Sciencemadness Discussion Board

Fischer esterification equilibrium

metalayer - 16-4-2015 at 02:44

Is it always necessary to use a molar excess of the alcohol? The alcohol has the same solubility profile as my ester product, so I would like to avoid this if possible. I'm also a cheap bastard so I would like to avoid buying a Dean Stark trap, will 4A molecular sieves do the job for removing water? Reaction will produce 6.8g of water according to my calculation so 100g of 4A seemed ok with H2SO4 catalyst.

C7H6O3 + C8H15NO = C15H19NO3 + H2O

Finally does anyone have any general/lab technique tips for a first-timer regarding this reaction? thnx



[Edited on 16-4-2015 by metalayer]

Chemosynthesis - 16-4-2015 at 03:02

Quote: Originally posted by metalayer  
Is it always necessary to use a molar excess of the alcohol? The alcohol has the same solubility profile as my ester product, so I would like to avoid this if possible. I'm also a cheap bastard so I would like to avoid buying a Dean Stark trap, will 4A molecular sieves do the job for removing water? Reaction will produce 6.8g of water according to my calculation so 100g of 4A seemed ok with H2SO4 catalyst.

C7H6O3 + C8H15NO = C15H19NO3 + H2O

Finally does anyone have any general/lab technique tips for a first-timer regarding this reaction? thnx

Thank you for adding the empirical formulas at least. Why not post a reference and explicit substrates, which is common practice for this subsection?

We're not too stupid to figure out you're asking about 3',4'-Methylenedioxy-α-pyrrolidinobutyrophenone (MDPBP), the pharmacologically-active substituted cathinone.

Also, there is an unfathomable amount of resources on esterification out there. Why not show us what work you've done so far? Right now it sounds like you're just asking us to help you make drugs.
Edit- without really putting the effort in the chemistry, that is.

[Edited on 16-4-2015 by Chemosynthesis]

metalayer - 16-4-2015 at 04:18

Quote:

We're not too stupid to figure out you're asking about 3',4'-Methylenedioxy-α-pyrrolidinobutyrophenone (MDPBP), the pharmacologically-active substituted cathinone.


That isn't my product and your comment is a bit unfair IMO. I am not asking to be spoonfed on how to make illegal drugs, my questions were specific.

The substrate is pseudotropine, with salicylic acid. I have found no references for this anywhere.

Chemosynthesis - 16-4-2015 at 05:03

Quote: Originally posted by metalayer  


That isn't my product and your comment is a bit unfair IMO. I am not asking to be spoonfed on how to make illegal drugs, my questions were specific.

The substrate is pseudotropine, with salicylic acid. I have found no references for this anywhere.


Ah, that does make some sense with your previous posts. Well, when you make a post without any substrates or references, then edit in an empirical formula that corresponds to a drug rather than a name, it's a reasonable suspicion even if the functional groups don't match (since you admit being new and don't appear to have looked at anything such as RSC Adv., 2014,4, 8675-8681).

You don't necessarily need a reference for the specific procedure to talk through it if you don't have one. What references do you have on Fischer esterification in general? Did you look up Tetrahedron (1998) Vol. 54, Iss. 14, 2 p3631–3644?

2) http://hdl.handle.net/2381/30038
3) http://ro.uow.edu.au/theses/1109
Might not exactly be what you wanted, but might explain some things.

[Edited on 16-4-2015 by Chemosynthesis]

metalayer - 12-5-2015 at 17:32

Thanks for the journals. I have a clearer idea of the procedure now.

I intend to use a 3x molar excess of the acid, remove water using molecular sieves in the flask and also in a improvised Hempel column and catalyse with a stoichiometric quantity of H2SO4. 5 hour reflux.

Workup should be simple, wash reaction solution with saturated NaHCO3 to remove residual salicylic acid and then extract with 5% HCL, evaporate to get the HCL salt. However I am concerned about possible ester hydrolysis when forming the HCL salt. Should I be, and how can i prevent this?

Fischer Esterification - Bunch of Questions!!!

