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What I want to do in this video
is think about what type

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of reaction we might have if
we have ingredients very

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similar to what we saw
in the last video.

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But instead of our nucleophile
or our base being methoxide,

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it's going to be something
slightly more involved.

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So it's still going to have the
O minus, but it's going to

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be bonded to a carbon, which is
then bonded to three methyl

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groups: CH3, CH3, CH3,
just like that.

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So we don't have methoxide
anymore.

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We have this thing
right over here.

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So just like before, we have
the exact same solvent.

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We have dimethylformamide.

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It's an aprotic solvent.

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That by itself would put us in
the Sn2 or E2 direction.

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But now we don't have
methoxide anymore.

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Methoxide was both a strong
base, very strong base.

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It's also a very small molecule,
and so it can really

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get in there and react
with the substrate.

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So it's also a strong
nucleophile.

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Now, this more bulky molecule,
it is still a strong base.

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It is still an extremely
strong base.

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But now it's this big,
bulky molecule.

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It would actually have trouble
getting in to react with your

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substrate, so it is no longer
a good nucleophile.

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This is not a good
nucleophile.

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So by making the base more, I
guess, bulky, it's now-- or I

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guess you could also call it the
nucleophile or the thing

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that would act as a nucleophile,
more bulky.

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It is no longer a strong
nucleophile, so it would no

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longer be good for
an Sn2 reaction.

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So just by changing the base a
little bit or the nucleophile

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a little bit, now this
one would go

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strictly in the E2 direction.

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So we wouldn't see anything like
this in the last video.

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We would only see something
like this.

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And obviously, the base in this
example is no longer just

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a methoxide.

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It looks like this.

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Let me clear it.

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Let me do my best to clear it.

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Edit, clear.

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Let me clear it over
here as well.

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So now instead of just being
bonded to a methyl group, it's

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bonded to a carbon.

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It's bonded to a carbon that's
bonded to three methyl groups.

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So CH3, CH3, CH3, or you could
call this a tert-butyl group;

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this whole thing over here.

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So that's a carbon bonded to
a CH3, a CH3 and a CH3.

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So the reaction occurs just like
what we saw in the last

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video, except this base is this
big, old, bulky thing,

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but it can still act as a strong
base, so it still nabs

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the hydrogen or really
just the proton.

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The hydrogen's electron that
was bonded now goes to the

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alpha carbon.

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The alpha carbon will then lose
an electron to the bromo

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group and that becomes bromide,
so the same exact

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mechanism, different base.

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But that base is now not a good
nucleophile, so you won't

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see Sn2 occurring at all.

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You will only see E2.

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