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MCAT
Course: MCAT > Unit 9
Lesson 16: Carboxylic acid derivatives- Carboxylic acid derivatives questions
- Nomenclature and properties of acyl (acid) halides and acid anhydrides
- Nomenclature and properties of esters
- Nomenclature and properties of amides
- Reactivity of carboxylic acid derivatives
- Nucleophilic acyl substitution
- Acid-catalyzed ester hydrolysis
- Acid and base-catalyzed hydrolysis of amides
- Beta-lactam antibiotics
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Acid-catalyzed ester hydrolysis
The mechanism for the acid-catalyzed hydrolysis of esters (and transesterification). Created by Jay.
Want to join the conversation?
- at the end of the video, why is the methyl ester group a better leaving group than the butyl ester group?
are smaller leaving groups better leaving groups? i would have thought the major product would have been with the methyl group over the butyl group?(3 votes)- The conditions in the solution would push the reaction forward. If there is an excess of butanol in acid while there is only a limited amount of ester, you would see transesterification occur.(3 votes)
- What's the best way to remember these reactions? I keep forgetting the products. Do we just need to memorize this? Please help!(2 votes)
- For MCAT purposes, just know what the final product is. You MIGHT get one question asking about reaction mechanisms on the test. Not worth all the stress to have a better chance at answering one question correctly. Focus your time elsewhere.(4 votes)
- Atto the rest of the video, why does the water or alcohol not add across the double bond? 8:37(2 votes)
- That could definitely occur, especially in acid. He simply doesn't mention it because he is focusing on the reactivity of the ester in this video.(1 vote)
- Hi guys...
at the mechanism of acid promoted ester hydrolysis... why you did not protonate oxygen atom that is next to R prime and formed an alcohol and left as a good L.G then molecule of water came and nucleophilically attacked carbonyl carbon then we deprotonated the oxygen atom and finally we have formed our carboxylic acid...??
instead of this long mechanism...?
IS my view false or true..?(2 votes)- The pi electrons in the carbonyl bond are more distortable towards the protonated and positively charged oxygen (think of resonance structures) than those in the sigma bond of C-OR'. Thus protonation of the carbonyl oxygen makes the carbonyl carbon more electrophilic than protonation of the other oxygen, thus promoting the attack by the nucleophile.(2 votes)
- we can do it by strong or weak acid....?(2 votes)
- A strong acid is used, such as hydrochloric acid or sulfuric acid(2 votes)
- From my understanding, an alcohol is a bad LG. Why is it a LG at? 5:28(2 votes)
- Yeah I'm not sure. And earlier at, OH2+ had the opportunity to leave as a water molecule (an excellent leaving group) but this did not happen. 2:44(2 votes)
- I can´t understand to which video he refers in the beginning.(2 votes)
- At, why didn't the water molecule which was acting as a base attack the OH group instead of the alkoxy group (ie: OR)? Wouldn't water function as a better leaving group? 4:00(2 votes)
- During transesterification, why wouldn't the product be a mixture of the butyl acrylate and the methyl acrylate? The mechanism atis going to protonate the esteric oxygen, but if your nucleophile is an alcohol, you'll have two esteric oxygens to choose from. How would you isolate only the butyl acrylate? 3:32(2 votes)
- You would have a mixture. You would probably have to use distillation or gas chromatography to seperate your products. But pushing reaction conditions to favor transesterification will increase your yield of the desired product.(1 vote)
- atwhy doesn't the ether diol stay like that as all the charges are now neutral? 3:40(1 vote)
- The diol is still in an acid solution.