Sniffity - 10-8-2015 at 21:46

Hey everyone,

I was wondering if I could get some help with a Fischer Esterification. I've read alot about it, but I'm really trying to get my procedure perfect before hitting the lab, as in to avoid accidents/mess ups.

I know these are ALOT of questions, it's, believe it or not, what I couldn't find on the net after long research. Mostly just little points/looking for some ideas from people here.

So.... INCOMING GIANT WALL OF TEXT!

Reaction Mechanism
First, I'm curious about one step in the reaction mechanism... When I look it up, it's quite simple to understand, seems like a common reaction, however one thing's bugging me:
After the attack on the carbonyl oxygen by the proton; the oxygen gains a formal charge of +1
(having 3 bonds and a free electron pair: 6-(3+2)=5).

This strutcture is shown to resonate with another in which the carbonyl carbon gains the formal charge of +1. However, how does this make sense? The carbonyl carbon, upon resonance, has 3 bonds and one free electron pain, giving it a total of 5 assigned electrons: 4-5 = -1. Shouldn't the formal charge on this carbonyl carbon be -1 then?

I think my mistake is in thinking the carbonyl carbon has a free electron pair, when in reality it doesn't.

I'll link the reaction mechanism I'm referring to:
https://upload.wikimedia.org/wikipedia/commons/thumb/d/d7/Fi...

Synthesis

Now, on to the actual experimental question: I understand that there's equilibrium involved in the reaction, and I need to shift it to the right by applying Le Chatelier's Principle. I was just wondering about this:

1.-Most lab guides/papers I've read suggest using an excess of the alcohol to drive the equilibrium to the right. Why is the alcohol always the one used in excess? Is it just because it's easier to remove unreacted alcohol at the end? Can acid be used in excess?

2.-How much excess alcohol should be used? I was thinking of using a 4x molar excess of alcohol (as in 5 moles moles of alcohol per mole of acid). Would this be enough, or overkill?

3.-As for the catalyst, is sulphuric acid always used? Can another acid be used? I read that HCl might just do the trick, but doesn't work as good as sulphuric. Any ideas?

4.- How do I determine the amount of sulphuric acid catalyst to add to the reaction mixture? Some procedures suggest using the acid to remove water from the reaction mixture, thus they use an amount which would normally be considered excessive for a catalyst. I usually tend to use 1-2% of mole total as catalyst. Would this be enough?

5.- This might seem like a stupid question: But how exactly is it that sulphuric acid removes water from the reaction mixture?

6.- Is reflux absolutely necessary? Any other methods to carry this out?

7.- Is there anyway to test if reflux is complete? A procedure I looked up used NH4OH to test for the carboxylic acid. Added a few drops of reaction mixture to a test tube, added half a mL of NH4OH, half a mL of water, pipette off supernatant to another test tube and added HCl to induce crystallization of the carboxylic acid. Why would HCl induce crystallization of the carboxylic acid? Any other ways to test for reflux completetion?

8.- Any other ways to shift the equilibrium to the right? Any ideas?


9.- How do you go about removing the alcohol after the reaction is complete? I was thinking about distillation, would this be okay?

Okay! That was quite alot of questions! Thanks in advance to anyone who takes the time to provide an opinion!!!

Doctor Cat - 10-8-2015 at 22:24

generally a 5 molar excess is just good.
For the acid, HCl can be used of course, but since generally you use it as a solution it will decrease yield because of, well, water... I once made an esterification with phosphoric acid and it worked just as good as sulphuric... only problem is that the reaction time increases way too much
1-2% Is what I usually use
it doesn't, you need to find a way to remove that water, for example with azeotropic distillation, a soxhlet extractor filled with a desiccator, etc.
Not strictly... there are a few low boiling point esters that can be directly made just by distilling a mix of acid-alcohol... some even catalyze their own reaction (as formic acid in synthesising methyl formate if I remember correctly) Yes, tons but Fischer esterification is the easiest... other methods involve reacting alkali acid salts with haloalkanes, etc.
Finding a way to remove a product... typically, water... as I said before some ester allow you to remove them by distillation during reaction in good yields
fractional distillation or simple distillation is good enough

The trick is, the less water, the higher yield


Also... may I ask what ester specifically are you trying to synthesise?

