Alcohol properties

WEBVTT
Kind: captions
Language: en

00:00:00.610 --> 00:00:02.120
Let's think a little bit
about some of the

00:00:02.120 --> 00:00:03.820
properties of alcohol.

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So the general formula for an
alcohol we saw is some type of

00:00:07.260 --> 00:00:12.050
group or chain of carbons bonded
to an oxygen, bonded to

00:00:12.050 --> 00:00:13.210
a hydrogen.

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And of course, the oxygen
will have two lone

00:00:15.980 --> 00:00:17.220
pairs just like that.

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Let's compare this to water.

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So water just looks like this.

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You have a hydrogen bonded to
an oxygen, bonded to another

00:00:24.260 --> 00:00:27.190
hydrogen with two lone pairs.

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Now in the case of water,
the oxygen is much more

00:00:30.960 --> 00:00:33.090
electronegative than the
hydrogen, so it hogs the

00:00:33.090 --> 00:00:34.350
electrons towards it.

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So you have a partial negative
charge at the oxygen end.

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Then you have partial positive
charges at the hydrogen ends.

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That's what allows oxygen to
kind of-- or sorry-- that's

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what allows water to bond to
itself or to have not a

00:00:54.680 --> 00:00:56.910
ridiculously low
boiling point.

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So let me show this.

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Let me copy and paste this.

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We've seen all this before
in regular chemistry.

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So copy and paste.

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So let me draw some more
water molecules here.

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Let me draw another water
molecule here.

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So you see water because the
oxygen end has a partial

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negative charge and the hydrogen
ends have partial

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positive charges, the oxygen of
one water molecule will be

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attracted to the hydrogen of
another water molecule.

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And we've seen this before.

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This we call hydrogen bonding.

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So that right there is
hydrogen bonding.

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The exact same thing can happen
with alcohols, although

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alcohols really only have
the partial positive

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charge on the hydrogen.

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We don't know exactly what's
going on here.

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We probably have carbons
bonded to the oxygen.

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And with the carbons, they're
reasonably electronegative.

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They're not going to have their
electrons hogged as much

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as a hydrogen would.

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So in the case of an alcohol--
let me draw.

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Instead of having this R for
radical there, let me make it

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a little bit more concrete.

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Let me draw an actual alcohol.

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So an actual alcohol.

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Maybe we have methanol.

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Maybe we have methanol that
would look like that.

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It has a hydrogen
right over here.

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Oxygen is much more
electronegative than the

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hydrogen, so you have a partial
negative charge there.

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And then you have a partial
positive charge there.

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So it too, because of these
hydrogen bonds, it will have a

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reasonable boiling point.

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It won't just turn immediately
into the gaseous state.

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It would actually try to
bond to each other.

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Let me copy and paste that.

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So it can also form the
hydrogen bonds.

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Although they won't to be quite
as strong as what you

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see in water.

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And that's why something like
methanol actually has a lower

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boiling point than water.

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It's easy to make it boil.

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It's easier to make these bonds
break apart because you

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don't have as much of the
hydrogen bonding.

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So this is an example
of hydrogen

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bonding with methanol.

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Now because methanol can have
hydrogen bonding and it has

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this slight polarity to it and
water obviously has hydrogen

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bonding, methanol is actually
miscible in water.

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And all that means is that it's
soluble in water in any

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proportion.

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No matter how much methanol
or how much water

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you have, it is soluble.

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So if I were to draw some
methanol molecules-- actually,

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maybe this is the water
right here.

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So if you draw a methanol
molecule right there, that

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would have a hydrogen bond
right over there.

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If I were to draw another
methanol molecule maybe right

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over here, you would have
another hydrogen bond right

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over there.

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And that's what allows
methanol to

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be soluble in water.

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Now, as this chain grows, or
if you have alcohols with

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longer radical chains, then
they become less and less

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soluble in water.

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But their boiling points
actually do go up.

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And let's think about
why that is.

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So if I have something like--
let me do butanol.

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So butanol's going to
have 4 carbons.

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So it's going to be H3C, H3--
let me just draw it like H3C,

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CH2, Ch2, CH-- let me
do it like this.

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H2C.

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Then that carbon, that last
carbon right there is going to

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be bonded to the oxygen.

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It's going to be bonded
to an oxygen, which

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is bonded to a hydrogen.

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Now, when you have a situation
like this, the oxygen will

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have a partial negative
charge.

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The hydrogen will still have
a partial positive charge.

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Just like we saw up
here with both the

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water and the methanol.

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But now you have this big thing

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here that has no polarity.

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So this part of the alcohol is
not going to be soluble in

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water, and it's going to make it
harder for this part to be

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soluble over here.

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So this right here
is less soluble.

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This is less soluble.

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It'd still be a little
bit soluble.

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So if you have some oxygen here,
you will still have a

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little bit of the hydrogen
bonding.

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You still will have a little
bit of the hydrogen

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bonding going on.

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But this part is kind of-- you
can imagine it's almost-- it

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doesn't want to dissolve
with the water.

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It is non-polar.

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You could actually, for
example, butanol in

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particular, it actually
is soluble in water.

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But not in any proportion.

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So methanol is miscible.

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Let me write this.

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This is a new word.

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I don't think I've ever used
it before in the context of

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the organic chemistry videos.

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So methanol is-- let me write
that in a brighter color since

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it's a new word.

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Methanol is miscible, which
just means soluble in any

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proportion.

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So I don't care what percent
is methanol,

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what percent is water.

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The methanol will dissolve
into the water in any

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proportion.

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If you look at butanol, it
is soluble but not in any

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proportion.

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If you had a ton of butanol,
some of it would not dissolve

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in the water.

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So this is soluble.

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So the butanol right here
is soluble, but

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not miscible in water.

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If you have too much of the
butanol, all of a sudden, some

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of it will not actually be
able to be dissolved.

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If this was a decanol or
something with a really long

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carbon chain, then of
course, it's going

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to be very non soluble.

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You might be able to get a
couple of molecules in the

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water, but most of them
will not dissolve.

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Now the other reason-- I
hinted-- look, you know the

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reason why the alcohols have a
reasonable-- not too low of a

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boiling point is that
they're able to do

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this hydrogen bonding.

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But you would say well, look.

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You know, these longer carbon
chains, these are going to

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have less of the hydrogen
bonding going on.

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Maybe these would have
lower boiling points.

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But actually, the longer the
chain gets, these actually

00:06:54.680 --> 00:06:56.210
have higher boiling points.

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And that's because
these chains can

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interact with each other.

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So the longer the chain, so
longer R or the longer R

00:07:05.930 --> 00:07:08.800
chain, I guess, I could say,
we could say the higher the

00:07:08.800 --> 00:07:10.620
boiling point in an alcohol.

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Higher boiling point.

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It's harder.

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You have to put more heat into
the system or the temperature

00:07:16.840 --> 00:07:18.860
has to be higher for the
things to break apart.

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And that's because this is one
decanol molecule here, another

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decanol molecule might
look like this.

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Maybe it might look like this.

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You have an oxygen and a
hydrogen and then you have

00:07:29.460 --> 00:07:31.420
your carbons.

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So you have your CH,
your CH2, CH2, H3C.

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So you have this other
butanol here.

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And what the interaction between
these two chains are--

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these are the van
der Waal forces.

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So even though they have
no [INAUDIBLE],

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so these guys are going to have
some polar interactions.

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They're going to have the
hydrogen bonding.

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We've seen that multiple
times already.

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But these long chains, they're
going to have the London

00:07:57.250 --> 00:08:00.020
dispersion forces, which are a
subset of van der Waal forces.

00:08:00.020 --> 00:08:02.840
Where even though they're
neutral, every now and then,

00:08:02.840 --> 00:08:05.790
one of these might become
slightly negative on one side.

00:08:05.790 --> 00:08:09.410
So you might have
a very temporary

00:08:09.410 --> 00:08:10.670
partial negative charge.

00:08:10.670 --> 00:08:13.370
And that's just because
of the randomness of

00:08:13.370 --> 00:08:14.720
how electrons move.

00:08:14.720 --> 00:08:18.110
On this side of the molecule,
all of a sudden, you might

00:08:18.110 --> 00:08:19.630
have more electrons
over there.

00:08:19.630 --> 00:08:21.460
So you have a partial
negative charge.

00:08:21.460 --> 00:08:24.220
And because of that, you're
going to have-- the electrons

00:08:24.220 --> 00:08:26.395
over here, they're not going
to want to be there.

00:08:26.395 --> 00:08:28.210
So you're going to want to have
a partial positive charge

00:08:28.210 --> 00:08:30.570
there and you're going to have
a very temporary interaction.

00:08:30.570 --> 00:08:31.910
That's a very weak force.

00:08:31.910 --> 00:08:33.880
Much weaker than
hydrogen bonds.

00:08:33.880 --> 00:08:36.720
But as these chains get longer
and longer, as they possibly

00:08:36.720 --> 00:08:38.840
even get intertwined with each
other and get close to each

00:08:38.840 --> 00:08:42.230
other, these London dispersion
forces or van der Waal forces

00:08:42.230 --> 00:08:43.220
are going to keep propagating.

00:08:43.220 --> 00:08:44.970
So all of a sudden, maybe these
guys are going to be

00:08:44.970 --> 00:08:46.900
attracted to each other and
that's going to disappear.

00:08:46.900 --> 00:08:48.790
Than these guys are going be
attracted to each other and

00:08:48.790 --> 00:08:50.000
then that's going
to disappear.

00:08:50.000 --> 00:08:51.800
And then these are going to be
attracted to each other and

00:08:51.800 --> 00:08:53.000
then that's going
to disappear.

00:08:53.000 --> 00:08:54.940
And so you can imagine, the
longer the chain, the more of

00:08:54.940 --> 00:08:57.290
these type of interactions
you're going to have. The more

00:08:57.290 --> 00:08:58.970
attracted they're going
to be to each other.

00:08:58.970 --> 00:09:01.720
And it's going to be harder to
break them apart, higher

00:09:01.720 --> 00:09:02.680
boiling point.

00:09:02.680 --> 00:09:04.810
So those are just kind of the
two big takeaways on the

00:09:04.810 --> 00:09:08.350
properties of alcohols.

00:09:08.350 --> 00:09:11.170
Especially smaller chained
alcohols are soluble in water.

00:09:11.170 --> 00:09:13.610
The very small ones are
completely miscible.

00:09:13.610 --> 00:09:16.026
And the longer the chain you
have, the harder it is to

00:09:16.026 --> 00:09:16.870
dissolve in water.

00:09:16.870 --> 00:09:19.130
But also, the higher
the boiling point.

00:09:19.130 --> 00:09:20.940
The harder it is to break them
apart because you have these

00:09:20.940 --> 00:09:23.180
London dispersion forces.


E2 E1 Sn2 Sn1 Reactions Example 3

WEBVTT
Kind: captions
Language: en

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Let's think about what type of
reaction might occur if we

00:00:03.640 --> 00:00:05.990
have this molecule
right over here.

00:00:05.990 --> 00:00:07.600
I won't go through the
trouble of naming it.

00:00:07.600 --> 00:00:09.590
It would take up too much
time in this video.

00:00:09.590 --> 00:00:11.680
But it's dissolved
in methanol.

00:00:14.350 --> 00:00:16.110
When we talk about what type of
reactions, we're going to

00:00:16.110 --> 00:00:22.230
pick between Sn2, Sn1,
E2 and E1 reactions.

00:00:22.230 --> 00:00:24.590
Now, maybe the first place to
start or the place I like to

00:00:24.590 --> 00:00:26.800
start is to just look at
the solvent itself.

