Let's think about what might happen if we had a solution of
this carboxylic acid here.
We might as well name it just to get some practice.
We have one, two, three, four, five, six, seven carbons.
So this is heptan-, and then we don't write heptane,
because this is a carboxylic acid.
It is heptanoic acid.
So let's see what happens if we have heptanoic acid
reacting with-- this is one, two carbons, and then it has
an OH group, so this is ethanol.
That's what the OH group does.
It makes this an alcohol.
And it's in the presence of a sulfuric acid catalyst. This
right here is sulfuric acid, one of the stronger acids.
I'll actually draw its structure, because I always
find it frustrating when people just write just the
formula here without the actual structure, because the
structure actually shows you why it's so acidic.
Sulfur has six valence electrons, just like oxygen.
So it has a double bond to an oxygen, another double bond to
an oxygen, and then it has a single bond to an OH group,
and then it has another single bond to an OH group.
And notice, it has one, two, three, four,
five, six valence electrons.
Now, the reason why this is such a strong acid, is that if
either of these oxygens take the electron from this proton,
so actually give away the proton to the solution,
there's a ton of resonance structures here.
And maybe I'll make a whole video on sulfuric acid, just
to show you all of the resonance structures.
But, in general, whenever you see a reaction when they say
it's catalyzed by an acid, all you have to do is realize it's
just going to make the surrounding solution a lot
more acidic, just a ton more acidic.
Maybe we're in a solution of ethanol, and if we are in a
solution of ethanol, it'll just add protons to the
ethanol itself.
So you can imagine this guy right over here.
Let me draw the sulfuric acid a little bit differently.
Let me draw these oxygen-hydrogen bonds, so you
have this oxygen and it is bonded to a hydrogen there.
This is floating around the solution.
You have your ethanol that looks like this, so two
carbons and then bonded to an oxygen, and then that oxygen
is bonded to a hydrogen.
The oxygen has two lone pairs, just like that.
And so this guy really is good at getting rid of the protons.
So you have a situation where this electron can be taken
back by this oxygen, and then it can actually be given here,
and there's all these resonance structures.
But it's just very good at taking that electron.
And that can happen at the exact same time that one of
the ethanols capture the hydrogen proton, at the exact
same time that this oxygen captures that hydrogen proton.
And if you just look at this part right here, this will
just result in-- this part alone will just result in a
ton of having these protonated ethanols flying around.
Actually, let me draw it over here.
I want to make sure I'm using my
screen real estate properly.
So this will result in a ton of these protonated ethanols.
So one, two carbons, it has this original
hydrogen over here.
But now it has one electron in that original pair, and it
gave the other electron to this hydrogen, so it gave the
other electron to that hydrogen nucleus.
Let me do it in that same color.
So it took a proton.
It gave away an electron, so it now has a positive charge.
And now what was a sulfuric acid now has a
negative charge over here.
So if I were to draw it, it would now look like this.
Plus sulfur, two double bonds, two oxygens.
You still have this OH group.
Actually, it could still donate.
This is still acidic, but now this oxygen right now gained
an electron.
It now has a negative charge.
Now, the whole reason I did this is I wanted to give you a
tangible sense of what sulfuric acid looks like and
why it's acidic.
But really, you just have to kind of internalize that
you're just going to have a bunch of hydrogen protons
floating around.
They could be attached to an ethanol.
If this was a water solution, they could create hydronium,
so you have a bunch of hydrogen protons floating
around that will catalyze this reaction.
They will be used to facilitate the reaction we're
going to explore, and then they will be let go.
So hopefully, this right here, by having this right here, if
I start involving some of these protonated ethanols in
our reaction, you won't view that as a huge stretch of the
imagination, because they would have gotten protonated
by the sulfuric acid.
Or if I just actually grab protons in our reaction, that
actually might make it a little bit simpler.
So let's start with the actual reaction.
So let me read redraw my heptanoic acid.
So I have one, two, three, four, five, six, seven
carbons, double bond to an oxygen, and then we have an OH
group right over here.
Now, the first step of this reaction, this oxygen right
here, we have all these protons floating around, very
acidic environment.
We have sulfuric acid there just giving protons away to
the ethanol or to other things.
