Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: anilin on April 13, 2009, 12:25:39 PM
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Why do acetyl chloride and acetic anhydride react with water at room temp, whilst ethyl acetate and acetamide do not?
thanks
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It's a combination of factors...resonance stabilization of the starting material, leaving group ability, and autocatalysis (look up if you're not familiar with it).
Ponder those ideas with respect to the reaction and see if you can figure out how they apply/explain the reactivity..
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It's a combination of factors...resonance stabilization of the starting material, leaving group ability, and autocatalysis (look up if you're not familiar with it).
Ponder those ideas with respect to the reaction and see if you can figure out how they apply/explain the reactivity..
the amin in acetamide is a better leaving group than the hydroxyl group in acetic anhydrate.
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The leaving group for acetic anhydride is not a hydroxyl (OH).
Could you show the structures of the leaving groups in each case? Just to make sure we're on the same page.
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The leaving group for acetic anhydride is not a hydroxyl (OH).
Could you show the structures of the leaving groups in each case? Just to make sure we're on the same page.
you are right, I was thinking of the structure of acetic acid. How about acetamide? It has a NH2 which could act as a leaving group. How does resonance stabilization improve water hydrolyses?
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I didn't say resonance stabilization "improves" hydrolysis, just that it is a factor. Some of the starting carbonyl's have better resonance stabilization than others. Would that make them more or less reactive toward nucleophilic attack?
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I didn't say resonance stabilization "improves" hydrolysis, just that it is a factor. Some of the starting carbonyl's have better resonance stabilization than others. Would that make them more or less reactive toward nucleophilic attack?
since neucleophiles are electron rich specious then they seek a positive charge. therefore resonance stabilization will reduce neuclophilic attack.
Now there is a resonance stabilization in all of these four componds and acetamide has a leaving group as well. From what you are saying I understand that resonance stabilization and the presence of a leaving group are factors that help hydrolysis. But I don’t understand why 1- resonance stabilization (distribution of -ive charge) helps neucleophilic attack 2- acetemide has a good leaving group but it is not hydrolyzed.
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Resonance stabilization of the carbonyl makes it less reactive toward nucleophilic attack. And yes, you can draw a resonance structure where a lone pair on the leaving group donates to the partially positive carbon for all carboxylic acid derivatives. However, this resonance structure contributes more to the hybrid for an amide than it does for any of the other carboxylic acid derivatives. This can be used explain why the C=O stretch of an amide occurs at a lower frequency than a ketone, and why in the proton NMR of N,N-dimethylformamide, you see signals for both N-methyl groups (slow rotation around the OC-N bond due to resonance). As such, the amide is the least reactive toward nucleophiles.
As far as quality of leaving groups go, compare pKa's of the conjugate acids. The stronger the conjugate acid, the more stable the leaving group is. pKa's: HCl = -7; carboxylic acid = ~4-5; alcohol = ~16-18; Amine = ~35-40. -NH2 is not that good a leaving group.
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Resonance stabilization of the carbonyl makes it less reactive toward nucleophilic attack. And yes, you can draw a resonance structure where a lone pair on the leaving group donates to the partially positive carbon for all carboxylic acid derivatives. However, this resonance structure contributes more to the hybrid for an amide than it does for any of the other carboxylic acid derivatives. This can be used explain why the C=O stretch of an amide occurs at a lower frequency than a ketone, and why in the proton NMR of N,N-dimethylformamide, you see signals for both N-methyl groups (slow rotation around the OC-N bond due to resonance). As such, the amide is the least reactive toward nucleophiles.
As far as quality of leaving groups go, compare pKa's of the conjugate acids. The stronger the conjugate acid, the more stable the leaving group is. pKa's: HCl = -7; carboxylic acid = ~4-5; alcohol = ~16-18; Amine = ~35-40. -NH2 is not that good a leaving group.
Thanks for your explanation but I'd like to know what you mean by "this resonance structure contributes more to the hybrid for an amide than it does for any of the other carboxylic acid derivatives". Contributing more to the hybrid for an amide, why? And why is it important in here.
Are these statements correct:
1-hydrolysis of acetyl chloride is because of the Cl ( a good leaving group).
2-acetamide is not readily hydrolyzed because NH2 is not a good leaving group.
3-as far as resonance stabilization, there is more resonance in acetic anhydride than ethyl acetate. therefore acetic anhydride is a more stable compound than ethyl acetate. but why is it more readily hydrolyzed than ethyl acetate?
thanks
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Why do acetyl chloride and acetic anhydride react with water at room temp, whilst ethyl acetate and acetamide do not?
thanks
AcOCl (Acetyl Chloride) Because the bond between Cl and the electron withdrawn C is very weak Cl atom is easily reduced by H-OH's H+ to HCl while the carbon atom of the AcO=C- is easily oxidated by OH- forming AcOH.