00:00:26.800 --> 00:00:28.680
And when we're trying to decide
what type of reaction

00:00:28.680 --> 00:00:31.630
will occur, the important thing
to think about is, is

00:00:31.630 --> 00:00:34.370
the solvent protic or aprotic?

00:00:34.370 --> 00:00:35.810
And if you look at this
solvent right

00:00:35.810 --> 00:00:37.410
here, this is methanol.

00:00:37.410 --> 00:00:40.420
It is protic.

00:00:40.420 --> 00:00:44.150
And in case you don't remember
what protic means, it means

00:00:44.150 --> 00:00:46.980
that there are protons flying
around in the solvent, that

00:00:46.980 --> 00:00:49.910
they can kind of go loose and
jump around from one molecule

00:00:49.910 --> 00:00:50.590
to another.

00:00:50.590 --> 00:00:54.680
And the reason why I know that
methanol is protic is because

00:00:54.680 --> 00:00:57.850
you have hydrogen
bonded to a very

00:00:57.850 --> 00:01:01.070
electronegative atom in oxygen.

00:01:01.070 --> 00:01:04.090
So every now and then, in one of
the methanol molecules, the

00:01:04.090 --> 00:01:10.840
oxygen can steal hydrogen's
electron.

00:01:10.840 --> 00:01:13.150
And then the hydrogen itself,
without the electron, the

00:01:13.150 --> 00:01:15.990
hydrogen proton will be flying
around because it

00:01:15.990 --> 00:01:17.140
doesn't have a neutron.

00:01:17.140 --> 00:01:19.970
So this is a protic solvent.

00:01:19.970 --> 00:01:22.460
Now, you might say, well does
anything with a hydrogen,

00:01:22.460 --> 00:01:23.280
would that be protic?

00:01:23.280 --> 00:01:24.100
And the answer is no.

00:01:24.100 --> 00:01:25.810
If you have a bunch of hydrogens
bonded to just

00:01:25.810 --> 00:01:27.250
carbons, that is not protic.

00:01:27.250 --> 00:01:29.650
Carbon is not so electronegative
that it could

00:01:29.650 --> 00:01:32.190
steal a hydrogen's electron
and have the hydrogens

00:01:32.190 --> 00:01:32.900
floating around.

00:01:32.900 --> 00:01:35.670
So a big giveaway is hydrogen
bonded to a very

00:01:35.670 --> 00:01:38.390
electronegative atom
like oxygen.

00:01:38.390 --> 00:01:39.780
So this is protic.

00:01:39.780 --> 00:01:42.130
And when we think about protic
out of all of the reactions we

00:01:42.130 --> 00:01:49.260
studied, that favors-- well,
even a better way to think

00:01:49.260 --> 00:01:50.990
about it is it disfavors.

00:01:50.990 --> 00:01:59.570
So it tells us that it's
unlikely to have an Sn2 or an

00:01:59.570 --> 00:02:00.730
E2 reaction.

00:02:00.730 --> 00:02:04.070
And the logic there is an Sn2
reaction needs a strong

00:02:04.070 --> 00:02:05.050
nucleophile.

00:02:05.050 --> 00:02:08.250
An E2 reaction needs
a strong base.

00:02:08.250 --> 00:02:11.720
Now, if you have protons flying
around, the nucleophile

00:02:11.720 --> 00:02:15.610
or the base is likely to
react with the proton.

00:02:15.610 --> 00:02:17.620
It would not be likely
to react with

00:02:17.620 --> 00:02:18.960
the substrate itself.

00:02:18.960 --> 00:02:22.170
So a protic solution, you're
unlikely to have an Sn2 or E2.

00:02:22.170 --> 00:02:31.060
What you are likely to have is
an Sn1 or an E1 reaction.

00:02:31.060 --> 00:02:33.970
Both of these need the leaving
group to leave on its own, and

00:02:33.970 --> 00:02:37.150
actually, having protons around
might help to stabilize

00:02:37.150 --> 00:02:39.070
the leaving group
to some degree.

00:02:39.070 --> 00:02:43.060
So it makes Sn2, E2 unlikely,
Sn1, E1 a little more likely.

00:02:43.060 --> 00:02:45.430
So far, these are our
good candidates.

00:02:45.430 --> 00:02:47.400
Now, the next thing to think
about is to just look at the

00:02:47.400 --> 00:02:50.360
leaving group itself,
or see if there is

00:02:50.360 --> 00:02:51.960
even a leaving group.

00:02:51.960 --> 00:02:54.805
And over here, everything we see
on this molecule is either

00:02:54.805 --> 00:03:00.070
a carbon or a hydrogen, except
for this iodine right here.

00:03:02.660 --> 00:03:07.320
And we know that iodide is
a good leaving group.

00:03:07.320 --> 00:03:11.580
Well, a good leaving group, it
does not make it any less

00:03:11.580 --> 00:03:13.350
likely that you'd
have Sn2 or E2.

00:03:13.350 --> 00:03:15.880
Both of those can do well with
a good leaving group.

00:03:15.880 --> 00:03:20.080
But it's a necessary requirement
for an Sn1 or an

00:03:20.080 --> 00:03:20.910
E1 reaction.

00:03:20.910 --> 00:03:23.450
Remember, an Sn1 and E1, in both
of them, the first step

00:03:23.450 --> 00:03:26.170
is that the leaving group
leaves on its own.

00:03:26.170 --> 00:03:28.180
That is the rate-determining
step.

00:03:28.180 --> 00:03:30.895
So that's a requirement
for Sn1 or E1, so

00:03:30.895 --> 00:03:32.460
it still looks likely.

00:03:32.460 --> 00:03:34.320
We haven't seen anything that
would make us think that we

00:03:34.320 --> 00:03:36.550
wouldn't have an Sn1 or E1.

00:03:36.550 --> 00:03:37.900
This is a good leaving group.

00:03:37.900 --> 00:03:38.520
Let me write it here.

00:03:38.520 --> 00:03:43.740
Good leaving group.

00:03:43.740 --> 00:03:47.060
Now, the last thing that we can
think about right here is

00:03:47.060 --> 00:03:49.330
the carbon that we might
be leaving from.

00:03:49.330 --> 00:03:52.000
So far, everything is pointing
in the Sn1 and E1 direction,

00:03:52.000 --> 00:03:55.200
and kind of the final thing
is when this leaving group

00:03:55.200 --> 00:03:57.990
leaves, it's going to form a
carbocation from the carbon

00:03:57.990 --> 00:03:59.250
that it's bonded to right now.

00:03:59.250 --> 00:04:01.770
And in order for that
carbocation to be reasonably

00:04:01.770 --> 00:04:05.010
stable, at minimum it should
be a secondary carbocation

00:04:05.010 --> 00:04:07.140
bonded to at least two carbons,
but ideally, it would

00:04:07.140 --> 00:04:08.430
be bonded to three carbons.

00:04:08.430 --> 00:04:10.470
It would be a tertiary carbon.

00:04:10.470 --> 00:04:16.490
Now, the carbon that the leaving
group is bonded to is

00:04:16.490 --> 00:04:17.630
a tertiary carbon.

00:04:17.630 --> 00:04:21.399
It's bonded to one, two,
three carbons.

00:04:21.399 --> 00:04:25.110
so it is a tertiary carbon.

00:04:25.110 --> 00:04:28.240
It can actually be a
stable carbocation.

00:04:28.240 --> 00:04:30.820
It's a tertiary carbon,
which it once again

00:04:30.820 --> 00:04:32.440
favors Sn1 and E1.

00:04:32.440 --> 00:04:34.950
So all of the clues here
tell us that Sn1 and E1

00:04:34.950 --> 00:04:35.970
are going to happen.

00:04:35.970 --> 00:04:37.520
And actually, they'll
both happen.

00:04:37.520 --> 00:04:41.190
So let's think about
the mechanism.

00:04:41.190 --> 00:04:44.010
In the very first step, the
leaving group leaves in either

00:04:44.010 --> 00:04:45.420
one of these reactions.

00:04:45.420 --> 00:04:47.800
So if we look at the iodine,
it already has

00:04:47.800 --> 00:04:49.080
seven valence electrons.

00:04:49.080 --> 00:04:54.410
One, two, three, four,
five, six, seven.

00:04:54.410 --> 00:04:57.250
And in the first step, the
rate-determining step of the

00:04:57.250 --> 00:05:02.440
Sn1 or E1 reaction, the iodine's
going to nab an

00:05:02.440 --> 00:05:03.510
electron off the carbon.

00:05:03.510 --> 00:05:08.060
I'll do that electron in
green right over there.

00:05:08.060 --> 00:05:12.200
That's going to get nabbed onto
iodine to make iodide,

00:05:12.200 --> 00:05:14.390
and then that carbon is going to
lose an electron and become

00:05:14.390 --> 00:05:15.570
a carbocation.

00:05:15.570 --> 00:05:17.790
So after that very first
step, we have something

00:05:17.790 --> 00:05:19.460
that looks like this.

00:05:19.460 --> 00:05:21.760
So that's our molecule.

00:05:21.760 --> 00:05:24.070
Just so it's clear what we did,
this arrow I'll do all

00:05:24.070 --> 00:05:26.750
the way over here, so
it's clear that

00:05:26.750 --> 00:05:29.320
this is our next step.

00:05:29.320 --> 00:05:30.740
And what's happened here?

00:05:30.740 --> 00:05:34.730
Our tertiary carbon has
lost its electron.

00:05:34.730 --> 00:05:38.520
So now this carbon right here
has a positive charge.

00:05:38.520 --> 00:05:41.190
It's a tertiary carbocation.

00:05:41.190 --> 00:05:43.610
And now the iodine has
become iodide.

00:05:43.610 --> 00:05:47.430
It has left the molecule.

00:05:47.430 --> 00:05:51.420
So it had its original seven
valence electrons: one, two,

00:05:51.420 --> 00:05:54.200
three, four, five, six, seven.

00:05:54.200 --> 00:05:58.250
It nabbed one more electron
from the carbon and now it

00:05:58.250 --> 00:06:00.880
is-- I wanted to do
that in green.

00:06:00.880 --> 00:06:05.080
It nabbed one more electron from
the carbon, now it is the

00:06:05.080 --> 00:06:10.550
iodide anion.

00:06:10.550 --> 00:06:12.510
So this step right
here is common to

00:06:12.510 --> 00:06:14.940
both Sn1 and E1 reaction.

00:06:14.940 --> 00:06:17.250
The leaving group
has to leave.

00:06:17.250 --> 00:06:20.170
Now, after this, they
start to diverge.

00:06:20.170 --> 00:06:23.520
In an Sn1, the leaving group
essentially gets substituted

00:06:23.520 --> 00:06:25.620
with a weak nucleophile.

00:06:25.620 --> 00:06:29.250
In an E1, a weak base strips off
one of the beta hydrogens

00:06:29.250 --> 00:06:30.420
and forms an alkene.

00:06:30.420 --> 00:06:33.850
So let's do them separately.

00:06:33.850 --> 00:06:35.220
So over here, I'm going
to do the Sn1.

00:06:38.120 --> 00:06:43.280
And on the right-hand side,
I will do the E1 reaction.

00:06:43.280 --> 00:06:46.250
So let me start over here.

00:06:46.250 --> 00:06:49.180
So the Sn1 is starting over
here at this step.

00:06:49.180 --> 00:06:51.290
I'll just redo this
step over here.

00:06:51.290 --> 00:06:53.980
So this has a positive charge.

