09-01-01 09:30
No 209251
Hi everyone. I have yet another hairbrain idea question that Im sure has been thought of b4 but hey, you never know...
I know there has been talk of taking 3,4 Methylenedioxy-1-Bromobenzene and making a Grignard adduct from it with Magnesium, then reacting this with 1-Bromoacetone to produce MDP-2Pdirectly.
I am wondering...If one could react the MDP-Br mentioned above, WITHOUT making the grignard, directly with acetone in the presence of a weak base, like sodium carbonate or acetate maybe. I would hope it would proceed like thus:
The Br attacks a hydrogen on the acetone, producing HBr and MDP-2P. There are only two "sets" of hydrogens for it to attack anyway: either of the methyl groups at position 1 or 3 of acetone, and obviously its symmetrical, and either will produce the same product. (MDP-2P)
Now I know that HBr has a tendency to really mess with the methylenedioxy ring, thats what the weak base is for, to neutralize the acid, but not a strong base like NaOH because you dont wanna risk it reacting directly with the MDP-Br to produce a phenol.
I just wonder if this little shortcut could be taken or will it spell utter disaster and a polymerized goop in the flask? Any and all thoughts on the subject are much appreciated. And btw, if anyone is wondering, I will answer in advance: YES, I AM trying to cut a corner in this synthesis, to NOT have to make the grignard or the Bromoacetone derivative would be wonderful. This would simplify a already known reaction, making it more valuable. Anyway, enough of my rambling and on with the discussion! Thanx again.
Of COURSE we don't know what we're doing! That's why it's called research! (boom...) :)
(Newbee)
09-01-01 17:46
No 209296
../rhodium /enolate
understanding is everything
(Newbee)
09-01-01 21:59
No 209330
Thank you. I had looked at that enolate synth many a time b4 but I completely forgot about it this time. Man, Drone and OP are into some serious heavy shit!
But thats really cool. To be honest, I was actually hoping for just a simple, dirty free radical coupling between benzene radical and acetone radical, making the P-2P derivative. The radicals would be produced from HBr elimination between the starting reagents. (Bromobenzene and Acetone.) I know thats a rather dirty and problem prone mechanism, but hey! You can't ask for anyhing simpler. 
Just heat and pray for product instead of goop...Thanx again, I really appreciate your unput.Of COURSE we don't know what we're doing! That's why it's called research! (boom...) :)
(Stoni's sexual toy)
09-02-01 02:34
No 209385
There won't be any reaction at all.
(Hive Bee)
09-02-01 02:40
No 209386
Due to what? Why won't this work? SOmething to do with the MDP-Br?
The acetone enolate will form, the DMSO will promote the benzyme formation.
Please explain yourself.
(Hive Bee)
09-04-01 07:40
No 209901
Better I think to take your MDP-Br, make the aryl Lithium via low temp halogen-metal-exchange, and add that to propylene oxide, with a dash of a Cu reagent present, to make the alcohol, then oxidize to the ketone with say, Swern conditions. Or make the mesylate, and displace with MeAm.
(Chief Bee)
09-04-01 11:47
No 209948
Is the lithium exchange that favorable to making the grignard and reacting that with propylene oxide, so that it warrants the use of dry ice and expensive/pyrophoric BuLi?
(Hive Bee)
09-06-01 01:54
No 210498
Os/Rhod? Which reaction won't work? "There wont be any reaction at all." The enolate, or the proposed process? I assume the enolate P2P synth works, or it wouldnt be on Rhodiums page. As for the proposed reaction, why wont it work? Please tell me, as I am very interested in your opinions and knowledge. Thank you, and I am always willing to learn.
Have fun and stay free everyone.
Of COURSE we don't know what we're doing! That's why it's called research! (boom...) :)