I have attached a paper below which I will be using as my main reference.
I've occasionally looked at this reaction but had not had any way to attempt it, no materials, etc. It appears to be very commonly utilized when it
comes to producing MDxx precursors.
In this paper, the diagram shows that the base deprotanates the hydroxyl groups and the dihalomethane swings in closing the methylenedioxy ring,
causing the halogen molecule to be ejected.
What happens to the bromine(or other halogen)? Is there a reason why it doesn't brominate a spot on the benzene ring? Bromination reactions are
commonly performed by adding elemental bromine to the substrate, so, in the case of a methylenation reaction, what happens to the halogen? Does it
react with the base, sodium hydroxide, to form hypobromite? If so why is this reaction more likely to occur than a bromination of the aromatic ring?
This procedure is commonly performed and halogenation does not seem to ever be reported as an issue, so what's happening here?
A mixture of water (20 cm3), dibromomethane(.15 mole) and Adogen 464 (1 mmole) was vigorously stirred and heated to reflux.
FireLion3 - 19-1-2014 at 13:28
I have indeed read the paper..... probably 10x+... that portion you quoted tells me nothing about what happens to the halogen. Does the halogen break
off prematurely and react with the base? Is that the purpose of refluxing before adding in the catechol? If this is the case this wouldn't explain how
other reactions work successfully in reverse (refluxing the catechol with the PTC first then adding the dihalomethane) DraconicAcid - 19-1-2014 at 13:57
If an anionic nucleophile reacts with a haloalkane, the halogen comes off as halide ion.FireLion3 - 19-1-2014 at 14:07
If an anionic nucleophile reacts with a haloalkane, the halogen comes off as halide ion.
Thank you! Sorry for this daft question but what is the inevitably fate of that halide ion? It has to find itself reacting with some other molecule
besides the freshly methylenated molecule, which is?DraconicAcid - 19-1-2014 at 14:18
I'm not bothering to read the reference (my computer's downloading things awfully slow), but it depends on what your base is. If you're deprotonating
the phenol with sodium hydroxide sodium carbonate, or sodium ethoxide, then the bromine in your dibromomethane will end up as sodium bromide.
Potassium hydroxide? Potassium bromide.
If you're using ammonia or (more likely) a bulky alkylamine, you'll get ammonium bromide or an alkylammonium bromide.FireLion3 - 19-1-2014 at 15:28
Thanks... that is what I was assuming, though initially I wasn't sure if the halogen would evaporate into the air or react. Thanks!smaerd - 20-1-2014 at 07:51
There's a pretty big difference between elemental/diatomic bromine and the bromide ion. For some reason I thought you were implying that they used
diatomic bromine in the reaction and was wondering why it didn't go by way of E.A.S. Yea you can look at it like a williamson ether synthesis that
happens once intermolecularly and then once intramolecularly.
For example here's a similar reaction,(not my own diagram taken from cliffsnotes.com)
Notice in this example NaCl is also a product. If you think about halogenation reactions Na+ and Cl- do not halogenate aromatic rings directly. There
are likely ways to generate say Cl2 or in your example Br2 in situ but the conditions called for in the paper don't point to
that.