The acid protonates one of the OH groups, and the water leaves to form an even more stable carbonyl group.(3 votes)
Video transcript
Voiceover: In the video
on Fischer esterification we saw that if we took a
carboxylic and alcohol, in an acid-catalyzed reaction, we produced an ester, and
we also produced water. Our goal in that video was
to make more of our ester, so we shifted the
equilibrium to the right, to make more of our product. In this video, we're talking
about the reverse reaction; we're going to talk
about ester hydrolysis. So if we increase the
concentration of water, that would shift the
equilibrium back to the left, and we would hydrolyze our ester, and turn it into our alcohol
and our carboxylic acid. So, it's important to think about what bond we're going to break; you can see that we're going
to break this bond, in here, and this oxygen, and this
R prime group turn into our alcohol, and so we'll
see that in our mechanism. So let's go ahead, and look at the details of the mechanism, where we're starting with an ester, and we're
going to, first, think about what else is present. One solution, H two O, and H plus, would give us H three O plus, and I'm going to go ahead and draw in the hydronium ion, over here,
so this is H three O plus. The first step of the
mechanism is to protonate the carbon EEL oxygen, so
this lone pair of electrons picks up a proton from hydronium and let's go ahead and show
the protonated carbon EEL, so now we have our oxygen,
it has been protonated, so it has a plus one formal
charge, so let's show those electrons, so these
electrons right here, in magenta, pick up this proton
to form this bond right here, and this activates our
carbon EEL, so we've seen this in earlier videos,
so when you think about a resonance structure,
that actually makes this carbon more positive,
it's more electrophilic, and therefore, that carbon is going to react with a nucleophile
in our next step, and our nucleophile here is water, so water's gonna function
as a nucleophile, the nucleophile attacks our electrophile, and that pushes these electrons
off, onto this oxygen, so when we show the bond
now, between the oxygen and that carbon, let
me go ahead and draw in the rest of this: this would
be a plus one formal charge on this oxygen, and let me
highlight those electrons here, so these electrons in blue,
are going to form the bond between this carbon and this oxygen. We still have an oxygen
over here, on the left, and now, that oxygen has
two lone pairs of electrons, so let me go ahead and
show those electrons here, so let me make them green, so these electrons right
here, in green, move off onto here, and are now a
lone pair on that oxygen, we still have an oxygen
bonded to this carbon, and an R prime group. The next step, we need to deprotonate; we need to get rid of that
plus one formal charge, and so, a molecule of
water can come along, and this time, function as
a base, so water's gonna function as a base,
gonna take this proton, leaving these electrons
behind, on that oxygen, so let's get some space down here, to show the deprotonation,
so we would now have this carbon, with an OH on the
left, and after we deprotonate, we're also gonna have an OH on the right, so let me go ahead and draw
in that OH, on the right, so, showing those electrons,
let's make those red, so these electrons in
here, are gonna move off onto this oxygen, and then
drawing in everything else, we have our R group on the left, and we have OR prime on the right, and I'm gonna go ahead and
put in lone pairs of electrons on this oxygen, because in the next step in the mechanism, we're going
to protonate that oxygen. So hydronium is present,
so H three O plus, so I'm gonna go ahead
and draw that in here, so H three O Plus, this
oxygen is gonna pick up a proton from hydronium,
leaving these electrons behind, and so we're gonna protonate that oxygen, and the reason why we
protonate that oxygen is it turns it into a
better leaving group, so let me go ahead and draw in everything, so we have our OH groups at the top, we have our R group on the
left, we have our oxygen, which has been protonated
now, so it's gonna have a plus one formal charge,
so plus one formal charge on this oxygen, let's
show those electrons, so let's make those blue,
here, so I'm saying that this lone pair is gonna take this proton, forming this bond, right here, and if you look closely,
you now have alcohol hiding as a leaving group,
because in the next step of the mechanism, we're going
to reform our carbon EEL, and alcohol is going to
leave, so if these electrons move into here, to reform our carbon EEL, that's too many bonds to
carbon, and so these electrons are gonna come off, onto the oxygen, and so here's the bond-breaking step, where the alcohol leaves,
and so let's go ahead and draw our product,
so we're going to form our carbon EEL, and this
oxygen is going to have a plus one formal charge
now, and we have an R group, and, over here, would be
an OH, because the alcohol is going to leave, so I'm gonna go ahead, and draw in the alcohol here, so OR prime, with now two lone pairs of
electrons, so let's follow some electrons, so
these electrons in blue, are the same ones as blue
up here, and let's make these electrons in here red,
so these electrons in red are going to come off, onto
this oxygen, so we lose our alcohol at this step,
