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Current time:0:00Total duration:3:06

Video transcript

what I want to do with this video is think about what type of reaction we might have we have ingredients very similar to what we saw in the last video but instead of our nucleophile or our base being meth X mythic side it's going to be something slightly more involved so it's still going to have the o- but it's going to be bonded to a carbon which is then bonded to three methyl groups ch3 ch3 ch3 just like that so we don't have methoxide anymore we don't have methoxide anymore we have this thing right over here so just like before we have the exact same solvent we have dimethyl formamide it's an a protic solvent that by itself would push it put us in the sn2 or e2 direction but now we don't have methoxide anymore methoxide was both a strong base very strong base and it's also a very small molecule at that and so it can really get in there and react with the substrate so it was also strong nucleophile now this more bulky molecule it is still a strong base it is still an extremely strong base but now it's this big bulky molecule it would actually have trouble getting in to react with your substrate so it is no longer a good nucleophile this is not a good nucleophile this is not a good nucleophile so by making the base more I guess bulky it's now and or I guess you could also call it the nucleophile or the thing that would act as a nucleophile more bulky it is no longer a strong nucleophile so it would no longer be good for an sn2 reaction so just by changing the base a little bit or the nucleophile a little bit now this one would go strictly in the e2 direction e to direction so we wouldn't see anything we wouldn't see anything like this in the last video we would only see something like this but and obviously the the base in this in this example is no longer just the methoxide it looks like this it looks like this let me clear it do my best to clear it edit clear let me clear it over here as well let me clear it over here as well so now instead of just being bonded to a methyl group it's bonded to a carbon it's bonded to a carbon that's bonded to three methyl groups so ch3 ch3 ch3 or you could call this a tert-butyl group this whole thing over here so that's a carbon bonded to a ch3 ch3 and a ch3 so the reaction occurs just like what we saw in the last video except this base is this big ol bulky thing but it can still act as a strong base so still NAB's the hydrogen or really just the proton the hydrogen's electron that was bonded now goes to the alpha carbon the alpha carbon will then use lose an electron to the bromo group and that becomes bromide so the same exact mechanism different base but that base is now not a good nucleophile so you won't see sn2 occurring at all you will only see e 2