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.ausetute.com.au/titrcalc.html)
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 (http://www.nitrate.com/ntk204b.pdf)
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/scihist/avogadro.htm (http://gemini.tntech.edu/~tfurtsch/scihist/avogadro.htm)
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_weight (http://en.wikipedia.org/wiki/Atomic_weight)
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/acidsbases.html (http://ptcl.chem.ox.ac.uk/MSDS/acidsbases.html)
.
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--
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 (https://www.thevespiary.org/talk/index.php?topic=11143.msg53325300#msg53325300)
(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-methylenedioxyphenyl)butane hydrochloride (METHYL-J or MBDB)
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.
No, you cannot use the procedure outlined in Post 533253 (https://www.thevespiary.org/talk/index.php?topic=11143.msg53325300#msg53325300)
(starlight: "there is probably a better way to do this", Newbee Forum)
Only these are proven to work satisfactorily: Post 536518 (https://www.thevespiary.org/talk/index.php?topic=11143.msg53651800#msg53651800)
(Rhodium: "Piperonal & 1-Nitropropane", Newbee Forum)
Thank you Rhodium for clearing that up for me.
Could one use Toulene or Xylene instead of benzene in the Post 536518 (https://www.thevespiary.org/talk/index.php?topic=11143.msg53651800#msg53651800)
(Rhodium: "Piperonal & 1-Nitropropane", Newbee Forum) method?
In the preparation of 3,4-Methylenedioxyphenyl-2-nitro-1-butene found here Post 103792 (https://www.thevespiary.org/talk/index.php?topic=7493.msg10379200#msg10379200)
(Rhodium: "3,4-Methylenedioxyphenyl-2-nitro-1-butene", Chemistry Discourse) Could one use Ammonium acetate instead of n-butylamine as the catalyst?
If so what would be the ratio?