For each of the following molecules, predict the product that ...

Question

For each of the following molecules, predict the product that would form upon reaction of a single equivalent of a strong base.

(e)

Accepted Answer

Hey everyone. And welcome back to another video Jody and I obtained in the reaction of the following molecule with one equivalent of a strong base. We're given the structure of four nitro once you dial because we are using only one equivalent of a strong base and we have two acidic protons, right. Those are the protons that are bonded to the oxygen atom because oxygen is much more electron negative than hydrogen. So those bonds will be the most polarized and therefore, they would have the most acidic protons. So we have two of them, let's label one of them in blue and another one in red, there are two alcohol groups and we can remove any of the two protons because we only have one equivalent of the base. The protein that will be removed will correspond to the more acidic proton. Now, we want to determine which one of them is more acidic. Instead that we have to consider the conjugate base resultant from the de protonation of each hydrogen starting with the blue hydrogen. If we remove it, we will get the conjugate base of the given compound. So we are just going to redraw it, we have Benze and ring, we are going to add our nitro group. Now we are going to leave one of the o age groups as it is. And for the other one, we will remove hydrogen so that we end up with oxide and ion. Let's recall that the more stable the conjugate base is the stronger the acid. So we have to consider resonance, right? Because we have our negative charge and oxygen bonds the benzene ring. So let's start drawing our resonance forms. Now, the workforce res form that we can get here is simply by de localizing the lone pair on oxygen into a pipe bond, followed by the cleavage of the adjacent pipe bond. This is how we get our second resonance form because the initial one is the first one. So we can simply draw out the result on second resonance form just as the arrow suggests, we're going to break, we're actually going to first, we introduce a pie bond and then because we have broken a pipe on within the ring, now we will have a lone pair on carbon. Well, then we have our second resins form. Now let's think about the third one. We can de localize the result on lone pa and carbon and then we can break the adjacent pipe on which will give us the third resonance structure. Let's go ahead and it just as the arrow suggests, we're creating a pi bond. And now we have a negative charge on carbon bonnets to the nitro group. That would be our third resonance form. Let's think about the next one. We can de localize the lone pair on carbon and the ring and break the other pipe bond that would give us the fourth resonance structure. OK. So let's go ahead and enjoy it. OK. So we got our fourth one. Let's go ahead and label it as the fourth rein structure. And now if we carefully think about the third one, and if we expand the nitro group, we will notice that nitrogen can also participate in resonances, right. So specifically, let's go ahead and expand nitrogen, oxygen bonds. Nitrogen will have a formal positive charge. One of the oxygen atoms has a formal negative charge. And we can essentially notice that there's another pathway for resonances in which that same loan pair can be de localized into a pi bond both by the cleavage of the nitrogen oxygen pi bond. This will give us an additional resonance structure. So let's go ahead and add it. Now, we noticed that we are forming carbon nitrogen double bond and the two oxygen atoms will be single bonded. So we have a formal negative charge on one of the oxygen atoms, nitrogen will get a formal positive charge because the cell has four bonds. So that will be our final fifth resonance structure. Now, our goal is to essentially perform very similar steps, but we are going to remove the proton from the O H group, which is to the nitro group, we're going to leave the other one as it is. And if we remove that proton, we will get the conjugate base, our P oxide and ion, let's draw our N 02 group that will be immediately our very first resonance structure. And now we can start de localizing the electrons. So we are going to de localize the loan pair on the negatively charged oxygen followed by the cleavage of the pi bond that would give us our second resonance structure. So let's go ahead and draw it. We can simply break the pipe bomb control another pipe bo for oxygen. And now we understand that we have a negative charge on carbon that gives us our second resonance structure. Let's go ahead and draw the third one which can be achieved by de localizing the lone parent carbon and breaking the adjacent pie bond that will give us structure number three. And you may notice that we are getting it by adding a pi bond breaking the other one so that we end up with a negative charge on carbon at a different position. And there is one more that we can control, we can also shift the lone pair to form a pi bond by breaking the final pie bond within the ring. And that will give us our fourth structure. Let's go ahead and try it. So now we have a double bonnet, a different position and our negatively charged carbon. So that's our fourth resonance structure. And essentially, if we go forward, we will just rebuild the same structure that we started off with. So now we have to compare the stability of our resonance structures, right? Because the more stable the conjugate base, the stronger the asset going back to the resonance forms of the conjugate base upon the removal of the proton from the a wage group that is pirate to the nitro group. We notice that we have a total of five resonance forms. However, what is more important is that structure one is a significant contributor because we have a negative charge on a more electron negative atom structure. Two has a negative charge on carbon. So it's not so significant. Same thing for structure number three four. And now the fifth one is also significant because we have a negative charge on oxygen. So we have two significant resonance contributors and a total of five resonance forms for the O H group at the meta position. We noticed that we have a total of four residents contributors. One of them has a negative charge on oxygen everywhere else, we have a negative charge on carbon. So we only have one significant rises contributor. So we can conclude that the O H group at the power position produces a more acidic proton. OK. We already know the reasons that's because there are more resonance forms and they are more significant because we have two resonance forms with the negative charge on oxygen versus one where the meta O H. And therefore, because the power hydrogen is more acidic, that means one equivalent of base will prefer to detonate the O age group at the power position because it has a more acidic and more easily removable hydrogen. Therefore, if we want to draw our final answer, we can just detonate the O H group at the power position, which corresponds to our welfare rein structure that we drew previously. Thank you for watching.

For each of the following molecules, predict the product that would form upon reaction of a single equivalent of a strong base.
(e)

Key Concepts

Strong Bases

Acid-Base Reactions

Functional Groups

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