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- 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.