laczlacylaci

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The key for this question wasn't that clear for me. I have a separation scheme in my mind:

your reactants are: amide, amine, anhydride.
you react it with NaOH
The amide could turn to a carboxylic acid
The amine could attack the anhydride to form a carboxylic acid and an amide.

Can you elaborate what happens in A?
 

bobeanie95

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In A, you're reacting a strong base with amide, amine and carboxylic anhydride. The base won't react with the amine or the amide (you can't have addition elimination reaction since the nitrogen group is a poor leaving group). Since the reaction will then be basic both the amine and amide will be in the unprotonated form. Meanwhile, the OH- acts as good nucleophile that can attack the anhydride to form a carboxylate (negative charge).

If you add ether, you will have two layers: water and ether. The amine and amide will be in the ether since it doesn't have a charge, while the carboxylate will be in water. So when you evaporate ether, you remain with the amide & amine.
 

theonlytycrane

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^ great answer.

Be careful about amide -> carboxylic acid. It won't happen- amide is stable! Think peptide bond stability.
 
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^ great answer.

Be careful about amide -> carboxylic acid. It won't happen- amide is stable! Think peptide bond stability.
Be careful about being so extreme (won't happen). It can and does happen, it's just that the amide is the least reactive of the COOH derivatives, partly due to the amine's poor ability as a leaving group. This makes the reaction require special circumstances to pull off.
 

aldol16

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Be careful about being so extreme (won't happen). It can and does happen, it's just that the amide is the least reactive of the COOH derivatives, partly due to the amine's poor ability as a leaving group. This makes the reaction require special circumstances to pull off.
Those special circumstances being strong acid or base, which is what we have here.
 
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Those special circumstances being strong acid or base, which is what we have here.
Which is why I cautioned against such strong language (i.e. "It won't happen..."). Those kinds of absolutes can lead to mistakes, especially in a relatively straightforward area like O-chem where so many people (myself included until I learned better) simply memorize.

For my MCAT, I am focusing on learning/understanding. Any tips for Biochem?
 

bobeanie95

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Those special circumstances being strong acid or base, which is what we have here.
The reaction will happen but not in a high enough yield without heat, so that's how you can still obtain the amide from the ether layer
 
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aldol16

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The reaction will happen but not in a high enough yield without heat, so that's how you can still obtain the amide from the ether layer
Depends on the reaction. I would avoid making blanket statements like that. It depends on the specific nature of the amine. I know that we usually teach in undergraduate organic chemistry that heat is required for amide hydrolysis but once you think about it, that cannot apply to all cases. The reason you need heat is to overcome the activation barrier for the reaction but if the bonds broken are weaker (which depends on the electronics of the specific amide), then the transition state becomes early and low-lying, which could bring the reaction within the room temperature limit (again, depends on the exact reaction).

But yes, in this case for the MCAT, one can assume the simplest case.
 

bobeanie95

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Depends on the reaction. I would avoid making blanket statements like that. It depends on the specific nature of the amine. I know that we usually teach in undergraduate organic chemistry that heat is required for amide hydrolysis but once you think about it, that cannot apply to all cases. The reason you need heat is to overcome the activation barrier for the reaction but if the bonds broken are weaker (which depends on the electronics of the specific amide), then the transition state becomes early and low-lying, which could bring the reaction within the room temperature limit (again, depends on the exact reaction).

But yes, in this case for the MCAT, one can assume the simplest case.
I was referring to this question specifically. But yes I agree with you. It really depends on the reactants. The amide could have additional functional groups which could make it a better leaving group and lower the activation energy.
 
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