The benzene ring alters the reactivity of a neighboring group ...

Question

The benzene ring alters the reactivity of a neighboring group in the benzylic position much as a double bond alters the reactivity of groups in the allylic position.

Benzylic cations, anions, and radicals are all more stable than simple alkyl intermediates.

c. Which of the following reactions will have the faster rate and give the better yield? Use a drawing of the transition state to explain your answer.

Accepted Answer

Hey everyone and welcome back in today's video, we are going to solve the following problem. Ben's Illich parents and aunts and radicals are all more stable compared to simple alcohol intermediates. Which of the following reactions will give the better yield and have the foster. Wait, explain your answer using a drawing of the transition state. And if you look at the two reactions, they're very similar nature. Except from the fact that the main difference is that the second reaction has the benzene rainy instead of cyclo hexane ring. Now, clearly we have a primary substrate. In each case we can label this as a primary substrate and this as a primary substrate because the leaving group which is chlorine is bonded to a primary carbon atom, this carbon atom and this carbon atom, they both have one car, one neighboring carbon atom. Right? So in each case we have a primary substrate sodium as a non spectator. So we have a negatively charged with oxide in each case as our nuclear file and our solvents are exactly the same. They're both ethanol solvents. So we can conclude that because we have a primary substrate, a strong nuclear file because it's charged. Right? Once again. So they miss a non spectator. This must be an essence a reaction in each case. So we can conclude that each reaction follows an SNC pathway. So this tells us a lot, we have a reference point for this reaction. And because in essence the reaction rate depends on multiple factors such as the nature of the leaving group. Right? The nature of the substrate. It's not really what we're going to consider here because we have the same leaving group chlorine, right? The same nature of substrate, primary substrate. Right? The same solvent, ethanol in each case and the same nuclear file as well with oxide. So we are going to exclude the standard factors whenever we are comparing the rate of an SN two reaction and we're going to think about more supple reasons, recall that SN two reaction rate heavily depends on the stability of the transition state. Right? The more stable the transition state is the foster the rate of an SN two reaction. And this is really useful for us because if you look at the problem, it says that Ben Zilla, Karen's and ions and radicals are all more stable. Meaning if we have benzene ring, it implies that the transition state is expected to be more stable than for the reaction. Which doesn't have benzene ring. Right here, we will have our Ben's Ilic transition state. So we already know that it's very likely that reaction number two is faster than the first one but we will look at it more in depth by drawing the transition state. First of all, if you look at reaction number one and then proceeds to reaction number two, we made troll the transition state for number one. So simply, if you draw that cyclo hexane ring, you made your carbon atom two hydrogen atoms and now the transition state essentially implies that you're forming a new bond and breaking another one, right? This is how the reaction takes place. We have S oxide, O E C. Right? With a negative charge, it's attacking this carbon atom. So we can essentially draw that bond formation. There will be a bond formation between oxygen and carbon. And simultaneously we would be breaking that carbon chlorine bond. Okay, we understand that chlorine is partially negative because it's more electro negative than carbon. And here oxygen is also partially negative because it's bonded to carbon and oxygen is more electro negative. So that makes sense. Right. This is our transition state which shows how we are forming a new bond and breaking an old bond. So this is our transition state for number one and for number two, also, we can essentially indicate that this is a transition state by adding brackets and a transition state sign. And now for number two, if we think about it, we can draw a similarly looking structure. But now because we have our benzene ring, we want to include these P orbital's that are for principal er, to our ring, at each carbon atom, we have those P orbital's. Alright, so they're perpendicular to the plane. And now, as we said before, this is our carbon atom, we have two hydrogen atoms. And now, when you think about it in this transition state, if you look at the previous catch pretty much you will see that this bond formation and bond cleavage essentially represents another p orbital. That's how we can think about it. Right? The high electron density over here and here during the bond formation and bond cleavage. It may be simulated as though it was an actual P orbital due to the fact that there is high electron density over here in the transition state. So, you may simply think about it by drawing another p orbital, right? Even though it's not technically correct, but this is the fact that there's high electron density. There is that bond formation and bond cleavage. You can visualize that way during the transition state. So you're forming one bond and losing the other. Right? So this is where your chlorine is partially negative. And here you are forming carbon oxygen bond. Okay, so once again, as we said, there will be this bond cleavage and bomb formation. Now, when you look at this transition state, let's label this as our transition state. As we previously assad essence reaction rate heavily depends on the stability of the transition state. And if you look at the two transition states, the main reason is that when we are thinking about this high electron density over here, you may notice that it's overlapping with the p orbital's of benzene ring. And due to that overlap, essentially, we have stabilization of the transition state, recall that a p orbital overlap leads to a higher stability. Okay, and just as we said before, this high electron density resembles a p orbital, right? And essentially it overlaps with the adjacent p orbital of the benzene ring. And that overlap. A higher stability is achieved. And therefore, Transition State No two would be more stable. So we can say this state is more stable due to the overlap of p orbital's with our high electron density of the bond that is being formed and the bond that is being broken and this one is less stable Because we don't have that stabilization. Right? There are no P orbital's that would be overlapping with our electron density from the two bonds. So that would be it. We can essentially say that reaction number two would have a foster rate and a higher yield for the reaction because the transition state for this essence reaction is more stable than the transition state for reaction number one. And the reason for that is because the electron density within the bond form than the bond broken, overlaps with the p orbital's of the benzene ring and that stabilizes the transition state. We may conclude that the correct answer choice to this multiple choice question is C. And we may also able the required transition state for this multiple choice question. The transition state required is the transition state for reaction number two. Because this is where we are actually explaining the phenomenon in terms of the p orbital of benzene. Thank you for watching

Key Concepts

Benzylic Position

Stability of Intermediates

Transition State Theory

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