Main content
Organic chemistry
Course: Organic chemistry > Unit 11
Lesson 2: Formation of carboxylic acid derivativesFischer esterification
Fischer esterification of carboxylic acids. Created by Sal Khan.
Want to join the conversation?
- .... hang on, didn't Sal say that the ethanol was already protonated by the sulfuric acid? Therefore, how can we have an unprotonated ethanol now floating about, which can stage a nucleophilic attack on the carbonyl carbon? Sulfuric acid is so strong that I thought all the ethanol would have been protonated by now? 7:01(6 votes)
- Remember that sulfuric acid is a catalyst and is present in small amounts. It protonates every ethanol molecule it can until it is all used up. But ethanol is the solvent and is present in much larger quantities (the excess reactant, if you like), so there is lots of unprotonated ethanol still remaining in the mixture.(7 votes)
- Ok this may sound stupid, but why does the protonated oxygen located on top lose a proton while the other (right hand one) gains one and acts as a leaving group? wouldn't it be possible for the 2nd (top) oxygen to accept a proton as well and act as the 2nd leaving water group?(1 vote)
- Exactly. The two oxygen atoms are equivalent, so, half the time the top oxygen will be protonated to form water as a leaving group. The other half of the time the other oxygen atom will be protonated. Both pathways are equally likely.(9 votes)
- you made the ester so what happened to agua(4 votes)
- I am actually looking at an updated organic textbook right now and I believeshould be be two different steps. Deprotanation occurs after the leaving group leaves hope that helps all you guys understand the mechanism better but thanks Kahn for doing most of it right! 13:53(3 votes)
- No. Actually you are mistaken. You are a step premature....there must be a proton transfer between OH and OR before the leaving group leaves. He has the correct mechanism...I have no idea what you're specifically referring to.(2 votes)
- Why would the oxygen form a bond with hydrogen and become positive? Isn't it more stable if it stays neutral?(2 votes)
- I assume you're asking about what Sal shows aroundor 3:00? 6:00
Yes, but you have to consider what else is present in the reaction — in this case there is a strong acid (sulfuric acid), which will force a proton onto the oxygen.
In other words, protonated ethanol and protonated heptanoic acid are more stable than sulfuric acid.
pKa tables are an important reference for figuring out where protons will end up — for example:
http://faculty.chemeketa.edu/jcammack/Chem%20Handouts/Acid%20Base%20pKa/pka%20March.jpg
ADDENDUM:
Found a pKa table that includes protonated carboxylic acids here:
https://www2.onu.edu/~b-myers/organic/2511_Files/Chapter3-pKa%20table.pdf(3 votes)
- Since this is in equilibrium, what would promote an Ester product? or on the other hand the Acid product?(1 vote)
- Right. A Fischer esterification is an equilibrium reaction.
RCOOH + R'OH ⇌ RCOOR' + H₂O
Think of Le Châtelier's Principle if you want to push the position of equilibrium in one direction or the other.
To favour ester formation, you could
• Use a large excess of the cheaper reactant (usually the alcohol).
• Remove the water as it is formed (look up "Dean-Stark apparatus").
To favour conversion of ester to acid,
• Use an excess of water.
• (Better) Use an excess of NaOH or KOH. The strong base will react completely with the acid as it forms and drive the hydrolysis to completion. This is called a saponification reaction.(5 votes)
- what are other names for this type of reaction besides esterification?(2 votes)
- there is no other bc this is a general idea (using an c acid to make an ester)... the exact opp of this is ester hydrolysis(2 votes)
- [Refering to the last step(s) of the mechanism] In my orgo class we learned that the lone pairs on the protonated oxygen would reform the C=O, become positive, and kick off H2O as the leaving group, and then in another step be deprotanated (this keeps the O from having a negative charge in an acidic solution). In this video they do all of that in one step. Is one way more thermodynamically favored or are both acceptable mechanistic steps?(2 votes)
- For the catalysts, can we only use Sulfuric acid? or are there other acids that can be utilized too? and like are there any difference between different acids used? like are there any advantages?(1 vote)
- For many acid catalyzed reactions, like Fischer esterfication, we can use just any general acid to do so. However it's more advantageous to use sulfuric acid because it's a strong acid and dissociates more completely. Additionally sulfuric acid is widely available since it's a common chemical used in industry. But for this reaction any other strong acid like hydrochloric or nitric acid would have also worked.
Certain reactions do require a specific acid for catalysis, but reactions which accept general acids are much more common.
Hope that helps.(2 votes)
- How is ethyl heptanoate (with a C-double-bond-O) differentiated from an ether that contains a seven-carbon chain and an ethyl group connected by an Oxygen?(1 vote)
- Take an infrared spectrum of each. The ester will show a strong C=O absorption at 1735-1750 cm⁻¹.(2 votes)
Video transcript
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.