jtkelectroman - 30-4-2007 at 12:22
Well I no that this is no big deal but I just made an ampoule of bromine last night. There is a picture of it below and yes it is a tad bit fuzzy but
oh well. BTW When I made the bromine last night I also made concentrated bromine water which I added salicylic acid too and the orange color
dissapeared. Does anyone know what position Br2 adds to on the benzene ring for salycilic acid. I read something about how it can actually
decarboxylate the acid when it adds to the ring.
not_important - 30-4-2007 at 20:06
In water, Br2 will substitute phenol at the 2, 4, and 6 positions; tri-substitution is just about the rule. In your case that works out to 3 and 5,
unless it does decarboxylate in which case you're back to 2,4,6 tri-Br phenol.
Nicodem - 30-4-2007 at 21:09
Monobromination of salicylic acid yields 5-bromosalicylic acid. You do not get much ipso substitution (in this case substitution of -COOH
with -Br) or bromination at position 3. One recrystallization is enough to get pure 5-bromosalicylic acid. Of course the amount of side products
depends also on the bromination method and reagent.
jtkelectroman - 1-5-2007 at 10:53
I just used Br2/H20 solution and added salicylic acid to it. The solution is now clear with a nice aromatic smell to it.