Sciencemadness Discussion Board

Chloromethylation vs. formylation

turd - 17-4-2006 at 02:33

Hello everyone,

which route do you think is better for making a beta-phenethylamine out of a phenol-ether:
1) chloromethylation followed by nucleophilic substitution with a cyanide and reduction
or
2) formylation with Zn(CN)2 followed by condensation with MeNO2 and reduction,
the phenol-ether being the limiting reagent?

Honestly, I don't like the sound of any of the two. Both are 3-steps routes, both are probably low yielding and both involve continuous production of gaseous HCl, which always gives me a headache.

All in all I'm leaning more towards the first route, because Organikum says that the benzonitrile can be made with the crude benzylchloride, making the first two steps more or less a one-pot reaction, and the reduction of the nitrile needs only 2 H2-equivalents versus 4 H2-equivalents for the reduction of the nitrostyrene.

Thanks for any suggestions.

formylation

SecretSquirrel - 17-4-2006 at 04:57

This is my first post, so I hope I'm not breaking any rules.

If you have phenol ether (ie 4-methoxyphenol) go for Riemer-Tiemann formylation. It is easier and more eco-friendly than chloromethylation and nitrile formation..

turd - 17-4-2006 at 05:31

Hi,

unfortunately I only have the ether of a phenol, not an ether and a phenol. And no POCl3 for a Vilsmeier-formylation either. :(

Sergei_Eisenstein - 17-4-2006 at 06:36

Quote:
Originally posted by SecretSquirrel
This is my first post, so I hope I'm not breaking any rules.

If you have phenol ether (ie 4-methoxyphenol) go for Riemer-Tiemann formylation. It is easier and more eco-friendly than chloromethylation and nitrile formation..



Reimer-Tiemann ecofriendly? That's not how I'd define a method that involves the generation of dichlorocarbene, but opinions might differ.

My suggestion is you follow Organikum's advise. That is how they produce phenethylamine and 3,4-dimethoxyphenethylamine industrially (i.e. arene > benzylchloride > arylacetonitrile > phenethylamine). Cyanides are not as dangerous as they want you to believe, as long as you know what you're doing.

SecretSquirrel - 17-4-2006 at 10:11

Quote:
Reimer-Tiemann ecofriendly? That's not how I'd define a method that involves the generation of dichlorocarbene, but opinions might differ.


Sorry. I didn't know that dichlorocarbene is formed during reaction. I guess you could use paraformaldehyde/Mg methoxide formylation instead. That procedure is known to give good yields on 4-methoxyphenol.

Regarding chloromethylation: I remember a conversation at the old Hive, when someone said the main problem with this method is significant amount of bis-chloromethyl compound. I'll try to find that particular thread in the archives and post it if anyone wants.

Nicodem - 17-4-2006 at 11:03

SecretSquirrel, please read the reply from Turd where he already told you that what he has is not a phenol but a phenol ether.

Biarylmethylenic compounds only became a serious side product when:
a.) the aromatic substrate is very nucleophylic (phenols, trimethoxybenzenes etc.) which normaly can't be chloromethylated;
b.) when the reaction is done in the presence of Lewis acid catalysts on substrates that are already reactive enough and don't need catalysis (like trying to chloromethylate anisole with ZnCl2 or H2SO4 catalysis, for example);
c.) when the reaction on a reactive substrate is performed in homogenous media (like 1,4-dimethoxybenzene in AcOH as solvent).

You just have to choose the right conditions depending on what you want to chloromethylate. If you would, just for example, want to chloromethylate a phenol ether like 1,4-dimethoxybenzene, you have to avoid catalysts (b) and use a biphasic system (c) and it will work (indeed there is a reference somewhere giving a 50% yield after recrystalization).

SecretSquirrel - 18-4-2006 at 01:06

I wasn't talking about biarylmethylenic compounds, but rather about bis-chloromethyl compounds. For example: if you chloromethylate 1,4-dimethoxybenzene, there will be some 2,5-bis-(chloromethyl)-1,4-dimethoxybenzene formed as a byproduct. How do you separate those two compounds, or even better, how do you prevent the formation of bis-chloromethylated compounds? I remember Rhodium once posted a procedure for chloro-/bromomethylation which gave almost no bis-chloromethyl product, but I can't find it right now.

