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.
b. Predict the mononitration products of the following compounds
(iv)

b. Predict the mononitration products of the following compounds

(iv)

Accepted Answer

Everyone and welcome back in today's video, we are going to solve the following problem by fennel is an aromatic compound having two benzene rings joined by a single bond. The site of an electro aromatic reaction is usually the one that is on the more activated ring. The other wing and other substitutions if present also affect the reaction. Regional chemistry predicted a major product of a mono chlorination reaction for the following compound. Just as the problem suggests, we have our by fennel because we understand that there are two benzene rings. Let's suppose that this is our first one and this is our second one. And you may clearly notice that they are bonded by this single bond. Okay, so this is our buy fennel structure and we want to understand what kind of product can we get if we use a mono chlorination reaction, recall that whenever you're dealing with manure chlorination, You essentially want to make sure that you're using chlorine in the presence of a Lewis asset. Iran three chloride is typically used along with chlorine. So based on this problem, our goal is to first of all understand whether electro chromatic substitution will take place at ring number one or at ring number two. And to do that, we want to identify our substitutions. So let's look at ring number one, it has one substitution. Right? This is just the bond connecting to benzene rings and then this is our substitution. We are going to go along our bond and we are going to identify the first atom. Okay, so here we have our first atom. That's our siege to group and we can identify this as an L kill substitution. Right? Because that's our saturated hydrocarbon chain. So that's our L kill substitution. On the other hand, if we look at ring number two, we notice that there's again one substitution. Of course you may notice that it's cyclic. But that's not a problem. We are only interested in the first atom. So if we go along that bond, we notice that this is oxygen. And because oxygen is bonded to two carbon atoms, you can think of this substitution corresponding to an ether functional group. Right? So we have two types of substitutions based on the hint. And the problem electrolytic aromatic substitution takes place at a more activated rings. So we have to think about the nature of these two substitutions. So let's recall that al kill groups are activating. The reason why they're activating is because alcohol groups can essentially supply electron density towards our ring and increase the electron density on the ring to increase the rate of an electro aromatic substitution. So, L kill groups are activating or basically increasing the rate of electro aromatic substitution. And because the other group is an ether functional group, we may also recall that they are activating as well. So both of them are activating. We essentially want to understand which one of the two is more activating. Now let's think about it. So the alcohol group essentially increases the electron density across our ring based on the inductive effect. While the ether group can actually supply a lone pair, right, We can notice that oxygen has two lone pairs. We can use one of them to show how this lone pair can be de localized into the ring system. Notice how we can create a pi bon here, meaning we're going to break the adjacent pi bon and we are going to transfer a lone pair of electrons from the original pi bond onto this carbon and so on. We are going to have several residence forms. So due to the fact that this lone pair of oxygen is the localized, right? Essentially, it allows us to increase the electron density as worth opposition. So that's our worth opposition. That's our other work opposition. And this is our power position. Right? So the electron density is significantly increased at these positions due to the fact that there's a lone pair on oxygen. And therefore we may conclude from here that ether is strong activating our rings stronger than an alcohol group. So let's see that, let's say that ether is more activating than an l kill group due to the fact that there's an actual lone pair rather than these carbon carbon electrons that work via an inductive effects. So that's definitely stronger because ethers more activating according to the context of the problem, substitution will take place at the ring, which has a more activating group. So, essentially that would mean the substitution will take place at ring number two, that's it. So all that we have to do is just think about which position we are going to choose. We call that activating groups increase electron density at Ortho and power positions. So we have to arte positions one power position. Keep in mind that para is the major product. Okay. The reason why power would produce the major product is because of a slight Sterritt, hinderance effect, right? Or the position is closer to our substitution. So this side is slightly hysterically hinder power position is more accessible and therefore we expect to have our substitution at the power position. Activating groups increased electron density of water power positions. Power position produces the major product. We understand that ether is more activating and we have all the information. We need to draw our product because this is our part of our opposition. We are going to replace this hygiene with chlorine and that's it. We are done. We can essentially draw our product. So let's draw our 6th member wing. That's Benzene Ring # one then we can connect another benzine ring. Right? That's our by final. Now we control the remaining part. That's our oxygen. Okay. And now we understand that the substitution takes place in the power position. So we're going to replace that implicit hygiene with chlorine and that will be our final product. So let's label it as our final product and based on the identity of our final product. We may conclude that the correct answer to the multiple choice question is B. However, if you have any questions, don't have states lever comment down below and thank you for watching.

Key Concepts

Biphenyl Structure

Electrophilic Aromatic Substitution (EAS)

Ortho/Para Directing Effects

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