Replacing halogens with hydrogens

[Melvin]

I'm new to organic chemistry, so bear with me here...

 

Say you had a halogenated organic compound; how would you replace the halogen with hydrogen?

[Fuzzwood]

Depends on the position of the halogen and the structure of the alkane.

 

t-butylchloride is "easy" as you can replace the chlorine with hydroxide and then reduce it to t-butane.

[UC]

LiAlH4 reduces primary and secondary alkyl halides to the alkanes, and in some cases tertiary alkyl halides.

 

assuming the structure is relatively simple and insensitive, you could dehydrohalogenate the tertiary alkyl halide and then hydrogenate it over Pd/C. (palladium on carbon)

 

Googling around also found me some stuff on tributyltin hydride with catalytic InCl3, which seems to occur via a radical mechanism.

[Melvin]

Hmm...interesting...thank you both for your responses.

 

Fswd, would a strong base (like NaOH) be able to replace chlorine with hydroxide?

 

Also, just for curiosity's sake, would electrolysis be able to reduce alkyl halides as well?

[John Cuthber]

Classically you convert it to a grignard reagent then add water.

Doesn't work with fluorides.

[UC]

Classically you convert it to a grignard reagent then add water.

Doesn't work with fluorides.

 

I can't believe I forgot this one. This isn't tolerant of beta-haloethers though, which undergo elimination (see the Boord olefin synthesis), or of hydroxy groups, or other non-LiAlH4 responsive functional groups that do react with grignards.

[Melvin]

Classically you convert it to a grignard reagent then add water.

Doesn't work with fluorides.

 

Hmm...what about compounds with multiple halogen groups attached? Can more than one magnesium be attached to one molecule?

 

I can't believe I forgot this one. This isn't tolerant of beta-haloethers though, which undergo elimination (see the Boord olefin synthesis), or of hydroxy groups, or other non-LiAlH4 responsive functional groups that do react with grignards.

 

Hmm...so if the compound has both hydroxyl groups and halogen groups, you would have to use the other methods you mentioned (LiAlH4, dehydrohalogenation and Pd/C, etc.)?

[John Cuthber]

The grignard reaction is elegant, simple, and damn near useless. It is mesed up by just about any other functional group on the molecule

[Theophrastus]

Originally posted by Melvin

Fswd, would a strong base (like NaOH) be able to replace chlorine with hydroxide?

 

Theoretically yes, but that's only if there aren't any other functional groups, or radical bonding sites, that get in the way. I suppose it would work with something simple like 1- chlorobutane, or something of the like though, and then as said before, you can reduce the alcohol, to ethane.

 

Doing a bit of searching, I found something similar; playing off of Markinov's Rule, and reversing the process. For example, one example I found was vinyl bromide, which when reacted with hydroiodic acid, breaks the central double bond, to form 1- bromo 1- iodoethane. This is then reduced with silver oxide to acetaldehyde and silver bromide and iodide salts.

 

[ce] CH2=CHBr + HI -> CH3-CHBrI [/ce]

 

[ce] CH3-CHBrI + Ag2O -> CH3CH=O + AgBr + AgI [/ce]

 

Theoretically, you could then go further, treating the aldehyde with a strong base, like [ce] NaOH [/ce] or [ce] KOH [/ce], to produce ethanol and potassium acetate, due to reaction of the acetic acid product, with the basic potassium hydroxide. You could then reduce the ethanol to ethane, but yeah, it's definitely a rather timely process.

 

[ce] CH3CH=O + KOH -> CH3CH2OH + CH3COOK [/ce]

Edited August 6, 2009 by Theophrastus

[Melvin]

This is then reduced with silver chloride to acetaldehyde and silver bromide and iodide salts.

 

[ce] CH2=CHBr + HI -> CH3-CHBrI [/ce]

 

[ce] CH3-CHBrI + Ag2O -> CH3CH=O + AgBr + AgI [/ce]

 

 

You mean silver oxide? That's what the reaction says, not silver chloride

 

Wouldn't the ethanol and acetic acid react to form ethyl acetate?

[Theophrastus]

Originally posted by Melvin

You mean silver oxide? That's what the reaction says, not silver chloride

 

Wouldn't the ethanol and acetic acid react to form ethyl acetate?

 

Happy now?

 

(yeah, I screwed; you're right that under acidic conditions, the products would react (they don't react on their own otherwise, due to the low acid ionisation constant of acetic acid), but, what if things are to be left basic? )

 

ps: Examine slight editting, made of my initial post

 

---

Merged post follows:

Consecutive posts merged

 

[edit]

 

I was actually doing a smidge of research, and I found another substitute reaction, to turn the aldehyde into a primary alcohol, that's actually also pretty viable for the home chemist, given the availability of the compounds described (isopropyl alcohol, and a source of aluminum). Personally, I haven't tried it myself (as I have only just heard of it), but given the simplicity, I should go about doing so relatively soon (granted, my free time, has been rather limited this summer).