Nicodem - 18-4-2006 at 11:18

The chloromethyl group is moderately ring deactivating by inductive effect so there is little or no bischloromethyl product formed when the the ratio of formaldehyde vs. the substrate is aproximately 1:1.
This is nearly not a problem when compared to the formation of biarylmethylenic compounds which in some cases are the major and not just the side product.
Quote:

I remember Rhodium once posted a procedure for chloro-/bromomethylation which gave almost no bis-chloromethyl product, but I can't find it right now.


I think you mean this.

[Edited on 18-4-2006 by Nicodem]

SecretSquirrel - 18-4-2006 at 11:32

Aha.:) Thank you for the explanation and the link to the old post.

turd - 18-4-2006 at 13:20

Thanks everybody, this thread was most insightful.

Suddenly Gattermann doesn't sound all that bad anymore. :/

Organikum gives a 65% yield for 3,4-dimethoxy-benzylchloride, by slow addition of paraformaldehyde to veratrol in chloro-benzene while saturating with dry HCl. Is the ortho-methoxy group in the 2,5-dimethoxy-compound the killer?

turd - 23-4-2006 at 02:39

Damn.

I tried the chloromethylation procedure given in Organikum with hydroquinone dimethyl ether on a 50mmol scale and the residue looks and smells exactly like unreacted HDME. :(

I probably shouldn't have cooled it to 5°C! :(

Madandcrazy - 23-4-2006 at 08:13

Interested,
maybe bromomethylation is better to handle and have the same ;) effect and have perhabs importance for reactions between different halogens and further syntheses.

3,4-dimethoxy-benzylchloride
(CH3O)2-C6H4CH2Cl
-->
(CH2Br)2-C6H4CHBrCl

turd - 24-4-2006 at 13:31

Quote:
3,4-dimethoxy-benzylchloride
(CH3O)2-C6H4CH2Cl
-->
(CH2Br)2-C6H4CHBrCl


I'm not sure I understand what you mean. Br being a better leaving group than Cl, wouldn't bromomethylation give even more of the diaryl-compound?

Anyway, 1H-NMR says 10% chloride, 12% dimer and 78% unreacted HQDME. If I'm reading it right, that is. :(

turd - 2-5-2006 at 11:36

Cool, the procedure Nicodem posted seems to work. :) I hope it's true that there is no formation of bis-chloromethyl compounds, because then yield is really decent. Unfortunately the peak of the diarylmethane is on top of one of the -O-CH3 peaks, but there doesn't seem to be much of it.

Nicodem - 3-5-2006 at 10:00

If you recrystalized the product then you removed most of the biarylmethylenic sideproducts.
The Ar-CH2-Ar methylene should show an NMR singlet at about 4.8ppm, somewhat upfield compared to the Ar-CH2-Cl singlet (about 4.6ppm). This way you can check the presence of this sideproduct. The two MeO in the 2,5-dimethoxybenzylchloride are not equivalent and thus should be two separate singlets.

turd - 3-5-2006 at 12:33

No, I didn't recrystallise it - I plan to use the crude product in the next step. I just distilled of the solvent and some unreacted reactant. I'm still undecided on whether I should use acetone, DMF or DMSO as solvent, but I think I will go with the acetone.

The CH2-Cl singlet is at 4.65ppm (of course I forgot to turn on the ppm display :/).

Just as you say, there's two MeO-singlets, at 3.8 and 3.9ppm respectively. On top of the 3.9ppm one I thought that there is the diaryl-peak, but it's hard to tell. Why do you think that it should come at 4.8ppm? There indeed is a very small peak at 4.85ppm, which I can't explain...

turd - 4-5-2006 at 08:47

Well, wouldn't the singlet at 4.85ppm be from the benzyl alcohol?

Nicodem - 4-5-2006 at 11:43

No benzyl acohols should form during chloromethylations. The Ar-CH2-OH singlet in 2,5-dimethoxybenzyl alcohol is at 4.606 ppm and note that there should also be the -OH hydrogen at 3.06 ppm (check SDBS for the full spectra of 2,5-dimethoxybenzyl alcohol).
How much does it integrates anyway? What is the ratio with the methylene of the benzyl chloride?

turd - 6-5-2006 at 01:02

Quote:
No benzyl acohols should form during chloromethylations.

