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Current time:0:00Total duration:13:21

Elimination vs substitution: tertiary substrate

Video transcript

let's look at elimination versus substitution for a tertiary substrate for this reaction we have a tertiary alkyl halide and we know that a tertiary alkyl halide will form a tertiary carbocation which is a stable carbo cation and therefore an sn1 reaction is possible an sn2 reaction is not possible because this tertiary alkyl halide has too much steric hindrance to undergo an sn2 mechanism and E one mechanism is also possible because an e1 mechanism requires a stable carbo cation and an e2 reaction is also possible we have three choices for a tertiary substrate next you want to look at your reagent here we have sodium chloride and we know that the chloride ion is a very weak base and it's going to function only as a nucleophile so if we have a nucleophile here we're going to go with substitution reactions so an sn1 reaction must be the case for this example and we know in an sn1 mechanism first step should be loss of our leaving group for this tertiary alkyl halide so we form the iodide anion and we take a bond away from this carbon in red so we form a tertiary carbo cations when we draw this in we have our six number ring we have our methyl group and the carbon in red is this one it gets a plus 1 formal charge so a positive charge on this carbon once we've formed our carbo cation in the next step of our mechanism our nucleophile attacks and our nucleophile is the chloride ion so lone pair of electrons on the chlorine are going to form a bond with that carbon in red and we end up with our product let me draw in our ring we have a methyl group and we have a bond to a chlorine now so let me highlight a lone pair of electrons in magenta form this bond between the chlorine and the carbon in red we don't have any stereochemistry to worry about here because we don't have any chiral centers so this was an sn1 mechanism first we look at our substrate and we can see that this is a tertiary alkyl halides the carbon that is directly bonded to our halogen is bonded to three other carbons so a tertiary alkyl halide is two sterically hindered to undergo an sn2 reaction so immediately we know that that is out and when we look at our reagent this is the calcium a charge of butoxide so let me draw out the structure for that we would have a negative one formal charge on this oxygen and this potassium would be a plus one charge so potassium tert-butoxide is two sterically hindered to function as a nucleophile so instantly we know that sn1 is out and potassium tert-butoxide is a strong base so we know that e 1 is out and that leaves an e2 mechanism for this reaction so let me draw the alkyl halides over here again and put these in here and we know that the carbon that's bonded to our halogen is our alpha carbon and the beta carbons are bonded to that so this would be a beta carbon this would be a beta carbon and this would be a beta carbon let's say our sterically hindered strong base comes along and takes a proton from this beta carbon so our base takes this proton these electrons move into here at the same time these electrons come off onto the bromine to form the bromide ion and we will get one product for this reaction so let me draw in here we would now have a double bond and let me show those electrons so electrons in magenta moved into here to form our double bonds now let's say we took a proton from a different beta carbon so let me redraw let me just redraw this so I'll draw the alkyl halide in here like this and again we know that this is the alpha carbon and let's take a proton from this beta carbon now so I'll draw in approach on and just think about our base coming along so our base coming along with a negative charge taking this proton these lecture I'll move into here at the same time these electrons come off onto the bromine so that's our e 2 mechanism so I'll draw the resulting alkene there's a Delvon in here this time the electrons in magenta moved into here to form our double bonds when we look at our products we know that the one on the right is actually more stable because the one on the right is a trisubstituted alkene whereas the one on the left is only a dive substituted alkene but when you're dealing with a sterically hindered base like potassium tert-butoxide it turns out that the dye substitute the less substituted alkene is the major product and that's just because our base is so sterically hindered also notice if our base came along and took a proton from this beta carbon we would get the same product as right here there's another tertiary substrate so we know an sn2 mechanism is out and when we look at our reagent which is sodium ethoxide sodium 1 is a plus 1 formal charge and the oxygen has a negative 1 formal charge the methoxide ion we saw an earlier video it's a strong nucleophile and a strong base and whenever you think strong base think e 2 reaction so the e2 reaction is going to dominate here and an e2 reaction means the methoxide ion is going to function as a base and take a proton from our alkyl halide so next we analyze our alkyl halide the carbon bonded to our halogen is our alpha carbon and the carbon is directly bonded to the alpha carbon are the beta carbons here the beta carbons are identical so I'm going to say that's our beta carbon and let's draw in one beta hydrogen then the Fox ID ion comes along so I'll draw that in so we have a negative negative one formal charge on our oxygen this is going to take this proton at the same time these electrons move in here to form our double bond and these electrons come off to form our iodide ion has our leaving group so