 

Anyhow, previously, we stripped the 1- bromo 1-iodoethane, of halogens, producing acetaldehyde. This acetaldehyde, is then placed in a solution of isopropyl alcohol (pure "rubbing alcohol, which can be obtained by adding salt, to an isopropyl alcohol solution (rubbing alcohol), resulting in the precipitation of an alcohol layer). Aluminum isopropylate is then added. This acts as a catalyst, in the following reaction:

 

[ce] CH3CH=O + Al(OiPr)3 + iPrOH -> CH3CH2OH + (CH3)2C=O + Al(OiPr) [/ce]

 

As for making the catalytic aluminum isopropylate, this can easily be done, by heating an aluminum plate, then adding it to isopropyl alcohol. Of course, you ought to use small amounts of iodine, or a globule of mercury, to initially break through the hydroxide layer, protecting the pure aluminum metal. As for the reaction with the alcohol, I would guess it'ld be quite vigorous, though I haven't tried it before, and thus cannot say, what extra percautions, in addition to the basic gloves, lab coat, and goggles, one would have to take. (I personally recommend trying it on an extremely small scale, in a test tube, just to test the nature of it, before proceeding to proper synthesis)

 

[ce] 2Al + 6CH(CH3)2OH -> 2Al(OCH(CH3)2)3 + 3H2 [/ce]

 

Here's a brief description of the actual reaction process...

 

http://en.wikipedia.org/wiki/Meerwein%E2%80%93Ponndorf%E2%80%93Verley_reduction

 

Cheers

Edited August 6, 2009 by Theophrastus
Consecutive posts merged.

[Melvin]

Very interesting infro on the aluminum isopropylate

 

Theoretically, you could then go further, treating the aldehyde with a strong base, like [ce] NaOH [/ce] or [ce] KOH [/ce], to produce ethanol and potassium acetate, due to reaction of the acetic acid product, with the basic potassium hydroxide. You could then reduce the ethanol to ethane, but yeah, it's definitely a rather timely process.

 

[ce] CH3CH=O + KOH -> CH3CH2OH + CH3COOK [/ce]

 

So, you're saying that acetaldehyde can oxidize itself to acetate, while simultaneously being reduced to ethanol? That seems very strange to me

[Theophrastus]

So, you're saying that acetaldehyde can oxidize itself to acetate, while simultaneously being reduced to ethanol? That seems very strange to me

 

Not quite. You see, this is what happens, the potassium hydroxide donates a hydroxyl group to the aldehyde, as such, the bond structure of the aldehyde shifts, and the oxygen that was formerly double bonded, is only single bonded, due to the bonding of the OH. As such, the oxygen is negatively charged.

 

By nature, this configuration, as an end product is unstable, and this intermediary gives off an oxygen, to oxidise present aldehyde, generating a carboxylic acid, while the first aldehyde/ former intermediary, having lost an oxygen, and gained a hydroxyl group, is now an alcohol.

 

Furthermore, given the basic environment, the potassium hydroxide, reacts with the acid, to form sodium acetate. I understand the confusion; it is a bit of a non- standard reaction really. Here's a link with pretty pictures, which I think would greatly help...

 

http://en.wikipedia.org/wiki/Cannizzaro_reaction

[UC]

Not quite. You see, this is what happens, the potassium hydroxide donates a hydroxyl group to the aldehyde, as such, the bond structure of the aldehyde shifts, and the oxygen that was formerly double bonded, is only single bonded, due to the bonding of the OH. As such, the oxygen is negatively charged.

 

By nature, this configuration, as an end product is unstable, and this intermediary gives off an oxygen, to oxidise present aldehyde, generating a carboxylic acid, while the first aldehyde/ former intermediary, having lost an oxygen, and gained a hydroxyl group, is now an alcohol.

 

Furthermore, given the basic environment, the potassium hydroxide, reacts with the acid, to form sodium acetate. I understand the confusion; it is a bit of a non- standard reaction really. Here's a link with pretty pictures, which I think would greatly help...

 

http://en.wikipedia.org/wiki/Cannizzaro_reaction

 

The Cannizarro reaction is only applicable to aldehydes that cannot undergo keto-enol tautomerism. The best known of these are benzaldehyde and formaldehyde. With acetaldehyde you will mainly get aldol reaction and aldol condensation products plus an enjoyable amount of tar.

