Markovnikov's Rule and Carbocations. Figuring out which addition reaction is more likely. Created by Sal Khan.
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- Could anyone explain me in detail why are tertiary carbocations the most stable while primary carbocations are least stable?(8 votes)
- Alkyl groups are electron donating groups. If they donate some electron density to the carbocation, this reduces its positive charge and lowers the energy (stabilizes) the cation.
A 1,1-dimethylethyl cation (3°), for example, has three methyl groups attached to the cationic carbon. An ethyl cation (1°) has only one methyl group on the cationic carbon. So 3° is more stable than 1°.(12 votes)
- At3:10how does Bromine atom with 8 e-'s attain an oxidation state of -1. Going by the Octet rule, shouldn't it be neutral??(3 votes)
- Even with 8e- it's not neutral. It has 7e- in the it last orbital on neutral state. Remember, overall numbers of protons in the nucleus against the overall numbers of electrons after you add the 1e- to form bromide and will get -1 oxidation state.(12 votes)
- What does Markovnikov's rule state?(5 votes)
- When you add HX (X could be a Cl, Br, OH, etc.) across a double bond, the major product will be the one where the carbon with more carbons bonded to it gets the X and the other carbon gets the H.
The explanation is that the double bond picks up the H and then the more stable carbocation is formed which will be the one with more carbons bonded to it(7 votes)
- Hi, could someone explain how Markovnikov's rule applies to the reaction between ethene and a hydrogen halide? I'm just a bit confused because ethene is symmetrical so wouldn't both isomers be of equal predominance? Thanks(3 votes)
- The most general statement of Markonvikov's rule is that when an unsymmetrical reagent adds to an unsymmetrical alkene, the more positive part of the agent goes to the carbon that has more hydrogen atoms. Since ethene is a symmetrical alkene, Markovnikov’s rule does not apply.
There is only one product from the addition of HX to ethene.(5 votes)
- sal i think that when we use HBr it follows the anti-markovnikov's rule isn't it??
so i believe that for this reaction the primary carbocation molecule is more stable..
This is an exeption of Markovnikoff's rule..
please correct me if i am wrong..
- no, anti-markovnikov's rule is observed in the presence of a peroxide. Unless u have a peroxide , markovnikov's rule will be followed(4 votes)
- Is it also possible to consider the Newman projections of the products? The first one has no gauche interactions while the second one does? And gauche interactions have higher energy state, therefore more instability?(3 votes)
- If by second one, you mean the 1-bromo-pentane, then not necessarily. Looking down the C2-C3 bond, we can arrange the bromine in the anti staggered conformation (pointing in the opposite direction) of the propyl group, so it wouldn't have any gauche interactions. The other staggered conformation would have a gauche interaction, but it would not be the most stable conformation.
The first one (2-bromo-pentane) also wouldn't have any gauche interactions as you pointed out because the Br and propyl groups are both connected to C3. Great question!(1 vote)
- in my textbook it's written that H+ acts an electrophile( but an elctrophile accepts a pair of electrons ) here the H+ snatches one elctron from secondary carbon ( it returns to its neutral state)and then forms a covalent bond with the primary carbon atom . how is this equal to accepting an electron pair(it only accepts one electron and the other in covalent bonding) then how is it an electrophile?(2 votes)
- Despite how Sal draws and explains it, the H+ isn’t accepting only one electron, it is accepting a pair of electrons.(3 votes)
- what is the difference between a free radical and a carbocation/anion?(1 vote)
- A free radical has no charge and has an orbital with one electron.
A carbocation has a positive charge and a vacant p orbital (no electrons) on a carbon atom.
A carbanion has a negative charge and has an orbital with two electrons (a lone pair) on a carbon atom.(5 votes)
- Around5:20Sal said that the experiment would result in more 2-bromopentane than 1-bromopentane. However, my professor emphasized that Markovnikov's rule is never broken, as in there won't be any 1-bromopentane at all. Who's right here?(1 vote)
- Not hating on your professor, but I think Sal gets a gold star for this one. To my knowledge, both products should form, but 2-bromopentane will be the major product.
Perhaps your professor's use of the term never broken could be interpreted to say: "Yes, 2-bromopentane will always be the major product, even though some 1-bromopentane will form."
Regardless, don't feel shy to ask your professor directly!(4 votes)
- During this mechanism, does the hybridization of the C's change? Thanks!(2 votes)
- Yes it does.
At the beginning, both of the Cs are sp2.
Once the H has bonded to one of the Cs, that C becomes sp3, but the carbocation C is sp2.