00:06:53.980 --> 00:06:55.390
That has a positive
charge here.

00:06:55.390 --> 00:06:59.290
The iodide has left.

00:06:59.290 --> 00:07:01.960
I don't have to draw all its
valence electrons anymore.

00:07:01.960 --> 00:07:03.520
And what's going
to happen next?

00:07:03.520 --> 00:07:05.380
We're going to get substituted
with the weak base, and the

00:07:05.380 --> 00:07:08.440
weak base here is actually
the methanol.

00:07:08.440 --> 00:07:10.320
The weak base here
is the methanol.

00:07:10.320 --> 00:07:12.190
So let me draw some
methanol here.

00:07:17.770 --> 00:07:22.810
It's got two unbonded pairs of
electrons and one of them,

00:07:22.810 --> 00:07:24.000
it's a weak base.

00:07:24.000 --> 00:07:26.400
It was willing to give
an electron.

00:07:26.400 --> 00:07:29.460
It has a partial negative
charge over here because

00:07:29.460 --> 00:07:31.280
oxygen is electronegative, but
it doesn't have a full

00:07:31.280 --> 00:07:33.460
negative charge, so it's not
a strong nucleophile.

00:07:33.460 --> 00:07:36.500
But it can donate an electron
to this carbocation, and

00:07:36.500 --> 00:07:38.480
that's what is going
to happen.

00:07:38.480 --> 00:07:42.210
It will donate an electron
to this carbocation.

00:07:42.210 --> 00:07:48.420
And then after that happens,
it will look like this.

00:07:48.420 --> 00:07:51.690
That's our original molecule.

00:07:51.690 --> 00:07:55.100
Now this magenta electron
has been donated to the

00:07:55.100 --> 00:07:56.370
carbocation.

00:07:56.370 --> 00:08:00.090
The other end of it is this blue
electron right here on

00:08:00.090 --> 00:08:01.080
the oxygen.

00:08:01.080 --> 00:08:02.770
It is now bonded.

00:08:02.770 --> 00:08:03.880
That is our oxygen.

00:08:03.880 --> 00:08:07.570
Here's that other pair of
electrons on that oxygen, and

00:08:07.570 --> 00:08:13.130
it is bonded to a hydrogen
and a methyl group.

00:08:13.130 --> 00:08:15.930
And then the last step of this
is another weak base might be

00:08:15.930 --> 00:08:19.150
able to come and nab off the
hydrogen proton right there.

00:08:19.150 --> 00:08:21.230
Oh, I want to be very
clear here.

00:08:21.230 --> 00:08:22.670
The oxygen was neutral.

00:08:22.670 --> 00:08:24.100
The methanol here is neutral.

00:08:24.100 --> 00:08:28.040
It is giving away an electron
to the carbocation.

00:08:28.040 --> 00:08:30.750
The carbocation had a positive
charge because it had lost it

00:08:30.750 --> 00:08:31.320
originally.

00:08:31.320 --> 00:08:32.450
Now it gets an electron back.

00:08:32.450 --> 00:08:33.990
It becomes neutral.

00:08:33.990 --> 00:08:37.230
The methanol, on the other hand,
was neutral, gives away

00:08:37.230 --> 00:08:42.210
an electron, so now it becomes--
it now is positive.

00:08:42.210 --> 00:08:45.040
So now you might have
another methanol.

00:08:45.040 --> 00:08:47.810
You might have another methanol
molecule sitting out

00:08:47.810 --> 00:08:54.570
here someplace that might also
nab the proton off of this

00:08:54.570 --> 00:08:56.520
positive ion.

00:08:56.520 --> 00:08:59.090
So this one right here,
it would nab it or it

00:08:59.090 --> 00:08:59.860
would bond with it.

00:08:59.860 --> 00:09:03.850
It would give the electron to
the hydrogen proton, really.

00:09:03.850 --> 00:09:08.110
The hydrogen's electron gets
nabbed by the oxygen, and so

00:09:08.110 --> 00:09:10.570
then that becomes neutral.

00:09:10.570 --> 00:09:14.920
So in the final step, it'll
all look like this.

00:09:14.920 --> 00:09:16.370
We have that over here.

00:09:16.370 --> 00:09:19.530
The methanol that had originally
bonded has lost its

00:09:19.530 --> 00:09:21.890
hydrogen, so it looks
like this.

00:09:21.890 --> 00:09:27.140
We just have the oxygen
and the CH3 there.

00:09:27.140 --> 00:09:30.830
It is now neutral because it
gained an electron when that

00:09:30.830 --> 00:09:32.920
hydrogen proton was nabbed.

00:09:32.920 --> 00:09:34.540
So if you wanted to draw it,
it has actually those two

00:09:34.540 --> 00:09:36.980
extra electrons,
just like that.

00:09:36.980 --> 00:09:39.100
And if you want to draw this
last methanol, it's now a

00:09:39.100 --> 00:09:42.080
positive cation, so it
looks like this.

00:09:42.080 --> 00:09:49.530
So it's OH, CH3, H, and then
it has unbonded pair right

00:09:49.530 --> 00:09:51.910
there, and now this has
a positive charge.

00:09:51.910 --> 00:09:54.120
So that was the Sn1 reaction.

00:09:54.120 --> 00:09:56.310
Now, the other reaction that's
going to occur is the E1.

00:09:58.850 --> 00:10:02.365
Once again, our first
step-- nope, I

00:10:02.365 --> 00:10:03.450
didn't want to do that.

00:10:03.450 --> 00:10:06.780
Our first step looks
like that.

00:10:06.780 --> 00:10:08.220
So that is our first step.

00:10:08.220 --> 00:10:09.390
Let me get everything
straight.

00:10:09.390 --> 00:10:11.550
So the leaving group had left.

00:10:11.550 --> 00:10:14.580
So in each situation, the
leaving group had left.

00:10:14.580 --> 00:10:15.990
Our iodide is up here.

00:10:19.010 --> 00:10:22.510
And in an E1 reaction, you
don't get substituted.

00:10:22.510 --> 00:10:26.120
What happens is one of the beta
carbons gets a hydrogen

00:10:26.120 --> 00:10:28.320
swiped off of it
by a weak base.

00:10:28.320 --> 00:10:30.410
Now, let's think about what
a beta carbon is.

00:10:30.410 --> 00:10:33.250
The alpha carbon is the
carbocation carbon.

00:10:33.250 --> 00:10:34.130
That's right over there.

00:10:34.130 --> 00:10:35.480
That's the alpha carbon.

00:10:35.480 --> 00:10:37.150
Beta carbons are one
carbon away.

00:10:37.150 --> 00:10:40.750
So this is a beta carbon, this
is a beta carbon, and then

00:10:40.750 --> 00:10:42.410
that is a beta carbon.

00:10:42.410 --> 00:10:44.930
Now, this carbon over here is
not bonded to any hydrogens.

00:10:44.930 --> 00:10:46.470
It's only bonded to
other carbons.

00:10:46.470 --> 00:10:49.280
So that one cannot lose
any hydrogens.

00:10:49.280 --> 00:10:51.180
And then we have to pick between
this carbon that's

00:10:51.180 --> 00:10:54.030
bonded to three hydrogens and
this carbon that's actually

00:10:54.030 --> 00:10:55.050
bonded to two hydrogens.

00:10:55.050 --> 00:10:57.530
I didn't draw it before, but
it's implicitly there.

00:10:57.530 --> 00:10:58.960
It's bonded to two hydrogens.

00:10:58.960 --> 00:11:03.860
Now Zaitsev's rule tells us that
the dominating product is

00:11:03.860 --> 00:11:07.390
going to be produced when the
carbon that has less hydrogens

00:11:07.390 --> 00:11:08.400
loses a hydrogen.

00:11:08.400 --> 00:11:10.820
So out of this, this carbon
has two hydrogens,

00:11:10.820 --> 00:11:12.030
this one has three.

00:11:12.030 --> 00:11:14.300
So the one that's going to lose
the hydrogen to the base,

00:11:14.300 --> 00:11:16.440
or more likely to lose the
hydrogen to the base, is the

00:11:16.440 --> 00:11:18.870
one that has two hydrogens, not

00:11:18.870 --> 00:11:20.670
three, the fewer hydrogens.

00:11:20.670 --> 00:11:23.520
So our base in this case is once
again the methanol, acted

00:11:23.520 --> 00:11:26.900
as a nucleophile in the
Sn1, acted as a weak

00:11:26.900 --> 00:11:28.350
nucleophile in Sn1.

00:11:28.350 --> 00:11:30.700
Now it'll act as a weak base.

00:11:30.700 --> 00:11:35.090
So we have methanol
right over here.

00:11:35.090 --> 00:11:36.480
That's a hydrogen.

00:11:36.480 --> 00:11:40.280
CH3 has some electrons
right over here.

00:11:40.280 --> 00:11:42.330
It has a partial negative
charge.

00:11:42.330 --> 00:11:47.440
It will give one of the
electrons to the hydrogen,

00:11:47.440 --> 00:11:49.030
just to the hydrogen proton.

00:11:49.030 --> 00:11:51.250
The hydrogen's not going to
take its electron with it.

00:11:51.250 --> 00:11:57.770
That electron is then going to
be given to the carbocation to

00:11:57.770 --> 00:12:00.750
make it neutral, and it will
form a double bond.

00:12:00.750 --> 00:12:03.060
So after that happens, we
are left with something

00:12:03.060 --> 00:12:04.070
that looks like this.

00:12:04.070 --> 00:12:05.820
This was our original
molecule.

00:12:05.820 --> 00:12:13.010
Now, this carbon right
here in yellow, it

00:12:13.010 --> 00:12:14.090
has lost its hydrogen.

00:12:14.090 --> 00:12:15.760
This hydrogen back here
is still there.

00:12:15.760 --> 00:12:16.960
I could draw it if I like.

00:12:16.960 --> 00:12:17.480
I don't need to.

00:12:17.480 --> 00:12:19.070
It just makes the thing messy.

00:12:19.070 --> 00:12:21.930
But it has now formed a double
bond with the primary carbon.

00:12:21.930 --> 00:12:24.040
It has now formed
a double bond.

00:12:24.040 --> 00:12:28.190
That electron that had bonded
with the hydrogen was now

00:12:28.190 --> 00:12:29.820
given to the carbocation.

00:12:29.820 --> 00:12:31.520
It has a double bond
with that.

00:12:31.520 --> 00:12:36.880
And then you have your methanol
has now turned into a

00:12:36.880 --> 00:12:38.710
positive ion.

00:12:38.710 --> 00:12:43.810
It has now turned into
a positive ion: CH3.

00:12:43.810 --> 00:12:46.390
And now it has this bond.

00:12:46.390 --> 00:12:49.020
It has this bond with
the hydrogen.

00:12:49.020 --> 00:12:52.520
I'll even make that electron
that it gave away in magenta.

00:12:52.520 --> 00:12:58.150
And then it has that extra
lone pair of electrons.

00:12:58.150 --> 00:13:00.660
So, in this circumstance, we
looked at all of the clues.

00:13:00.660 --> 00:13:03.330
All of the clues were
against Sn2 and E2.

00:13:03.330 --> 00:13:04.860
They favored Sn1 and E1.

00:13:04.860 --> 00:13:07.080
So if you were to actually
make this, if you were to

00:13:07.080 --> 00:13:09.830
actually try to see what happens
is this reaction, you

00:13:09.830 --> 00:13:13.900
would get products of both Sn1
and E1 reactions, these

00:13:13.900 --> 00:13:17.270
products and these products
over here.


