The following compound reacts with a hot, concentrated solutio...

Question

The following compound reacts with a hot, concentrated solution of NaOH (in a sealed tube) to give a mixture of two products. Propose structures for these products, and give a mechanism to account for their formation.

Accepted Answer

Hey everyone and welcome back in today's video, we are going to solve another problem which reads when one chloral to methyl benzene is treated with sodium Amytal pneumonia, it undergoes nuclear filic aromatic substitution to form a mixture of two products. Right? The products and the mechanism that accounts for the formation of two products. We may begin solving this problem by drawing our substrate. And because we have a benzene derivative, we can just start by drawing the benzene ring. So that would be our six member ring With three double bonds. And then we see that there are two substitue ints position number one contains a chloral substitution isn't? So we can simply add our chloral substitution at position one and the adjacent position carbon # two has a metal group bonded to it. Now we have to understand that sodium a mite is a strong base. And because we're using a strong base in combination with a good leaving group, this allows us to perform a nuclear aromatic substitution on benzene. So, if we think about the beta hydrogen, notice that this is our alpha carbon, that has a good leaving group, right? And this is our beta carbon which contains a beta hydrogen that can be removed by a strong base. Keep in mind once again, that generally this doesn't take place unless we have a good leaving group adjacent or basically at the alpha position. So what's going to happen in the step is that we're I'm going to use our strong base, which is a mine and we're going to perform an elimination. So the negative charge on nitrogen essential attacks the beta proton and we are going to form a pi bond over here. So let's see how that works. We're going to form a pi bond followed by the loss of the leaving group, which is which essentially explains why this step is possible. Right? We have a good leaving group, a strong base and the elimination is favorable. So now in this step, we are forming a triple bond. If you notice, we can essentially draw our intermediate. We understand that between these two carbons. We end up with a triple bond, that would be the methyl group, the original double bonds. And it's also very important to understand that this intermediate is called ban sign because it has a triple bond in an original benzene ring. Right now, a triple bond in a ring causes a lot of strain. You may recall that triple bonds tends to be linear, but we cannot achieve that in a six member ring. Right? So that means benzene is really reactive. Now, the molecule of ammonia can essentially allow Ben Ben Zayn to restore its automaticity or basically to restore the sp two hybridized carbon atoms by a nuclear attack at any of the two positions. Which essentially explains why there are two possible products. So, if we if we think about the attack at the first position, we are using the molecule of ammonia, just as the problem suggests. And the reason why we're using it is because benzene is really reactive right? Because of that triple bond. So we don't need to use a negatively charged nuclear biologists efficiency is a molecular one. And we understand that there might be two pathways to attack the carbon within the triple bond. Either that this position or that position. So, if you look at the first pathway, the lone pair of nitrogen attacks the carbon atom, we're breaking the pi bond by transferring the pi electrons in the form of a lone pair onto this carbon. And let's see what we get for this blue pathway. So, first of all, let's draw our intermediate, that would be our six member ring. Now we have our double bond because we have broken the pi bon at this position, we have ammonia. So we can essentially expand that additional hydrogen atom because nitrogen has a formal positive charge. Now that would be NH two H which corresponds to NH three overall. And now it's very important to keep in mind that you also see negative charge of this position because you have transferred these pi electrons in the form of a lone pair. So now the world last step is the proton transfer step. Notice how nitrogen has an excess of protons. So this acidic proton can be easily transferred to our carbon, which will essentially allow us to get rid of both charges at once. So we essentially get our final product for this one. We control our benzene ring the double bonds And then we simply end up with an Amino group that be NH two. Okay. And we understand that the metal group and the amino group, they correspond to a one comma three dice substitute, advancing derivative. Right, So that would be our first product. And now let's think about the formation of the other one. Let's suppose that we are thinking about the attack at the other position. And let's use another color for that. Let's suppose we use black if we use our next pathway, the lone pair, nitrogen from ammonia can also attack this carbon atom, which will essentially mean that we need to break the pi bond by transferring these pi electrons onto different carbon atoms. So we will have a negative charge of this position. Let's see what we get. So that would be our six members doing notice how we have formed a double bond at this position, We have broken one of the two pi bonds. And because we are transferring the pi electrons onto this carbon, we now see a negative charge of this position. And also ammonia is bonded to this one. So we can essentially once again expand that acidic hydrogen because nitrogen will correspond to a form of positive charge since it has four bonds. And when we got our intermediate, we are thinking about it once again and we understand that there would be a proton transfer because we have a negatively charged carbon which can capture that proton from ammonia to eliminate the positive charge on nitrogen. So this would give us our next product. Let's draw The six member ring, our double bonds, the resultant amino group and the method group. Let's label our final products. This is our complete mechanism. This is product number one because you may notice it's one comma three di substituted. And this would be our second product because it's one committee that substituted. So because the positions, the relative positions of the two substitue ints are different. That implies that we indeed see two different products. So let's label the whole mechanism as our final answer because the problem was referring to the products and the mechanism. So that would be our final mechanism. Looking at the mechanism and at the products form, we may conclude that the correct answer choice This multiple choice question is d However, if you have any questions don't have States, Libya, comment down below in the comment section and thank you for watching

The following compound reacts with a hot, concentrated solution of NaOH (in a sealed tube) to give a mixture of two products. Propose structures for these products, and give a mechanism to account for their formation.

Key Concepts

Nucleophilic Substitution

Elimination Reactions

Reaction Mechanisms

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