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Elimination vs substitution: secondary substrate

Video transcript
- [Instructor] Let's look at elimination versus substitution for a secondary substrate. And these are harder than for a primary or tertiary substrate because all four of these are possible to start with. So, if we look at the structure of our substrate and we say it's secondary, we next need to look at the reagent. So, we have NaCl which we know is Na plus and Cl minus and the chloride anion functions only as a nucleophile. So, we would expect a substitution reaction, nucleophilic substitution. So, E1 and E2 are out. Between SN1 and SN2 with the secondary substrate, we're not sure until we look at the solvent and DMSO is a polar aprotic solvent, which we saw in an earlier video, favors an SN2 mechanism. So, SN1 is out and we're gonna think about our chloride anion functioning as a nucleophile. So, let me draw it in over here. So, this is with a negative one formal charge. And an SN2 mechanism are nucleophile attacks the same time we get loss of a leaving group and our nucleophile is going to attack this carbon in red. So, we're gonna form a bond between the chlorine and this carbon in red and when the nucleophile attacks, we also get loss of our leaving group. So, these electrons come off onto the oxygen and we know that tosylate is a good leaving group. So, when we draw our product, let's draw this in here, and the carbon in red is this one, we know an SN2 mechanism means inversion of configuration. The nucleophile has to attack from the side opposite of the leaving group. So, we had a wedge here for our leaving groups, so that means we're gonna have a dash for our chlorines. We're gonna put the chlorine right here and that's the product of our SN2 reaction. For our next problem, we have a secondary alkyl halide. So, just looking at our reactions, we can't really rule any out here. So, all four are possible, until we look at our reagent. Now, we saw in an earlier video, that DBN is a strong base, it does not act like a nucleophile. So SN1 and SN2 are out. And a strong base means an E2 reaction. So, E1 is out. Now that we know we're doing an E2 mechanism, let's analyze the structure of our alkyl halide. The carbon that's directly bonded to our halogen is our alpha carbon and the carbons directly bonded to the alpha carbon are the beta carbons. So, I'll just do the beta carbon on the right since they are the same essentially. And we know that our base is gonna take a proton from that beta carbon. So, let me just draw in a hydrogen here. And DBN is a neutral base, so I'll just draw a generic base here. Our base is going to take this proton at the same time these electrons move in to form a double bond and these electrons come off to form our bromide anion. So, our final product is an alkyne and our electrons in magenta in here moved in to form our double bond. For our next problem, we have another secondary alkyl halide, so right now all four of these are possible until we look at our reagent which is sodium hydroxide, Na plus, OH minus, and we know that the hydroxide ion can function as a strong nucleophile or a strong base. So a strong nucleophile makes us think an SN2 reaction and not an SN1. The strong base makes us think about an E2 reaction and not an E1 reaction. Since we have heat, heat favors an elimination reaction over a substitution, so E2 should be the major reaction here. So, when we analyze our alkyl halide, the carbon bonded to the halogen is our alpha carbon and the carbons directly bonded to that would be our beta carbons. So, we have two beta carbons here and let me number this ring. I'm gonna say the alpha carbon is carbon one, I'm gonna go round clockwise, so that's one, two, three, carbon four, carbon five and then carbon six. And next we're going to translate this to our chair confirmation over here. So, carbon one would be this carbon and then carbon two would be this one. This'll be carbon three, four, five and six. The bromine is coming out at us in space at carbon one which means it's going up. So, if I look at carbon one, we would have the bromine going up, which would be up axial. At carbon two, I have a methyl group going away from me in space, so that's going down, so at carbon two we must have a methyl group going down which makes it down axial. So, we care about carbon two. Let me highlight these again. So, we care about carbon two which is a beta carbon. We also care about carbon six which is another beta carbon. So, let's put in the hydrogens on those beta carbons. At carbon two, we would have a hydrogen that's up equatorial and at carbon six we would have a hydrogen that's down axial and one that is up equatorial. So, when we think about our E2 mechanism, we know our strong base is going to take a proton and that proton must be antiperiplanar to our halogen. So, our halogen, let me highlight our halogen here which is bromine, that is in the axial position, so we need to take a proton that is antiperiplanar to that bromine, so that carbon two, and let's look at carbon two first. At carbon two I do not have a hydrogen that's antiperiplanar to my halogen but I do have one at carbon six. It's the one that is down axial. So, our base is gonna take that proton, so let's draw in the hydroxide ion which is a strong base and the hydroxide ion is going to take this proton and then these electrons