Comparing E2 E1 Sn2 Sn1 Reactions

WEBVTT
Kind: captions
Language: en

00:00:00.530 --> 00:00:02.730
What I want to do in this video
is to try to figure out

00:00:02.730 --> 00:00:06.470
what type of reaction or
reactions might occur if we

00:00:06.470 --> 00:00:07.150
have-- what is this?

00:00:07.150 --> 00:00:09.710
One, two, three, four,
five It's in a cycle.

00:00:09.710 --> 00:00:12.205
This is bromocyclopentane.

00:00:12.205 --> 00:00:18.560
If we have some
bromocyclopentane dissolved

00:00:18.560 --> 00:00:21.380
and our solvent is
dimethylformamide.

00:00:21.380 --> 00:00:23.060
Sometimes you'll see that
just written as DMF.

00:00:23.060 --> 00:00:26.160
And I've actually drawn the
formula for it here, so we can

00:00:26.160 --> 00:00:28.470
think about what type
of a solvent it is.

00:00:28.470 --> 00:00:32.890
And also, in our solution, we
have the methoxide ion.

00:00:32.890 --> 00:00:35.340
So we also have the methoxide
ion right here.

00:00:35.340 --> 00:00:39.250
So let's think about what type
of reaction might occur.

00:00:39.250 --> 00:00:41.400
And just to narrow things down,
we'll think about it in

00:00:41.400 --> 00:00:44.380
the context of the last four
types of reactions

00:00:44.380 --> 00:00:45.850
we've looked at.

00:00:45.850 --> 00:00:51.480
So this might be an Sn2
reaction, an Sn1 reaction, an

00:00:51.480 --> 00:00:54.950
E2 reaction, or an
E1 reaction.

00:00:54.950 --> 00:00:56.520
We're going to look at all the
clues and figure out what's

00:00:56.520 --> 00:00:59.060
likely to occur, and then
actually draw the mechanism

00:00:59.060 --> 00:01:01.560
for it occurring.

00:01:01.560 --> 00:01:05.050
Now, the first thing since they
gave us the solvent and

00:01:05.050 --> 00:01:07.160
other things that are in the
solvent, let's think about how

00:01:07.160 --> 00:01:08.670
those might affect
the reaction.

00:01:08.670 --> 00:01:11.960
So if we look at this solvent
right here, whenever you look

00:01:11.960 --> 00:01:13.560
at any of these reactions, when
you look at the solvent,

00:01:13.560 --> 00:01:14.330
you just want to
think about it.

00:01:14.330 --> 00:01:15.650
Is it protic or not?

00:01:15.650 --> 00:01:18.230
And protic means that it has
hydrogens that can kind of be

00:01:18.230 --> 00:01:21.120
released or that their electrons
could be nabbed off

00:01:21.120 --> 00:01:22.970
and these protons could
just float around.

00:01:22.970 --> 00:01:25.610
And if we look over here, we do
have hydrogens, but all of

00:01:25.610 --> 00:01:26.990
the hydrogens are bonded
to carbon.

00:01:26.990 --> 00:01:30.910
And carbon is unlikely to just
steal a hydrogen's electrons

00:01:30.910 --> 00:01:32.170
and let the hydrogen
float around.

00:01:32.170 --> 00:01:34.080
Carbon is not that
electronegative.

00:01:34.080 --> 00:01:37.000
If you had hydrogens bonded
to an oxygen, that'd be a

00:01:37.000 --> 00:01:37.650
different question.

00:01:37.650 --> 00:01:40.610
Then you would have
a protic solvent.

00:01:40.610 --> 00:01:43.180
But in this case, all the
hydrogens bonded to carbons,

00:01:43.180 --> 00:01:45.470
not likely to get their
electrons nabbed off and float

00:01:45.470 --> 00:01:46.870
around as free protons.

00:01:46.870 --> 00:01:48.350
So this is an aprotic solvent.

00:01:54.180 --> 00:01:57.630
Now, we've gone over this a
little bit with Sn2 and Sn1,

00:01:57.630 --> 00:01:59.650
but the same idea applies.

00:01:59.650 --> 00:02:07.020
In order to have an Sn2 or an E2
reaction, you have to have

00:02:07.020 --> 00:02:09.669
either a strong nucleophile or
a strong base, and the same

00:02:09.669 --> 00:02:12.160
thing could actually be both,
although they're not always

00:02:12.160 --> 00:02:12.600
correlated.

00:02:12.600 --> 00:02:13.940
We've seen that before.

00:02:13.940 --> 00:02:17.790
Now, if you had a protic
solvent, it would stabilize

00:02:17.790 --> 00:02:21.390
the strong base or the
strong nucleophile.

00:02:21.390 --> 00:02:23.080
The protons would
react with them.

00:02:23.080 --> 00:02:25.450
They would take the electrons
from that strong base or that

00:02:25.450 --> 00:02:26.580
strong nucleophile.

00:02:26.580 --> 00:02:29.250
So in order to have an Sn2 or
an E2, you have to have no

00:02:29.250 --> 00:02:32.610
protons flying around, so you
need an aprotic solvent.

00:02:32.610 --> 00:02:40.580
So this aprotic solvent will
favor Sn2 or an E2 reaction.

00:02:40.580 --> 00:02:45.670
Now, so our mind is already
thinking in Sn2 or E2, let's

00:02:45.670 --> 00:02:47.320
think about the reactants
themselves.

00:02:47.320 --> 00:02:49.990
So over here, we have
the methoxide ion.

00:02:54.460 --> 00:02:57.180
And let's think about whether
it's a strong or weak.

00:02:57.180 --> 00:02:59.370
Let's think about it first as a
strong or weak nucleophile.

00:02:59.370 --> 00:03:02.610
It's actually a pretty
strong nucleophile.

00:03:02.610 --> 00:03:04.010
It is a strong nucleophile.

00:03:08.470 --> 00:03:13.090
So that would put us in the
direction of an Sn1.

00:03:13.090 --> 00:03:14.280
So we have two data points.

00:03:14.280 --> 00:03:15.670
I'm sorry, for an Sn2.

00:03:15.670 --> 00:03:18.510
We have two data points for Sn2
because remember, it has

00:03:18.510 --> 00:03:20.500
to just kind of go in
there and be active.

00:03:20.500 --> 00:03:22.640
It's not too big of a molecule,
so it's not going to

00:03:22.640 --> 00:03:24.000
be hindered.

00:03:24.000 --> 00:03:27.390
But it's also an extremely
strong base, even stronger

00:03:27.390 --> 00:03:28.880
than hydroxide.

00:03:28.880 --> 00:03:38.920
So it's also an extremely strong
base, which might lead

00:03:38.920 --> 00:03:40.590
us or that does imply
that we're going

00:03:40.590 --> 00:03:43.680
to have an E2 reaction.

00:03:43.680 --> 00:03:47.640
Now, the last thing we need to
think about is the carbon

00:03:47.640 --> 00:03:49.660
where the leaving group
might leave from.

00:03:49.660 --> 00:03:52.230
And immediately, when you look
at the bromocyclopentane,

00:03:52.230 --> 00:03:54.590
there's only one functional
group attached to the chain,

00:03:54.590 --> 00:03:58.350
and that is the bromo group
right here, right there.

00:03:58.350 --> 00:03:59.760
It is attached to this carbon.

00:03:59.760 --> 00:04:01.860
We could call that the
alpha carbon, and it

00:04:01.860 --> 00:04:03.150
is a secondary carbon.

00:04:03.150 --> 00:04:05.100
This carbon right here
is bonded to

00:04:05.100 --> 00:04:07.655
one, two other carbons.

00:04:10.350 --> 00:04:13.660
This alpha carbon-- let
me write it this way.

00:04:13.660 --> 00:04:23.040
This alpha carbon is a secondary
carbon, and that

00:04:23.040 --> 00:04:25.540
kind of makes it neutral
in this mix.

00:04:25.540 --> 00:04:29.190
If it was a methyl or
primary carbon, it

00:04:29.190 --> 00:04:32.740
would favor Sn2, actually.

00:04:32.740 --> 00:04:36.540
I mean methyl, the only thing
you could have is an Sn2.

00:04:36.540 --> 00:04:39.970
And if it was a tertiary carbon,
it would favor Sn1 or

00:04:39.970 --> 00:04:43.500
E1 because it would favor
a stable carbocation.

00:04:43.500 --> 00:04:45.770
The leaving group could just
leave. And if this guy was

00:04:45.770 --> 00:04:49.210
bonded to another carbon,
it would be very stable.

00:04:49.210 --> 00:04:51.300
But in this situation, it's
a secondary carbon

00:04:51.300 --> 00:04:52.250
bonded to two carbons.

00:04:52.250 --> 00:04:54.460
It's a little bit neutral.

00:04:54.460 --> 00:04:56.750
Any of these reactions
might occur.

00:04:56.750 --> 00:04:59.000
When we look at all of the other
data points, they're

00:04:59.000 --> 00:05:01.010
pointing at both Sn2 or E2.

00:05:01.010 --> 00:05:03.430
We have a strong
nucleophile/base.

00:05:03.430 --> 00:05:04.770
We have an aprotic solvent.

00:05:04.770 --> 00:05:08.110
It's going to be Sn2
or an E2 reaction.

00:05:08.110 --> 00:05:10.000
So let's actually draw
the reactions.

00:05:10.000 --> 00:05:15.860
Let me do the Sn2 first. So
let me do it in orange.

00:05:15.860 --> 00:05:19.770
So if we were to have
an Sn2 reaction, let

00:05:19.770 --> 00:05:21.020
me redraw the molecule.

00:05:23.190 --> 00:05:26.630
Let me draw the cyclopentane
part.

00:05:26.630 --> 00:05:28.350
I want to make sure-- let me
draw it the same way I had it

00:05:28.350 --> 00:05:29.570
drawn up there.

00:05:29.570 --> 00:05:32.750
So the pentagon is
facing upwards.

00:05:32.750 --> 00:05:35.790
And then we have our bromo
group right there.

00:05:35.790 --> 00:05:41.910
So we have our methoxide
ion right over here.

00:05:41.910 --> 00:05:45.980
So CH3O minus.

00:05:45.980 --> 00:05:48.410
Or another way we could view
it is that this oxygen has

00:05:48.410 --> 00:05:52.740
one, two, three, four, five,
six, seven valence electrons

00:05:52.740 --> 00:05:54.440
with a negative charge.

00:05:54.440 --> 00:06:02.070
One of these electrons right
over here, this can attack the

00:06:02.070 --> 00:06:04.410
substrate right over
there, that carbon.

00:06:04.410 --> 00:06:09.960
Right when that happens,
simultaneously this bromine is

00:06:09.960 --> 00:06:13.460
going to be able to
nab an electron

00:06:13.460 --> 00:06:14.880
from that same carbon.

00:06:14.880 --> 00:06:18.350
And then we are going to be left
with-- the bromine now

00:06:18.350 --> 00:06:20.070
becomes the bromide anion.

00:06:22.900 --> 00:06:25.620
It had one, two, three,
four, five, six,

00:06:25.620 --> 00:06:27.670
seven valence electrons.

00:06:27.670 --> 00:06:30.730
One, two, three, four,
five, six, seven.

00:06:30.730 --> 00:06:35.000
Now it nabbed one more electron,
making it bromide.

00:06:35.000 --> 00:06:37.180
Now it has a negative charge.

00:06:37.180 --> 00:06:42.300
And if we were to draw
the chain, it

00:06:42.300 --> 00:06:43.550
would look like this.

