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Current time:0:00Total duration:12:19

Elimination vs substitution: secondary substrate

Video transcript

let's look at eliminations 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 a secondary substrate we're not sure until we look at the solvent and DMSO is a polar a protic solvent which we saw in an earlier video favors an sn2 mechanism so sn1 is out and we're going to think about our chloride anion functioning as a nucleophile let me draw it in over here so this is with a negative one formal charge an sn2 mechanism our nucleophile attacks at the same time we get lost of a leaving group and our nucleophile is going to attack this carbon in red so we're going to form a bond between the chlorine and this carbon in red and with the nucleophile attacks we also get lost of our leaving group so these electrons come off onto the oxygen and we know that tousle eight is a good leaving group so when we draw our products 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 on the side opposite of the leaving group we had a wedge here for our leaving group that means we're going to have a dash for our chlorines we're going to 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 and we saw in an earlier video that dbm is a strong base does not act like a nucleophile so sn1 and sn2 are outs and a strong base means an e2 reaction Zoey one is out now that we know we're doing an e2 mechanism let's analyze the structure of our alkyl halides the carbon that's directly bonded to our halogen is our alpha carbon and the carbon is directly bonded to the alpha carbon are the beta carbon so I'll just do the beta carbon on the right since they are the same essentially and we know that our base is going to take a proton from that beta carbon let me just draw in the hydrogen here and vbn 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 products is an alkene and our electrons in magenta and here moved in to form our double bonds 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 o h- 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 heats heat favors and elimination reaction over a substitution so e2 should be the major reaction here when we analyze our alkyl halides the carbon bonded to the halogen is our alpha carbon and the carbon is directly bonded to that would be our beta carbon so we have two beta carbons here and let me number this ring I'm going to say the Alpha carbon is carbon one I'm going to go around clockwise so that's one two three carbon four carbon five and then at carbon six and next we're going to translate this to our chair conformation over here so carbon one would be this carbon and then carbon two would be this one to 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 at carbon one we would have the bromine going up which would be up axial a carbon - I have a methyl group going away from me in space that's going it down so at carbon - we must have a methyl group going down which makes it down axial so we care about carbon - let me highlight B's again so we care about carbon - which is a beta carbon we also care about carbon six which is another beta carbon so let's put in two hydrogen's 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 up equatorial so when we think about our e two mechanism we know our strong base is going to take a proton and that proton must be anti-periplanar 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 anti-periplanar to that bromine so at carbon 2 once again carbon 2 first at carbon 2 I do not have a hydrogen that's anti-periplanar to my halogen but I do have one at carbon six it's the one that is down axial so our base is going to take that proton 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 going to 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 1 and carbon 6 so that means a double bond forms in here and then at carbon 2 we still have a methyl group going away from us in space let me draw that end like that so the electrons in red hard to see what if you think about these electrons in red back here are going to move in to form our double bonds between what I've labeled as carbon 1 and carbon 6 when we label those again here so one and carbon six again not I you pack 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 we saw a lot of these types of problems in the videos on elimination reaction so it's not going to be sn1 or sn2 and we don't have a strong base so don't think e to 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 oxygens and now our oxygen has two bonds to hydrogen one lone pair of electrons and a +1 formal charge on the oxygen so this lone pair of electrons on the oxygen picked up a proton from phosphoric acid it's you form a dis 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 carbo cation let's draw in our secondary carbo cation and the carbon in red is this one and that carbon would have a +1 formal charge when we draw on a plus 1 formal charge here and now we have water which can function as a weak base in our e1 reaction and take a proton from the carbon next to our carbon with the positive charge so let's say this carbon right here it has two hydrogen's 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 cyclohexene 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 halides 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 carbo-cation it so let's draw the carbo cation 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 +1 formal charge and let's draw our carbo cation so we have our benzene ring here I'll put in my PI electrons and the carbon in red is this one so that carbon gets a +1 formal charge this is a secondary carbo cation but it's also a benzylic carbo cation so the positive charge is actually delocalized because of the pi electrons on the ring so this is more stable than most secondary carbo cations next if we're taking an sn1 type mechanism this would be our electrophile our carbo cation 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 mechanism since i cover this in a 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 de'longhi 2 and 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 we have planar geometry in our carbo cation our nucleophile can attack from either side and we're going to 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 carbo cation II one is possible because we have a carbo cation 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 benzene ring so I'll draw that in and then we would have and we have a double bond so another possibility is e 1 mechanism