[Edited on 10-08-2015 by Doctor Cat]

gdflp - 10-8-2015 at 22:57

Mechanism question - You're adding an additional electron pair from nowhere. In the middle resonance structure, the carbon simply has three bonds and no lone pairs, giving it a formal charge of +1. To try to solve these problems, draw in all of the electron pairs and use arrows to signify where they move. This makes it easier to keep track of all the electrons and helps prevent confusions like this.

Reaction questions -
1. Yes, the acid can be used in excess. The reason that the alcohol is typically used in excess is that, in general, alcohols have lower melting points and boiling points than similar corresponding carboxylic acids. This means that the alcohol can be removed under vacuum rather easily for short chain alcohols, while removing carboxylic acids could be quite a hassle. Especially if the acid is a solid at room temp, workup of the reaction is more difficult.

2. A large excess of alcohol is typically used. 5x molar excess is on the low end, cheap alcohols such as methanol and ethanol are typically used in a 10x molar excess to increase the yields, but most of this can be recovered, depending on the method of workup. Yields might be lower when a 5x molar excess is used, but it should still work quite well. Look at Vogel's Practical Organic Chemistry in the forum library, procedures for the synthesis of many esters can be found(ignore the mentions of using carbon tetrachloride).

3. Any strong acid will work. Sulfuric acid is the most commonly used for several reasons. Firstly, it is rather cheap and readily available, and it can be purchased in a nearly anhydrous form. Secondly, as you mentioned, it does help to increase the yield slightly by sequestering some of the water. Hydrochloric acid will work, but the concentrated acid is only a 37% aqueous solution and thus contributes a significant amount of water to the reaction mixture, which hinders the reaction due to Le Châtelier's Principle. p-Toluenesulphonic acid is a strong organic acid that is often used as a catalyst for this reaction, it is also a solid at room temp and thus easily weighed out. Hydrogen chloride is also used sometimes, but instead of using an aqueous solution, anhydrous HCl is bubbled directly into the reaction mixture, thus preventing the introduction of unnecessary water.

4. See Vogel for proper amounts. I believe 5 mol% is the standard amount, but I'm not sure and I can't check Vogel on this device.

5. The sulfuric acid irreversibly protonates the water since it is a strong acid. The protonated oxygen is no longer nucleophilic and thus attack of the electrophilic carbonyl cannot occur.

6. Reflux is not necessary for this reaction. The purpose of refluxing the reaction is to increase the rate at which the reaction comes to equilibrium. If the mixture is refluxed, the reaction will typically reach equilibrium within several hours depending on the reactants. If the mixture is left to stand at room temperature, reaching equilibrium can take weeks.

7. That procedure is one of the only ways to test if there is still significant amounts of the carboxylic acid remaining. The reason it works is that while many(any with chain of 6+ carbons) carboxylic acids are are quite soluble in alcohols, they are nearly insoluble in water. The initial reaction with ammonia forms the ammonium salt of the carboxylic acid, which is quite soluble in water. This causes it to migrate from the organic ester alcohol phase to the aqueous phase which is drawn off. Addition of hydrochloric acid acidifies the mixture, regenerating the carboxylic acid which is insoluble. In some cases, depending on the acid, instead of a precipitate you may see a second layer form if the acid is a liquid at room temperature e.g. hexanoic acid.

A more complex way to determine if the reaction is done is to azeotropically remove water during the reaction and collect it in a Dean-Stark trap or similar. This removes water from the reaction and helps to drive it to completion, while at the same time, makes it easier to determine how much water has been produced. If the theoretical quantity of water is collected, the reaction is complete.