00:06:46.460 --> 00:06:48.195
Well, we could draw it on this,
and I might as well draw

00:06:48.195 --> 00:06:50.250
it on this side, just so it's
attacking from the other side.

00:06:50.250 --> 00:06:53.870
This is the chiral substrate,
so we don't have to be too

00:06:53.870 --> 00:06:56.550
particular about how we draw the
connections to the carbon.

00:06:56.550 --> 00:06:57.850
We're not actually even
showing anything

00:06:57.850 --> 00:06:59.210
popping in or out.

00:06:59.210 --> 00:07:03.470
But we would have the methoxide
ion, where now it's

00:07:03.470 --> 00:07:05.940
bonded, so it's no longer
an ion, so it's

00:07:05.940 --> 00:07:11.450
OCH3, just like that.

00:07:11.450 --> 00:07:12.920
It has bonded to this carbon.

00:07:12.920 --> 00:07:15.660
Obviously, implicitly this
carbon had another hydrogen

00:07:15.660 --> 00:07:17.150
that we are not showing.

00:07:17.150 --> 00:07:20.800
Just that quickly, that
was the Sn2 reaction.

00:07:20.800 --> 00:07:21.720
That is the mechanism.

00:07:21.720 --> 00:07:23.360
Now let's think about what
the E2 reaction is.

00:07:23.360 --> 00:07:25.780
To do the E2 properly, to give
it justice, we're going to

00:07:25.780 --> 00:07:27.540
have to draw some of
the hydrogens.

00:07:27.540 --> 00:07:32.030
So on the E2 reaction, let
me draw that in blue.

00:07:38.830 --> 00:07:41.320
Let me draw the cyclopentane
part.

00:07:41.320 --> 00:07:43.140
Let me draw it big.

00:07:43.140 --> 00:07:46.200
Actually, over here, it's less
important to draw it too big.

00:07:46.200 --> 00:07:49.890
So let me draw the pentagon.

00:07:49.890 --> 00:07:52.130
The pentagon just like that.

00:07:52.130 --> 00:07:57.660
That is the bromine, three,
four, five, six, and then it

00:07:57.660 --> 00:08:00.590
has a seventh valence electron
right over here.

00:08:00.590 --> 00:08:03.080
This is the alpha carbon.

00:08:03.080 --> 00:08:06.050
That right there is
the alpha carbon.

00:08:06.050 --> 00:08:08.020
And then there are
two beta carbons.

00:08:08.020 --> 00:08:11.340
There are two beta carbons right
over there and there.

00:08:11.340 --> 00:08:13.970
They each have two hydrogens
on them.

00:08:13.970 --> 00:08:15.390
They each have two hydrogens.

00:08:15.390 --> 00:08:17.390
I know it's becoming a
little hard to read.

00:08:17.390 --> 00:08:19.210
They each have two hydrogens
on them.

00:08:19.210 --> 00:08:23.450
And in an E2 reaction, the
strong base will react-- let

00:08:23.450 --> 00:08:24.620
me make it a little
cleaner than that.

00:08:24.620 --> 00:08:27.370
Let me get rid of the beta.

00:08:27.370 --> 00:08:29.070
The beta makes it's
a little dirty.

00:08:29.070 --> 00:08:32.440
OK, so they each have two
hydrogens on them.

00:08:35.890 --> 00:08:39.580
Now in an E2 reaction, the
strong base-- over here, the

00:08:39.580 --> 00:08:42.320
methoxide ion was acting as
a strong nucleophile.

00:08:42.320 --> 00:08:44.790
In E2, it's going to act
as a strong base.

00:08:44.790 --> 00:08:47.470
It's going to nab off a hydrogen
off of one of the

00:08:47.470 --> 00:08:48.500
beta carbons.

00:08:48.500 --> 00:08:49.800
And you might want to
say, OK, which one?

00:08:49.800 --> 00:08:51.730
Let's look at Zaitsev's rule.

00:08:51.730 --> 00:08:52.520
It doesn't matter.

00:08:52.520 --> 00:08:53.670
These are symmetric.

00:08:53.670 --> 00:08:56.550
They are both bonded to
two other carbons.

00:08:56.550 --> 00:08:58.430
They both are bonded to the
same number of hydrogens.

00:08:58.430 --> 00:08:59.270
It doesn't matter.

00:08:59.270 --> 00:09:01.060
It's actually going to be random
which one, and you

00:09:01.060 --> 00:09:02.710
actually won't be able
tell the difference

00:09:02.710 --> 00:09:03.990
because it's symmetric.

00:09:03.990 --> 00:09:06.260
So let's just draw
it like this.

00:09:06.260 --> 00:09:08.740
Let me draw the methoxide ion.

00:09:08.740 --> 00:09:11.860
One, two, three, four--
or anion, maybe I

00:09:11.860 --> 00:09:13.710
should say-- five, six.

00:09:13.710 --> 00:09:16.970
And then it has one
bond to the CH3.

00:09:16.970 --> 00:09:21.440
It has a negative charge, very,
very, very strong base.

00:09:21.440 --> 00:09:24.980
It can go over here and nab
the hydrogen and leave

00:09:24.980 --> 00:09:28.310
hydrogen's electron behind.

00:09:28.310 --> 00:09:29.870
Maybe I'll take a color.

00:09:29.870 --> 00:09:34.160
This electron can be given to
the hydrogen so that it forms

00:09:34.160 --> 00:09:35.030
a bond with it.

00:09:35.030 --> 00:09:38.220
Hydrogen's electron-- let
me do this in a suitably

00:09:38.220 --> 00:09:39.490
different color.

00:09:39.490 --> 00:09:44.180
Hydrogen's electron that is
sitting right over there can

00:09:44.180 --> 00:09:46.940
now be given to the
alpha carbon.

00:09:46.940 --> 00:09:48.940
It can now be given to
the alpha carbon to

00:09:48.940 --> 00:09:50.170
form a double bond.

00:09:50.170 --> 00:09:53.430
And now that the alpha carbon is
getting that electron, now

00:09:53.430 --> 00:09:56.280
the bromo group can leave. It's
a decent leaving group.

00:09:56.280 --> 00:09:57.840
And that was another thing that
we should think about in

00:09:57.840 --> 00:09:58.250
our equation.

00:09:58.250 --> 00:10:00.180
But a good leaving group
actually favors all of the

00:10:00.180 --> 00:10:04.140
reactions: Sn2, E2, Sn1, E1.

00:10:04.140 --> 00:10:06.800
And so the carbon's getting
the electron, and then the

00:10:06.800 --> 00:10:12.150
bromine can then take this
carbon's electron.

00:10:12.150 --> 00:10:14.610
And just in one step that's
what's distinctive about the

00:10:14.610 --> 00:10:16.470
E2 and the Sn2 reactions.

00:10:16.470 --> 00:10:17.910
All of the reactions
are involved in the

00:10:17.910 --> 00:10:20.850
rate-determining step and there
really is only one step.

00:10:20.850 --> 00:10:23.850
Just like that, after that
happens, what we're left with

00:10:23.850 --> 00:10:27.380
is the methoxide anion
takes the hydrogen,

00:10:27.380 --> 00:10:29.330
so it becomes methanol.

00:10:29.330 --> 00:10:30.450
Let me draw that.

00:10:30.450 --> 00:10:33.510
So it becomes methanol.

00:10:33.510 --> 00:10:39.150
So it had one, two, three, four
and then five, that's

00:10:39.150 --> 00:10:41.280
this one right there, but then
this guy goes and bonds with

00:10:41.280 --> 00:10:42.540
the hydrogen.

00:10:42.540 --> 00:10:46.570
This guy goes and bonds with the
hydrogen, just like that,

00:10:46.570 --> 00:10:50.580
and hydrogen leaves its
electron behind.

00:10:50.580 --> 00:10:53.580
Let me draw the cyclopentane
part now.

00:10:53.580 --> 00:10:57.470
And so the cyclopentane looked
like this before, if I just

00:10:57.470 --> 00:10:59.770
focus on the ring.

00:10:59.770 --> 00:11:04.700
Now, this guy was bonded
to a hydrogen.

00:11:04.700 --> 00:11:07.540
He was bonded to this hydrogen
over here, but now that

00:11:07.540 --> 00:11:10.470
electron is going to be used to
form a bond with this alpha

00:11:10.470 --> 00:11:11.390
carbon right over here.

00:11:11.390 --> 00:11:12.640
Let me draw the alpha carbon.

00:11:12.640 --> 00:11:15.040
And the alpha carbon is
right over there.

00:11:15.040 --> 00:11:16.680
Obviously, implicitly at
every one of these

00:11:16.680 --> 00:11:17.970
edges, we have a carbon.

00:11:17.970 --> 00:11:19.670
But now, a double bond
is going to form

00:11:19.670 --> 00:11:21.350
with that alpha carbon.

00:11:21.350 --> 00:11:23.980
We could just draw it like
that, a double bond.

00:11:23.980 --> 00:11:25.620
Obviously, there's another
carbon here.

00:11:25.620 --> 00:11:27.200
I could write up another
carbon over there.

00:11:27.200 --> 00:11:29.070
And now this double
bond will form.

00:11:29.070 --> 00:11:31.280
And now the bromide has left.

00:11:31.280 --> 00:11:33.700
It's taken an electron with it
from that carbon now that the

00:11:33.700 --> 00:11:34.530
carbon doesn't need it.

00:11:34.530 --> 00:11:37.010
It was already starting to
hog it because it's so

00:11:37.010 --> 00:11:38.550
electronegative.

00:11:38.550 --> 00:11:39.500
So that's bromine.

00:11:39.500 --> 00:11:41.250
It takes that orange electron.

00:11:43.750 --> 00:11:46.040
Now, it is bromide.

00:11:46.040 --> 00:11:47.080
And we're done.

00:11:47.080 --> 00:11:50.030
And so just to go back to the
original question here, which

00:11:50.030 --> 00:11:52.750
reaction is likely to occur
or which mechanism?

00:11:52.750 --> 00:11:55.510
It's actually both Sn2 and E2.

00:11:55.510 --> 00:11:57.960
You would see a mix of both of
these occurring because you

00:11:57.960 --> 00:11:59.520
have all of the environmental
factors that

00:11:59.520 --> 00:12:00.800
would enable both.

00:12:00.800 --> 00:12:03.130
And so you would have both
of these mechanisms.

00:12:03.130 --> 00:12:04.960
Here's the-- let me
separate them out.

00:12:04.960 --> 00:12:07.360
Here's the Sn2 reaction.

00:12:07.360 --> 00:12:09.650
You would have the Sn2 reaction
occurring in your

00:12:09.650 --> 00:12:12.690
whatever, your vial, or your
pot, or whatever you're making

00:12:12.690 --> 00:12:17.180
all of this stuff occur in, and
you would also have your

00:12:17.180 --> 00:12:18.320
E2 reaction.

00:12:18.320 --> 00:12:21.960
So you would see some of all of
these, some of all of those

00:12:21.960 --> 00:12:25.780
products and these products
right over there.


E2 reactions

WEBVTT
Kind: captions
Language: en

00:00:00.570 --> 00:00:03.470
Let's try to come up with the
reaction of when we have this

00:00:03.470 --> 00:00:10.120
molecule right here reacting
with sodium methoxide in a

00:00:10.120 --> 00:00:13.595
methanol solution, or with
methanol as the solvent.

00:00:17.170 --> 00:00:19.030
Just so we get a little practice
with naming, let's

00:00:19.030 --> 00:00:21.710
see, this is one, two,
three, four carbons.

