Show how you would accomplish the following synthetic transformat...

Question

Show how you would accomplish the following synthetic transformations. You may use any necessary reagents. (a) N-ethylbenzamide --> benzylethylamine (b) ethyl benzoate --> N-ethylbenzamide

Accepted Answer

Hey everyone. Let's do this problem and says propose a synthetic approach for the following conversion using any needed re agents. So first in our reaction a we have a secondary hmeid, right, we have a nitrogen attached to two carbon groups next to that carbon you'll group And we are converting that to a secondary amine. The nitrogen has two carbon groups again. And for reaction be we have an ester with the formula R. O. R. And we are forming a secondary A meid. Okay, so starting with reaction A. The reduction of the meid can be done by an isolation reduction using lithium aluminum hydride known as L A. H, which is L I A L H four. And that will give the corresponding amine. So the overall reaction for this can be written as our mind, lithium aluminum hydride in a second step of an acid work up to form our mean. And the mechanism is a little bit complicated. There's sort of a long version and a short version. I'll show you a shorter version. And your professor might not ask you to know the mechanism. It might they might just want you to know that you can perform this reduction but I'll include it. If not, you could skip ahead to part B. So starting with our reactant, not our product, we have the hydrogen atom of the L. A. H. That is going to attack the carbon in the carbon L group. And that will give us the oxygen atom with three lone pairs and a negative charge. And now we've added the hydrogen here to that same carbon. Now the aluminum with only three hydrogen will be added to that oxygen. Right? The negative oxygen will attack that aluminum and form a complex that looks like this. Okay. And now the lone pair on the nitrogen will swing down to form an m mean a double bond between the carbon and the nitrogen kicking off the oxygen and everything connected to it. Okay, So we're losing L. I. To A L H 30. Sorry, it should just be L. I. And now we are left with a double bond and a positive charge on our nitrogen. And similar to the first step, we have that hydrogen adding itself to that carbon which will move these electrons back onto the nitrogen. And that will give us are I mean, Alright, so that is the mechanism let's go to reaction to or be rather. So we have the ester being converted to a secondary a meid and we can react the Esther with a primary amine in order to form the secondary of mind. So let's show that reaction. So we're starting with our five member carbon ring in our Esther forming our a meid. So you want to react with a primary amine and the carbon group attached to that Amin should be the carbon group on the end of the mind. So we have a methyl group here. So that means we're going to react with C H three and H two. And this is done with heat. Okay, so let me show you the mechanism quickly for this. So we have our Esther and the lone pair on our amine will attack that carbon in the carbon yield group attaching itself. So now the oxygen will have three lone pairs and we have R O S L R O methyl rather and are a mean that we just added and the nitrogen has a positive charge. So now notice in our final product are nitrogen only has one hydrogen. So since this nitrogen has a positive charge we can remove that hydrogen with more of our amine. Alright. And then the last step, I'll show it all. In one step. We have these electrons of the oxygen moving down and kicking off our O methyl group and that forms our product. Right? So if we have those electrons come down, we form our carbon group kicks off our O methyl groups. So we're left with our N our hydrogen and our CH three group. Okay, So these are the correct reactions which leaves B2, B the correct answer to this problem. That's it for this one. I hope this video is helpful and thanks for watching

Show how you would accomplish the following synthetic transformations. You may use any necessary reagents. (a) N-ethylbenzamide --> benzylethylamine (b) ethyl benzoate --> N-ethylbenzamide

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