Cou
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Amounts of sulfuric acid and excess reactant in fischer esterificatoin procedures
* moles, name, and (ratio to limiting reactant)
In nurdrage's ethyl propionate video:
* 0.267 mol propionic acid (1)
* 0.856 mol ethanol (3.21)
* 0.0187 mol sulfuric acid (0.07)
in nile red's methyl salicylate video:
* 0.362 mol salicylic acid (1)
* 6.10 mol methanol (6.10) (solvent)
* 0.746 mol sulfurid acid (2.06 ... more than one of the reactants???)
in my textbook's isopentyl acetate procedure:
* 0.0230 mol isopentyl alcohol (1)
* 0.0632 mol acetic acid (2.75)
* 0.00933 mol sulfuric acid (0.4)
it seems that in general, you should use either the carboxylic acid or the alcohol in three fold molar excess (whichever is cheaper per mole), unless
the alcohol is used as a solvent.
but the amount of sulfuric acid varies wildly per procedure. nurdrage used a tiny amount, nile red used a huge amount, and my textbook was in between.
what's going on here? how am i gonna decide how much sulfuric acid to use? i plan to do a lot of fischer esterifications soon, but i don't know
enough about catalyst kinetics to decide how the amount of catalyst affects reaction rate (haven't taken physical chemistry yet)
on another note, i know molecular sieves can improve yield of fischer esterifications, but it gets more complicated when there isn't an azeotrope
between the alcohol and water (e.g. methanol and water). with making methyl acetate, i think it could work because methyl acetate and methanol form a
low-boiling azeotrope, so eventually an azeotrope would come up as the reaction progresses. but you could also find a ternary azeotrope with water and
methanol (e.g. toluene)
[Edited on 10-12-2019 by Cou]
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Tsjerk
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Did you take the work-up of the product in account? I would hate a precursor in my mixture I can't separate.. Rater a lower yield over a twenty cents
precursor.
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wg48temp9
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I am guessing that Nile red was concerned that the methanol was not anhydrous and or that the large excess of methanol would react with part of the
sulphuric acid .
On a vaguely related subject: are binary azeotropes always unaffected by a third component even if no other azeotropes (binary or ternary) can form
with the third component ?
Can the azeotrope be considered as a component of an ideal mixture so Raoults law is applicable ?
I am wg48 but not on my usual pc hence the temp handle.
Thank goodness for Fleming and the fungi.
Old codger' lives matters, wear a mask and help save them.
Be aware of demagoguery, keep your frontal lobes fully engaged.
I don't know who invented mRNA vaccines but they should get a fancy medal and I hope they made a shed load of money from it.
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Cou
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Using a large excess of one reactant in Fischer esterification is only necessary if one reactant is cheap, and the other is expensive. E.g. using a
cheap industrial chemical like ethanpl to form an ester with a valuable small amount of some synthetic carboxylic acid
In that case, using a large molar excess of the cheap reagent increases the percentage of expensive reagant that becomes ester. As you use an
increasingly huge amount of cheap alcohol, the percentage of carboxylic acid that becomes an ester approaches 100% at the cost of more alcohol being
unreacted. Equilibrium constant of esterification is about 1
If both reactants are similar price (e.g. methanol and acetic acid have similar prices per mole for me, both being cheap industrial chemicals), then
it's fine to use an equal amount of both. Using one in excess wont decrease total unreacted reactants
[Edited on 11-12-2019 by Cou]
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Dr.Bob
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Most of the time for a Fischer the alcohol (if it is a simple alkyl one, like methanol or ethanol) is used as the solvent, thus in a large excess, and
the carboxlic acid is just added as the limiting reagent. The amount of sulfuric or other acid (HCl, tosyl acid, or other strong acids mostly all
work) depends on how fast you want to go, the acid cost, disposal costs, but since the acid is catalytic, it really does not matter, and in many
cases, only a very small amount of the acid is ever dissolved, thus able to do anything. For expense acids or alcohols, Fischer is not as common, as
it works best with an excess of the cheaper or smaller MW reagent. I use a bit of acetyl chloride in methanol to make HCl often to do the reaction,
as everything is volatile, and the alcohol and HCl can all be evaporated to provide the crude ester, this works great for amino acids to ester HCl
salts.
In most cases the amount of sulfuric acid can be much lower if you are willing to patiently distill the ester slowly, down to a few drops in a big
flask, stirring can matter much more than amount, as the acid tends to settle to the bottom of the flask, where it does little.
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Amos
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Different alcohols, acids, and reaction conditions should also caution the concentration of sulfuric acid in some procedures. Larger alcohols can be
readily dehydrated by sulfuric acid at high enough temperature, and some acids may also be sulfonated or dehydrated. The dreadful odor of pentene
drifting out of an amyl acetate synthesis taught me (and my immediate family) this lesson. There's no one-size-fits-all ratio.
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Cou
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Quote: Originally posted by Amos  | | Different alcohols, acids, and reaction conditions should also caution the concentration of sulfuric acid in some procedures. Larger alcohols can be
readily dehydrated by sulfuric acid at high enough temperature, and some acids may also be sulfonated or dehydrated. The dreadful odor of pentene
drifting out of an amyl acetate synthesis taught me (and my immediate family) this lesson. There's no one-size-fits-all ratio. |
XD
also, when making smaller esters such as methyl acetate, ethyl acetate, with fischer esterification. I think i figured out why distillation is
performed after reflux and before workup. Small esters have a high solubility in polar solvents, so you won't get any layer if you add water
immediately after reflux. but Small esters often form low boiling azeotropes with the alcohol and water, that have a higher percentage of the ester.
Those come out of reaction mixture and leave behind most polar solvent, so a layer can form on addition of small amount of water
[Edited on 13-12-2019 by Cou]
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