Predict the mononitration products of the following compounds....

Question

Predict the mononitration products of the following compounds.

a. o-nitrotoluene

b. m-chlorotoluene

c. o-bromobenzoic acid

Accepted Answer

Hey everyone and welcome back in today's video, we are going to solve the following problem. What will be the product or products when each of the following compounds is mona and I traded and we are given three different compounds. 13 dimethyl benzene, salicylic acid, one bromo three chloral benzine. Let's start with part A. And we are going to throw benzene. We have one comma three that metal benzene. We may start by throwing our benzene ring positions one and three. They contain two metal substitue ints. And we are going to intentionally toward these metal substitutions in different colors. We will soon understand why we're doing this. But for now let's add one methyl group at position number one. And if we go counterclockwise, that would be our position number three. And we are going to add another metal group over here in blue. So now according to the problem, we want to perform a mon on it trey shin reaction essentially. That means we are going to use nitric asset in the presence of sulfuric asset. We call that in a mono nitrate in reaction. We are generating an electro file in the form of N O T. Plus. So our electro file would be N O T. Plus and it is going to replace one of the hydrogen atoms on the ring. Right? So we simply want to understand the Ortho power directors and activating deactivating groups. First of all notice how these two groups are l kill groups. So we can essentially label that in part a. We have alcohol substitute prints which are activating as well as Ortho and para directors. Right? So we're going to identify the positions for our works and power carbon atoms on our benzene ring. Starting with Our first metal substitution, 10 blue. So first of all we have to our two positions over here and then a power position over here. So these are our three positions for which will undergo mono nitrate ation with respect to that method group and moving on to the green one, we notice that there are two or three positions and one power position. You may notice how the two alcohol suspicions are directing to exactly the same positions on the benzene ring. Now one thing to understand is that these two positions were basically or so with respect to the left metal group and worked with respect to the right metal group. Their equivalent because of the line of symmetry passing through the ring. So we actually have two unique positions. We may essentially forget about this one and just concentrate on this one because the two are exactly the same. Now we have two possibilities for our products. Let's think about the minor and major products. First of all, we are going to produce an electrical aromatic substitution at this carbon atoms. So this would give us our first compound. So we're going to add our two methods substitutions and then both of them are directing to worth opposition. So we are just adding an N. 02 group here which replaces hydrogen plus another product. As we already said, we are going to choose one of these two equivalent positions. And we get our second product let's throw it a bit more carefully. So that would be our six member ring. Now we are adding our pie pons. There are these two original metal groups and we understand that we can add our nitro group over here to identify which one of the two years the major product. We have to recall that the stereo hinderance effect is not so important because method groups are not that bulky. Instead, it's actually the fact that the two method groups are directing at the same Ortho position. Notice how this position is worth with respect to the blue metal and worth with respect to the green one. But this position is Ortho with respect to the green one, but Sarah with respect to the blue one. So generally whenever we are thinking about electrical aromatic substitution, if stare ignorance is not such an important effect, we want to make sure that the two substations are pointing at the same position. So it's Ortho Ortho with respect to each and Ortho para with respect to each meaning the common position is Ortho and this is where substitution is preferred. So this will be the major product and the other one will be the minor one because the two substitutions are pointing at work and power positions respectively. And these are different with respect to each substitution. So we got our two products and we are ready to move on now for B we have salicylic acid and we just need to know how the structure looks like. We are going to draw benzene we can add our car bookstore gas group and at the adjacent carbon atom we simply have an alcohol group. So that's our salicylic acid. And once again we can actually label our substations in different colors. And let's not think about the whole substation. Let's just add that bon in blue for the curb oxalic acid substitute. And then let's suppose that the other one, the alcohol group will be drawn in green. Okay so we have our substitutes. We need to think about the Ortho and Para Directors activating and deactivating groups. So we're going to add nitric asset in the presence of sulfuric acid. Let's recall that here we have two substitutes. One of them is highly activating. That's our oh age group because it has a lone pair of electrons which can be de