Meltonium
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Deciding Time for Reflux
Today I tried doing a simple experiment. I wanted an ester of propionic acid so that I could follow in NurdRage's footsteps and make some 3-ethyl,
propan-3-ol for use as a catalyst to make sodium from the magnesium-reduction method.
On hand I had propionic acid (conc.) and 2-chloropropane. I converted the propionic acid into sodium propionate using sodium hydroxide. After
isolating, recrystalizing and drying, I had 18.5 grams.
I then put the sodium propionate into a roundbottom flask along with 100 mL of 91% isopropyl alcohol and 17.5 mL of 2-chloropropane. This makes for
stoichiometric amounts of the reactants. (It's also all the 2-chloropropane I had.)
I planed to reflux these for 1 hour and then attempt to collect my product, but I decided to continue refluxing for 5 hours total since this would
follow an Sn2 mechanism and it is acting on a secondary carbon.
After the five hours, I still had a lot of sodium propionate left over, but I decided to go ahead and try and recover some product. I added a five
times the amount of reaction mixture in water and stirred until the sodium propionate had dissolved. I expected there to be some oily substance on the
top of the water, but there was none. I'm afraid that I did not get enough product at all. My next step is to try and wash the solution with DCM to
try and recover anything.
But the question that comes up is: how long did I need to reflux?
Are reflux times determined experimentally or is there a general rule to follow on the time?
I have heard that the reaction speed doubles for every 10 or so degrees C you raise the temperature, but I don't know how that helps me with this.
"If at first you don't succeed, fail, fail again."
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Tsjerk
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The mechanism by which the reaction proceeds is not determined by how long you reflux, quite the opposite.
Time for reflux is determined experimentally, for example by TLC determination of one of the starting compounds.
Why do you think the solid in your reaction is sodium propionate? Couldn't it be sodium chloride?
How soluble is your product in water/isopropylalcohol?
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Meltonium
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Oh I didn't mean to imply that I thought the longer reflux time would make it an Sn2 mechanism, rather, I chose a longer reflux because it was Sn2.
I just thought the solid was sodium propionate rather than sodium chloride because I always expect failure, haha. It definitely could be sodium
chloride.
According to PubChem, the solubility of isopropyl propionate is 5.95 mg/mL at 25C. I don't know how soluble it is in isopropanol, but I would expect
it to be miscible. Because the solubilty in water is so low (relatively) I expected that after adding ~500mL of water for it to separate out.
Thank you for the answer to my question.
"If at first you don't succeed, fail, fail again."
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Reboot
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Quote: | But the question that comes up is: how long did I need to reflux?
Are reflux times determined experimentally or is there a general rule to follow on the time? |
Experimentally, but in my experience energetically favorable reactions tend to go largely to completion fairly quickly (within a few hours anyway.)
Quote: | I have heard that the reaction speed doubles for every 10 or so degrees C you raise the temperature, but I don't know how that helps me with this.
|
My guess is that it's not relevant. Broadly speaking, raising temp can provide more activation energy (beyond simply hoping to speed things up), so
getting the mix hotter is indeed sometimes the answer to driving a reaction. But, it also promotes side reactions, so it can also be the fast lane
to a flask full of waste products. :-) In this case, if I wanted to get a higher temp I would probably go with a pressure vessel.
It's a bit unusual to see a reflux with a solvent that's well above the BP of one of the reagents. Under some conditions, it might have been
possible that you boiled off your chloropropane before it had a chance to react (or even after it did react, driving the reaction backwards by
reducing reagent concentration over time.)
Or maybe the basic alkoxide was able to reduce the chloropropane to propene, in which case you definitely lost it to the atmosphere.
Knowing the reflux temp might shed light on it. If the end reaction mix boils at a temp near that of isopropanol (instead of chloropropane) then
you've lost your reagent (as opposed to it simply not having reacted yet.)
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Tsjerk
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Quote: Originally posted by Reboot ![](images/xpblue/lastpost.gif) | Quote: | But the question that comes up is: how long did I need to reflux?
Are reflux times determined experimentally or is there a general rule to follow on the time? |
Experimentally, but in my experience energetically favorable reactions tend to go largely to completion fairly quickly (within a few hours anyway.)
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This is bullocks, you can't just state "a few hours", times differ from seconds to days, weeks are not unheard of.