 

Salt will not generate anhydrous isopropanol. It will get you to 91% and no better. Take the 91%, shake with 10% of it's weight in NaOH flakes, remove the aqueous layer. Repeat with another small amount of NaOH, and distill to give an effectively anhydrous product. If you want better, use a little sodium metal and redistill in flame-dried glass.

 

I can dig up the reference if you'd like.

 

Also, proper nomenclature is aluminum isopropoxide.

Edited August 12, 2009 by UC

[John Cuthber]

I don't think the Canizzaro reaction works with acetaldehyde. Formaldehyde is fine, but with acetaldehyde you deprotonate the alpha carbon atom then the anion atacks the carbonyl of another acetaldehyde.

Something like this.

http://en.wikipedia.org/wiki/Aldol_condensation

Then you get further reaction to form polymeric sludge.

[Theophrastus]

Salt will not generate anhydrous isopropanol. It will get you to 91% and no better.

 

Well yeah, water would screw with yield, resulting in some incompletion, and thus acetone formation, but I suppose the reaction will still occur to a reasonable extent, right?

 

[edit] Damn, I just thought of something, incompletion would mean there is an equilibrium present. $@! That could be screwy, very screwy.

 

As for aldol condensation, yeah, I failed in that respect, though out of interest, what sort of factors determine, whether the shift will occur?

 

[edit]

 

Nevermind, I just looked at the structural formulas of acetaldehyde and formaldehyde. If it is so, wouldn't propionaldehyde, be able to participate in the canizzaro reaction?

[UC]

Well yeah, water would screw with yield, resulting in some incompletion, and thus acetone formation, but I suppose the reaction will still occur to a reasonable extent, right?

 

No. You won't make any aluminum isopropoxide. Instead you'll make aluminum hydroxide until you run out of water, but Al(OH)3 is voluminous, so you probably would have sludge as your reaction mixture.

 

 

Nevermind, I just looked at the structural formulas of acetaldehyde and formaldehyde. If it is so, wouldn't propionaldehyde, be able to participate in the canizzaro reaction?

 

Look at the drawing below. From left to right, I have drawn formaldehyde, benzaldehyde, neopentaldehyde, acetaldehyde, propionaldehyde, and cyclohexanal (formylcyclohexane).

 

The alpha carbon is the carbon adjacent to the aldehyde carbonyl carbon. If the alpha carbon has no hydrogens (alpha-hydrogens) on it, the cannizzaro reaction will occur, with no condensation products.

 

For formaldehyde, there is no alpha carbon at all, so it's good to go for cannizzaro.

For benzaldehyde, the alpha-carbon is part of the aromatic ring, and has no hydrogens.

For neopentaldehyde, the alpha-carbon is quaternary and has no hydrogens.

 

Acetaldehyde, propionaldehyde, and cyclohexanal all have at least one alpha hydrogen, and are therefore capable of undergoing keto-enol tautomerism. Thus, they are capable of forming enolates, which participate in the aldol reaction/condensation.

 

There is ALWAYS an equilibrium present, but sometimes it lies so far to one side of the reaction that it effectively goes to completion. Aluminum isopropoxide is picky about which things it reduces. I'm not sure it would even work on acetaldehyde, but some research on standard enthalpies of formation should tell you the answer.

Edited August 12, 2009 by UC

[Theophrastus]

Holy crap, alpha hydrogens!

 

lovely.

 

Thanks for the help. By the way, how'ld you make those images? Do you have some sort of custom program, to model structural formulas and whatnot? It could be damned useful to be honest (implying its taken on the free economy model in one way or another; I don't have much money to spend).

[UC]

lovely.

 

Thanks for the help. By the way, how'ld you make those images? Do you have some sort of custom program, to model structural formulas and whatnot? It could be damned useful to be honest (implying its taken on the free economy model in one way or another; I don't have much money to spend).

 

http://www.scienceforums.net/forum/showthread.php?t=31089

 

I had been using it long before coming to this forum. To make forum-friendly images, you need to save as a bitmap and then convert to a JPEG in another program, which drastically reduces the file size.

 

I actually find it more usable than chemdraw, which costs @$400 IIRC. Of course, it does lack some of the advanced stuff chemdraw has, but for anything you'll need to post here, it's just fine.

[Melvin]

One thing I've really wondered about is whether electrolysis can be used to replace halogens with hydrogens. I've seen references to it on other forums and websites:

 

http://www.sciencemadness.org/talk/viewthread.php?tid=11450

 

It is mentioned from the sixth post onward.

 

http://81.207.88.128/science/chem/exps/precision_electrolysis/index_acetate.html

 

This is a little different, in the case of the acetate being converted to ethane.