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E2 E1 Sn2 Sn1 reactions example 2

E2 E1 Sn2 Sn1 Reactions Example 2. Created by Sal Khan.

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Video transcript

What I want to do in this video is think about what type of reaction we might have if we have ingredients very similar to what we saw in the last video. But instead of our nucleophile or our base being methoxide, it's going to be something slightly more involved. So it's still going to have the O minus, 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 have this thing right over here. So just like before, we have the exact same solvent. We have dimethylformamide. It's an aprotic solvent. That by itself would 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. It's also a very small molecule, and so it can really get in there and react with the substrate. So it's also a 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. So by making the base more, I guess, bulky, it's now-- 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. So we wouldn't see anything like this in the last video. We would only see something like this. And obviously, the base in this example is no longer just a methoxide. It looks like this. Let me clear it. Let me do my best to clear it. Edit, clear. 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, a CH3 and a CH3. So the reaction occurs just like what we saw in the last video, except this base is this big, old, bulky thing, but it can still act as a strong base, so it still nabs 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 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 E2.