Sciencemadness Discussion Board

Hydrogenation of isopropanol

Wurfgurf - 7-1-2018 at 06:19

Hi, ok so I've searched the net far and wide on this subject and all I can find is oxidation of said alcohol or reduction of acetone to said alcohol. The reason I'm proposing the question I will be is because isopropanol is a perfect, cheap, available-to-the-public solvent that I just so happen to have on hand. However, if it gets reduced I'll have a hell of a mess on my hands, maybe. So my question is this: what is the end product of isopropanol after it has been reduced? I understand from reading past posts that there are some really powerful minds in here and I humbly and RESPECTFULLY remind you that I am not the most advanced in chemistry just yet (not a chemjedi). Thanks for your answers to the ones who have the answer to this question.

unionised - 7-1-2018 at 06:31

"what is the end product of isopropanol after it has been reduced? "
Exhaustive hydrogenation of IPA will give methane + water.
I'm not sure why anyone would bother






[Edited on 7-1-18 by unionised]

Wurfgurf - 7-1-2018 at 06:35

I'm not TRYING to reduce this alcohol, it's the SOLVENT of choice, perhaps, for the substance I am reducing. I just needed to make sure I won't have a, um, conglomerate of some kind. But methane and water? THAT would be acceptable.

Wurfgurf - 7-1-2018 at 06:37

Maybe I should ask this, what would you get if it reduces only one time, when the hydroxyl is just an h? Or does that not happen..?

SWIM - 7-1-2018 at 07:59

It can reduce to propane as a first step.

However there are reductive reactions which couple molecules too, so it's not possible to predict anything about your reduction or its side reactions without knowing something about it.

What reductants?
What else is in the pot?
Questions like that are vital.

Reboot - 7-1-2018 at 09:26

A reducing agent might strip off the (weakly) acidic hydrogen of the alcohol group, giving you isopropoxide. Aluminum isopropoxide can be produced that way if memory serves (from aluminum amalgam.) If the reducing agent is exceptionally strong, it might further reduce it to propane. Reductions beyond that would require fairly exotic conditions I think (like high temp thermal cracking over a catalyst.)

Generally IPA isn't very reactive (especially if kept cold), but there are a lot of options for polar aprotic solvents if you're concerned about it. Some things have unusual sources. For instance, where I live a premium brand of plumber's PVC primer is based on THF, and DMSO is sold by the health nuts as a skin ointment for treating arthritis.

Wurfgurf - 7-1-2018 at 18:13

Sorry it took me so long, long day.. Anyways, the reductamt of choice, only because I discovered it's massive abundance is anhydrous ammonia, coupled with an alkali metal, typically the lightest or second lightest as either are abundant as well. The substrate would be vanilla, or even Benzyl alcohol, but I may play with menthol or some other aromatic thing. Typically the substance I aim to reduce is extracted in isopropanol, some of them can really be dried as they evaporate as well. Thanks.. I had another post but it seems to have disappeared ...?.idk thanks guys, u all answer at light speed here

[Edited on 8-1-2018 by Wurfgurf]

Texium - 7-1-2018 at 19:05

Sounds like you're being cagey. You won't say exactly what the substrate is, don't seem to have an understanding of the chemistry behind what it is you're trying to do, and won't even refer to lithium or sodium by name. I strongly suspect cookery.

NEMO-Chemistry - 8-1-2018 at 03:01

Quote: Originally posted by Texium (zts16)  
Sounds like you're being cagey. You won't say exactly what the substrate is, don't seem to have an understanding of the chemistry behind what it is you're trying to do, and won't even refer to lithium or sodium by name. I strongly suspect cookery.

To be fair he did say he has another post wiped, this has happened to me a couple of times lately.

Hopefully no cookery as this sounds interesting.

off topic

Why the name change ZT? i like the new one but its confusing :D

Wurfgurf - 8-1-2018 at 09:48

How do you not see that I said I am reducing VANIILLIN, BENZYL ALCOHOL, MENTHOL, BENZENE (maybe), and more to the point, NO "cookery" as you call it. I apologize if I confused you with my intentions, referring to the alkali metals lithium and sodium as "the lighter two" or some such. I was just being a nerd, trying too hard I guess, nerdness. I am definitely new to chemistry in the proper sense. I am teaching myself thru experience, reading, watching. But as you may know, the hands on approach is the best way to truly understand what is happening. Thanks to those of u who gave me answers to the question. And thanks to those who let me understand how not to word things that might get me in trouble. I do not break the law.

