic rings with electron withdrawing groups resist Friedel Crafts alkylation reactions.) If a secondary or even better a tertiary alcohol were used, the rate limiting loss of water would be faster and alcohols can be used for Friedel Crafts alkylations.
If the protonated ethanol does not react further, then a slower and less thermodynamically favored protonation of the COOH group can take place. Once this happens, neutral ethanol can add to a protonated COOH group. The intermediate of that reaction has three similarly basic oxygen atoms§ and the final product of the addition is determined by le Chatelier's principle. If water is removed, the equilibrium favors the ester and if an excess of water is present the equilibrium favors a carboxylic acid.
§You must know the mechanism of Fischer esterification/acid catalyzed ester hydrolysis for this to make sense.
You will notice the citation of zaphraud succeeds with ethanol by heating the reaction. Presumably, heating increasing the rate of water loss.
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well I was just studying for the Dental Admissions Test and that was a question but I think the responses that I obtained were a little bit more complicated than I hoped for but thank you very much..=D