Elimination of methyl groups from Toluene

[aommaster]

Hey guys!

 

I'm wondering whether or not it is possible to be able to eliminate the methyl group from toluene by using a relatively low amount of energy/reagents.

 

I am well aware that this reaction is not favourable.

 

The overall reaction I'm looking for would be:

Methyl Benzene ->Benzene + CH4 (I'm guessing the other H would be obtained from solution)

 

I'm considering the probability that the reaction may be impossible due to the fact that you'll need to have a CH3 leaving group.

 

I'm also thinking that a Hoffman degradation wouldn't work because you need to use Amines?

 

Thanks

[YT2095]

IIRC, the easiest way is to convert it to benzoic acid, and then reduce that to the benzene.

I don`t rem the Exact synth, but I think the 1`st part involves KMnO4 and H2SO4 with the methlybenzene to form the benzoic acid.

 

the Reduction part I don`t quite remember, but there was something about mixing it with Hydroxide and destructive distilation.

there is another that uses Hydrides also.

 

BTW, Long time no see dude! , how goes?

[aommaster]

Hey YT!

 

I've been busy at university. I'm double majoring in Chemical Engineering and Chemistry and doing a minor in Math... so, I have a lot less free time than I had

 

The conversion to Benzoic acid makes sense, but I'm not exactly sure how the redcution would work.

 

You could use a variety of substances for the reduction (Sodium Borohydride, or Lithium Aluminium Hydirde would also work). But as I see it, you'd compeltely reduce the benzoic acid to Benzyl (or phenyl) methanol, instead of removing the carbon group from it.

 

Unless I've gotten the mechanism completely wrong...

[John Cuthber]

It's not a reduction, it's a decarboxylation. The classic method is to heat with soda lime.

RCOOH--> RH +CO2

 

 

In principle, there's another way, react it with a strong lewis acid catalyst and a little of a methyl halide like CH3Br and AlBr3

You get a series of friedel crafts type reactions which are reversible. If you fractionally distill the mixture slowly you should get benzene and a whole bunch of dimethyl and trimethyl benzenes etc left behind. I don't know if this has ever been used as a synthetic method.

[aommaster]

Thanks John!

 

Would you happen to know the mechanism for the decarboxylation?

[Ozone]

I too am interested in this (the soda lime is no where near basic enough to abstract a proton from toluene, unless I am mistaken?). Although not the requested, mechanism, here's one I had from a while back. IIRC, the double bond and the "six-membered" intermediate are a couple of things to look for; they make the reaction take place under much milder conditions (nevermind, in this case, the re-aromatization driving the thing). I think it may be relevant.

 

For other examples of this sort of thing, check out the Strecker degradation whereby, in the presence of a reducing sugar (50 g/100 or so) amino acids are decarboxylated to yield the aldehyde, ammonia and CO2. For example, phenylalanine yields phenylacetaldehyde. The reaction proceeds first via the Schiff base with the carbohydrate. See also Maillard reaction and Amadori rearrangement. I have made this reaction work at 80°C.

 

Cheers,

 

O3

[aommaster]

I believe Hoffman degradations also reduce carbon chains, but I believe you need to have amide groups in.

[John Cuthber]

"I too am interested in this (the soda lime is no where near basic enough to abstract a proton from toluene, unless I am mistaken?). "

So what? the reaction doesn't involve toluene, its the decarboxylation of benzoic acid that gives benzene.

 

The hofman degradation would convert benzamide to aniline, potentially a useful reaction, but not for making benzene unless you want the trouble of diazotizing it the reducing that .

[YT2095]

decarboxylation.

 

Fantastic! that`s exactly the word I was after but just couldn`t remember, Cheerz

[Fuzzwood]

I too am interested in this (the soda lime is no where near basic enough to abstract a proton from toluene, unless I am mistaken?). Although not the requested, mechanism, here's one I had from a while back. IIRC, the double bond and the "six-membered" intermediate are a couple of things to look for; they make the reaction take place under much milder conditions (nevermind, in this case, the re-aromatization driving the thing). I think it may be relevant.

 

For other examples of this sort of thing, check out the Strecker degradation whereby, in the presence of a reducing sugar (50 g/100 or so) amino acids are decarboxylated to yield the aldehyde, ammonia and CO2. For example, phenylalanine yields phenylacetaldehyde. The reaction proceeds first via the Schiff base with the carbohydrate. See also Maillard reaction and Amadori rearrangement. I have made this reaction work at 80°C.

 

Cheers,

 

O3

 

Nono, its oxidized to benzoic acid first (potassium permanganate is a good option), THEN decarboxylize it.