FlaskBreaker
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1-Propanol from 2-Propanol
Yesterday I made some 99+% isopropanol by the salting out method. Now that I have some anhydrous (or close to it) isopropanol, I was thinking, how
could I convert this to n-propanol?
I had the idea of reacting the 2-propanol with hydrobromic acid to make 2-bromopropane, then adding back the OH group with sodium hydroxide. However,
I'm not sure if this will actually work. If I did add back the OH group, it seems like it would just recreate the isopropanol.
Does anyone have any ideas?
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Nicodem
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Thread Moved
2-2-2018 at 09:32 |
JJay
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It wouldn't be easy. You might dehydrate it to propene with high temperature sulfuric acid or something and then hydroborate and oxidize the propene.
I've been taught that it is possible to do anti-Markovnikov addition to alkenes with hydrogen iodide in the presence of a peroxide catalyst, but I'm
not really sure what the reaction conditions would be and suspect that they might require high pressures and dangerous catalysts.
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LearnedAmateur
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Yeah, 2-bromopropane will undergo nucleophilic substitution with the hydroxide, I don’t think there’s some magic way to just move a hydroxyl group
around a molecule, at least not without exotic reagents. The best/easiest method I can think of is what JJay recommended.
I found this interesting article to explain anti-Markovnikov addition to (eventually) make propan-1-ol in high yield, although it states why only HBr
can be used so you’ll end up with 1-bromopropane, plus smaller iodoalkanes are quite unstable and must be used quickly unlike the other haloalkanes
because I- is such a good leaving group.
https://chem.libretexts.org/Core/Organic_Chemistry/Alkenes/R...
In chemistry, sometimes the solution is the problem.
It’s been a while, but I’m not dead! Updated 7/1/2020. Shout out to Aga, we got along well.
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clearly_not_atara
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Basically it's impossible. The reaction JJay is thinking of is the addition of bromine radicals to alkenes, usually a system like HBr in CCl4 with
benzoyl peroxide catalyst. Not high temperature, but it requires rigorously dry conditions, moreso than most amateurs can achieve.
More practical might be to oxidize all the way to bromoacetone and convert this to acetol via the acetate; acetol is then reduced to propanol with
hydrazine in ethylene glycol.
[Edited on 04-20-1969 by clearly_not_atara]
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LearnedAmateur
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Maybe so, it is achievable for even an amateur but bromoacetone is a terrible lachrymator and quite toxic. It was known as ‘BA’ or ‘B-Stoff’
during the First World War, used as an early chemical weapon and high concentrations can damage the respiratory tract - not something many a chemist
would want floating around, like phosgene or NO2.
In chemistry, sometimes the solution is the problem.
It’s been a while, but I’m not dead! Updated 7/1/2020. Shout out to Aga, we got along well.
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Texium
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Quote: Originally posted by JJay  | | I've been taught that it is possible to do anti-Markovnikov addition to alkenes with hydrogen iodide in the presence of a peroxide catalyst, but I'm
not really sure what the reaction conditions would be and suspect that they might require high pressures and dangerous catalysts.
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HBr will do the same with a catalytic amount of peroxide present, to form Br•
I think hydrogen peroxide would work, but a relatively benign organic peroxide like benzoyl peroxide would work better. You'd need to heat it gently.
I can't find much on how to practically carry out this reaction, but you'd probably want to bubble the alkene into the HBr and peroxide mixture in a
suitable solvent (DCM, perhaps?). High pressures would not be required though.
Edit: Sniped by CNA who clearly is more familiar with the practical carrying-out of this reaction than I am. My guessing wasn't too bad though 
[Edited on 2-2-2018 by Texium (zts16)]
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JJay
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| Quote: Originally posted by LearnedAmateur | Yeah, 2-bromopropane will undergo nucleophilic substitution with the hydroxide, I don’t think there’s some magic way to just move a hydroxyl group
around a molecule, at least not without exotic reagents. The best/easiest method I can think of is what JJay recommended.
I found this interesting article to explain anti-Markovnikov addition to (eventually) make propan-1-ol in high yield, although it states why only HBr
can be used so you’ll end up with 1-bromopropane, plus smaller iodoalkanes are quite unstable and must be used quickly unlike the other haloalkanes
because I- is such a good leaving group.
https://chem.libretexts.org/Core/Organic_Chemistry/Alkenes/R... |
I couldn't make sense of that description, but this one is pretty easy to read: http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_o...
It would appear that it is very hard to add hydrogen iodide and hydrogen chloride to a double bond by anti-Markovnikov addition.
[Edited on 2-2-2018 by JJay]
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clearly_not_atara
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| Quote: Originally posted by Texium (zts16) |
Edit: Sniped by CNA who clearly is more familiar with the practical carrying-out of this reaction than I am. My guessing wasn't too bad though
[Edited on 2-2-2018 by Texium (zts16)] |
I must credit most of my knowledge of this reaction to a discussion I had two years ago when we were discussing the functionalization of styrene by
this method. Nicodem explains the reaction conditions and pitfalls pretty well.
http://www.sciencemadness.org/talk/viewthread.php?tid=22421&...
[Edited on 2-2-2018 by clearly_not_atara]
[Edited on 04-20-1969 by clearly_not_atara]
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