Reactions at the benzylic position
Free radical bromination, nucleophilic substitution, oxidation. Created by Jay.
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- Do you have any videos on organometallics (excluding grinards) like the suzuki, stille, ring opening, cross coupling etc?(21 votes)
- At3:39, I understand the losing a Hydrogen on the benzylic position can be resonance stabilized, while on any other carbon on the akyl group can't. But why not just losing a Hydrogen on the ring itself, as it can be resonance stabilized in the exact same way?(7 votes)
- I think the difference is that the ring has aromatic stabilization, while the benzylic position does not. This means that removing a benzylic hydrogen results in a smaller energy gain (and thus requires less energy) that removing a ring hydrogen, since the latter destroys the aromaticity of the ring.(2 votes)
- Why does the hydrogen turn into a radical at2:43?(3 votes)
- This is a free radical reaction. NBS stands for N-bromosuccinimide. In the initiating step, the NBS loses the N-bromo atom. leaving behind a succinimidyl radical (S·):
NBS → S· + ·Br
It is the S· that removes the magenta hydrogen to form succinimide (SH):
S· + C6H5CH2CH2H3 → SH + C6H5CH(·)CH2CH3
Then C6H5CH(·)CH2CH3 + NBS → C6H5CHBrCH2CH3 + S·
and so on.(9 votes)
- Do you have any videos explaining the free radical substitution reaction with NBS on allylic positions?(4 votes)
- Jay says at4:20that substitutions at the benzylic position can be either SN1 or SN2 but the german Wikipedia article on nucleophilic Substitution states that SN2 reactions can never occur there.
On an Exercise i did with pretty much the same reagents from5:35I also thought that it would be SN2 but the solution said SN1.
So i guess Jay is wrong, but why can't SN2 occur here? Because even the primary carbokation is stabilized enough by the Benzene ring?(2 votes)
- 1° benzylic halides typically react via an SN2 pathway,.
2° and 3° benzylic halides typically react via an SN1 pathway, via the resonance stabilised carbocation.(3 votes)
- Is there a way to add Chlorine to the benzylic position instead of bromine?(2 votes)
- Since Chlorine is a halogen it can be added in the same way. It would just have a different rate of reaction due to the difference in electronegativity.(2 votes)
- why doesn't and electrophilic aromatic substitution take place in this case ?(1 vote)
- Because you are using a nucleophile. Electrophilic substitution requires an electrophile.(2 votes)
- At about5:20when substitution occurs at primary carbon via SN2, why it has to be the benzylic carbon? It doesn't seem like there's a radical that would need to be resonance stabilized, since the mechanism is concerted.(1 vote)
- Doesn't KMnO4-KOH also give benzoic acid as product on reaction with any type alkyl benzene?(1 vote)
- It does for all alkyl groups with a benzylic hydrogen. t-butyl groups do not react.(1 vote)
- Dear Khan Academy Community,
At4:40, how would the reaction be affected if instead of Sodium Hydroxide we used something much more prominent like sunlight?
So, Benzyl bromide + Sunlight = What? A reaction does take place, right?(1 vote)
Reactions that occur at the benzylic position are very important for synthesis problems. So let's look at a few. We'll start with the free radical bromination of alkyl benzenes. And so here is my alkyl benzene, so a benzene ring, and I have an alkyl group attached to that. So this is a carbon. And on that carbon, this is the benzylic position. So this is will be a benzylic hydrogen attached to that carbon, which is necessary for this reaction. And then there are two other things attached to that carbon here. So if we add some NBS and some heat-- and this is a free radical mechanism, so usually you add something to initiate this, like peroxide-- you can see that we get a bromine added on to the benzylic position in the place of that hydrogen there. So let's look at a quick reaction. So if I took propylbenzene-- so I'm going to go ahead and draw out propylbenzene-- so three carbons coming off like that. And to it, I had some NBS and some heat, and you could do this in a solvent like carbon tetrachloride and some peroxide here, add in some peroxide to start your radical mechanism. And we have three carbons here. So there's one, two, three carbons for a propylbenzene on our alkyl group. But the only one that is at the benzylic position is, of course, this one right here. So that's where we're going to end up adding our bromine. So when I draw the product, I go ahead and I still have all three of those carbons, but the bromine adds exclusively to the benzylic position right here like that. And if I look at the hydrogens that I have attached to this carbon-- so we started off with two hydrogens attached to that carbon, and we'd left with only one of those hydrogens. So let me just go ahead and highlight the hydrogen in magenta here. So if we go back up here to this general reaction, so this benzylic hydrogen, I could say that it's either one of these two. I'm going to go ahead and say that it's this one right here. And so the other hydrogen is attached to that carbon, and it is still left for our final product. So the bromine took the place of one of those benzylic hydrogens there. Now reason why the reaction occurs only at the benzylic position is because in this mechanism you form a radical. So let me just go ahead and draw what the radical would look like for this reaction here. So a radical for the reaction of propylbenzene, so we have our benzene ring. And we already lost the hydrogen in magenta, so now we're left with just the hydrogen in blue here like that. And we're going to have an unpaired electron right here on this carbon. And because this radical, this electron, is right next to this benzene ring, we can draw some resonance structures. So I could take this one electron and show it moving into here with my fish-hook arrow. And then from this pi bond, I could take one of those electrons and move it into here. And the other one could come off onto this carbon. So for a resonance structure here, I can draw my ring, and I can show these pi bonds over here. Now