Reaction of an alkene with a hydrogen halide, converting the double bond to a halogenated single bond. Created by Jay.
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- How do you know when to do a hydride shift? Couldn't you of just moved the Cl where the positive charge is in the second hexane?6:28(7 votes)
- You want the most stable carbocation - meaning you want that positive charge to be on a carbon connected to as many other carbons as possible, or to be resonance stabilized. That hydride shift happens to move the positive charge onto the most stable carbocation (from the secondary to the tertiary carbon).... Tertiary > Secondary > Primary being the order of preference there.(14 votes)
- From3:28to4:30, based on Markovnikovs rule, would the tertiary carbocation be the major, and the secondary carbocation be the minor?(5 votes)
- I've understood that Markovnikov's underlines that the carbocation forms always to the most stable Carbon - while the elektrofile (H+) goes where the most hydrogens are.
However according to Zaitsev's rule, in elimination you would get Major and Minor (usually Cis-Trans) products.
Then again, I'm not a master in chemistry yet :)(6 votes)
- In another video we saw a similar reaction with vinyl chloride and HCl only the resulting reaction made a polymer. Couldn't a similar thing take place here whereby the positive carbocation steals an electron from ethylene the way the hydrogen ion did and start a polymer?(7 votes)
- In the polymerisation video there was a Cl attached to the carbocation stabilising it with +R effect [ sharing its lone pairs with carbocation ] which is greater then the -I effect caused by the Cl atom . Since the resultant carbocation is stable the pi electron pair can be delocalised and form bond with previous carbon and so on....Hope you understood my english(0 votes)
- At03:36, Jay says that the tertiary carbocation will form faster than the secondary carbocation. What is the reason for this? Is the activation energy for the formation of tertiary carbocation less than that of secondary carbocation... Or is it just because tertiary carbocations are more stable (which does not make any sense to me).(2 votes)
- You're right that the stability of the carbocation itself cannot explain the rate difference, since we need to compare activation energies; however, the stability of the carbocations does come into play indirectly. According to Hammond's postulate, the transition state involved in carbocation formation (endothermic) will resemble the carbocation itself. Thus, when we are forming a more stable carbocation, we would expect the transition state leading to that carbocation to be more stable, and thus the activation energy to be lower.
For your example: Since tertiary carbocations are more stable than secondary carbocations, the transition state to form a tertiary carbocation will be lower in energy. This lower energy transition state translates to a lower activation energy, and thus a faster rate of formation compared to secondary carbocations.
(See http://en.wikipedia.org/wiki/Hammond's_postulate#The_Transition_States_for_SN1_Reaction for more about Hammond's postulate)(5 votes)
- at about4:00, isn't there a hydrogen missing from the aromatic ring? Or did he just leave it out?(2 votes)
- There is no aromatic ring.
Remember that in bond-line structures you do not show the H atoms on C unless you want to draw attention to them.
He just didn't bother to show the missing H atoms.(3 votes)
- At6:15, what happened to the original hydrogen that got attached to the right of the double bond? Does the hydride shift not occur with an existing hydrogen on the ring?(2 votes)
- The original H atom from H-Cl got attached to C-2 of the ethenyl group and stayed there. It didn't move.
The H atom on C-1 of the ring then shifted to C-1 (the positively charged carbon) of the side-chain. So yes, the hydride shift does occur with an existing hydrogen on the ring.(3 votes)
- At1:22, the nucleophile attacks the electrophile, creating the product that has the H and X on the same side (syn addition), but couldn't the bond rotate, leaving the positive charge on the bottom side? This would create an anti-addition.(2 votes)
- I see where you might think that it is a syn-addition, but Jay never mentions it because there is no stereochemistry in his simple line drawing. Jay could've drawn the H and X adding in any position (between the methyls, on bottom, or "anti"). The + charge is on the entire carbon atom, never on just one side (that is an important concept).
Remember, when the carbocation is formed, it has trigonal planar geometry with an empty p orbital. This means the halide can attack from either side of the plane (possibly producing enantiomers if chirality centers are formed). There will be regioselectivity (Markovnikov) if the molecule is not symmetrical. Keep watching, Jay goes in to greater detail.(3 votes)
- At6:26, I don't understand why that tertiary carbon has a positive charge.. Didn't it lose a proton? Therefore shouldn't it be negatively charged?(2 votes)
- No it lost a hydride, the H took both electrons that were in the C-H bond.