00:00:21.710 --> 00:00:25.600
So it has but- as a prefix and
no double bond or triple

00:00:25.600 --> 00:00:27.890
bonds, so it's butane.

00:00:27.890 --> 00:00:29.470
And we have a chloro
group here.

00:00:29.470 --> 00:00:30.990
So if we start numbering
at the side

00:00:30.990 --> 00:00:32.630
closest to it, one, two.

00:00:32.630 --> 00:00:33.880
So it's 2-chlorobutane.

00:00:39.000 --> 00:00:42.220
Let's think about what
might happen here.

00:00:42.220 --> 00:00:47.420
The first thing-- let
me just redraw the

00:00:47.420 --> 00:00:48.430
molecule right there.

00:00:48.430 --> 00:00:52.760
The first thing you need to
realize is this sodium

00:00:52.760 --> 00:00:55.960
methoxide is a salt.

00:00:55.960 --> 00:00:57.870
When it's not dissolved,
it's made up of a

00:00:57.870 --> 00:01:00.650
positive sodium cation.

00:01:00.650 --> 00:01:01.660
Let me draw them right here.

00:01:01.660 --> 00:01:08.580
It's a positive sodium
cation and a

00:01:08.580 --> 00:01:11.860
negative methoxide anion.

00:01:11.860 --> 00:01:14.060
Let me draw the methoxide
part right here.

00:01:19.000 --> 00:01:22.380
It normally would have just
two pairs, but now we have

00:01:22.380 --> 00:01:23.360
three pairs here.

00:01:23.360 --> 00:01:26.276
The oxygen has an
extra electron.

00:01:29.380 --> 00:01:30.715
Actually, let me draw
another electron

00:01:30.715 --> 00:01:32.450
as the extra electron.

00:01:32.450 --> 00:01:35.050
This'll be useful for
our mechanism.

00:01:35.050 --> 00:01:36.330
It could be any of them.

00:01:36.330 --> 00:01:37.930
The oxygen has an
extra electron.

00:01:37.930 --> 00:01:39.320
It has a negative charge.

00:01:39.320 --> 00:01:41.720
In solid form, when they're not
dissolved, they had formed

00:01:41.720 --> 00:01:45.080
an ionic bond and they form
a crystal-like structure.

00:01:45.080 --> 00:01:48.470
It's a salt, but when you
dissolve it in something, in a

00:01:48.470 --> 00:01:52.130
solution, in this case, we have
methanol as the solution,

00:01:52.130 --> 00:01:54.170
they will disassociate
from each other.

00:01:54.170 --> 00:01:57.940
And what you have right here is
this methoxide right here.

00:01:57.940 --> 00:02:02.040
This is a very, very,
very strong base.

00:02:02.040 --> 00:02:09.729
This right here is
a strong base.

00:02:09.729 --> 00:02:11.970
And if you use the Lewis
definition of a base, that

00:02:11.970 --> 00:02:17.650
means it really, really, really
wants to give away this

00:02:17.650 --> 00:02:19.880
electron to something else.

00:02:19.880 --> 00:02:23.570
If you use the Bronsted-Lowry
definition, this means that it

00:02:23.570 --> 00:02:26.900
really, really, really wants
to take a proton off of

00:02:26.900 --> 00:02:28.030
something else.

00:02:28.030 --> 00:02:31.160
In this situation, that is
exactly what it will do.

00:02:31.160 --> 00:02:37.440
I'll actually give you the most
likely reaction to occur

00:02:37.440 --> 00:02:40.020
here, and we'll talk about other
reactions, and why this

00:02:40.020 --> 00:02:43.660
is the most likely reaction
in future videos.

00:02:43.660 --> 00:02:45.310
So it wants to nab a proton.

00:02:45.310 --> 00:02:46.500
It is a strong base.

00:02:46.500 --> 00:02:50.040
It wants to give away
this electron.

00:02:50.040 --> 00:02:57.590
Let's say that it gives away
this electron to this hydrogen

00:02:57.590 --> 00:02:58.880
right over here.

00:02:58.880 --> 00:03:01.340
Now, this hydrogen already
had an electron.

00:03:01.340 --> 00:03:06.780
If it's getting an electron from
the methoxide anion, the

00:03:06.780 --> 00:03:11.410
methoxide base, then it can give
away its electron to the

00:03:11.410 --> 00:03:13.360
rest of the molecule.

00:03:13.360 --> 00:03:16.970
It can give away this electron
to the rest of the molecule.

00:03:16.970 --> 00:03:19.010
Now, carbon won't need
the electron.

00:03:19.010 --> 00:03:21.350
Carbon doesn't want to have
a negative charge.

00:03:21.350 --> 00:03:26.920
Maybe simultaneously that
electron goes to that carbon

00:03:26.920 --> 00:03:27.810
right over there.

00:03:27.810 --> 00:03:29.900
But once again, this carbon
doesn't want it.

00:03:29.900 --> 00:03:32.920
It already has four bonds, but
what we see is we have this

00:03:32.920 --> 00:03:33.990
chloro here.

00:03:33.990 --> 00:03:38.160
This is a highly electronegative
group.

00:03:38.160 --> 00:03:40.540
The chlorine is very
electronegative, so the whole

00:03:40.540 --> 00:03:43.500
time, the chlorine was already
tugging on this

00:03:43.500 --> 00:03:45.200
electron right here.

00:03:45.200 --> 00:03:48.310
Now, all of a sudden, it's all
happening simultaneously, when

00:03:48.310 --> 00:03:52.680
an electron becomes available
to this carbon, this carbon

00:03:52.680 --> 00:03:54.240
doesn't need this electron
anymore.

00:03:54.240 --> 00:03:58.630
The chlorine already wanted it,
so now the chlorine can

00:03:58.630 --> 00:03:59.700
take the electron.

00:03:59.700 --> 00:04:02.550
And just like that, in exactly
one step, let's think about

00:04:02.550 --> 00:04:03.080
what happened.

00:04:03.080 --> 00:04:09.630
If all of these simultaneous
reactions occurred, what are

00:04:09.630 --> 00:04:12.280
we left with?

00:04:12.280 --> 00:04:18.170
Let me redraw my two
chlorobutanes.

00:04:18.170 --> 00:04:19.839
I'm going to have to
change it now.

00:04:19.839 --> 00:04:24.180
So now, the chlorine
has disappeared.

00:04:24.180 --> 00:04:27.000
So we clear it.

00:04:27.000 --> 00:04:27.780
That's disappeared.

00:04:27.780 --> 00:04:31.700
The chlorine has now left.

00:04:31.700 --> 00:04:34.000
This chlorine is now up here.

00:04:34.000 --> 00:04:34.700
It has left.

00:04:34.700 --> 00:04:37.360
It had this electron right over
there, which is right

00:04:37.360 --> 00:04:38.450
over there.

00:04:38.450 --> 00:04:41.070
The other electron it was paired
with that was forming a

00:04:41.070 --> 00:04:43.025
bond with is now also
on the chlorine.

00:04:46.080 --> 00:04:50.710
It becomes a chloride anion.

00:04:50.710 --> 00:04:52.780
Let me draw the rest of
the valence electrons.

00:04:52.780 --> 00:04:56.210
One, two, three, four, five,
six, and it has the seventh

00:04:56.210 --> 00:04:57.040
one right here.

00:04:57.040 --> 00:05:01.670
One, two, three, four, five,
six, seven, eight, so it now

00:05:01.670 --> 00:05:04.370
has a negative charge.

00:05:04.370 --> 00:05:09.810
This electron up here-- let me
clear this part out as well.

00:05:15.700 --> 00:05:19.590
Now, this electron right here,
this magenta electron, this is

00:05:19.590 --> 00:05:21.560
now given to this carbon.

00:05:21.560 --> 00:05:23.030
Let me draw it here.

00:05:23.030 --> 00:05:26.000
That magenta electron is now
given to that carbon.

00:05:26.000 --> 00:05:28.170
And if we look at the other
end, the one that it was

00:05:28.170 --> 00:05:29.780
paired with, that it was bonded
with, it still is

00:05:29.780 --> 00:05:33.190
bonded with it, so that green
electron is now still on that

00:05:33.190 --> 00:05:35.250
carbon and now they
are bonded.

00:05:35.250 --> 00:05:36.705
Now, they're forming
a double bond.

00:05:39.550 --> 00:05:43.190
This all of the sudden has
become an alkene and now the

00:05:43.190 --> 00:05:44.960
methoxide took the hydrogen.

00:05:47.560 --> 00:05:48.810
Let me redraw the methoxide.

00:05:51.240 --> 00:05:59.670
OCH3, the oxygen had one,
two, three, four, five.

00:05:59.670 --> 00:06:02.190
It had six.

00:06:02.190 --> 00:06:06.370
Actually, it had seven valence
electrons, one with the carbon

00:06:06.370 --> 00:06:08.470
and then all of these
six unpaired ones.

00:06:08.470 --> 00:06:10.170
Neutral oxygen has six.

00:06:10.170 --> 00:06:12.880
But now it gave one of them
away to a hydrogen.

00:06:12.880 --> 00:06:21.900
It gave that green electron
there to the hydrogen-- I'll

00:06:21.900 --> 00:06:25.240
make this hydrogen the same
color-- to this hydrogen right

00:06:25.240 --> 00:06:26.890
over here, so now
this hydrogen is

00:06:26.890 --> 00:06:28.350
now bonded with it.

00:06:31.750 --> 00:06:32.930
So what are we left with?

00:06:32.930 --> 00:06:38.710
We have a chloride anion,
so this is chloride.

00:06:41.340 --> 00:06:43.200
We now have methanol.

00:06:43.200 --> 00:06:44.770
This was a strong base.

00:06:44.770 --> 00:06:47.840
Now it has become its
conjugate acid.

00:06:47.840 --> 00:06:57.900
So now we have methanol, which
is the same as the solvent, so

00:06:57.900 --> 00:06:59.660
it's now mixed in.

00:06:59.660 --> 00:07:02.720
And now we're left off
with one, two,

00:07:02.720 --> 00:07:05.330
three, four, still four.

00:07:05.330 --> 00:07:12.550
We still have four carbons,
but now it's an alkene.

00:07:12.550 --> 00:07:16.400
We have a double bond, so we
could call this but-2-ene, or

00:07:16.400 --> 00:07:19.310
sometimes called 2-butene.

00:07:19.310 --> 00:07:20.570
Let's think about what
happened here.

00:07:20.570 --> 00:07:22.370
It all happened simultaneously.

00:07:22.370 --> 00:07:23.920
Both reactants were present.

00:07:23.920 --> 00:07:25.750
There was actually only one
step so this is the

00:07:25.750 --> 00:07:27.300
rate-determining step.

00:07:27.300 --> 00:07:28.610
If we would try to name
it, it would probably

00:07:28.610 --> 00:07:30.170
have a 2 in it someplace.

00:07:30.170 --> 00:07:32.360
And something got eliminated.

00:07:32.360 --> 00:07:37.660
The chloride, or I guess you
could call it the chloro

00:07:37.660 --> 00:07:39.120
group, got eliminated.

00:07:39.120 --> 00:07:39.610
It left.

00:07:39.610 --> 00:07:41.780
It was a leaving group
in this situation.

00:07:41.780 --> 00:07:51.350
So this was eliminated, and this
type of reaction where

00:07:51.350 --> 00:07:55.680
something is eliminated and
both of the reactants are

00:07:55.680 --> 00:07:58.110
participating in the
rate-determining step, and we

00:07:58.110 --> 00:08:00.640
only had one step here so that
was the rate-determining step,

00:08:00.640 --> 00:08:02.400
is called an E2 reaction.

