Reductions...

[Rhodium]

The reduction of 2C-B is generally very low-yielding, due to dimer formation and other side-reactions. Without distillation you have no idea how much 2C-H you have in the oil you want to brominate. And purification after the bromination is even more tedious than doing it before (as well as the usual lowered yield), and you have to purify it by distillation (or possible column chromatography) at SOME point in the process, or you will never get a pure product.

[Tricky]

Question about distillation is alarmin' me too.
Is it possible to treat w/o it?
Here

https://www.thevespiary.org/rhodium/Rhodium/chemistry/2cb-lah.html

is write-up of purification w/o distilling, as far as I unerstand. Realy it's wrong?  

Quote: "Then 20ml 4N NaOH was added slowly. Each portion was allowed to react before the next was added. 60ml water was added and the reaction mixture stirred for an hour or so until it was almost white and granular. The mixture was filtered through a porcelain filter and the filter cake was extracted twice with 50 ml portions of THF in a separate beaker. The organic solvents was combined and evaporated under vacuum. The residue was dissolved in water/HCl (pH=1). The aqueous phase was washed with 3x50 ml CH2Cl2. The organic phase was re-extracted with 50ml HCl/water. Then the combined aqueous phase was made basic with NaOH (pH=14) and extracted with 4x50ml CH2Cl2. The organic phase was dried with K2CO3 and evaporated. The residue (2C-H) was only slightly colored and weighed 16.98g (75%).

A bit more cost effective then earlier methods using LAH I dare say!

According to the original JMC article it is also possible to directly dry the THF solution after filtering off the inorganic lithium/aluminum salts, evaporate the solvent and distill directly to get the freebase. About the same amount of work, and you'll get an even cleaner product."
Pliz calm me. It's very urgent problem to me just now. And I hate vacuum distilling.
Wanna make laugh the GOD? - Tell him about your plans.

[Rhodium]

Personally, I think that either the 2C-H freebase should be distilled at that point, or the freebase 2C-B should be after the bromination. If you would do a TLC of that 2C-H, I don't think it would show a single spot, rather 3:1 2C-H/dimer or something. I'd be very glad if you could prove me wrong here.

You won't die off from impure 2C-B, but I don't understand people going through a whole synthesis like that, and then skimp on the purification, ingesting a tan product of unknown purity - you will never know what the compound really should be like if you would have made it properly. And giving an impure product to other people, that is outright nasty.

[Tricky]

Thanx for advise, Rhod, I don't skimp on the purification, but vacuum distilling is really serious problem for me, so thought that simply acid/base/DCM purification of 2CH will give me product with valid purify to other synthetic steps and finally 2CB will not be very revolting... 
Wanna make laugh the GOD? - Tell him about your plans.

[Rhodium]

Make the HBr salt, it can at least be recrystallized to aid in your vacuum distillation-free purification.

[Tricky]

Subj.  
Wanna make laugh the GOD? - Tell him about your plans.

[scram]

"I think LiAlH4 will usually work in place of NaBH4"

What would be the molar amounts of LiAlH4 to Ketone in place of Al? Does MeOH pose any oxidizing threat to this reducing agent?

[Osmium]

> Does MeOH pose any oxidizing threat to this reducing agent?

Yes, that's another way to express it  

LAH + MeOH = Boom!

I'm not fat just horizontally disproportionate.

[spectralmagic]

As fascinating as all this has been...  I'm still confused about the P2NPol -> PPA thing...  will I get amphetamine instead, or does that outcome not pose a risk with the reducing agents I've mentioned?

I tells ya, it's enough to drive one mad!

Perhaps in a few weeks I'll report on SWISM's results using various methods...  he doesn't have the theory (or knowledge of first principles) to work from, so all he's got is brute force (try, try again).  So, brute force it is then!
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