That carbon now has 3 bonds and 0 lone pairs
Formal charge = valence electrons - lone pair electrons - bonds
4 - 0 - 3 = +1(2 votes)
- what is the difference between base and nuleophile?(2 votes)
- The species is called a nucleophile when it attacks a carbon atom and a base when it attacks a hydrogen atom.(1 vote)
Here's the general reaction for a hydrohalogenation. You have an alkene, and you react that with a hydrogen halide. And the hydrogen adds to one set of your double bond, and the halogen adds to the other set of your double bond. Let's look at the mechanism for this reaction. So here we have our alkene and our hydrogen halide. Think something like hydrochloric acid. So a strong acid donates protons in solution. So the hydrogen halide's going to function as an acid. The alkene's going to function as a base. The electrons in this pi bond here are going to take this proton and leave these two electrons behind on your halogen. So let's go ahead and draw the results of that acid-base reaction. So I'm going to say the proton adds to the carbon on the left. And the carbon on the right over here used to have four bonds to it. Now it has only three bonds to it, which means it's positively charged and is a carbocation. So it wants electrons. It wants to get an octet of electrons. So it loves electrons. It's an electrophile. We have our halogen over here, which had three lone pairs of electrons. It got one more lone pair for a total of eight electrons around it, which gives it a negative 1 formal charge, which means that it likes nuclei. It is a nucleophile. So our nucleophile is going to attack our electrophile like that and form a new bond to give us our alkyl halide as our product. Let's follow some of these electrons through our mechanism. So the electrons in this pi bond here, these are the same electrons that form this bond right here with the proton. And let's follow these electrons in here. So the electrons in here are the ones that ended up on our halogen. And I happened to choose these very same electrons to form the bonds between the carbon and the halogen like that. So make sure to be able to follow electrons in mechanisms. Let's look at an actual problem, and let's follow the mechanism through. So let's look at this reaction here. This is my alkene. And I'm going to react this alkene with hydrochloric acid. First step of the mechanism-- the pi electrons function as a base and take this proton here. Kick these electrons off onto your chlorine. Now, we have a problem-- which side do I add my proton to? I have two options. I could add the proton to the top carbon. So let's go ahead and do that. Let's add the proton to the top carbon there and see what we get. So if I add that proton there to the top carbon-- here's my methyl group. Here is my proton that I added. Well, I just took a bond away from this carbon. So this is where my carbocation is going to be. What kind of carbocation is that? That carbon is connected to two other carbons. So that is, of course, a secondary carbocation. Let's see what would happen if I add the proton on to the other carbon. So what would I get if I add the proton on to that carbon right there? So my methyl group is still here. I add my proton on to that bottom carbon now. Now, this is my carbocation. That is my positively charged carbon right there. What kind of carbocation is that? Well, that positively charged carbon is bonded to one, two, three other carbons. So this is actually a tertiary carbocation. We know that a tertiary carbocation is more stable than a secondary carbocation. So the tertiary carbocation is going to form faster than the secondary carbocation. What happens in the second step of the mechanism? Well, our chlorine ends up being negatively charged. Right? It is going to function as a nucleophile. This lone pair of electrons is going to attack this carbon and form our product. All right. So let's go ahead and draw our product here. Our product is going to have a methyl group and then a chlorine attached to that carbon. So the halogens adds to the more substituted carbon. And this is called Markovnikov's rule. So let's go ahead and write that. See if we can spell Markovnikov. The halogen adds the more substituted carbon. And the reason it does that is because the more substituted carbon is the one that was the more stable carbocation in the mechanism. So let's do another mechanism here. Whenever you have a carbocation present, you could have rearrangement. So let's do one where there's some rearrangement. So let's start out with this as our alkene and react that with hydrochloric acid once again. First steps-- pi electrons function as a base. These electrons kick off onto your chlorine. So which side do we add the proton to? Right? We could add the proton to the left side of the double bond. We could add the proton to the right side of the double bond. Well, we know that we want to form the most stable carbocation that we can. So it makes sense to add the proton to the right side of the double bond right here because that's going to give us this as a carbocation. What kind of carbocation is that? So let's identify this carbon as the one that has our positive charge. That carbon is bonded to two other carbons. So it is a secondary carbocation. If we had added on the proton to the left side of the double bond, we would have a primary carbocation here. So a secondary carbocation is more stable. Can we form a tertiary carbocation? Because we know tertiary carbocations are even more stable than secondary carbocations. And of course, we can. There's a hydrogen attached to this carbon. And we saw-- in our earlier video on carbocations and rearrangements-- we could get a hydride shift here. All right. So