in this video we're going to focus on sn2 intramolecular reactions now first let's talk about the difference between an inter molecular reaction and an intra molecular reaction so in an intermolecular reaction the reaction is between two different molecules for instance let's say if we have hydroxide reacting with methyl bromide this is an inter molecular sn2 reaction Now with an intramolecular s into reaction you have both the nucleophile and the leaving group in the same substrate which is what we're going to consider in this example so in the first step hydride which is a strong base is going to remove the acidic proton in the oh group so we're going to get an alkoxide ion now notice that we have both the nucleophile and the leaving group all in the same molecule so this is going to be where we have an intra-molecular reaction where the molecule reacts with itself so the nucleophile is going to attack this carbon kicking out the leaving group whenever you have these intramolecular reactions you're going to get a ring so if we count the carbon atoms and the oxygen we could say this is number one two three four five six so we're going to get a six-membered ring so what we have here is a cyclic ether let me draw that better so that's the answer for this problem so that is an intramolecular sn2 reaction now let's try another example problem feel free to pause the video and work on this problem we're going to react with sodium hydride again now just like before the hydride ion is going to participate in the acid-base reaction removing a proton and we're going to get an alkoxide ion which is also part of an alkyl halide now with the other reactant we're going to get the same thing hydride can remove a hydrogen put in the negative charge on that oxygen but notice that the stereochemistry is different what do you think is going to happen here because we have a nucleophile and the leaving group all in the same molecule now if you recall in order for the sn2 reaction to work the nucleophile must approach the substrate from the back so if bromine is in the front the nucleophile must be coming in the back it's not going to come and do a frontside attack because hydroxide will be repelled by The partially negatively charged bromide atom so the front side attack won't work notice that this oxygen and the bromine they're locked in the same position they're both in the front so this negatively charged oxygen it's not going to attack this carbon because in order for it to attack it it's approaching it from the front so this is going to stop here now for the other one the bromines in the back the oxygens in the front so for this one we can get an intramolecular sn2 reaction this can attack this carbon kicking out the BR and what we're going to get is a special kind of ether when you get a three-membered ether like this this is known as an epoxide but in order to get the epoxide these two groups have to be anti with respect to each other and then The Ether I mean the epoxide the carbon oxygen bonds will be on the same side because this is going to attack from the front kicking out the BR that's in the back so that's another example of an sn2 intermolecular reaction let's look at another example so here we have 1 5 bromo pentane and we're going to react with ammonia now this particular reaction can give us a mixture of products but we're going to focus on the formation of one product and what we need to do is we're going to propose a mechanism for the formation of this particular product even though we can get a lot of other products for this reaction so feel free to pause the video and propose a mechanism for the formation of this product so ammonia is a nucleophile it's also a weak base and what we have here is a primary alkyl halide so we're going to get an sn2 reaction the nucleophile is going to attack from the back kicking out the leaving group so we're going to get this so right now the nitrogen has three hydrogen atoms attached to it so it's going to have a positive formal charge because it has a total of four bonds now we need this nitrogen to behave as a nucleophile and attack this carbon so that we can close and form the ring before we can do that though we need to remove a hydrogen from this nitrogen and we could use another ammonia molecule to do that so now this nh2 is once again nucleophilic so at this point an intramolecular reaction is going to happen this nitrogen will attack the carbon kick out the leaving group so we can call this number one two three four five and six so we're going to get a six-membered ring with nitrogen being part of that ring but right now nitrogen has two hydrogen atoms attached to it and it has a positive formal charge so we need to use another ammonia molecule to remove this hydrogen I can definitely draw that better and this is going to give us our final product which is a cyclic Amine so that's how we can get this product from this particular alkyl halide now keep in mind there's a lot of other products that we can get for that reaction this is just one of the many products that we can get