are gonna move in to form a double bond at the same time we get these electrons coming off onto the bromine to form the bromide ion. So, let's draw the product for this reaction. We would have our ring and a double bond forms between carbon one and carbon six. So, that means a double bond forms in here. And then at carbon two, we still have a methyl group going away from us in space. So, let me draw that in like that. So, the electrons in red, hard to see, but if you think about these electrons in red back here, are gonna move in to form our double bond between what I've labeled as carbon one and carbon six. Let me label those again here. So, carbon one and carbon six. Again not IUPAC nomenclature just so we can think about our product compared to our starting material. So, this would be the major product of our reaction which is an E2 reaction. It would also be possible to get some products from an SN2 mechanism, but since heat is here, an elimination reaction is favored over a substitution. Next we have a secondary alcohol with phosphoric acid and heat. And we saw a lot of these types of problems in the videos on elimination reactions. So, it's not gonna be SN1 or SN2 and we don't have a strong base, so don't think E2, think E1. And our first step would be to protonate our alcohol to form a better leaving group. So phosphoric acid is a source of protons and we're going to protonate this oxygen for our first step. So, let's draw in our ring and we protonate our oxygen, so now our oxygen has two bonds to hydrogen, one lone pair of electrons and a plus one formal charge on the oxygen. So, this lone pair of electrons on the oxygen picked up a proton from phosphoric acid to form this bond. And now we have a better leaving group than the hydroxide ion. These electrons come off onto the oxygen and we remove a bond from this carbon in red which would give us a secondary carbocation. So, let's draw in our secondary carbocation and the carbon in red is this one and that carbon would have a plus one formal charge. So, let me draw in a plus one formal charge here. And now we have water which can function as a weak base in our E1 reaction and take a proton from a carbon next to our carbon with a positive charge. So, let's say this carbon right here. It has two hydrogens on it. I'll just draw one hydrogen in and water functions as a base, takes this proton and these electrons move in to form a double bond. So, let's draw our final product here. We would have a ring, we would have a double bond between these two carbons, so our electrons in, let's use magenta, electrons in magenta moved in to form our double bond. So, our product is cyclohexane. So, a secondary alcohol undergoes an E1 reaction if you use something like sulfuric acid or phosphoric acid and you heat it up. For this reaction we have this secondary alkyl halide reacting with an aqueous solution of formic acid. Formic acid is a weak nucleophile and water is a polar protic solvent. A weak nucleophile and a polar protic solvent should make us think about an SN1 type mechanism because water as a polar protic solvent can stabilize the formation of a carbocation. So, let's draw the carbocation that would result. These electrons would come off onto our bromine and we're taking a bond away from this carbon in red. So, the carbon in red gets a plus one formal charge and let's draw our carbocation. So, we have our benzine ring here. I'll put in my pi electrons and the carbon in red is this one, so that carbon gets a plus one formal charge. This is a secondary carbocation but it's also a benzylic carbocation. So, the positive charge is actually de-localized because of the pi electrons on the ring. So, this is more stable than most secondary carbocations. Next, if we're thinking an SN1 type mechanism, this would be our electrophile, our carbocation is our electrophile and our nucleophile would be formic acid. And we saw in an earlier video how the carbonyl oxygen is actually more nucleophilic than this oxygen. So, a lone pair of electrons on the carbonyl oxygen would attack our carbon in red. And we would end up with, let's go ahead and draw in the result of our nucleophilic attack, and I won't go through all the steps of the mechanisms since I cover this in great detail in an earlier video. So, this is from our SN1, SN2 final summary video. So, let me draw in what we would form in here. So, this would be carbon double bonded to an oxygen and this would be a hydrogen. So, this is our product and this carbon is a chiral center and because this is an SN1 type mechanism and we have planer geometry in our carbocation, our nucleophile can attack from either side and we're gonna end up with a mix of enantiomers. So again, for more details on this mechanism, I skipped a few steps here, please watch the SN1, SN2 final summary video. Next, let's think about what else could possibly happen. So, SN2 is out, we formed a carbocation, E1 is possible because we have a carbocation here and we also have a weak base present. So, our weak base could be something like water, and I'll just draw a generic base in here, and let's draw in a proton on this carbon. So, our base could take this proton here and these electrons would move in to form a double bond. So, another possible product, we would have our benzine ring, so I'll draw that in, and then we would have a double bond. So, another possibility is an E1 mechanism.