So this guy can grab a proton either directly from sulfuric
acid or maybe from one of the protonated
ethanols, either one.
So we could just draw it like this.
He just grabs a hydrogen proton.
The hydrogen proton might have had an electron associated
with it that would then go back to a sulfuric acid, but
just to make things simple, I'll just say grabs a proton
from something else.
So once he grabs that proton, then it looks like this.
So I'll draw my two, three, four, five, six, seven
carbons, double bond to an oxygen, single bond to an OH.
This oxygen had two lone pairs, but now one of the lone
pairs is broken up because it gave an electron to this
hydrogen proton right over there.
Hydrogen proton, it was positively charged.
It gained an electron.
Now it is neutral.
But this oxygen right over here gave away the electron,
so it is now positive.
Now, the next step, we have all this
ethanol floating around.
So let me introduce some ethanol, and I'll do this in a
different color.
So we have this ethanol floating around, and this is
one, two carbons.
This is our ethanol.
You have two electron pairs on that oxygen.
And then you could imagine, especially because this is now
a good leaving group, you can imagine that this acts as a
nucleophile.
It would do a nucleophile attack on this carbonyl carbon
right here.
And so what you could imagine is that this electron attacks,
or gets given to, this carbon right here on the carbonyl.
And at the exact same time that this happens, this oxygen
is positively charged.
It wants an electron.
It wants to get neutral again.
It's already hogging these electrons.
That's why you have a partial positive
charge on this carbon.
That's why this guy might be attracted to this.
We've see this in Sn2 reactions many, many, many,
many times already.
So this electron will be taken up by this
top oxygen over there.
So after that happens, what does our newly formed
molecule look like?
Let me just scroll down and have some clear space.
After that, our newly formed molecule will look like this.
We have two, three, four, five, six, seven carbons to
get to the carbonyl carbon, and now it will have a single
bond to that oxygen up there.
That oxygen had one lone pair already.
It had this bond to this hydrogen that it nabbed from
the solution, that proton that it nabbed from the solution,
and now it has another pair.
It had this electron that was participating in a double bond
with this carbonyl carbon, but now it took the other side of
that bond back.
So now it has that electron and the other one that it took
from the carbon, so it has two lone pairs again.
It had a positive charge, but now it took back an electron.
Now, it is neutral.
Now the rest of the molecule, we have this OH
group right over here.
We have that OH group right over there.
And now we have the actual ethanol that has attached
itself, so it is no longer ethanol.
It's attached itself to this larger molecule.
So this oxygen right over here, it still has one lone
pair, but the other lone pair has been broken up, and it has
given an electron to this carbon.
It is now bonded to the larger molecule, and then the rest of
it, you have these one, two carbons right there, and then
you have this hydrogen right over there.
Remember, we have a bunch of protons floating around.
Actually, I should make it clear.
These are all reversible reactions, so I actually
shouldn't even draw one-way arrows here.
Let me scratch that out.
Actually, delete that.
A better thing to do, instead of drawing these one-way
arrows is to show that the reaction could actually go in
either direction.
It's just as likely to go from here to here as it is from
there to there.
So let me draw that.
These are kind of in equilibrium with each other.
So you can imagine the next thing that could happen is
another oxygen could grab a proton from the medium.
And actually, before I do that, let me make sure this
guy can lose a proton to the medium.
Actually, this guy could take the proton from that guy, but
I won't do it that way.
So you could imagine a situation where this proton
just jumps off.
I should have pointed it out.
This guy gave away an electron to this carbonyl carbon.
So this right here has a positive charge.
In general, the oxygen, very electronegative, it's going to
be hogging the electrons of this proton, so it
can take them away.
So far, we gained a proton, and now we
can give back a proton.
So this electron right here can go back to this oxygen,
making it neutral.
And the proton can be picked up by anything.
It could be picked up by another molecule of heptanoic
acid through this first protonation we saw.
It could be picked up by another molecule of ethanol.
Let me draw it like that.
Let me draw it as getting picked up by another molecule
of ethanol.
It just gets thrown back in the solution.
So this guy gives an electron to the proton, but the end
result is the hydrogen leaves.
The electron goes back to the oxygen.
It now has a neutral charge.
And once again, these are all reversible reactions.
It can go in either direction.