Quote: | I have heard that the reaction speed doubles for every 10 or so degrees C you raise the temperature, but I don't know how that helps me with this.
|
Quote: Originally posted by Reboot ![](images/xpblue/lastpost.gif) |
My guess is that it's not relevant. Broadly speaking, raising temp can provide more activation energy (beyond simply hoping to speed things up), so
getting the mix hotter is indeed sometimes the answer to driving a reaction. But, it also promotes side reactions, so it can also be the fast lane
to a flask full of waste products. :-) In this case, if I wanted to get a higher temp I would probably go with a pressure vessel.
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Temperature and it accelerating reactions is very relevant. It goes exponentially up, so between room temp and isopropyl BP is a world of difference.
Quote: Originally posted by Reboot ![](images/xpblue/lastpost.gif) |
It's a bit unusual to see a reflux with a solvent that's well above the BP of one of the reagents. Under some conditions, it might have been
possible that you boiled off your chloropropane before it had a chance to react (or even after it did react, driving the reaction backwards by
reducing reagent concentration over time.)
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This reaction is not an equilibrium.
Quote: Originally posted by Reboot ![](images/xpblue/lastpost.gif) |
Or maybe the basic alkoxide was able to reduce the chloropropane to propene, in which case you definitely lost it to the atmosphere.
|
There is no alkoxide in this reaction, nor is there a reducing environment. Do you mean dehydrohalogenation?
Quote: Originally posted by Reboot ![](images/xpblue/lastpost.gif) |
Knowing the reflux temp might shed light on it. If the end reaction mix boils at a temp near that of isopropanol (instead of chloropropane) then
you've lost your reagent (as opposed to it simply not having reacted yet.) |
I don't think a 15%-ish solution of chloropropane is going to boil closer to the boiling point of chloropropane than the billing of isopropyl. So the
starting boiling point is probably quite high.
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AvBaeyer
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Your real problem is running this reaction in a polar protic solvent and worse, with water present. You are attempting an Sn2 reaction with a poor
nucleophile and a poorly reactive alkyl halide. The sodium propionate is very highly solvated in the reaction medium you used and thus has very highly
reduced nucleophilicity if any at all. If you really want to do an ester formation via alkylation you need to run the reaction in a polar aprotic
solvent such as DMF or acetonitrile. Consult Jerry March's text book for explanatory details and relevant references regarding alkylative
esterifications. As an aside, if you use an unreactive halide such as i-PrCl, an iodide catalyst often is used as well as the potassium salt of the
acid.
Can you dry your 91% i-PrOH to 99% or better? I recall there is good information on this process on the forum. If so, then just do a standard acid
catalyzed esterification. It may take a while timewise (ie how long to reflux question) but I'll bet that you can find such a procedure for your
compound with a bit of careful searching.
AvB
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Tsjerk
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Ah, I missed that 9% water, that is going to give 1-propranol, which also liberates HCl, but that is buffered by the propionate.
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Meltonium
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Yeah, I thought that the solvent may be a hindrance when running the reaction, but in my organic chemistry class my professor told us that the solvent
didn't matter that much for a reaction. Now I know.
I have dried isopropanol on hand, but I didn't use it for this reaction because I didn't know that water was going to destroy my yield.
Previously, I tried making ethyl propionate through a classic Fischer Esterification following NurdRage's video, but I had little success. I only
tried this way because I had 2-chloropropane that I wanted to use up.
"If at first you don't succeed, fail, fail again."
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Quote: Originally posted by Tsjerk ![](images/xpblue/lastpost.gif) | The mechanism by which the reaction proceeds is not determined by how long you reflux, quite the opposite.
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Except sometimes.
https://chemistry.stackexchange.com/questions/31711/thermody...
Another important factor is practicality.
Checking the literature will find lots of references to "refluxed for 16 hours" which means "we set it going last thing, then came back in the morning
to see what had happened".
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aga
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Experimental experience - no substitute really, especially in an amateur setting.
E.g. you can calculate and measure as much as you like, then find that the fizzing only stops after addition of another 20% acid.
(impure reagents/duff acid/bum calculations)
How long to reflux for ? Er, whatever the procedure says, OR, for as long as it takes for something to happen.
... or as long as the cooling water remains cool if you run out of ice.
... or until bedtime
... or until you get fed up waiting and want to see if it worked at alll
I suppose a big difference between Amateur and Pro Chemistry is that a pro can spend $ on perfect reagents and follow a procedure and have it work 1st
time, and get paid.
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Tsjerk
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I was talking about time of reflux, not temperature of reaction. Temperature obviously influences the outcome of a reaction.
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