(b) When 3-bromopyridine is used in this reaction, stronger re...

Question

(b) When 3-bromopyridine is used in this reaction, stronger reaction conditions are required and a mixture of 3-aminopyridine and 4-aminopyridine results. Propose a mechanism to explain this curious result.

Accepted Answer

Welcome back everyone. Under stronger reaction conditions, the reaction between 3 chloroperidine and sodiumethylamide forms N ethyl purdene 3 amine and N Ethyl purdene 4 amine. Proposed a plausible mechanism that explains the formation of the products. So first of all, let's draw the starting materials. And we understand that the first one is 3 chloroperidine. We know that purdine consists of a 6 membered ring, 5 carbons and 1 nitrogen. So we're going to introduce purdine, and position number 3 contains a chlorosubstituent. The reactant that we are using is sodium ethyl amide. Now, sodium is an ion spectator, so we're just going to draw nitrogen bonded to an ethyl group with a negative charge, which means that there must be 1 hydrogen available. Right? We have our amide. Now, essentially, we are going to consider the elimination addition mechanism for this problem. We have to understand that there are 2 hydrogens available adjacent to the electron withdrawing choral group. So first of all, we are going to consider one of them, which will explain the formation of the first product. So let's expand that carbon hydrogen bond, and let's show the deprotonation step to form the benzine intermediate. Right? This is what we are going to get. As we can see, we are getting a triple bond and we know that benzyne intermediates are really reactive. So from here we have 2 pathways, right? We can attack the benzyne intermediate from two sides. Let's illustrate both pathways. First of all, let's introduce another equivalent of ethylamide. And let's use different colors to illustrate the resultant intermediates. First of all, we will choose a red color to illustrate the nucleophilic attack and the placement of the pi electrons in the form of a lone pair on the adjacent carbon. That's one pathway. Now if we choose a blue color, we can go in reverse. We can attack the other carbon and place a negative charge on the adjacent carbon. And we are going to get 2 possible structures. Let's go ahead and draw them out. So first of all, we can just introduce a 6 membered ring. Now, both of the structures will have a double bond instead of a triple bond, and we simply want to draw the resultant carbon nitrogen bond. We are going to have it at 2 different positions. So the first one, let's introduce nitrogen with an ethyl group, right? Nitrogen is still bonded to hydrogen. That's our first structure and we have to be cognizant that there is a negative charge formed on the adjacent carbon and the second intermediate is going to look very similar. But we are now adding nitrogen to a different position, position number 3. So nitrogen bonded to an ethyl group. There is still hydrogen available, and then we have a negative charge. So we have formed 2 carbanions. And now these carbanions will essentially be protonated to form the final product. Essentially, we can use the resultant amine. Previously, we have formed an amine, so we're just going to expand that nitrogen hydrogen bond. And lets illustrate the proton capture for each step. And this gives us our first two possible products. We are simply adding nitrogen I'm sorry, we are simply adding hydrogen and we are done. Now let's consider the same mechanism, but now we are going to think about the removal of the other hydrogen adjacent to the chloro substituent, right? So now we're going to follow the same strategy, but remove the other hydrogen. That is available to us. Nitrogen deprotonates the structure. We want to form the benzyne intermediate. Notice that essentially to do that, we want to have a different pibond arrangement, so let's shift our pibond slightly. And we can do that because essentially we have a structure with a with an aromatic ring. Right? It is resonance stabilized, so we are removing the hydrogen, forming a triple bond and this is followed by the loss of the leaving group. Another addition elimination step So now we are forming a triple bond, but now we have to understand that there's only a negative charge on nitrogen that will allow us to perform another nucleophilic attack step. Notice that we are forming an ethylpyridine 3 amine and an ethylperdine 4 amine. So as the problem suggests, we are not going to consider the attack at position number 2. Right? We are only going to consider the attack at position number 3. And this is based on the context of the problem. So we're attacking carbon at position number 3, forming a Carbanion Let's illustrate that specific carbanion. And introduce nitrogen to the structure. So we have our substituent. And of course, the final step is the protonation step. For the protonation step, we're going to use our amine just as before. So let's expand the nitrogen hydrogen bond, and let's capture that proton, which essentially gives us our product. And this product essentially corresponds to the previously drawn product as well. We have just shown an alternative pathway for the formation of that product. Thank you for watching.

(b) When 3-bromopyridine is used in this reaction, stronger reaction conditions are required and a mixture of 3-aminopyridine and 4-aminopyridine results. Propose a mechanism to explain this curious result.

Key Concepts

Nucleophilic Substitution

Aromaticity and Electrophilic Substitution

Regioselectivity

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