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
(vi)

b. Predict the mononitration products of the following compounds

(vi)

Accepted Answer

Hey 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 electrostatic aromatic reaction is usually the one that is on the more activated ring. The other ring and other substitutions at present also affect the reaction. Regional chemistry predict the major product of a monarch chlorination reaction for the following compound. And indeed, if we look at the given compound, it consists of two benzene rings bonded by a single bond and each ring contains its own sub stich. In first of all, if we look at the left hand side, we notice that the substitution bond, it's our ring is a carbon compound and in particular that's our key tone. If we look at the right hand side, the second ring contains and a wage group, that's our alcohol group. Right. We can essentially say that this is our alcohol group or simply speaking O. H. For complete clarity. And on the left hand side we can say that's our ceo, right? So that would be C. E. O double bond with an L. Kill substitution. So that be our metal group in this case. But let's just in general state that this is our. Okay, so now we see clearly what our submissions are. As the problem suggests, we want to understand at which position we are going to observe mono chlorination and we have to recall that the majority of the carbon ill groups are deactivating. So the group that is deactivating is our carbonell group. So we can say key tones are deactivating whenever you have an AL to hide or a key zone, that would be the activating group. However, we have one of them being an activating group and hide. Troxell is actually strongly activating because there's a lone parent oxygen which essentially can be de localizing into that system. Recall that whenever we have a lone parent oxygen, it can form a double bond, a disposition breaking this pi bon and essentially de localizing the lone parent of oxygen along the ring. So oh, ages strongly activating. As the problem suggests, the substitution will take place at a more activated ring. So because ohh activating and our carbon group is deactivating. We are going to choose the second ring since it has an activating substitute. Print. And because we're looking at a monoplane nation reaction, we just have to determine which position is going to be substituted. Now recall that not only a wages activating, it's actually increasing the electron density at Ortho and power oppositions so that the electrical substitution is expected to be observed at worth oppositions. So notice how with respect to the wage group we have to Ortho positions. One of them is already substituted. So you can only think about one so far and then we have our power our position. So we have decide which one of the two we are going to use for our monarch chlorination reaction. Now the major product whenever an Ortho and power products can be produced will be at a power positions. We are going to choose a power position due to the fact that the Ortho position is slightly starik lee hindered. So generally we are going to choose our power position to form our major product even though Ortho produces quite a significant amount of the product generally that's about 2 to 1 ratio. But the power product is going to dominate. So all that we have to do is just combine our logic. We understand that we are going to use our second ring for the substitution. We understand the reason why the para substituted product will dominate. And of course because high Troxell is activating and Ortho para directing, we just had to choose one of the two positions. And we are going with para due to the fact that hysteric hindrance is an effect here. So let's draw our final product. We understand that simply we need to draw our starting material. So we're going to draw our two benzene rings bonded by a single bond. Now let's draw our key tone than our high Troxel group and have some play. Let's find our power position. So that's Ortho meta para right, so we are going to add our chloral substitution at this position. Keep in mind that for this reaction whenever we want to perform on a chlorination, the set of regents required would be as follows, we're going to use Cl to write the atomic chlorine and then we of course, need to use our Lewis asset as a catalyst, which would be iron three chloride. So this would allow us to achieve a substitution at the power position. And we may label this as our final product because we were asked to draw the major product formed in this reaction. So let's label it in green and based on the answer that we have got the correct answer choice to this multiple choice question is D However, if you have any questions, don't hesitate to leave a comment down below in the comments section and thank you for watching.

Key Concepts

Electrophilic Aromatic Substitution (EAS)

Activating and Deactivating Groups

Ortho/Para-Directing Effects

Watch next