Once the Br has added to the carbocation, all Cs are sp3.(2 votes)
In the last video, we saw a potential mechanism where if we reacted hydrogen bromide with this alkene right over here, that we could essentially add. We had the addition of this halide to what started as an alkene, and then it ended up as 2-bromo-pentane, as an alkane. But when we did that, we made a somewhat arbitrary decision, or I didn't explain why we made the decision. We said, look, this hydrogen is going to be partially positive, because this guy's so electronegative, and maybe when it's partially positive, it'll be attracted. Maybe it'll just bump in just the right way into one of these carbons. It'll maybe swipe its electron. We somewhat arbitrarily in the last video decided that it would swipe this guy's electron. But you could just as easily imagine a world where it swipes an electron from this guy. So let's draw a mechanism for that and just think about which one is more likely to actually happen. So what happens? So once again, this guy-- let me draw all of his valence electrons, so this is the bromine. One, two, three, four, five, six, seven valence electrons. You have the hydrogen. I'll do it in the same color. The hydrogen has its electron right there. This is partially positive and this is partially negative. The hydrogen might want to swipe one of these electrons away. Let's do it from this guy right here. So he has this electron right over here, so the other side of that bond, it goes to the hydrogen when the hydrogen goes near it, or maybe it's attracted to it. And when it goes to hydrogen, then the hydrogen lets go of the electron that the bromine wanted all along, because it's so electronegative. So then that electron goes to the bromine. So after we do that, what will it look like? What will be the next step in our reaction? And it will look fundamentally different than this right over here. So now what happens? So we have a carbon bonded to two hydrogens, and it only has a single bond to the other carbon, which is bonded to the original hydrogen right over there. Let me write my hydrogens a little bit-- actually, let me write this whole thing a little bit neater. So you have your carbon bonded to a hydrogen and another hydrogen, and now it only has a single bond to this carbon right here, which is bonded to a hydrogen and then the rest of the chain. Let me just draw the rest of the chain right here. And now this electron went to the hydrogen. The other electron that it was paired with is still with this carbon, so now this carbon is now bonded to that hydrogen over there. So this blue electron is now with the hydrogen. Let me draw. So the blue electron that was here has now gone over to the orange hydrogen. Let me draw it a little bit neater than that. It has now gone over to this hydrogen right over there. And then the hydrogen lost its electron to the bromine. So the bromine originally had seven valence electrons: one, two, three, four, five, six, seven. And then it nabbed an extra electron from the hydrogen, so now it will have a negative charge. It is a negative ion. It is bromide. It's a bromide anion, I guess you could call it. And since this guy lost an electron, he had four valence electrons, lost one to the hydrogen, he now has a positive charge. He's a carbocation. So notice the difference. Before, this guy lost the electron, and so the hydrogen bonded to this carbon. In this situation, this guy lost the electron, so the hydrogen bonded to the other carbon. And so you can imagine, from here, something very similar happens as what happened in the first video, but now it happens to this carbon right over there. So let's do that. This carbon is positive. The bromide ion is obviously negative, so maybe he'll want to swipe his electron away. So this electron then goes to the carbocation and then it will form a bond. This green will go to the carbocation and then this purple one still stays with the bromine. And so they'll have a bond. They're paired up, you can imagine it. So then we're left with, we have a carbon. We have our original hydrogens. We have this carbon, that hydrogen, the rest of the chain: CH2, CH3. And then you have this hydrogen right here that it bonded to. That was our first step. And now the bromine has bonded to this carbon right over here. The bromine has bonded over to that carbon right over there. And we're done! This is another possible mechanism. This one we ended up with 2-bromopentane, right? Because it's on the number two carbon. Here we have 1-bromo-pentane. One, two, three, four, five. Still five carbons. It's just the bromine is attached to the one carbon here, attached to the two carbon here. So we now need to think about it, because on a first cut, these both seemed like reasonable mechanisms. But if you did it experimentally, you would see that this is the one that you'd really observe. I actually haven't done this exact experiment, so I don't know the proportions. But you're going to observe this one disproportionately. The great majority of the products that you see are going to be this one, not that one. And so the question is, well, you know, they both seem like reasonable things to do up here. Why is this one so much more likely to happen than that one? It all comes from something called Markovnikov's rule. And there's a couple of ways to think about it. When Markovnikov thought it up, or he observed it, it seemed to work. They weren't 100% sure about why it worked. We can think a little bit about why it worked. So Markovnikov's rule, a couple of ways you can think about it. You can think of it as the thing that already has more hydrogens is more likely to get more hydrogens, so that's what happened here. This thing had more hydrogens on it than the right carbon right here. This right carbon had a hydrogen, but it had some other alkyl group attached to it. And so the thing that had more hydrogens ended up with the hydrogen, and then the thing that had more groups, this character right here had more groups, right? He had one group over here. This carbon over here had no groups. He ended up with the bromine. So the thing that has more hydrogens ends up with more hydrogens. The thing that has more groups ends up with more groups. So I guess you kind of go more in the direction that you are going in. But that still is just a rule, so why does that make sense? It starts to make sense when you think about that in both mechanisms, we had to have a carbocation. We talked about it in the last video. We had a carbocation right over there. This is the left carbon being a carbocation. This is the right carbon being a carbocation. And Markovnikov's rule all comes from which carbocation is more stable, which one has a lower energy level. It turns out that the carbocation that is a bonded to more electron-rich molecules or atoms is going to be more stable. You can imagine it has more things that, look, it's positive, but it has more carbons around it so it can share some of those electrons. The electron clouds will help it out a little bit to be a little bit more stable. This one right here is only bonded to one other carbon, so not as much sharing. This is bonded to two. So, in general, when you're only bonded to one other carbon, you're called a primary carbon. And if you're carbocation, this is a primary carbocation right here. This guy is bonded to two carbons, so he would be called a secondary carbon. Since it's a carbocation, it's a secondary carbocation, so this right here is secondary. So a secondary carbocation is more stable than a primary. And actually a tertiary, if you had another carbon group here or something else that had a lot of electrons around it, that would be even more stable. So bonded to three things, more stable than two things. And when I say two things, two things other than hydrogen, and then, more stable than one. So Markovnikov's rule all is a byproduct of the fact that this carbocation is more stable than this one over here. That's because it's secondary versus primary. Because it's secondary, it can borrow electrons from some of its friends. It has more neighbors to borrow electrons than this one. And since this is more stable, this is more likely to happen. This is a more likely intermediate to have. This is a less likely outcome to have in general. And that's why you're more likely to get to this left product, the 2-bromo-pentane than the 1-bromo-pentane.