8. Search the forum. Fischer esterification is a popular organic reaction for beginners and has thus been discussed in a multitude of threads. Essentially any method to irreversibly remove water from the reaction mixture will shift the equilibrium, such as appropriately sized molecular sieves, a Dean-Stark trap, and many others(I recall seeing a prep in OrgSyn using a Soxhlet extractor, I don't remember the details though). A traditional esterification with a 10x molar excess of alcohol, long reflux times, and careful workup will typically yield 80-90% without any attempt to remove water from the reaction.

9. Distillation is alright, but you run the risk of hydrolyzing the ester as the reaction is at the same temperature that it was during the refluxing period, but the amount of alcohol is diminishing, shifting the equilibrium away from the ester. The best methods are to use vacuum distillation, which greatly lowers the boiling point of the alcohol and thus the rate of the reaction shifting the equilibrium, or to directly quench the chilled reaction into cold water or sodium bicarbonate solution, the alcohol then migrates to the aqueous layer after several washings.

[Edited on 8-11-2015 by gdflp]

Sniffity - 11-8-2015 at 01:16

First of all, thank both of you Doctor Cat and gdflp for your amazing replies. Amazingly helpful, thank you very much!

@ Doctor Cat I'll probably be doing something involving acetic acid, as I have tons of it readily available. Most likely iso-propyl acetate.

@ gdflp Thank you very much for being so thorough and answering all my questions. Much, much appreciated, friend. I will certainly check out that Vogel. Thanks alot for poinitng me in that direction! :)

Would you mind clarifying a few, few (only two this time!) little things? :o

1.- Okay, so the alcohol is easier to remove, got that. But say I have a large, large, large excess of acid available (not theoretical, I actually do have loads of it sitting around) and I wanted to the reaction with an excess of acid instead of alcohol. For simplicity's sake, I'll asume that all of the alcohol will be consumed in the reaction, and I'll be left with quite a mess of carboxylic acid.

I was thinking of removing it by distilling the ester off, after workup. Do you think this would work if we asume the ester has a lower boiling point than that of the acid?

7.- So I can safely asume that the NH4OH can be replaced by any other base which will produce a soluble carboxylic acid salt, correct? The procedure I watched called for testing every 30 minutes.

Ashamed as I am to say this: I've never actually done this in any of my previous reflux condensations for other reactions. I'll asume I just quickly remove the condenser from the RBF, pipette some of the reaction mixture, and put condenser back on, with no need to turn heat off or do anything else out of the ordinary, correct?

Doctor Cat - 11-8-2015 at 04:49

1.- We use exces reactant due to Le-chatelier principle so we can drive the reaction to the products... Theoretically you can use excess acid but temperature for reflux will be higher and general work up will be more... messy? than if you use excess alcohol...
when neutralizing a bunch of acetate salt will precipitate out and you will have a lot of mechanical loss

I think distillation may work but I would go for azeotropic distillation in that case... (Make sure your ester is lower in its boiling point than your acid and your alcohol otherwise you will be distilling just IPA in this case) also adding a solvent will also decrease your mechanical loss due to volume increase
also I do believe that your yield will decrease as well due to ester hydrolysis
You can stop and restart the reaction as much as you want... is not, generally, very specific on that

I made Isopropyl acetate some time ago... It is generally relatively generous about yield... in a very very crude set up using phosphoric acid as catalyst, refluxing for 1/3 of the time recommended and being just absolutely lazy about watching the reaction I got a 35-40% Yield of relatively pure ester



[Edited on 10-08-2015 by Doctor Cat]

[Edited on 10-08-2015 by Doctor Cat]

gdflp - 11-8-2015 at 07:53

1. If you're using acetic acid, then you shouldn't have any issue using an excess of acid, it only becomes an issue if the acid is a solid at room temperature. Vogel has a method for the exact reaction you are looking to do on page 383, preparation of isopropyl acetate using an excess of acetic acid and has an alternative workup procedure which may be of interest as it doesn't require a fractional distillation. Alternatively, to workup the reaction , you could also use a distillation. If you're interested in isopropyl acetate, this boils at a lower temperature than acetic acid by about 26°C so you should be able to get a decent separation with a simple distillation, I would use a fractionating column if you have one though. Isopropyl acetate forms a low boiling azeotrope with both isopropanol and water. I can't find any info about an azeotrope of isopropyl acetate and acetic acid, but I don't believe one exists. Therefore, you can distill the ester along with any water and excess alcohol, then wash as normal.