00:08:07.050 --> 00:08:09.336
And once again, E stands
for elimination.

00:08:12.000 --> 00:08:16.740
And the 2 stands for that both
reactants were involved in the

00:08:16.740 --> 00:08:17.790
rate-determining step.

00:08:17.790 --> 00:08:20.680
We had one reactant
and two reactants.

00:08:20.680 --> 00:08:23.410
They were both involved with
the rate-determining step.

00:08:23.410 --> 00:08:25.290
E2, just like Sn2.

00:08:25.290 --> 00:08:27.390
It's obviously not the same
reaction, but in Sn2, we had

00:08:27.390 --> 00:08:28.290
substitution.

00:08:28.290 --> 00:08:31.420
The leaving group left,
nucleophile came.

00:08:31.420 --> 00:08:34.360
It was substitution using
a nucleophile so that's

00:08:34.360 --> 00:08:35.440
the S and the n.

00:08:35.440 --> 00:08:37.309
And then the 2 was we
had both reactants.

00:08:37.309 --> 00:08:39.799
It was all happening
simultaneously.

00:08:39.799 --> 00:08:40.370
I'll leave you there.

00:08:40.370 --> 00:08:43.390
We're going to go into more
detail on the different types

00:08:43.390 --> 00:08:45.760
of each reactions, which ones
are favored, and then we'll

00:08:45.760 --> 00:08:46.950
talk a little bit about E1.

00:08:46.950 --> 00:08:48.820
You could maybe guess what
that's going to be based on

00:08:48.820 --> 00:08:50.920
our experience with
Sn2 and Sn1.


Nucleophilicity (Nucleophile Strength)

WEBVTT
Kind: captions
Language: en

00:00:00.400 --> 00:00:02.450
What I want to do with this
video is talk about

00:00:02.450 --> 00:00:03.700
nucleophilicity.

00:00:14.180 --> 00:00:17.270
This is really just how good of
a nucleophile something is.

00:00:17.270 --> 00:00:20.180
Or I'll just make up a
definition right now: the

00:00:20.180 --> 00:00:35.680
ability for an atom slash ion
slash molecule to act as a

00:00:35.680 --> 00:00:51.220
nucleophile, or to give away
extra electrons and bond with

00:00:51.220 --> 00:00:53.930
a nucleus or with
something else.

00:00:53.930 --> 00:00:55.180
I'll say with a nucleus.

00:01:01.490 --> 00:01:03.420
I want to say with a nucleus,
because that's what

00:01:03.420 --> 00:01:04.480
nucleophilicity is saying.

00:01:04.480 --> 00:01:07.450
It loves nucleuses, especially
positive ones, that it can

00:01:07.450 --> 00:01:09.420
give its extra electrons
to it.

00:01:09.420 --> 00:01:13.150
Now, as a first cut, if you
want to identify a good

00:01:13.150 --> 00:01:16.660
nucleophile, it should have
extra electrons to give away.

00:01:32.160 --> 00:01:35.510
The best things that have extra
electrons to give away

00:01:35.510 --> 00:01:40.750
are negative ions or anions, so
just at a very high level,

00:01:40.750 --> 00:01:44.770
something like the
fluoride anion.

00:01:44.770 --> 00:01:48.590
Normally, fluorine has seven
valence electrons: one, two,

00:01:48.590 --> 00:01:53.130
three, four, five, six, seven,
but it's so electronegative,

00:01:53.130 --> 00:01:55.700
it might be able swipe off
another electron from

00:01:55.700 --> 00:01:59.880
something else and then it
becomes the fluoride anion.

00:01:59.880 --> 00:02:03.650
Then it becomes the fluoride
anion with a negative charge.

00:02:03.650 --> 00:02:05.150
You can do that for all
of the halides.

00:02:05.150 --> 00:02:06.145
You could do that
for chlorine.

00:02:06.145 --> 00:02:07.860
It can become chloride.

00:02:07.860 --> 00:02:09.210
Bromine can be bromide.

00:02:09.210 --> 00:02:11.360
Iodine could be iodide.

00:02:11.360 --> 00:02:13.190
Let me do iodine, too.

00:02:13.190 --> 00:02:15.060
Iodine, once again,
it's a halide.

00:02:15.060 --> 00:02:16.590
It has seven valence
electrons.

00:02:16.590 --> 00:02:19.980
It has many, many more electrons
than fluorine, but

00:02:19.980 --> 00:02:21.250
if you just look at
its valence shell,

00:02:21.250 --> 00:02:23.730
it has seven electrons.

00:02:23.730 --> 00:02:26.850
And then it is also reasonably
electronegative, not as

00:02:26.850 --> 00:02:28.220
electronegative as fluorine.

00:02:28.220 --> 00:02:32.060
Remember, the trend goes like
this from the bottom left to

00:02:32.060 --> 00:02:32.740
the top right.

00:02:32.740 --> 00:02:36.930
So fluorine is extremely
electronegative, but iodine is

00:02:36.930 --> 00:02:39.110
still pretty electronegative.

00:02:39.110 --> 00:02:43.490
It is a halogen so it also might
be able to swipe off an

00:02:43.490 --> 00:02:45.560
electron from someone else
and become iodide.

00:02:48.250 --> 00:02:52.380
In general, things with extra
electrons, lone pairs of

00:02:52.380 --> 00:02:54.820
electrons, and especially a
negative charge, are going to

00:02:54.820 --> 00:02:59.880
be pretty good nucleophiles.

00:02:59.880 --> 00:03:01.570
Another example that's
not a halide is the

00:03:01.570 --> 00:03:05.340
hydroxide anion, so OH.

00:03:05.340 --> 00:03:09.420
This is an example of something
that is a molecule.

00:03:09.420 --> 00:03:12.770
Maybe traditionally water
would look like this.

00:03:12.770 --> 00:03:17.430
Traditionally, this is just
neutral water, and oxygen has

00:03:17.430 --> 00:03:19.600
two lone pairs like that,
but oxygen is pretty

00:03:19.600 --> 00:03:20.650
electronegative.

00:03:20.650 --> 00:03:23.530
It is already kind of taking the
electrons away from one of

00:03:23.530 --> 00:03:26.210
the hydrogens, and at some
point, it might just take it

00:03:26.210 --> 00:03:31.900
altogether and then you have
the hydroxide anion.

00:03:31.900 --> 00:03:33.540
This would look like this.

00:03:33.540 --> 00:03:36.370
You have your original two lone
pairs, just like that,

00:03:36.370 --> 00:03:39.010
and then you have this pair
that's going to be taken.

00:03:39.010 --> 00:03:40.100
It already had that electron.

00:03:40.100 --> 00:03:42.490
It takes that electron from the
hydrogen, so now it has

00:03:42.490 --> 00:03:44.000
two more electrons.

00:03:44.000 --> 00:03:45.920
Let me color code it so
you see what it took.

00:03:45.920 --> 00:03:49.350
It took this electron from the
hydrogen and now this also has

00:03:49.350 --> 00:03:50.990
a negative charge.

00:03:50.990 --> 00:03:55.280
That's also a reasonably
good nucleophile.

00:03:55.280 --> 00:03:56.815
And, of course, you have
your hydrogen now.

00:03:56.815 --> 00:03:58.770
It lost its one electron.

00:03:58.770 --> 00:04:01.320
It only has a proton
in its nucleus.

00:04:01.320 --> 00:04:03.580
Whenever you see H plus, this
is really just a proton.

00:04:03.580 --> 00:04:05.580
There's nothing else
to that hydrogen.

00:04:05.580 --> 00:04:06.730
So that's hydroxide.

00:04:06.730 --> 00:04:11.100
These are all reasonably good
nucleophiles in that they have

00:04:11.100 --> 00:04:12.020
something to give away.

00:04:12.020 --> 00:04:14.000
They have extra charge.

00:04:14.000 --> 00:04:17.209
Now, what I want to do is think
about between these, how

00:04:17.209 --> 00:04:19.620
do you think about what's going
to be a stronger or

00:04:19.620 --> 00:04:21.180
weaker nucleophile?

00:04:21.180 --> 00:04:25.020
And here, it becomes a little
bit more nuanced.

00:04:25.020 --> 00:04:27.090
What we're going to do is
differentiate what happens in

00:04:27.090 --> 00:04:31.980
a protic solvent versus what
happens in an aprotic solvent.

00:04:31.980 --> 00:04:33.310
Let me write down.

00:04:33.310 --> 00:04:35.510
Let me start with a
protic solvent.

00:04:40.400 --> 00:04:43.050
I'll make two columns here,
protic solvent and then we'll

00:04:43.050 --> 00:04:54.030
do aprotic solvent
right over here.

00:04:54.030 --> 00:04:56.830
Once again, these are fancy
words, but they mean something

00:04:56.830 --> 00:04:57.680
pretty simple.

00:04:57.680 --> 00:05:01.850
Protic solvent is something that
has hydrogens that can be

00:05:01.850 --> 00:05:04.820
taken away or might have free
protons floating around.

00:05:04.820 --> 00:05:07.060
An example of a protic
solvent is water

00:05:07.060 --> 00:05:08.800
or really any alcohol.

00:05:08.800 --> 00:05:12.880
Water is the simplest example
or maybe the most common.

00:05:12.880 --> 00:05:15.620
The reason why you might have
protons floating around is

00:05:15.620 --> 00:05:18.020
exactly this reaction
I showed right here.

00:05:18.020 --> 00:05:21.430
Maybe every now and then a
hydroxide anion forms. Even

00:05:21.430 --> 00:05:26.330
more likely, maybe a water takes
a hydrogen from one of

00:05:26.330 --> 00:05:27.910
the other waters.

00:05:27.910 --> 00:05:29.790
One of the water molecules takes
hydrogen from one of the

00:05:29.790 --> 00:05:33.490
other water molecules and
becomes a hydronium, where

00:05:33.490 --> 00:05:37.665
it's not a proton necessarily,
but it's an oxygen.

00:05:41.930 --> 00:05:44.980
Let's say if you start with
water and one of these

00:05:44.980 --> 00:05:48.350
electrons were to be given to
some proton floating around,

00:05:48.350 --> 00:05:50.900
it would look like this.

00:05:50.900 --> 00:05:53.000
And it has a positive
charge and then this

00:05:53.000 --> 00:05:56.410
proton is very available.

00:05:56.410 --> 00:05:58.520
You can almost imagine it's
almost floating around because

00:05:58.520 --> 00:06:01.400
that oxygen really wants to
take back that electron.

00:06:01.400 --> 00:06:03.420
So protic solvent is water.

00:06:03.420 --> 00:06:05.360
In water, you might see a little
bit of hydroxide, a

00:06:05.360 --> 00:06:07.450
little bit of proton, a little
bit a hydronium.

00:06:07.450 --> 00:06:09.810
You see all of it in there, but
the bottom line is that

00:06:09.810 --> 00:06:13.005
there are protons that can
react with other things.

00:06:15.520 --> 00:06:19.070
Let me clear this away so that
I have some real estate.

00:06:19.070 --> 00:06:21.490
Let me just write down water.

00:06:21.490 --> 00:06:24.400
Water is a protic solvent.

00:06:24.400 --> 00:06:25.700
Now, wait.

00:06:25.700 --> 00:06:27.290
Is that always the case?

00:06:27.290 --> 00:06:28.510
It seems like hydrogens
are everywhere.

00:06:28.510 --> 00:06:29.790
Well, no, it's not
always the case.

00:06:29.790 --> 00:06:32.270
Let me show you an
aprotic solvent.

