E2 E1 Sn2 Sn1 Reactions Example 2

E2 E1 Sn2 Sn1 Reactions Example 2

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.