7. Yes, any water soluble base should work, don't use concentrated solutions though, 5% NaOH is commonly used. Testing every 30 minutes is overkill, I would wait 2-3 hours before you begin testing at all, then test every hour. To remove a test sample, turn off the heat and allow the reaction mixture to cool to below reflux before removing the condenser, otherwise the room will fill with acetic acid vapors.

[Edited on 8-11-2015 by gdflp]

Texium - 11-8-2015 at 07:59

Quote: Originally posted by Sniffity  
7.- So I can safely asume that the NH4OH can be replaced by any other base which will produce a soluble carboxylic acid salt, correct? The procedure I watched called for testing every 30 minutes.
No, actually it can't be substituted, at least not if you're thinking of a strong base such as NaOH or KOH. The presence of alkali hydroxides will cause the ester to rapidly hydrolyze, destroying your product.

[Edit] And even if you're just doing it as a test on a small sample of your reaction mixture, it would still cause a false positive since more carboxylic acid would be formed with the hydrolysis.

[Edited on 8-11-2015 by zts16]

gdflp - 11-8-2015 at 08:06

Quote: Originally posted by zts16  
Quote: Originally posted by Sniffity  
7.- So I can safely asume that the NH4OH can be replaced by any other base which will produce a soluble carboxylic acid salt, correct? The procedure I watched called for testing every 30 minutes.
No, actually it can't be substituted, at least not if you're thinking of a strong base such as NaOH or KOH. The presence of alkali hydroxides will cause the ester to rapidly hydrolyze, destroying your product.

[Edit] And even if you're just doing it as a test on a small sample of your reaction mixture, it would still cause a false positive since more carboxylic acid would be formed with the hydrolysis.

[Edited on 8-11-2015 by zts16]


If you use a dilute solution of an alkali hydroxide, no appreciable amount of ester hydrolysis should occur.

Doctor Cat - 11-8-2015 at 14:52

Here is an article that describes the synthesis of isopropyl acetate by using reactive distillation. They got yields of 98, 76,74 and 48% the only problem is that you will have to stop the reaction several times to "refill" the mixture... It is still a very good procedure.

Semibatch Reactive Distillation for Isopropyl Acetate Synthesis. Ind. Eng. Chem. Res. 2011, 50, 1272–1277

Link: http://pubs.acs.org/doi/pdf/10.1021/ie100354x

Cou - 6-11-2015 at 19:13

Quote: Originally posted by Doctor Cat  
generally a 5 molar excess is just good.
For the acid, HCl can be used of course, but since generally you use it as a solution it will decrease yield because of, well, water... I once made an esterification with phosphoric acid and it worked just as good as sulphuric... only problem is that the reaction time increases way too much
1-2% Is what I usually use
it doesn't, you need to find a way to remove that water, for example with azeotropic distillation, a soxhlet extractor filled with a desiccator, etc.
Not strictly... there are a few low boiling point esters that can be directly made just by distilling a mix of acid-alcohol... some even catalyze their own reaction (as formic acid in synthesising methyl formate if I remember correctly) Yes, tons but Fischer esterification is the easiest... other methods involve reacting alkali acid salts with haloalkanes, etc.
Finding a way to remove a product... typically, water... as I said before some ester allow you to remove them by distillation during reaction in good yields
fractional distillation or simple distillation is good enough

The trick is, the less water, the higher yield


Also... may I ask what ester specifically are you trying to synthesise?

[Edited on 10-08-2015 by Doctor Cat]

Thanks a lot, I've been needing a lot of help with fischer esterification too.

EDIT: Sorry about the necro post, I thought the previous post was on october 8, not august 10...

[Edited on 7-11-2015 by Cou]