08-25-04 04:48
No 527195
Okay fresh off my posts about FLEA I am now looking at MBDB. Not much is written on this compound, and I again have a couple questions. Please be paitient with me.
Shulgin writes: The mixture was occasional swirled, and cooled externally as needed to keep the temperature below 50 °C. After the reduction was completed (no metallic aluminum remaining, only gray sludge), it was filtered and the residues washed with MeOH. The combined filtrate and washes were stripped of organic volatiles under vacuum, the residue treated with 100 mL Et2O, and this was extracted with 2x50 mL 3 N HCl.
Could Toulene or DCM be used instead of Et2O?
Also, what exactly is 3 N HCL?
(Stranger)
08-25-04 06:13
No 527206
Sounds like just a basic extraction, you could probably use any Non-Polar. It seems like the solvents role in this synth is just to take up the residue and hold it for a few minutes, until you extract with HCL. Im sure others would agree wiyh me.
P.S. 3 N HCL denotes a certain concentration of HCL, I dont know which concentration, but the imformation could be easily found if you just do some looking, or maybe someone here knows. Anyone?
(Hive Addict)
08-25-04 06:16
No 527207
Could Toulene or DCM be used instead of Et2O?
Yes, use Toluene.
Also, what exactly is 3 N HCL?
http://www.ausetute.com.au/titrcalc.html
http://www.geocities.com/torhavn/citizen
(Quasiquerumodo)
08-25-04 09:52
No 527230
3N HCl = 3M HCl = 3 mol/l HCl
(but 3N H2SO4 = 1.5M H2SO4 = 1.5 mol/l H2SO4)
if you don't know _that_, you shouldn't make drugs, because you can't assess
how clean your product is.
He always lied while on the earth and now he's lying in it
(Stranger)
08-26-04 05:29
No 527398
I found this information on the preparatoin of 3 N HCl
Prepare 3 N HCl by adding 15 ml concentrated HCl to 45 ml di-water. Mix.
http://www.nitrate.com/ntk204b.pdf
(Hive Addict)
08-26-04 17:08
No 527506
3 N means : normality, the concentration of protons (hydrogen ions) in the solution is 3 mol / L
what is a mol ? 1 mol = 6.022 . 1023 molecules. http://gemini.tntech.edu/~tfurtsch/scihi
Now we need to find out, how many protons HCl donates to the solution. HCl __> H+ + Cl-
As you can see 1 mol of hydrogen chloride gives 1 mol of protons in solution (dissociation). Therefore a 3N solution is 3M. H2SO4 gives 2 protons, therefore a 3N solution of sulfuric acid would be 1.5M.
What does 3M mean. M is for molarity and specifies the concentration of a compound by mol / L. Just like normality but now it is not restricted to only protons (H+) anymore.
How to prepare a 3M soln. of HCl ?
What we need to find out is how much of a solution, of which is known that 36% of the weight are HCl, needs to be mixed with water to leave a concentration of 3 mol / L.
We've got the weight m , we need the amount of substance n.
M = m / n ;; molecular weight = mass / amount of substance. The question is how many grams do 6.022 . 1023 molecules (1 mol) weight ?
If you put amount of substance as 1, the molecular weight can be calculated from the atomic weights of its constituents, found in the periodic table of the elements.
H has an atomic weight of 1.0079 u
Cl has an atomic weight of 35.453 u
http://en.wikipedia.org/wiki/Atomic_weig
So one molecule of HCl weights 36,4609 u and thus one mol = 36,4609 g
Yeah, back to the solution we wanted to prepare.
3 mols will weight 109.3827 g. This much is in (109.3827 g / 36) . 100 = 303.84 g of a 36% HCl (the strength of concentrated HCl). To get the volume of those 303.84 g we need the density (which is dependant on temperature and usually given for 20°C if not specified differently). That is 1.18 g/mL. http://ptcl.chem.ox.ac.uk/MSDS/acidsbase
So the volume will be 303.84 g / 1.18 g/mL = 257,49 mL.
We're almost there. Since we want a concentration of 3 mol / L those 257,49 mL need to be mixed with water, until a total volume of 1000 mL is reached.
BTW: this also works exactly the same way with bases like NaOH, where N (normality) would specify the concentration of hydroxide ions (OH-) in the solution as mol / L.
mixing 15mL 36% HCl with 45mL should give a 2,913 N solution, with 38% HCl it would be
3,075 N. So it sounds about right.
Hope that helped anyone...
--psyloxy--
(Stranger)
10-20-04 06:10
No 536728
So then, this is how I see this synth working out.