I can show a double bond into here. There's still a hydrogen attached at this point. And then I have my other carbons like that. And then there is an electron out on this carbon. So the benzylic radical is resonance stabilized. And if you formed a radical at the other two carbons on the rings-- not on the ring, on the akyl group right here-- these are not resonance stabilized, because they're not right next to the ring. And so that's just a little bit of insight as to why the reaction occurs only at the benzylic position. It's because of resonance stabilization of the benzylic radical. So let's look at another type of reaction. And this, of course, would be a substitution reaction of a benzylic halide. So I put a bromine, here but you can imagine a different halogen. And if you add a nucleophile in a solvent, depending on the classification of your alkyl halide portion of the molecule and what solvent you use, the mechanism could be either SN1 or SN2. But of course, the end result would be for your nucleophile to substitute in for your halogen at the benzylic position. So if we were to look at a reaction here-- so let me just go ahead and show one. So let's say we started with benzyl bromide here. So this is our benzyl bromide molecule. And to benzyl bromide, we add some sodium hydroxide, so I have Na plus and OH minus. I'm going to go ahead and draw my lone pairs on my hydroxide anion, which is going to function as a nucleophile. So this is our nucleophile. And if I look at the structure of this alkyl halide over here-- so this carbon that's bonded to my bromine is bonded to one other carbon, so that's primary. And if we're talking about a primary alkyl halide, decreased steric hindrance, and so your mechanism would be SN2. So if you think about an SN2 type mechanism for this, which is a concerted mechanism, your nucleophile is going to attack this carbon at the same time that these electrons kick off onto your bromine to form the bromide anion as a relatively stable leaving group. And so this is a concerted mechanism. Everything happens at the same time in an SN2 type mechanism. And we end up with an OH replacing our Br. So we form benzyl alcohol for this reaction. So that's SN2. If you thought about an SN1 type reaction, let's go ahead and think about that. If I started with a tertiary alkyl halide-- so something like this-- so if I were to analyze that, put my lone pairs of electrons on this bromine, the carbon that is attached to my bromine-- is attached to one, two, and three other carbons. So therefore, it's tertiary. So you could think about an SN1 type of mechanism. And we know in an SN1 type mechanism-- let me go ahead and just write that here. So in an SN1 type mechanism, your first step is dissociation. So you're going to get these electrons in here kicking off onto your halogen onto your bromine to form your bromide anion. And you would, therefore, be taking a bond away from this carbon here at the benzylic position. So let me go ahead and draw the resulting carbocation. So I have my ring. I have these two methyl groups coming off of that carbon. And that is where I have a plus 1 formal charge. So I have a benzylic carbocation here, which is resonance stabilized. So I could think about these electrons moving into here, and I could go ahead and draw a resonance structure. So for this resonance structure, these pi electrons are over here. Now, there's a double bond between those two carbons. And let me show those electrons here. So the electrons in blue have moved out to here to form a pi bond, took a bond away from this carbon. So that carbon gets our plus 1 formal charge. And you could, of course, continue. But the point is that the presence of our benzene ring allows for resonance stabilization of our benzylic carbocation, which means that the substitution is going to occur at the benzylic position, because of that resonance stabilization here. So let's look at one more reaction. And this is oxidation of alkyl benzenes. And so here, I have my alkyl benzene. So once again, carbon with a benzylic hydrogen and then two other things attached to my carbon here. If I added some sodium dichromate source of protons, so something like sulfuric acid, and I heat up my reaction, I can oxidize my alkyl side chain to a carboxylic acid functional group over here. So this would, of course, be the benzoic acid molecule. You could also do this oxidation with something like permanganate and some heat as well. So that's another possibility. Let's go ahead and do an example. Let's start with-- let's do butylbenzene this time. So if I have butylbenzene, so four carbons, for my alkyl group here. And instead of rewriting all that stuff, I'm just going to put some ditto marks here. So we add some sodium dichromate source of proton, sulfuric acid, and some heat. We're going to oxidize our side chain. And it doesn't really matter the length of the alkyl group you have. In this case we have four carbons. We're still going to get a carboxylic acid. The reaction is still going to occur at the benzylic position. And we get a carboxylic acid here. Again, a complicated mechanism, so I won't go into it. But again the length of carbons that you have, it doesn't matter. So you could add on a bunch more, and you would still get this as your products. What about if you try to do this reaction with something like tert-Butylbenzene. So instead of butylbenzene, now we have tert-Butylbenzene like that. And we went ahead and added sodium dichromate and sulfuric acid and heat again. We heated it up. We wouldn't get any kind of reaction. So no reaction this time. And that's because we don't have any benzylic hydrogens present. So we back here to our generic reaction. There's a benzylic hydrogen. If I go down to here, here's a methyl group, here's a methyl group, here's a methyl group. And then this is a carbon. So we don't have any benzylic hydrogens. There are no hydrogens attached to this carbon here, which is necessary for the reaction to occur. And so we see no reaction in that case. So just keep these reactions in mind when you're doing synthesis problems, because you will see that they are very useful.