Amide Formation - Video Tutorials & Practice Problems
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concept
Amide Formation
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In this video, we're gonna talk about amide formation. Now, under this type of reaction, a carbolic acid and an amine undergo a condensation reaction to form an amide. Now recall a condensation reaction results in the loss of water to do this. The carbolic acid loses its oh group ooh portion and you mean nitrogen loses its one hydrogen. OK. So for this to work, the nitrogen needs to possess at least one hydrogen. Now, here for this to start, we need an H plus catalyst. OK. So it jumps, starts this whole process. If we take a look here at a typical amide formation general reaction, we have our carboy acid and remember squiggly line means it's connected to something else which we don't care about at this moment. And then we have our mean which again, it also has a squiggly line, meaning it's connected to something else. Again, we're not concerned with that. We're focusing on the carbolic acid portion and the amine portion, we're gonna lose water. So the oh from the carbolic acid and the h from the amine, we lose those to produce water. This carbon needs to make up for the bond it just lost. And so do this nitrogen. So what they do is they bond to each other. And in that way, we create in a mind, remember, an amide is when we have a carbonyl connected, single bonded to a nitrogen group, right? So this would be how we form an amide through a condensation reaction between a carbolic acid and an amine.
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example
Amide Formation Example
Video duration:
48s
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Determine the amide product form when four methyl Pentin OIC acid reacts with butty amine. Now, we know that the Carboy acid and the amine will undergo a condensation reaction where we lose the oh from the carbolic acid. And what are the hydrogens from the amine? What's left behind connects together to form our mind product? So here we have our carbon, we're making the amide connection. Now, this nitrogen loses one of its HS but it still has another one and it's still connected to this beautiful chain. So this would represent our m product again, formed through the condensation reaction of our starting carbolic acid and amine.
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Problem
Problem
Predict the amide product formed when 2,2-dimethylpropanoic acid reacts with dimethylamine.