I was worried because the reaction is done with an aqueous phase (albeit saturated with HCl) and during workup I washed it with cold H2O and cold NaHCO3 solution. But apparently benzylchlorides aren't as easily hydrolised as I feared.
Quote:
What is the ratio with the methylene of the benzyl chloride?

2:100 :P

If you compare the integral of the CH2-Cl peak with the one of the aromatics or the -O-CH3 peaks, then yield looks really good, but once you realise that the dimer can have up to two chloromethyl groups, that isn't necessarily true any more. :P

Anyway, I did a small scale test with KCN in acetone and all I got was unchanged reactant and crap. Seems like I'll have to convert KCN to NaCN (how!?) or to dry DMSO or DMF. Sigh.

Nicodem - 10-5-2006 at 09:51

Quote:
2:100

I thought it was something noteworthy. Was it even worth bothering me for a couple of mol% of an impurity?:D
Congratulations on the clean reaction.
Quote:

Anyway, I did a small scale test with KCN in acetone and all I got was unchanged reactant and crap. Seems like I'll have to convert KCN to NaCN (how!?) or to dry DMSO or DMF. Sigh.


Nucleophilic substitution of 2,5-dimethoxybenzyl chloride with NaCN in DMSO using the method described in [1] gives quantitative yield of the nitrile. This particular substrate was not checked in the mentioned paper but it was used and reported in [2] – which is the same paper where Rhodium found the chloromethylation procedure for 1,4-dimethoxybenzene that you used. KCN can also be used. The authors of [1] tried KCN only on two cases which gave a lower yield than NaCN, however they were substrates much, much less electrophilic than benzyl chlorides (1- and 2-chlorobutane). I assume that the yields will not be much lower on 2,5-dimethoxybenzyl chloride when using KCN instead NaCN (I guess it is only a question of solubility in DMSO). Keep also in mind that the reaction is exothermic and if the temperature rises the cyanide is a strong enough base to catalyse the classical oxidation of benzyl chlorides with DMSO to yield the benzaldehydes (this is otherwise done with NaHCO3/DMSO at 100°C).
Alternatively, a method using phase transfer catalysis should also give satisfactory results.


[1] J. Org. Chem., 26(6); 877-879 (1960) (attached)
[2] J. Org. Chem., 29(10); 2860-2864 (1964).

Attachment: jo01076a001.pdf (390kB)
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Some chloromethylation references

Vitus_Verdegast - 14-5-2006 at 06:15

Chloromethylation mechanism and orientation study:
Bull. Soc. Chim. Fr.[5] 6, 1025 (1939)

toluene -> p-methylbenzylchloride 82% yield
Bull. Soc. Chim. Fr. [4] 33, 313 (1923)

naphtalene -> 1-chloromethylnaphtalene 95% yield
Compt. Rend. 202, 73 (1936)

indane -> 5-chloromethylindane -> "indanylaldehyde"
J.A.C.S. 61, 1405 (1939)
Quote:

5-Chloromethylhydrindene.-
To a mixture containing 118 g. of hydrindene, 126 cc. of formaldehyde (30%), and 212 cc. of concentrated hydrochloric acid and held at a temperature of 60°C, was added 139 cc. of concentrated sulfuric acid over a period of seven hours. Stirring was continued for twenty hours and the reaction mixture worked up in the usual way. The yield is 95 g. (57%) ; b. p. 110-112°C (4mm.).

