Biphenyl is two benzene rings joined by a single bond. The sit...

Question

Biphenyl is two benzene rings joined by a single bond. The site of substitution for a biphenyl is determined by (1) which phenyl ring is more activated (or less deactivated), and (2) which position on that ring is most reactive, using the fact that a phenyl substituent is activating and ortho, para-directing.
a. Use resonance forms of a sigma complex to show why a phenyl substituent should be ortho, para-directing.
(i)

a. Use resonance forms of a sigma complex to show why a phenyl substituent should be ortho, para-directing.

(i)

Accepted Answer

Hey everyone and welcome back in today's video. We're going to solve the following problem. Buy fennel is an aromatic compound having two benzene rings joined by a single bond, the site of an electrical a traumatic reaction is usually the one that is on the more activated ring the other ring and other substitutions if present also affect the reaction. Regional chemistry for the following reaction use resonance forms of a sigma complex to explain the observed ritual chemistry. Now, we may notice that we are starting with our by final compound. Right? We have two benzene rings joined by a single bond. One of the rings. Let's essentially label it. S ring number one has no substitutions other than the fact that it's bonded to another ring. The second ring besides being bounded to ring, number one has a substitution and this substitution is our third beautiful group. Right? So that's our third beautiful substitution. Now, we may notice that we are essentially performing a nitrate in reaction because we're using nitric acid and combination with sulfuric acid. And you may essentially notice how we are adding our N. O. T substitute print at the power of position of the second ring. Let's understand, first of all why that happens. And then we will govern our explanation using a mechanism and the stability of the sigma complex. First of all, let's consider the first ring. The first ring only has one substitution. If we think about the second ring is our actual substations, we can say that ring number one has one substitution and that substitution is our final substitution. Now a final substitution is weekly activating. Let's recall that. And because it's weekly activating, we may understand that activating substitutions will essentially increase the electron density at Warsaw and power of positions. So we have this directing effect. What about the second one? Now, for the second ring, it essentially has two substations if we think about the first string as being a substitution. So first of all, let's think about the third beetle group, It has a third beetle group which is classified as an L. Kill group. And you may recall that it's also activating and because an L kill group is activating, we have similarly Ortho positions and power position being hire an electron density. And these positions we would expect to observe our electro filic aromatic substitution. Right? Because activating substitutions are Ortho and para directing. What about this ring? Now? It's weekly activating. We have another final substitution for the second one. So that's our final we already know that it's weekly activating. Right? So if it's weekly activating again, we would have Ortho. Right, so still these two positions, we may label another circle to indicate that this final substation is also worth a directing. So there are two arthur positions with respect to that ring and one power. So exactly same positions. Right? The two substations our third beetle group and our final ring, they're both directing at the same positions because we have a greater directing effect. Right? Because we have to substitute parents directing the same positions. This means the substitution will take place at this ring. Now the reason why we're choosing power position with respect to third beetle group is because that's what corresponds to a major product. And the main reason hysteric hindrance. Because third Beatle group is a hysterically hindered group or basically it's very bulky. We would not want to perform our electrical aromatic substitution at work propositions, but rather our power position is not as hysterically hindered. And that's why we cnooc at our opposition with respect to our third beetle group. So that makes sense. But now let's see the mechanism and let's understand why this is favorable thinking about the stability of the sigma complex in its residence forms. First of all, we are going to indicate the electro silicate text up. Or basically the attacks on the electro file. Right? So those would be our rings. Now, essentially we have our third beetle substitution. And now, as we said before, because we are generating our electro file, recall that the electro file generated here would be an OT with a positive charge on nitrogen. So that would be our electro file. And we already know that we are adding it at the power position with respect to Ter Beatles. We're going to use this bond to clearly show how the attack takes place. Now we are ready to draw an arrow, indicating that this is our actual attack on the electro file. So let's add our electro file to the structure. Okay, so let's retry that ring. Now, the next one and here we have our third beetle substitute. Right? And essentially you may notice that we are going to add an ot group at this position. And as a