so loss of our alcohol gives us this, over here,
so we're almost done with our reaction, so
this is really close to a carboxylic acid, all we would
have to do is deprotonate, so we could think about
another molecule of water coming along, and acting as a base, so water acts as a base, takes this proton, leaves
these electrons behind, and that, of course, gives
us our carboxylic acid. So, if I just go ahead
and draw in our R group, and then we have ROH, and
let's follow those electrons, so let's make those red as
well, so these electrons, right here, come off, and
we get our carboxylic acid as our product, so there's your mechanism for acid-catalyzed ester
hydrolysis, which produces a carboxylic acid, and
it produces an alcohol, and when we look at some
reactions in a second here, we're gonna think about
this part, right here, where we lose our alcohol
as one of our products, to also give us our carboxylic acid. So let's look at a reaction: So, over here on the left,
we have methyl salicylate, or oil of wintergreen, and
we're looking for an ester, because we have H two
O and H plus, giving us hydronium in solution,
and so this is the portion of the molecule that's going
to react, it's going to hydrolyze that ester, and
we know from the mechanism, this is the bond that's going
to break in our mechanism, and so, we can go ahead
and draw our products, so we're gonna break that
bond, and the whole left portion, so let me go ahead and draw that, this whole left portion here is going to form our carboxylic acid, so let's go ahead and draw
in our carboxylic acid, as one of our products,
so we have our ring we have our carboxylic
acid, so this OH on our carboxylic acid, this OH came from H two O in the mechanism, and then
we have this other OH, which produces, of course,
salicylic acid, and then we're thinking about our
alcohol product, let me go ahead and use green for
this, so this is going to leave, we can see that, after you
protonate, you get loss of that as your alcohol, and so
if we just add a proton onto this oxygen here,
we can see that we would form methanol, as our other
product, so if I go ahead and draw in the methanol here,
we have our two products: we have salicylic acid,
and we have methanol, so acid-catalyzed ester hydrolysis. This reaction is at
equilibrium, technically, and so you could do things
like push the equilibrium to the right, and if you remember from the Fischer esterification
video, this is what we use to make our wintergreen; we used methanol, and we used salicylic acid
to produce our wintergreen, and so everything depends
on reactions conditions, in terms of shifting the equilibrium. Let's look at another
practice problem here: So, once again, we have our
ester, and we have water, and I drew the water in a
little bit of a weird way, and I'll show you why in a second, so acid-catalyzed ester
hydrolysis, once again, we think about what bond is gonna break; this bond is gonna
break, and the left side is going to turn into our carboxylic acid, so we can go ahead and
draw our carboxylic acid, so we would have this
portion of the molecule, so our carboxylic acid,
like that, and once again, this OH, let me go ahead and highlight it, so this OH right here,
in our carboxylic acid came from our water
molecule, so we could think about losing a proton off
of water, and then we form our carboxylic acid on the
right, our other product, let's be consistent and
stick with green here, this is the portion that's
gonna turn into our alcohol, so we think about adding
a proton onto that oxygen, and, once again we have
methanol, as our other product, so, acid-catalyzed ester hydrolysis. Now the reason I drew the water molecule in a weird way, is
because, what would happen if, instead of this
hydrogen, what would happen if we had an R group, an alkyl group? Well, that would change this hydrogen into an alkyl group, and
then we would form an ester as our product, so we'd be
starting with one ester, and we'd be turning it into another ester, and that is called, "transesterification," so pretty much the same reaction that we've been talking
about, except we wouldn't get a carboxylic acid,
as one of our products; we would get another ester. And so, let's go ahead and
do this again, starting with the same ester, but
this time, we're gonna use butanol, instead of water. So, once again, we can think about losing, we're gonna break that
bond, and we're going to lose this proton, and
then we could stick those portions of the molecule together. So, if we stick this together
with this, we can see the identity of the other
ester that will form, so let's go ahead and draw in our product, so we would have an oxygen
here, and then we'd have four carbons: one, two, three, four; so these four carbons from
butanol, which form our ester. Our other product, once
again, we see this portion over here, that would give
us methanol, so once again, we have methanol as our
product, and even though everything is at equilibrium,
methanol has a very low boiling point, so we could
boil off the methanol, and shift the equilibrium to the right to make more of our product. So we can see, once gain,
we're starting with an ester of methanol, and we're converting it into an ester of butanol, simply by changing the
reaction conditions, in a transesterification reaction, so this is a pretty useful
reaction in industry.