But we're now over here, and let me redraw
my heptanoic acid.
Two, three, four, five, six, seven single bond now to the
oxygen, single bond to this OH.
And now we have this bond right over here to this now
deprotonated, what was ethanol, so you have O, and
then we have one, two carbons, just like that.
And this guy has a hydrogen attached to it.
And I won't even bother to draw this protonated ethanol.
The protons are flying around everywhere.
Now, the next step, this guy, this OH group, and especially
the oxygen in it, he could grab a proton from the
surrounding solution.
So he's got these two lone pairs.
He can grab a proton from one of these protonated ethanols
from the sulfuric acid, or maybe from one of these other
intermediate molecules, from anywhere.
That's the whole point of having this acid catalyst, so
he could donate an electron to a proton and then
form a bond with it.
I'll keep it in that color.
And so if that happened, then the next step in our
reaction-- and remember, these can all
go in either direction.
The next step in our reaction will look like this.
You would have your two, three, four, five, six, seven
carbons, single bond to an oxygen, and then you have your
bond to this oxygen, which is bound to two carbons, one, two
carbons, just like that.
And this guy on top is bonded to an oxygen.
He's got two lone pairs.
And this guy over here grabbed a proton.
He gave an electron to a hydrogen proton.
He had two, but now he gave one of them to this proton.
And so he gave it to that hydrogen.
That hydrogen is now neutral.
It gained an electron.
This oxygen is now positive, because it
gave away an electron.
It still has this other lone pair over here.
It is now positive, and frankly, it is now a good
leaving group.
So, in the next step, you can have someone else.
Remember, other people need protons
earlier in this reaction.
So this proton might get lost, maybe by one of the other
ethanol molecules, so let me draw that.
Let me do a color I haven't used yet.
I'll use orange.
So maybe one of the other ethanol molecules, or one of
the other intermediaries in this whole reaction.
So ethanol, I'll just do ethanol because
it's easier to draw.
It might give an electron to just the nucleus.
And then this guy can take the electron back.
This guy could take that hydrogen's electron back and
give it to this carbon that was the carbonyl carbon
several steps ago.
And then since he's got that electron, he can then give
back this electron to this, what was an OH group, but now
it's got another hydrogen there, so he can then give it
back to him.
And then the resulting products will look like this.
And this is all in equilibrium.
We now have two, three, four, five, six, seven carbons.
Now, this guy has a double bond again.
So you have an oxygen right there.
It is now a double bond.
I'll draw this newly formed double bond in magenta.
This guy has left as water, so I can just draw that, so now
you have this other OH group that is bonded to this
hydrogen over here.
He has now left as water.
And now you have this other thing over here.
You have what was that ethanol group has now attached itself.
That ethanol has lost its hydrogen.
It has attached itself to what was a carboxylic acid, so now
it looks like this.
It now is bonded to an oxygen and is bonded to one, two
carbons, just like that.
And this whole reaction that I showed you is called
esterification, one of those words that I
have trouble saying.
This one in particular is called the Fischer
esterification.
And he won the Nobel Prize in 1902 generally for his work in
organic chemistry.
And the reason why it's called that is we started with the
carboxylic acid.
We started with a heptanoic acid, and
now we have an ester.
An ester is something that, instead of an OH group like
you have in carboxylic acid, you have an OR.
You have an oxygen with an actual alkyl
group attached to it.
And this ester right here, and we'll probably talk more about
esters in future videos, this ester right here, just to give
you a hint of how to name it, you first start with the group
that's attached to this oxygen right here.
There are two carbons over there, so we'll use ethyl to
name that over there.
And then we have the rest of this, and we already know that
that is one, two, three, four, five, six, seven carbons,
including the carbon in the carbonyl group.
And so that is hepta-, heptan-.
And you would be tempted to call it heptanoic acid, but it
is no longer a carboxylic acid.
It is now an ester.
So you call it heptanoate.
And this is what tells you that you are
dealing with an ester.
And this tells you what's on the other side of the oxygen
in the ester.
This is telling you how many carbons you have attached to
the carbonyl chain of the actual ester.
Anyway, hopefully, you found that useful.
I just wanted introduce you to a well-known reaction
mechanism of creating an ester out a carboxylic acid.