Author Topic: A little 2C-I perhaps  (Read 11527 times)

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  • Guest
A little 2C-I perhaps
« on: November 26, 2002, 05:44:00 AM »
1 g 2C-H freebase (5,5 mmol)
1,8 g KI (6,5 mmol)
4 g Oxone (I belived this to be 6,5 mmol but it turned out to be 13 mmol of the active component)
15 ml MeOH

The potassium iodide was dissolved in 10 ml MeOH and this solution was then added to a solution of the 2C-H in 5 ml MeOH in a 50ml rb flask equipped with a magnetic stirbar. Oxone was then added in three portions during 10 minutes. The colorless solution turned ornage-brown immediately upon the addition of oxone. The temperature of the solution went up from 21°C to 32°C when the oxone addition was complete. This was then left to stir at room temp overnight.

Some eighteen hours later the color of the solution was the same as last night - orange-brown with some solids. The reaction mixture was poured into 25 ml water containing 100 mmol HOAc. This caused a color change, from orange-brown to deep purple. This ugly solution was then washed with two 50 ml portions DCM which removed all color from the aqueous phase. The aqueous phase was then made alkaline with NaOH, which gave a distinct yellow color to the solution, saturated with NaCl and extracted with 2x50 ml DCM. The combined DCM extracts was washed once with water and the solvent then removed in a rotovap. This gave a dark oil which was dissolved in 20 ml IPA and neutralized with dry HCl/IPA (4,5 mmol required).

The IPA solution was then put in a freezer for 30 minutes. Nothing happened even when scratching the walls with a glass rod. 10 ml IPA was removed in the rotovap and the residue put in the freezer again for 15 minutes. Still nothing!
The wall was scratched with the glass rod again, and white crystals begun to grow immediately. Within 1 minute the solution turned into a almost solid cake of white crystals. The crystals was washed with 2x15 ml EtOAc and dried to constant weight.

Yield 1,2 g of white crystals.

If all this is 2C-I it is then a yield of 63,6%. But this contains 2C-H as well. My HPLC is out of order right now so I can´t verify the ratios of 2C-I/2C-H. I guess bioassaying is the only honorable way to go at this point.


Catalytic hydrogenation freak


  • Guest
Yes we have activity
« Reply #1 on: November 26, 2002, 06:20:00 AM »
8 mg of the absolutely beautiful crystals was insufflated. Not by far as painful as 2C-B*HCl but not painless. 20 minutes later when working in the lab something is suddenly felt. It is like having something removed from my eyes. Everything is brighter and has more color. This will be a interesting day....  :)

Edit: Some two and a half hours later I can most definetly say that we have activity. It´s been a great day in the lab. Damn, the rotovap has some new nice colors  ;)  

Catalytic hydrogenation freak


  • Guest
« Reply #2 on: November 26, 2002, 03:47:00 PM »
Barium, you are THE MAN.




  • Guest
Five stars!
« Reply #3 on: November 26, 2002, 05:24:00 PM »
This must be the easiest and cheapest iodination of 2C-H this far. Now we just need to figure out the proper reaction conditions to fully iodinate the starting material so that we can isolate the target compound free from any unreacted 2C-H, fractional crystallization is hard to do right using very small amounts, and with unknown composition between 2CH/2CI (but that shouldn't be too much of a hassle, now when we have a proof-of-principle, thanks to you Barium!)


  • Guest
I think using excess iodinating reagent will be ...
« Reply #4 on: November 27, 2002, 02:20:00 AM »
I think using excess iodinating reagent will be the way to go. Due to the deactivating properties and the overall low reactivity of iodinating reagents diiodination can be ruled out to happen. So if that 2C-H is resisting and refuses to cooperate and react in a 1:1 ratio then overwhelm it with twice as much of the reagent. Beat the shit out of the stubborn redhead stepchild  ;D

I'm not fat just horizontally disproportionate.


  • Guest
I second that
« Reply #5 on: November 27, 2002, 04:14:00 AM »
One can´t whine about the iodinating system being expensive. So let´s hit it with a 1:2 ratio and see what happens.