Use this method from Rhodium - piperonal from pepper
Next, you make 1-(3,4-methylenedioxy)-2-nitrobutene using - Post 533253 (starlight: "there is probably a better way to do this", Newbee Forum)
Then reduce using Shulgin's method for Methyl-J. By using Al/Hg and MeAm.HCL being that you already have it for above process.
Extract and clean up to arrive at - 2-methylamino-1-(3,4-methylenedioxypheny
Is this a correct way to go about it, or am I seriuosly off-track? If I am off-track, please let me know where I am going wrong.
(Chief Bee)
10-20-04 09:07
No 536757
No, you cannot use the procedure outlined in Post 533253 (starlight: "there is probably a better way to do this", Newbee Forum)
Only these are proven to work satisfactorily: Post 536518 (Rhodium: "Piperonal & 1-Nitropropane", Newbee Forum)
The Hive - Clandestine Chemists Without Borders
(Stranger)
10-20-04 17:32
No 536807
Thank you Rhodium for clearing that up for me.
Could one use Toulene or Xylene instead of benzene in the Post 536518 (Rhodium: "Piperonal & 1-Nitropropane", Newbee Forum) method?
(Hive Bee)
10-20-04 18:41
No 536812
Yes, benzene is mostly replaced with toluene these days, due to its toxicity. It is just mentioned very often because it was the standard solvent in the earlier days of chemistry. Todays use of benzene is almost solely limited to applications in chromatography, and as reagent in certain syntheses. There is no case where you can't substitute benzene for toluene solvent AFAIK.
indole_amine
(Chief Bee)
10-21-04 00:44
No 536902
Post 454922 (Rhodium: "Always* use Toluene instead of Benzene", Newbee Forum)
The Hive - Clandestine Chemists Without Borders
(Stranger)
10-24-04 02:43
No 537377
Little by little I hope I am getting this.
Reading the synthesis of P2P on Rhodium, I saw this iron/acid reduction:
Reduction with Iron Powder
If phenyl-2-nitropropene is reduced by iron powder in an acidic medium (such as acetic acid or aqueous hydrochloric acid) the nitroalkene is reduced to the oxime, which is then hydrolyzed by the acid into the desired phenyl-2-propanone.
Phenyl-2-nitropropene (10 g, 61 mmol) was dissolved in 75 ml HOAc and slowly dripped into a refluxing slurry of Fe powder (32 g, 0.57 mol) in 140 ml HOAc. The mixture turned brownish and foamy, and the mixture was refluxed on low heat for 1.5 h. The reaction mixture was poured into 2000ml water, and was extracted with 3x100 ml CH2Cl2. The combined extracts was washed with 2x150 ml H2O and dried over MgSO4. The solvent was distilled off, and the residue distilled under vacuum to give phenyl-2-propanone in 75% yield.
Is this iron/acid reduction the same one that Shulgin talks about in the preparation of 1-(3,4-methylenedioxyphenyl)-2-butanone from 1-(3,4-methylenedioxyphenyl)-2-nitro-1-b
(can be found @ end of third paragraph for #94-J)
Could I also use Methyl Man's Al/Hg + Nitromethane method for the reduction of 1-(3,4-methylenedioxyphenyl)-2-nitro-1-b
(Hive Bee)
10-24-04 04:43
No 537416
about 1): Yes (without checking; but in PiHKAL, Shulgin definitely makes use of the same method to convert phenylnitroalkenes to their ketones)
about 2): no, MeNO2 Al/Hg is a reductive amination to arrive at methylated amines, read honey. The Al amalgam reduces the nitromethane to methylamine, which condenses with the ketone to give an intermediate methylimine. This is reduced to the methylamine (honey) by aluminum amalgam, too. Al/Hg has nothing in common with Fe/HCl, except both have great reducing power and therefore both reactions are reductions. But nitromethane has no use in Fe/HCl reductions.
(side note: maybe Al could be useful with Fe/HCl. Additional hydrogen maybe? Does anyone know?)
indole_amine
(Stranger)
10-29-04 04:48
No 538426
MBDB was, until recently, available on the mail order internet research chemical scene. I have heard a user of it describe the effects as a quote, "psychedelic quaalude." Others noted that MBDB lack all stimulant properties associated with the amphetamines and, albeit less so, with MDMA.
(Stranger)
11-06-04 04:08
No 539958
In the preparation of 3,4-Methylenedioxyphenyl-2-nitro-1-buten
If so what would be the ratio?