the proton and these two electrons here are hydride anion. And these two electrons are going to move over here, shift over one carbon, and form a new covalent bond. So what would we get if we get a hydride shift in our mechanism? Well, now our hydride has shifted over here to that carbon. This carbon no longer has a positive charge on it. We took a bond away from this carbon. So now, this is where our positive charge is. So we have a carbocation. How would we classify this carbocation? Well, one, two, three other carbons. So it's tertiary. It's more stable than our secondary carbocation. So in the final step of our mechanism, we had our chloride anion over here from the first step of our mechanism. So a chloride anion, negatively charged nucleophile file. So a nucleophilic attack on our carbocation. So right there. And we're going to form a bond between that halogen and that carbon. So our final product is going to end up with our cyclohexane ring. Let's do another one. So we have our cyclohexane ring like that. And then, we have our ethyl group attached to this carbon. And then, our chlorine attached to that carbon. Like that. So that's going to be our major product. All right. Let's look at the stereochemistry of this reaction really fast. So let's look at what happens if we react this alkene with hydrochloric acid. So what will we get it? All right. First step-- pi electrons take the proton. Kick the electrons off onto the chlorine. So once again, which side do we add our proton? So two possibilities. I could add to the left side of the double bond or add to the right side of the double bond. It makes sense to add it to the right side of the double bond because that gives us a more stable carbocation. So if I add it to the right side of the double bond, this carbon ends up being positively charged. What kind of a carbocation is that? That carbon is bonded to two other carbons. So it's a secondary carbocation here. All right. So we have a secondary carbocation. There's no kind of rearrangement that we could get here to get a tertiary carbocation. So a secondary carbocation is as stable as we're going to get. Now, carbocations are carbons with three bonds to them, meeting that carbon is sp2 hybridized. So let's redraw this carbocation here. So I'm going to say that this carbon right here represents my carbocation. What's bonded to it? Well, on the left side, there is an ethyl group-- CH2, CH3 right here. And I know that there's a methyl group bonded to it. I'm going to put the methyl group going back in space here. And then there's also a hydrogen attached to that carbon. We just didn't draw it in on our carbocation. So like this. I know this carbon is sp2 hybridized, meaning there's an untouched, unhybridized p orbital on this carbon. All right. So let me draw my p orbital in there like that. And let me go ahead and make sure that everyone realizes this is my carbocation. So sp2 hybridized carbon means the atoms bonded to that sp2 hybridized carbon are in the same plane. We could think about our carbocation being flat because of the trigonal planar geometry like that. So when our chloride anion comes along, it sees a flat sp2 carbocation here. So it's negatively charged. So when that chloride anion nucleophilic attacks our carbocation, it has two options. It could attack from the top of that plane. So let's go ahead and draw the product that would result if it attacked from the top of that plane. So let's see what we'd have here. We would have our carbon form the new bond with our chlorine like that. And then we have a CH2, CH3 over here on the left side. And we have a hydrogen coming out at us. And we have a methyl group going away from us. So that's one possibility. And what about if that chloride anion attacked from the bottom? So that's a possibility as well. So since it's a plane-- a flat sheet of paper-- 50% chance it attacks from the top, 50% chance it attacks from below. So what would we get if it attacks from below? Well, we would have a carbon forming a bond with that chlorine right here. And then we'd have an ethyl group over here on the left side-- CH2, CH3. We have a hydrogen coming out at us. And we would have a methyl group going away from us like that. So if you look at these products here, this is a chirality center. This carbon right here has four different groups attached to it. It's sp3 hybridized. Same with this group. So these two are actually mirror images of each other-- non-superimposable mirror images-- enantiomers of each other. Let's see if we can draw them in a dot structure that's more familiar to you here. So if you need a MolyMod set, go ahead and take it out right now. And put your eye right here and stare at your chirality center. So what enantiomer is this one over here? Well, here is my chirality center carbon. And if I were looking at it from this vantage point, there'd be an ethyl group here on the left-- CH2, CH3. On the right, there'd be a methyl group. And this hydrogen would be coming out at me in space. And the chlorine would be going away from me in space. So that is the enantiomer I'm looking at here. What about this one? So same thing. I put my eye right here. I stare at my chirality center. So here's my chirality center. I'd have an ethyl group on the left, a methyl group on the right. This time, though, the chlorine would be coming out at me. So the chlorine would be coming out at me like that. And I'd get a racemic mixture of my products. So be careful when you have carbocations in your mechanism. Think about the fact that they're trigonal planar. And think about to does this reaction form any chirality centers? If it forms chirality centers, you're going to get a racemic mixture for your products.