AcOAc (Acetic Anhydride) is a dequenching agent this means he is very prone to hydration. Because he is an ether of two Acetyl groups the ether bonds are very weak. When this molecule is met with an Acidic Hydrogen the Basic O traps that Hydrogen AcOHAc and that ether bond breaks and an Ac group is liberated to find an O or N.
EtOAC (Ethyl Acetate) This is a quite stable molecule so he wont react with H-OH as AcNH2 (Acetamide) is.
Hope these helps!
Lutesium...
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Cl atom is easily reduced by H-OH's H+ to HCl while the carbon atom of the AcO=C- is easily oxidated by OH- forming AcOH.
Acid base reactions are not redox reactions. C stays at same oxidation state (the acid oxidation state) and Cl also stays at same oxidation state (the chloride oxidation state)
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Thanks for your explanation but I'd like to know what you mean by "this resonance structure contributes more to the hybrid for an amide than it does for any of the other carboxylic acid derivatives". Contributing more to the hybrid for an amide, why? And why is it important in here.
As for "contributing to the hybrid" I mean that, when you have resonance structures that are not equivalent, the overall picture (hybrid) is a weighted average of the contributing canonical forms. Some canonical forms contribute more or less (lower weighting) than others based on separation of charge, full octets, electronegativity of atoms bearing charges, etc.
Between N & O, nitrogen is a better resonance donating group b/c N is less electronegative and holds on less tightly to it's lone pair of electrons. As such, in an amide the resonance structure where N donates a lone pair to the partial positive carbon is more significant (contributes more to the weighted average) than the analogous one in an ester.
Between Cl and the others, electronegativity doesn't explain all of it b/c O>Cl. There is also the difference in size between the 3p orbitals on Cl and 2p orbitals on carbon leading to less than ideal overlap.
Overall, the resonance stabilization increases (lowest to highest) in order of:
acid chloride < anhydride < ester < amide
Reactivity follows the opposite trend
Are these statements correct:
1-hydrolysis of acetyl chloride is because of the Cl ( a good leaving group).
2-acetamide is not readily hydrolyzed because NH2 is not a good leaving group.
I'd agree with both of those, but keep in mind the quality of the leaving group is most useful in determining which leaving group will be kicked out once a nucleophile attacks. How much resonance stabilization the starting material helps determine whether the weak nucleophile water will attack the carbonyl in the first place.
3-as far as resonance stabilization, there is more resonance in acetic anhydride than ethyl acetate. therefore acetic anhydride is a more stable compound than ethyl acetate. but why is it more readily hydrolyzed than ethyl acetate?
There is more resonance for the anhydride, but the middle oxygen of the anhydride has two carbonyl's that it can donate to...effectively halving it's donating ability to either one.
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Cl atom is easily reduced by H-OH's H+ to HCl while the carbon atom of the AcO=C- is easily oxidated by OH- forming AcOH.
Acid base reactions are not redox reactions. C stays at same oxidation state (the acid oxidation state) and Cl also stays at same oxidation state (the chloride oxidation state)
I know that but I couldn't find any other route to explain!
Thanks for the correction!
Lutesium...
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Thanks for your explanation but I'd like to know what you mean by "this resonance structure contributes more to the hybrid for an amide than it does for any of the other carboxylic acid derivatives". Contributing more to the hybrid for an amide, why? And why is it important in here.
As for "contributing to the hybrid" I mean that, when you have resonance structures that are not equivalent, the overall picture (hybrid) is a weighted average of the contributing canonical forms. Some canonical forms contribute more or less (lower weighting) than others based on separation of charge, full octets, electronegativity of atoms bearing charges, etc.
Between N & O, nitrogen is a better resonance donating group b/c N is less electronegative and holds on less tightly to it's lone pair of electrons. As such, in an amide the resonance structure where N donates a lone pair to the partial positive carbon is more significant (contributes more to the weighted average) than the analogous one in an ester.
Between Cl and the others, electronegativity doesn't explain all of it b/c O>Cl. There is also the difference in size between the 3p orbitals on Cl and 2p orbitals on carbon leading to less than ideal overlap.
Overall, the resonance stabilization increases (lowest to highest) in order of:
acid chloride < anhydride < ester < amide
Reactivity follows the opposite trend
Are these statements correct:
1-hydrolysis of acetyl chloride is because of the Cl ( a good leaving group).
2-acetamide is not readily hydrolyzed because NH2 is not a good leaving group.
I'd agree with both of those, but keep in mind the quality of the leaving group is most useful in determining which leaving group will be kicked out once a nucleophile attacks. How much resonance stabilization the starting material helps determine whether the weak nucleophile water will attack the carbonyl in the first place.
3-as far as resonance stabilization, there is more resonance in acetic anhydride than ethyl acetate. therefore acetic anhydride is a more stable compound than ethyl acetate. but why is it more readily hydrolyzed than ethyl acetate?
There is more resonance for the anhydride, but the middle oxygen of the anhydride has two carbonyl's that it can donate to...effectively halving it's donating ability to either one.
thanks a lot, it really helped.