Wurfgurf - 8-1-2018 at 12:36

I saw a video clip on YouTube from this one very smart guy, he's on here too I believe. Nilered, I salute you! Anyways, in one of his videos he makes cyclohexadiene, I believe from benzene. I intend to emulate soon. This is an intermediate for another project which I'll get into detail so you can see then. The days are very cold and my setup is outside. The irony is that the weather is better for working with liquified gases with boiling points below zero (-33.34 c) but as said previously, it's tough working in the super cold days, fingers go numb it's just not practical.. Anyways, I've said enough. Besides rocketry, chemistry has grown on me.

zed - 8-1-2018 at 16:28

Reducing Isopropyl Alcohol, is pretty difficult under normal conditions.

Makes a pretty good reducing agent though. Simply heating excess Isopropyl Alcohol and Benzaldehyde together, under some conditions, could be adequate to reduce the Benzaldehyde to Benzyl Alcohol. I've got a paper on the topic somewhere. Works surprisingly well.

happyfooddance - 9-1-2018 at 10:05

Quote: Originally posted by zed  
Reducing Isopropyl Alcohol, is pretty difficult under normal conditions.

Makes a pretty good reducing agent though. Simply heating excess Isopropyl Alcohol and Benzaldehyde together, under some conditions, could be adequate to reduce the Benzaldehyde to Benzyl Alcohol. I've got a paper on the topic somewhere. Works surprisingly well.


I would be interested in looking at such a paper, I will try to find it. I have read that IPA is used as a combination solvent and hydrogen source in some (obviously) high-pressure hydrogenations, but I am curious to what these "some conditions" that you mentioned actually are. I, too, have been using IPA as a preferred solvent because of it's price, low toxicity, and in general have been pleased with its performance, but am obviously interested in what its limits are.

Wurfgurf, why the interest in the flavor compounds? That is also where my chemistry passion takes me the most. What do you plan on reducing benzene to?

Edit: I did find a couple papers with said reductions. They used activated alumina under microwave radiation. Very high yields converting ketones and aldehydes to corresponding alcohols (high 80's, 90's).

[Edited on 1-9-2018 by happyfooddance]

Melgar - 9-1-2018 at 12:50

You'd get alkali isopropoxide salts, but they wouldn't really be "reduced", per se, you'd just be making the salt. As long as you used a huge excess of alkali metal to account for all the isopropanol, that would probably work, unless there's something I'm not taking into account. Alkali alkoxide salts tend to be reducing agents in their own right, and might reduce aldehydes to alcohols.

But unless you've got buckets of alkali metal, you really ought to evaporate some of that isopropanol off first, at least, or do a vacuum distillation.

chornedsnorkack - 9-1-2018 at 13:29

So the logic is like:
Step 1: a substrate is reduced with an excess of alkali metal (Na or Li)
Step 2: the reaction mixture is extracted with iPr-OH
What does the excess, so far unreacted alkali metal do to i-PrOH?
The answer seems to be that alkali metals do not reduce alcohols beyond reducing the acid hydrogen and forming an alkoxide.

zed - 9-1-2018 at 13:54

Nope, the substrate is reduced by isopropylalcohol.

http://www.organic-chemistry.org/namedreactions/meerwein-pon...

The reduction can be accomplished by passing the reactants through a heated tube. This is handy, as a small diameter heated tube, has no problem resisting high pressures if required.

http://pubs.acs.org/doi/abs/10.1021/jo035251f



[Edited on 9-1-2018 by zed]

[Edited on 9-1-2018 by zed]

[Edited on 10-1-2018 by zed]

MrHomeScientist - 10-1-2018 at 10:36

Quote: Originally posted by Wurfgurf  
How do you not see that I said I am reducing VANIILLIN, BENZYL ALCOHOL, MENTHOL, BENZENE (maybe), and more to the point, NO "cookery" as you call it. I apologize if I confused you with my intentions, referring to the alkali metals lithium and sodium as "the lighter two" or some such. I was just being a nerd, trying too hard I guess, nerdness. I am definitely new to chemistry in the proper sense. I am teaching myself thru experience, reading, watching. But as you may know, the hands on approach is the best way to truly understand what is happening. Thanks to those of u who gave me answers to the question. And thanks to those who let me understand how not to word things that might get me in trouble. I do not break the law.

No worries, we just get a lot of cookery garbage posts here so it's really on people's radar. Vague descriptions of reaction conditions, particularly involving lithium, anhydrous ammonia, red phosphorus, etc. are viewed with extra suspicion. There's no reason not to fully describe your reaction conditions, and more often than not you won't get good answers anyway unless people know everything that's going on in your setup. It's nothing against you; we're just trying to screen out the cooks. Stick around a while and you'll learn the forum culture. Welcome!