Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Shino on February 18, 2021, 08:28:53 PM
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Hi, I have some doubts about the reaction between an amise and an alcohol
Some textbooks say that there is no reaction because in the tethraedral intermediate -OR is a better leaving group than -NH2 ...so OR- is the leaving group and the amide is reformed
But other textbooks say that this is an equilibrium and this reaction can occour anyway
So where is the truth?
Thanks!!
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According to my textbook (Blackman's, 4th edition): "Amides do not react with alcohols under any experimental conditions," since alcohols are not strong enough nucleophiles to attack the carbonyl group of the amide.
Which textbooks are you using? Dated works may not be accurate.
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I am completely out of my element here, so I am not addressing details of the reaction itlsef, but:
Which textbooks are you using? Dated works may not be accurate.
These are rather basic things, I wouldn't be surprised if they were researched to death in XIX century, somehow I doubt OP uses THAT old textbooks.
In many cases trick is such handwavy explanations are all in some ways right and they don't contradict each other - even if the amide is reformed in most cases this can be still an equilibrium, just heavily shifted.
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I am completely out of my element here, so I am not addressing details of the reaction itlsef, but:
Which textbooks are you using? Dated works may not be accurate.
These are rather basic things, I wouldn't be surprised if they were researched to death in XIX century, somehow I doubt OP uses THAT old textbooks.
In many cases trick is such handwavy explanations are all in some ways right and they don't contradict each other - even if the amide is reformed in most cases this can be still an equilibrium, just heavily shifted.
Yes. Initially I wanted to address that a heavily shifted equilibrium could be possible. However, my textbook is extremely adamant that absolutely no reaction occurs, which seems strange. I honestly asked to make sure he was using actual textbooks and not just online resources...
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Depending what level of chemistry we are discussing now, there are surely methods to convert amides to esters:
Tetrahedron Letters 55, 50, 2014, 6935-6938
Nature 524, 79–83(2015)
Org. Lett. 2018, 20, 18, 5622–5625
and possibly others. Especialy the orglett paper is interesting IMO. The trick is to either cleave the C-N bond by transition metals or block the resonance stabilization and planarity by EWG groups like Boc or Tosyl
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Does water count as an alcohol? One can hydrolyze peptide bonds (6 M HCl, 110 °C, 24 hours) under forcing conditions.
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I think its different with water, you get an carboxylic acid, not an ester as product. It is hard believe that if you boil a amide with strong acid in n-butanol that you get 0% ester.
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Does water count as an alcohol?
Is formaldehyde a ketone?
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Meter,
If water can hydrolyze an amide, then why shouldn't propanol, for example, be able to bring about an analogous reaction under similar conditions?
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At the active site of a serine protease, an amide is first converted into an ester with a nucleophilic serine residue (Ser195 in chymotrypsin). Although hydrolysis is the second half of the overall reaction, the first half of the reaction forms an ester because the side chain of serine is -CH2OH.
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That was nice Babcock!
Its a special case but still proves that it can occur.
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Water is more nucleophilic than an alkyl alcohol Babcock.
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In Bruice Organic Chemistry 8th edition in chapter 15 she claims that an amide can be converted to an ester using alcohol and acid and heat. The driving force for the process is described as being the protonation of the amine leaving group. If it is very acidic and the amine leaving group is 100% protonated, then it cannot add back and form the more stable amide. Obviously this would be very slow because the amide is very weak electrophile and the alcohol is a very weak nucleophile.
She specifically points out that under basic conditions (alkoxide + amide) there would be no driving force to form the ester, and this version of the reaction is 100% impossible.
Perhaps it is completely impractical with low yield? Occasionally blatant falsehoods that only "sound" correct make it into these textbooks. But anybody using Bruice in Sophomore organic might be expected to know it.
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I think if its completely impossible or just gives low yield are two very different things.
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Hmmmm. Strong acids and alcohols.... What could go wrong?