SN1 and SN2 reactions of alcohols

Voiceover: Let's look at a few nucleophilic substitution reactions of alcohols. And I'm assuming that you have seen a SN-one and SN-two mechanism before. Let's start with a primary alcohol. So this is ethanol. It is a primary alcohol because the carbon bonded to the OH is bonded to one other carbon. And primary alcohols react with HBR to form an alkyl bromide to be a SN-two process. And so we have HBR a strong acid and we have ethanol which is going to function as a base, so actually the first step is to protonate the oxygen. So we got a proton transfer here. So lone pair of electrons in the oxygen pick up this proton, and these electrons are left behind on the bromine so we form our bromide anion here. So we got a bromide anion negative one formal charge. So these electrons in here in blue, lets say that these are these electrons right here on our bromide anion. So we are going to protonate the oxygen. So lets go ahead and show that oxygen being protonated here. So there is still one lone pair of electrons on this oxygen which gives this oxygen a plus one formal charge. So these electrons in magenta right here on this oxygen pick up this proton to form this bond right here. And so now we need to think about our SN-two type mechanisms. So in an SN-two type mechanism we need a nucleophile, and we need an electrophile, and we are also going to have our leaving group here. The reason why we protonate the oxygen is to form a much better leaving group. So if we just had a nucleophile attack ethanol right here we would have hydroxide as a leaving group. That is not a good leaving group. Here we would have water as our leaving group, which is what we are going to see in a second. And so that helps this process occur. So lets think about this oxygen here withdrawing some electron density from this carbon that I just marked in red. So this carbon in red is the electrophilic portion. So the nucleophile is going to attack the electrophile, so the bromide anion is going to function as a nucleophile and it's going to attack our electrophile, and it's going to form a bond between the bromine and the carbon. At the same time, these electrons in here are going to come off onto the oxgyen. So a SN-two type mechanism remember is concerted. So the bond that forms, this happens at the same time that this bond breaks and the electrons come off onto our leaving group. So when we draw our final product, we would form ethyl bromide or bromoethane. Lets put in our electrons right here. And so these electrons in blue formed a bond between the carbon and the bromine. So this is the carbon in the red down here, and so we form a bromoethane. We would also form water. So let me go ahead and draw in water. So H-two-0, which is a good leaving group because water is so stable. So we go ahead and think about these electrons in green here coming off onto the oxygen and that gives us water. So in an SN-two type mechanism sometimes you can think about stereochemistry, but not for this example. Because carbon in red right here, so this carbon in red right here is not a chiral center. There are two hydrogens attached to that carbon in red, and so this is our only product. This is the only thing that we have to think about. So bromoethane as our final result. Alright lets look at another SN-two reaction. This time we are doing a secondary alcohol. And so here we have our secondary alcohol. The carbon bonded to the OH is bonded to two other carbons, so a secondary alcohol. And we saw at an earlier video that alcohols will react with tosyl chloride and pyridine to form a tosylate. So lets go ahead a draw the product. The first step we form a tosylate. We retain the stereochemistry. So we have a wedge over here and we have a wedge down here as well. So we have an oxygen and then we will just put TS here for our abbreviation. So we talked in great detail about this in an earlier video. So once again, this gives us a better leaving group. So in the previous reaction we protonated the oxgyen to give us a much better leaving group, and here we have an excellent leaving group. So that's one of the reasons for using tosylates here. So in the second step we are going to add sodium bromide and we are going to get again a SN-two type mechanism. So a nucleophile is going to attack our electrophile. So we can identify our electrophile. It's the carbon bonded to the oxygen. So the oxygen is withdrawing some electron density from this carbon, so this the electrophilic portion of the molecule. And then once again our bromide anion is going to function as a nucleophile. So here is our bromide, and I am going to highlight this lone pair of electrons right here in blue. And so our bromide anion attacks our electrophilic carbon and forms a bond. At the same time, these electrons come off onto the oxygen. So once again a concerted SN-two type mechanism. And this time we do have to worry about stereochemistry. So we have this wedge in here. We have this wedge, so this part is coming out at us