result we're getting our positive charge over here. So we got our sigma complex and we simply want to think about the possible residence forms. Let's begin with the first one. So this is where residents starts. And we're going to show that first of all, there is a possibility to shift these pi electrons to form a carpool, karaoke, vanity, different position. So we're going to shift these by electrons forming a double bond over here and we will have a positive charge here. So let's add that. Our first string Now our 2nd Let's add our 3rd video group. We understand that. Now we have a positive charge in this position, there will be pi bond, the original pi bond and our Nrt group. So that's our nitro group. What would be the next possibility for us to de localize our pi electrons. So now we can think about this adjacent pi bon and we can essentially shift it over here so that we will get a positive charge at a different position specifically over here. That would be our made up position with respect to turbinado. So let's add this residence form. Now we have our original pi bond. We are going to add another one over here. And specifically here, we still have our energy group, our turbinado substation. And then we may notice how we are getting our positive charge at this position. That would be meta position with respect to turbinado. Okay, would be the other possibility to troy residents form. So we got our positive charge. We can essentially perform the following the localization first. Let's begin by shifting this pipe on over here. But if we only do that, we return to the previous structure, meaning we need to do something else. We are going to essentially shift this bond over here. Right? But this implies that we get the second, the first residence form that we got. So that's not sufficient for us. We need an additional shift. And this is where this rain comes in and that's why the sigma complex is really stable because this ring also participates in residence. Notice how we can use these pi electrons form a pi bond at this position and actually get our positive charge over here. So we are again the localizing our pi electrons to an even greater extent and this shows and extended stability of the sigma complex. So let's draw the result on residence form for this one. Okay, so now we have our pi bonds, the positive charge ends up at this carbon atom. And now looking at the shifts that we perform, we have our pi bonds at these positions. Now we are going to add our substations and that's it. We got our residence form, let's just keep in mind that we still need to make sure that we add a pi bond over here. So there's one more pie bon. Okay. And now we are ready to think about the next structure. We're going to notice that there is a possibility to shift this pi bond over here, right, Because we have a positive charge, so we can definitely do that. And if we do so we get our additional residence form, let's draw it. Okay, so let's add our substitutions. And now let's think about the positions of our pi bond. So these to remain the same. Now, we haven't affected this one. So we're going to add a double bond. But now we have our pi bond over here. So let's added the previous double bond. And now, because we have a loss of electrons at this position, there will be positive charge. Well done. So we got an additional residence form. And now we can think of this pipe bomb, we can essentially de localize these by elections over here and end up with a positive charge at the upper position. So we're going to do that as well, let's indicate resonance to our rings. But before that we can just show how the pi electrons are de localized over here to get a positive charge of this position. So we can throw our rings. Now let's add our pipe on here. We have our original pipe on and this one has been formed. So we have our positive charge over here because we had a loss of electrons or this carbon has lack of electrons and as a result we have our positive charge. Now, what would happen if we d localize these pi electrons? We would essentially form a positive charge at this carbon. And this is actually what we started off with. Right, of course, you may notice that the positions of the pi bonds will actually differ right? Because we have we we would essentially have pi bonds these positions instead of the original ones. But that's perfectly fine. It's ready sufficient to stop here. We can essentially notice that of course we would have a different distribution of pi bonds on this ring. But it's already sufficient to understand that we went back to the same atom and the positive charge would now be at the same carbon atom. So we may say that this is our final residence form because we are just getting back to the previous atom and we are going to get a positive charge, the same atom that we started off with and we can label these as our resonance forms. So, these would be our residence forms and based except from this one, right, because that's the actual attack on the electro file. So we have 123456 residence forms. and we may conclude that the correct answer to the multiple choice question is B. However, if you have any questions, don't hesitate to leave your comment down below. Thank you for watching.

Key Concepts

Biphenyl Structure

Electrophilic Aromatic Substitution (EAS)

Resonance and Directing Effects

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