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Author: Subject: Benzhydrol--->bromodiphenylmethane
Bitburger
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[*] posted on 13-9-2011 at 06:04
Benzhydrol--->bromodiphenylmethane


I try to prepare alpha-bromodiphenylmethane from the commercial available benzhydrol. My first attempt was done using NaBr and excess conc. phosphoric acid. This wasn't a good idea because of the facile formation of the corresponding polyphosphates which are formed at higher temperature and the formation of a brown tar-like substance which was also soluble in DCM.

Second attempt was done by using ether as a solvent, 36% sulphuric acid and NaBr. All reagents where used in stochiometric proportions and 200 mL ether was added so that the reaction itself was homogenous. Reflux it for 10 hours. Excess water (from sulphuric acid) was removed with dried magnesium sulphate. The ether was distilled over but than the mixture rapidly becomes darker and darker at higher temperatures. I stopped at 120°C and the residu was a blood red liquid with solid magnesium sulphate and maybe also some solid bromodiphenylmethane and benzhydrol..

The red resdual liquid created some fumes when exposed to air. So, it's hygroscopic and might react with water in the air.

The magnesium sulphate was separated out of the residual mixture by using toluene to wash my bromodiphenylmethane. The toluene had a brown colour and evaporation at room temperature couldn't be reached to obtain pure bromodiphenylmethane (mp.45°C, bp. 184°C 20mmHg)

I suggest that bromodiphenylmethane is a stong nucleophile itself and attacks the benzhydrol at higher temperatures, thus creating the corresponding ether:

What causes the brown colour in the bromination of benzhydrol?

benzhydrol side rxn.jpg - 14kB




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not_important
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[*] posted on 13-9-2011 at 06:38


As I recall benzyl alcohol will react with cold concentrated aqueous HCl, O'd expect similar from benzhydrol and HBr(aq). So skip the heating, perhaps make the HBr using H3PO4 then distilling off the constant boiling acid for use in the conversion.
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smuv
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[*] posted on 13-9-2011 at 21:53


My thoughts:

1.Don't do stochiometric proportions, should be an excess of HBr.
2.Ether is an iffy solvent, as it is cleaved by acid in the presence of good nulceophiles (bromide is good enough).
3. The side product you mention, even if it did form (which it certainly would be very minor) would react under the reaction conditions to form the desired product.
4. You have to understand that these are SN1 reactions. Pick conditions accordingly.


And I second everything not important said, but if HBr is not available I think that HBr generated in situ is viable.

Good Luck!

Edit: I think AcOH would be a reasonable solvent for this rxn. HBr can be made in situ in AcOH, by adding H2SO4 to a suspension of NaBr/KBr with stirring on an ice bath. Do not allow the temp to rise, or more Br2 will be formed. H3PO4 might do the same without any Br2 formation, I simply have never done it....

[Edited on 9-14-2011 by smuv]




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[*] posted on 14-9-2011 at 06:31


You haven't subjected the raction mixture to an aqueous workup to remove excess acid and bromide (it may not have gone to completion). Distillation to 120*C to remove the solvent is way over the top, and in combination with my previous point, allows for the formation of elemental bromine, or dehydration/oxidation of the organics by sulfuric acid or bisulfate. You havent stated the proportions used, but assuming you used 36% H2SO4 and 200ml ether was sufficient to make the reaction homogenous, I suspect it was rather small scale indeed (cf miscibility of water with ether).
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Bitburger
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[*] posted on 14-9-2011 at 09:51


I don't think that it is a good idea to use water to remove excess H2SO4 which didn't react. Sure, there is some sulphuric acid in the mixture at the end. But H2O and hydroxide ions should react quite easy with the bromodiphenylmethane to give (again) the corresponding alcohol...

Solid NaOH didn't react with the mixture.

The last part of the post makes sense. 120°C was way too much and causes idd the formation of bromine which explains the brown red colour.

I HAVE to repeat it again but now at lower temperatures. Maybe I can use a organic base instead to neutralize H2SO4 at the end?




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not_important
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[*] posted on 14-9-2011 at 21:53


Once again I'd say don't have H2SO4 in the reaction mix, if anything have an excess of the alkali bromide.

Supporting information that describes making various substituted a-chlorodiphenylmathanes. Could provide useful information.


[Edited on 15-9-2011 by not_important]

Attachment: f500845_s.pdf (306kB)
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