in space. And so the bromide anion has to attack from the opposite side. And so if it is attacking from the opposite side when you draw the final product, you would have to show this is a dash. So the bromide had to attack from the opposite side which gives us inversion of configuration. Inversion of absolute configuration here. So when you assign your absolute configuration, for your starting alcohol, this would be R. So our chiral center would be right here. So this carbon is chiral and so this carbon is chiral for our products, and for our product we would form the S in the [an-te-um] right here. So we formed S in the [an-te-um] row and we started with a R in the [an-te-um] row. So a SN-two type mechanism inversion of configuration because a nucleophile has to attack from the opposite side. So lets look at one more example here. Lets do a SN-one mechanism using a tertiary alcohol. So lets do that. So here we have tertbutyl alcohol or tertbutanol reacting with concentrated hydrochloric acid. So concentrated hydrochloric acid is going to function as an acid. Our alcohol is going to function as a base. And let me just highlight the fact that this is going to be a SN-one type mechanism because we have a tertiary alcohol. This carbon bonds to the OH bond to three other carbons. And so the alcohol functions as a base, and is protonated and we would form the chloride anion here. So lets go ahead and draw that in. So we would have the chloride anion, so negative one formal charge. And lets go ahead and put those electrons in blue like we did before. So these electrons in here, lets say that those are these electrons. So the chloride anion. We protonate the oxygen, so lets go ahead and draw that. So if we protonate the oxygen, now we have our tertbutyl group over here, protonate the oxygen so a plus one formal charge on this oxygen so lets go ahead and draw that in. So plus one formal charge on the oxygen. Electrons in magenta right here picked up this proton. Lets say it is this bond right here. So once again we have an excellent leaving group like we talked about before. Water is a good leaving group. And so in SN-one type mechanism, these electrons are going to come off onto the oxygen, which gives us water. It also gives us a carbo cation. A plus one formal charge on this carbon right here. This carbon red gets a plus one formal charge, so that's this carbon because it is losing a bond. So let me go ahead and highlight the electrons that it is losing. So these electrons in here are coming off onto the oxygen to form water. So we can go ahead and draw water over here. So we lose water at this point. So H-two-O, and the electrons in green come off onto this oxygen right here so we form H-two-O. Taking a bond away from the carbon in red, so we form a carbo cation. So this is a stable carbo cation. This is a tertiary carbo cation. So that is why this tertiary alcohol reacts via a SN-one type mechanism. The stability of the carbo cation. And so in our final step, we have the nucleophile is going to attack our electrophile. So the nucleophile attacks our electrophile. The chloride anion attacks our carbo cation, attacks that carbon there and so we form our final product, which is tertbutyl chloride. So let me go ahead and draw in these electrons and lets highlight some again. The electrons in blue. These electrons form the bond, So they bonded right here. And so let me go ahead and highlight this carbon in red. So this carbon in red is this one right here. So we form tertbutyl chloride and we lost water in the process. So this is a very easy reaction to do. It occurs at room temperature and just take tertbutanol and add some hydrochloric acid and just shake them together and you can form your final product this way. So thinking about stereochemistry and SN-one type mechanism, the carbon in red right here is not a chiral center and so we don't have to worry about what kind of stereochemical outcome that we would predict for the product. So this is our final product. There is no stereochemistry. Just to refresh your memory, for a SN-one type mechanism because this formation of this carbo cation, this carbon and the carbo cation is SP-two hybridized, and so it is planar. And so when we draw out that carbon in red here. Lets say that's that carbon in red. It's SP-two hybridized, which means that the carbons that are directly bonded to it lie on the same plane. So these carbons lie on the same plane. SP-two hybridized with a p orbital. So there is a p orbital. Lets sketch that in. So there is a plus one formal charge in this carbon. So when your nucleophile attacks, your nucleophile could attack from either side of that plane. So the nucleophile could attack from this side or it could attack from this side. So if your final product has a chiral center, you need to think about stereochemistry. But not in this case. In this case we don't have one. So we lucked out. This one was a little bit straightforward. So that's a couple of examples of SN-one and SN-two reactions of alcohols.