localized into the pi system. So in this case oh age is activating and it is also or so Para Director. So that's our first nails is recalled that activating groups generally are increasing the electron density on our ring and they are directing to work and power positions while the other group which is our core books telecast group is opposite, it's actually deactivating. And because it's deactivating just as the maturity of the carbonell groups, it is actually meta director. So let's identify our positions. Let's start with the Green one. That's our oh age group. We already said that it's worth though and power directing. Notice how this Ortho position is already occupied by the chromebooks telecast of the group. So we're not going to worry about it. We just have one Ortho position available and now we also have our power position over here. So these are two positions that can undergo substitution and that with respect to the car broke cilic cast group, we need to identify our meta position. So that's worth of meta. Right? So this power position is also meta to the car broke cilic cast group and similarly worth of meta. This is meta position with respect to the curb look silly cast group. Okay, so once again, just in the previous problem there directing at two exactly same sides. We need to think about the major and minor products. But first of all, let's draw the two possible products that we got over here for the first one. We're just adding our original substitue ints the car book silic acid group, the alcohol group. And let's suppose that we are going to start with this position. We said that the presuppositions are pointing at the power position with respect to wage. So we're going to add our nitro group over here, that's our first possible product and the second possible product would have N. O. T. Group. Ortho to the alcohol group. So we're going to add it at the Ortho position. That's N. OT. We simply want to understand which product is the predominant product. The first one is A major product and the 2nd is a minor product. The reason for that is it's simple because if you notice N OT group in each case it doesn't matter with respect to the curb oxalic acid group. So it looks like the curb oxalic acid group is not controlling anything. However, a wage group is a factor. Notice how in the major product, N. O. T. Group is as far as possible from a wage. So there are actually two reasons. One of them is the stereo hinderance effect. Right? If we added at the Ortho position, there's a hysteric entrance effect because the group is added to wage group, even though it's relatively small, that's a bit significant. So this also controls the fact that whenever we have fennel ring with a wage right? Benzene rings bonded to HHS final, The para substituted product is always the major one due to a minor strength endurance effect. And in addition to that, we also know that no two contains two oxygen atoms. So there's high electron density. We want to avoid these electronic interactions and repulsion between oxygen atoms. This is really significant when the two groups are adjacent to each other so the oxygen atoms can interact and we can experience the lone pair repulsion. We don't want that. So we're trying to make sure that our nitrile group is as far as possible from a wage. And to achieve that we're just going to add it on the opposite side of the ring. Notice how we added it, the power position. And this allows us to avoid sterile mandarins as well as electronic repulsion. And that's why the first structure is a major product and the second one is a minor one due to the fact that N. 02 is adjacent to O. H. Electronics and starik interactions essentially allows center center. This would be the minor product. Okay and now moving on to part C. We have one bromo 3 chloral benzine that we are part C. We are going to throw a benzene ring and we have to at our substations position number one will contain bromo group, let's add it in blue. So that would be Romo substitution. And then we need to add chloral at position three. The reaction conditions are exactly the same. We want to make sure that it's mona penetration. So we're going to use nitric acid in the presence of sulfuric acid. Now these are two halogen, they have lone pairs and we have to recall that halogen groups are activating. So going back to our description we can say that acts where access a halogen, they are activating because they can supply electron density to the rain deeds the fact that they have lone pairs and therefore that means they are worth. So para directors. Okay so let's see what positions we get for the directing effects. Starting with romaine in blue. There are two or three positions and one power position. So we have labeled these orto orto aura with respect to bro mean now with respect to chlorine we have to Ortho positions in green so we can say or so or so. And now there's one power position with respect to chlorine. Just as we saw before in the previous two examples the two substations are pointing exactly at the same site and we have to predict our products. Let's talk with the first one for the first one we are just going to use this position and we are going to form our mono nitrate in product. Notice how we will end up with three different products because the compound is not symmetrical. Right? So the three positions will allow us to get three different products. We are going to start by