5-Hydrindene Aldehyde.-
Forty-five grams of 5-chloromethylhydrindene and 36.0 g. of hexamethylenetetramine were refluxed in 1000 cc. of 60% ethanol for six hours. After evaporation of the excess ethanol the mixture was extracted with ether. Purification was readily accomplished by the formation of a sodium bisulfite addition product. The yield of aldehyde was 15 g.; b. p. 135-138°C (23 mm.). The structure was proved by the formation of the corresponding anil and carboxylic acid melting at 85- 86°C and 177°C, respectively.

http://rapidshare.de/files/20437485/ja01875a023.pdf.html

turd - 17-5-2006 at 09:21

Screw DMSO. Nasty, stinking, undistillable product mix. :P

But PTC rocks. I tried a procedure analogous to J.Chem.Soc.Perkin.Trans 1 2539(1991): 18-crown-6 and KCN in acetonitrile 24h at RT. The CH2-Cl singlet ist completely gone (not even the smallest residue) as is the singlet at 4.85ppm and there is a new singlet at about 3.68ppm, which should be the -CH2-CN signal. And of course still lots of crown ether. :P

But the MeO-peaks are now better resolved and there is definitely quite some secondary product.

I think I will try to distill it, but my vacuum will only allow me to distill it at about 180°C and I really fear that it will decompose. :( Maybe I should do distillation only after reduction. Sigh.

turd - 26-5-2006 at 23:13

Argh... NaBH4/Me3SiH does not work on 2,5-diMeO-benzonitrile, or only very slowly. After 12h I got only little conversion. :(

I have no way of making AlH3 (no 100% H2SO4 and no AlCl3) and no equipment for catalytic hydrogenation. LiAlH4 seems to give only mediocre yields (50%).

Maybe the route via benzaldehyde is easier after all. I'm thinking about making the Li-aryl and converting that to the benzaldehyde. But it's 1 more step. :/

Nicodem - 27-5-2006 at 02:44

Try with BH3 formed in situ, that is either NaBH4/I2 or NaBH4/CF3COOH in THF.
Should work. Of course, if you have sodium it is probably the simplest way to use Na/PrOH (or with other alcohols). You can also try the Zn-Ni couple that is supposed to work well on nitriles. NaBH4/CoCl2 might also work. But these things require experimentation as usually some methods simply don't work on just any substrate.
Let us know if you need any references.

turd - 4-6-2006 at 04:07

Thanks a lot you for your invaluable help, Nicodem.

I tried BH3*THF and it definitely worked. No NMR of the amine yet, but things look very good. The intermediate borazine (it is?) has its -CH2- groups at 2.75ppm (t) and at 3.60ppm (t, quite low field!) respectively. Yield was on the low side, but then I used the residue of the failed reduction and was completely freestyling (reflux for 1h, hydrolysis with 15%HCl, reflux for 30min, acid base extraction, distillation).

So there seems to be no reason why in-situ BH3 wouldn't work. I'm not going to try this in the near future, because I have enough material to keep me happy and there's already a bunch of other projects waiting to be realised. :)

All in all I think we can conclude that chloromethylation is definitely a viable route for people who don't have access to POCl3 or similar.

In a 3-neck flask equipped with a thermometer and a gass inlet tube, 13.8g (100mmol) HQDME was dissolved in 25ml C6H5Cl. After addition of 50ml 33% HCl, the reaction was saturated for 30min with HCl (made by dripping ~125ml H2SO4 on ~200g NaCl) at 10°C (cooling with an ice bath). 7.5g of formalin(37%) was slowly added over the course of 30min while maintaining the HCl flow and the temperature. The cooling bath was removed and the gas flow increased for another 1h. The temperature rose to 38°C. The organic phase was separated, washed with ice cold H2O and NaHCO3 solution, and dried over K2CO3. The solvent and unreacted HQDME were distilled off under vacuum after addition of a pinch of K2CO3. This step will need some optimisation (unreacted reagent vs. dimerisation).

The residue of the last step was dissolved in 60ml acetonitrile. 1.85g (7mmol) 18-crown-6 and 6.51g (100mmol) KCN were added and the resulting suspension stirred for 18h at RT. 400ml of DCM were added, the suspension was filtered and the filtrate washed with H2O (2x300ml). The organic phase was dried over MgSO4, the volatiles removed under vacuum (the DCM reused!) and the residue distilled. Yield from HQDME: ~50%.

As for the reduction step, there's a lot of alternatives:
Catalytic hydrogenation, LiAlH4/AlCl3, LiAlH4/H2SO4, Na/alcohol (sounds so simple!), BH3, in situ BH3 from NaBH4.