for our final products we're going to have a double bond here and let's change colors again the electrons in magenta moved into here to form our double bond since our beta carbons are all the same this is the only product for this reaction here's the same tertiary alkyl halide we saw in the previous problem so an sn2 reaction is out and when we analyze our reagents we know that water is a weak nucleophile and a weak base and since water is a weak base the e2 reaction is out that leaves the e1 reaction and the sn1 reaction which both proceed via a carbo cation so let's draw the carbo cation that we would form these electrons come off to form the iodide ion and we take a bond away from the carbon in red to form our tertiary carbo cation so our tertiary carbo cation has a plus 1 formal charge on our central carbon so this carbon in red if water acts as a weak base water can take a proton for one of the carbons next door to the carbon in red so on the draw on a hydrogen on this carbon and if water takes this proton these electrons would move into here to form a double bond so that's one of the possible products for this reaction so we form an alkene and our electrons in magenta moved in to here to form our double bonds so that's when water acts as a weak base and that would be an e1 mechanism we also have the possibility of an sn1 mechanism so let's draw our carbo cation again so here's our carbo cation our carbon in red is this one with a plus 1 formal charge and this time we're going to show water acting as a nucleophile let's draw in our water molecule with two lone pairs of electrons on the oxygen so the nucleophile attacks our electrophile and we form a bond between the oxygen and the carbon in red so let's draw that in here so now we have a bond between our oxygen and our carbon and red which is this one so a lone pair of electrons on our oxygen forms that bonds in here and our oxygen is still bonded to two other hydrogen's and still has one lone pair of electrons on the oxygen which gives the option a +1 formal charge and we would also need in our next steps to have an acid-base reaction another molecule of water could come along and take one of these protons so in the interest of time and space I'm just going to write minus h plus here we lose one of our protons and we form our products which would be a tertiary this tertiary alcohol so we have two possibilities for this for this reaction we can have either an e1 reaction which would give us this alkene or an sn1 reaction which was which would give us this alcohol for our last example let's look at this tertiary alcohol we know sn2 is out and we have a strong acid presence not a strong base so an e2 reaction is out and whenever you see a tertiary alcohol but something like sulfuric acid or phosphoric acid and heat think II one reaction so we need to form a carbo cation here but before we do that let's analyze the structure of our alcohol the carbon bonded to the O H is our alpha carbon and the carbon is directly bonded to that our our beta carbon so I'll call this one beta 1 this one beta 2 and this one at beta 3 we can't have loss of a leaving group right away because the hydroxide ion is a poor leaving group but our sulfuric acid is a source of protons so our first step is to protonate the alcohol so the lone pair of electrons on the oxygen pick up this proton and let's draw what we would form here so the oxygen would now will be bonded to two hydrogen's so has a lone pair of electrons and a plus one formal charge on the oxygen so these electrons here in magenta pick up this proton to form a bond and now we are ready for loss of a leaving group when these electrons come off onto the oxygen they would make water which we know is a stable molecule it's a good leaving group and now we're taking a bond away from this carbon in red so now we're going to form a carbo-cation so let's draw in our carbo cation so actually let me make this a better methyl group here and our carbon in red is this one so that gets a +1 formal charge and our e1 mechanism we know we're going to have a weak base come along and take a proton from one of the beta carbons so I'll go with beta 3 for this carbo cation here so I'll draw in a hydrogen coming off of this carbon we mark it in blue here this carbon in blue was this carbon on our starting alcohol so a weak base would have to be water we just formed water in our previous step so water comes along and takes that proton so let's draw in our lone pairs of electrons so we're going to take this proton and these electrons going to move in here to form our double bond and that would of course get rid of the formal charge on the carbon in red and give us our product which is an alkene let's draw that in here so our electrons in magenta moved into here to form our double bond that's a tri substituted alkene what about if we took a proton from one of the other beta carbon so going back over here to our picture of the tertiary alcohol beta 1 and beta 2 would actually both give us the same product but we'll just take one of them and first I'd have to redraw my tertiary carbo cation so here it is plus one formal charge on the carbon in red and I'll draw in a hydrogen on this carbon so water is going to function as a base let me draw in our water molecule here and it's going to take this proton and these electrons are going to move in to form our double bond so we draw in our products we're going to have a double bond here and our electrons in magenta would move in to here this is only a dive substituted alkene so this would be the minor product and the major product would be the trisubstituted alkene which is the more stable one so this is the major product for our II 1 reaction