drawing our original substitutions bonded to the benzene ring. So we have our bromo substitution chloral and now for the first one we are going to use this position or the with respect to bombing and add the nitro substitution. This is our first product. Now for the second structure we said that we can also use this or the position or that one with respect to chlorine. Let's suppose that we start with this one. So we are adding our N. O. T group adjacent to chlorine. And there's one more possibility we can actually add nitro group in between the two substitutions. So these are our three possible products. Now that we see our possible products, we are going to think about the major ones and minor ones. Now let's start with the first one or actually let's start with the third one. That will be easier for us to understand for comparison. We call how previously we said that when we had two metal groups over here and we added our end of two substitution, it was the major product because the two metal substitutions were pointing at the same words opposition. However, in this case it's not really true to think about this as the major product. There are two reasons one of them is that bromine and chlorine. They are significantly large atoms. They have three lone pairs each and those lone pairs are directly interacting with the energy group. So we have electronic interactions and that would cause a lot of propulsion and low stability of the product. So there is therapy entrance effect. We're adding an 02 group in between. And in addition to that we have high electronic repulsion is because N 02 group is so closely substitution is basically squeezed right? It's in between. So we can say that this one is least probable for our products because if you look at the other ones to group is quite far from glory in right? As far as possible. And here it's basic N OT group is as far as possible from roaming. So we have maximized our distance, meaning we have lowered our electronic repulsion. And for this one, N OT group is as close as possible to each substation simultaneously. So that will be our least probable product now between the two remaining ones. This is really matter which adam is bigger between bromine and chlorine, bromine is slightly larger than chlorine is because whenever you go down the group in the periodic table, atomic radius increases. Right? So grooming is larger. Now notice how energy group here is closer to booming which is larger. And other two group here is closer to chlorine which is smaller. So it might look like this would be the major product, but it's actually not the case. We call that booming is less electro negative. Chlorine is more electro negative. Okay, so because bruning is less electro negative, it's essentially withdrawing the electrons are basically whenever you're thinking about the booming carbon bond, it essentially implies that the bond is less polarized electrons are less polarized towards blooming. So g to the inductive effect, we can say that this position adjacent to booming or basically this carbon bromine bonds. Right? We can say that this position is more activated simply speaking, roaming is not withdrawing the electron density that much due to its lower electro negativity, but chlorine is right, it's more electro negative, meaning that this position is less activated than this position because roaming is less electro negative. And this is explained using the inductive effect. So, knowing that there's the major effect here which corresponds to the inductive effect, we can conclude that the first product would be the major product. So the first product is the major product And the 2nd is the minor one. Once again, now that we clearly see which one is the major and which one is the minor. We can just recall that as we said before, even though grooming is slightly larger than chlorine, it's actually the inductive fact that it's important here. We want to make sure that the halogen is not decreasing electron density significantly at the adjacent positions. And we want to make sure that it's not so electro negative because the more electro negative anonymous, the more polarized our bond becomes. And because pruning is less electro negative than chlorine, this position becomes more activated and the electron density at this position is larger than the electron position at that. Then the electron density at this position. And therefore this will be the major product. The second would be the minor one and the third one would be the least probable jesus terry hindrance as well as in electronic interactions. So now that we clearly see what our products are. Let's label our final answers starting with A. We have two products one major one minor. Moving on to B. We have two products one major, one minor, and finishing with C. We have three products one major, one minor and one least probable. Right? So these are our possible products based on the identity of the products. We can conclude that the correct answer to this multiple choice question is B. However, if you have any questions, don't hesitate to leave a comment down below. Thank you for watching.

Predict the mononitration products of the following compounds.
a. o-nitrotoluene
b. m-chlorotoluene
c. o-bromobenzoic acid

Key Concepts

Electrophilic Aromatic Substitution

Ortho/Para vs. Meta Directing Effects

Nitration Reaction Mechanism

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