00:06:32.270 --> 00:06:34.830
Diethyl ether looks like this.

00:06:41.050 --> 00:06:43.150
And just so you know the naming,
it's an ether because

00:06:43.150 --> 00:06:46.640
it has oxygen, and it's diethyl
because it has two

00:06:46.640 --> 00:06:47.300
ethyl groups.

00:06:47.300 --> 00:06:50.360
That's one ethyl group right
there and that is the second.

00:06:50.360 --> 00:06:53.050
So it's diethyl ether.

00:06:53.050 --> 00:06:56.230
Now, you might say, hey, this
guy's got hydrogens lying

00:06:56.230 --> 00:06:56.890
around as well.

00:06:56.890 --> 00:06:58.090
Maybe those can get released.

00:06:58.090 --> 00:07:01.270
But, no, these hydrogens are
bonded to the carbon and

00:07:01.270 --> 00:07:06.950
carbon is not anywhere near as
electronegative as oxygen.

00:07:06.950 --> 00:07:10.600
Carbon is unlikely to steal
these hydrogens' electrons and

00:07:10.600 --> 00:07:12.110
these hydrogens to be loose.

00:07:12.110 --> 00:07:14.870
If they were bonded to the
oxygen, that would have been a

00:07:14.870 --> 00:07:16.040
possibility.

00:07:16.040 --> 00:07:20.500
With water, you have
obviously H-O-H.

00:07:20.500 --> 00:07:27.400
In alcohols, you have some maybe
carbon chain bonded to

00:07:27.400 --> 00:07:29.540
an oxygen, which is then
bonded to a hydrogen.

00:07:29.540 --> 00:07:32.860
So in either of these cases,
in either water or alcohol,

00:07:32.860 --> 00:07:36.030
you have these hydrogens where
the electron might be taken by

00:07:36.030 --> 00:07:38.140
the oxygen because it's so
electronegative and then the

00:07:38.140 --> 00:07:39.270
hydrogen floats around.

00:07:39.270 --> 00:07:42.770
Anyway, that's a review of
protic versus aprotic.

00:07:42.770 --> 00:07:45.850
In a protic solvent-- and this
is actually a general rule of

00:07:45.850 --> 00:07:51.060
thumb-- if a nucleophile is
likely to react with its

00:07:51.060 --> 00:07:54.290
solvent, it will be bad at
being a nucleophile.

00:07:54.290 --> 00:07:54.950
Think about it.

00:07:54.950 --> 00:07:57.410
If it's reacting with the
solvent, it's not going to be

00:07:57.410 --> 00:07:58.320
able to do this.

00:07:58.320 --> 00:08:01.710
It's not going to be able to
give its electrons away to

00:08:01.710 --> 00:08:04.960
what it needs to give it away,
to maybe what we saw in an

00:08:04.960 --> 00:08:07.270
Sn2-type reaction.

00:08:07.270 --> 00:08:09.870
In a protic solvent, what
happens is that the things

00:08:09.870 --> 00:08:11.970
that are really electronegative
and really

00:08:11.970 --> 00:08:15.290
small, like a fluoride anion--
let me draw a fluoride anion.

00:08:19.520 --> 00:08:26.250
In a protic solvent, what's
going to happen is it's going

00:08:26.250 --> 00:08:29.050
to be blocked by
hydrogen bonds.

00:08:29.050 --> 00:08:30.660
It's very negative, right?

00:08:30.660 --> 00:08:31.940
It has a negative charge.

00:08:31.940 --> 00:08:33.700
And it's also tightly packed.

00:08:33.700 --> 00:08:36.909
As you can see right here,
its electrons are

00:08:36.909 --> 00:08:38.500
very close, tied in.

00:08:38.500 --> 00:08:42.150
It's a much smaller atom
or ion, in this case.

00:08:42.150 --> 00:08:45.000
If we looked at iodide,
iodide has 53

00:08:45.000 --> 00:08:48.290
electrons, many orbitals.

00:08:48.290 --> 00:08:49.940
Actually, iodide
would have 54.

00:08:49.940 --> 00:08:52.160
It would have the same
as iodine plus one.

00:08:52.160 --> 00:08:55.740
Fluoride will have 10 electrons,
nine from fluorine

00:08:55.740 --> 00:09:00.610
plus it gains another one, so
it's a much smaller atom.

00:09:00.610 --> 00:09:03.450
So when you have water hanging
around it, let's say you have

00:09:03.450 --> 00:09:04.740
something like water.

00:09:04.740 --> 00:09:06.170
That has a negative charge.

00:09:06.170 --> 00:09:11.800
Water is polar.

00:09:11.800 --> 00:09:14.695
Actually, both of these are
polar, so I should write down

00:09:14.695 --> 00:09:15.750
polar for both of these.

00:09:15.750 --> 00:09:18.770
This is a polar protic
solvent.

00:09:18.770 --> 00:09:21.270
This is a polar aprotic
solvent.

00:09:21.270 --> 00:09:23.620
In this case, water is still
more electronegative than the

00:09:23.620 --> 00:09:26.060
carbon, so it still has a
partial negative charge.

00:09:26.060 --> 00:09:27.930
These parts still have
a partial negative.

00:09:27.930 --> 00:09:30.200
Water still has a partial
negative charge.

00:09:30.200 --> 00:09:33.780
The hydrogen has a partial
positive charge so it is going

00:09:33.780 --> 00:09:35.970
to be attracted to
the fluorine.

00:09:35.970 --> 00:09:38.280
This is going to happen all
around the fluorine.

00:09:38.280 --> 00:09:40.900
And if these waters are
attracted to the fluorine in

00:09:40.900 --> 00:09:44.580
kind of forming a tight shell
around it, it makes it hard

00:09:44.580 --> 00:09:45.830
for fluorine to react.

00:09:52.870 --> 00:09:58.680
So it's a worse nucleophile
than, say, iodide or hydroxide

00:09:58.680 --> 00:10:00.700
in a polar protic solvent.

00:10:00.700 --> 00:10:02.390
Hydroxide has the same issue.

00:10:02.390 --> 00:10:05.030
It's still forming hydrogen
bonds, but if you wanted to

00:10:05.030 --> 00:10:10.800
compare them, iodide
is much bigger.

00:10:10.800 --> 00:10:12.630
Maybe I'll draw it like this.

00:10:12.630 --> 00:10:14.010
I'll draw its valence
shell like this.

00:10:14.010 --> 00:10:20.180
It's a much bigger ion.

00:10:20.180 --> 00:10:22.270
It has all these electrons
in here.

00:10:22.270 --> 00:10:26.310
And so, it still will form
hydrogen bonds with the water.

00:10:26.310 --> 00:10:29.070
It still will form hydrogen
bonds with the hydrogen end of

00:10:29.070 --> 00:10:31.510
the water because they're
partially positive, but it's

00:10:31.510 --> 00:10:34.255
going to be less
tightly packed.

00:10:34.255 --> 00:10:47.180
And on top of that, iodide is
more polarizable, which means

00:10:47.180 --> 00:10:50.500
that its electron cloud is so
big and the valence electrons

00:10:50.500 --> 00:10:53.820
are so far away from the nucleus
that they can be

00:10:53.820 --> 00:10:56.380
influenced by things and then
be more likely to react.

00:10:56.380 --> 00:11:00.350
So let's say this iodide is
getting close to a carbon that

00:11:00.350 --> 00:11:02.620
has a partial positive charge.

00:11:02.620 --> 00:11:09.190
So let's say carbon is attached
to a bromine and then

00:11:09.190 --> 00:11:12.210
it's attached to three
hydrogens.

00:11:12.210 --> 00:11:14.480
We've seen that this will have
a partial negative charge.

00:11:14.480 --> 00:11:17.450
It's more electronegative than
the carbon, which will have a

00:11:17.450 --> 00:11:19.420
partial positive charge.

00:11:19.420 --> 00:11:22.580
When this guy, this big guy with
the electrons really far

00:11:22.580 --> 00:11:28.040
away, gets close to this, more
of the electron cloud is going

00:11:28.040 --> 00:11:30.090
to be attracted to the partial
positive charge.

00:11:30.090 --> 00:11:33.450
It'll get distorted a little bit
and so it is more likely

00:11:33.450 --> 00:11:37.640
to react in a polar
protic solvent.

00:11:37.640 --> 00:11:40.450
Fluoride, on the other hand,
is very tightly packed,

00:11:40.450 --> 00:11:43.230
blocked by the hydrogen bonds.

00:11:43.230 --> 00:11:45.020
It's less likely to react.

00:11:45.020 --> 00:11:49.210
If you were to look at the
Periodic Table, if you look at

00:11:49.210 --> 00:11:57.410
just the halogens in a polar
protic solvent, the halides--

00:11:57.410 --> 00:12:00.420
this would be the ion version of
the halogens-- the halide's

00:12:00.420 --> 00:12:03.810
iodide will be the
best nucleophile.

00:12:03.810 --> 00:12:07.740
Fluoride will be the worst. So
in a polar protic solvent--

00:12:07.740 --> 00:12:21.090
let me write this down-- we
have a situation where the

00:12:21.090 --> 00:12:25.860
iodide is the best nucleophile,
followed by

00:12:25.860 --> 00:12:30.900
bromide, followed by chloride,
and then last

00:12:30.900 --> 00:12:33.920
of all is the fluoride.

00:12:33.920 --> 00:12:38.870
The exact opposite is true
in an aprotic solvent.

00:12:38.870 --> 00:12:42.720
In an aprotic solvent, the
fluoride, which is-- fluorine

00:12:42.720 --> 00:12:44.755
is far more electronegative.

00:12:44.755 --> 00:12:48.560
Fluoride is more basic.

00:12:48.560 --> 00:12:52.120
It will be more stable if it
is able to form a bond with

00:12:52.120 --> 00:12:54.030
something than iodide.

00:12:54.030 --> 00:12:56.210
Iodide is pretty stable.

00:12:56.210 --> 00:13:01.610
If you look at a hydrogen
iodide, it's actually a highly

00:13:01.610 --> 00:13:02.590
acidic molecule.

00:13:02.590 --> 00:13:05.200
So iodide itself, the conjugate
base of hydrogen

00:13:05.200 --> 00:13:09.600
iodide, is going to be
a very bad base.

00:13:09.600 --> 00:13:12.470
When you're dealing with an
aprotic solvent, you go in the

00:13:12.470 --> 00:13:14.390
direction of basicity.

00:13:14.390 --> 00:13:15.990
We're going to learn in the
next video that actually

00:13:15.990 --> 00:13:19.170
basicity and nucleophilicity are
related, but they aren't

00:13:19.170 --> 00:13:20.280
the same concept.

00:13:20.280 --> 00:13:23.110
We're going to talk about
that in a little bit.

00:13:23.110 --> 00:13:27.030
If you're in an aprotic solvent,
you're not reacting

00:13:27.030 --> 00:13:28.320
with the solvent as much.

00:13:28.320 --> 00:13:32.310
And then in this situation,
fluoride is actually the best

00:13:32.310 --> 00:13:37.920
nucleophile, followed by
chloride, followed by bromide,

00:13:37.920 --> 00:13:39.420
followed by iodide.

00:13:39.420 --> 00:13:42.150
So here, you're going in the
direction of basicity.

00:13:42.150 --> 00:13:47.860
This is the best. This is the
worse in an aprotic solvent.

00:13:47.860 --> 00:13:50.490
If it was in a protic solvent,
this is flipped around.

00:13:50.490 --> 00:13:54.110
This becomes the best and
this becomes the worst.