Catalytic hydrogenation freak


  • Guest
A little 2C-C perhaps
« Reply #6 on: November 28, 2002, 07:47:00 AM »
1 g 2C-H*HCl (4,6 mmol)
1,4 g Oxone (2,3 mmol)
10 ml MeOH
5 ml water

2C-H*HCl was dissolved in the mixture of water nad MeOH in a 50 ml rb flask equipped with a magnetic stirbar and the oxone added in three portions during 5 minutes. The solution gradually became more yellow during 10 minutes. No change in temperature was noted. After 30 minutes 25 mmol HOAc was added and the solution extracted with 2x15 ml DCM which removed most of the color. The acidic aqueous phase was then made strongly alkaline with NaOH, which caused the solution to become dark amber in color, then extracted with 2x15 ml DCM. Most of the color was now in the DCM extracts.

According to PIHKAL none of the hydrochlorides of the 4-halogenated 2,5-DMPEA`s are particualry soluble in water, while 2C-H*HCl is quite soluble. So I decided to perform a simple test to see if any 2C-C was made. To the DCM phase 20 ml conc HCl was added in one portion. Within 10 seconds the acidic aqueous phase was cloudy with white crystals. The phases was separated and 40 ml IPA was added to the aqueous phase and the crystals removed by filtration and dried to constant weight.

Yield 370 mg of, most likely, 2,5-dimethoxy-4-chlorophenethylamine hydrochloride (2C-C*HCl)

The yield is terrible low. But the theory seems to be correct. Next time I´ll try this without the MeOH and use a two-phase system consisting of water/DCM instead. Dissolve 2C-H*HCl in 10 ml water, add 10 ml DCM, then oxone in portions. When the reaction is over add NaOH until pH 12, separate and 20% HCl until pH 2-3 is reached (perhaps wash the DCM phase with aq NaHSO3 before acidification). Then it should just be a matter of remove the crystals, wash then a little and dry them to constant weight.

Catalytic hydrogenation freak


  • Guest
Where'd you get the 2c-h from???
« Reply #7 on: November 28, 2002, 09:54:00 AM »
By the way, you ARE the PIMP for the day...

(off to buy a hat so I can take it off).

"Small Minds Have Big Mouths."


  • Guest
2C-B.HCl is insanely insoluble, but I wouldn't ...
« Reply #8 on: November 28, 2002, 01:33:00 PM »
2C-B.HCl is insanely insoluble, but I wouldn't say that holds as true for 2C-I.HCl/2C-C.HCl - you lost a lot in the water above. Use your standard 5M HCl in IPA crystallization, and then recrystallize from acetonitrile to remove any unreacted 2C-H - check with TLC to find out if it is completely removed.


  • Guest
As I said, it was just a quick test to see if ...
« Reply #9 on: November 29, 2002, 06:59:00 AM »
As I said, it was just a quick test to see if 2C-C had been made.

Catalytic hydrogenation freak


  • Guest
Wich solvent system ?
« Reply #10 on: November 30, 2002, 12:27:00 PM »
Wich solvent system could be adequate to separate 2CI and 2CH ?


  • Guest
Solvent system
« Reply #11 on: December 01, 2002, 02:19:00 AM »
I would try with water, MeOH, EtOH or IPA. 2C-H.HCl is much more soluble in any of them than 2C-I.HCl

Catalytic hydrogenation freak


  • Guest
« Reply #12 on: December 02, 2002, 08:58:00 AM »
May be plain water is the better solvent so.


  • Guest
oxone calculations
« Reply #13 on: March 17, 2003, 04:18:00 PM »
"4 g Oxone (I belived this to be 6,5 mmol but it turned out to be 13 mmol of the active component)"

The patent however states the use of "oxone" so 4 gram of  oxone would make 6,5 mmol, thus your calculation should bee correct


  • Guest
« Reply #14 on: March 29, 2003, 12:26:00 AM »
How is 2C-C optimization going?

(just want to add that 2C-I.HCl tastes slightly sweet, that can bee diagnostical for some bees)  ;D

I could dissolve as most 20 mg 2C-I.HCl in 1 mL water at RT (20 mg/mL).

Barium, I forgot to say: YOU ARE THE BEST!