Making Anavar from m1t
- Thread starter 3Wheels
- Start date
Frankie xq
New member
i don't think so
Billy_Bathgate
Community Veteran, Chemistry Guru
Ya, but you need acess to a professional labs and chemicals that are not readily available to the public.
Billy_Bathgate
Community Veteran, Chemistry Guru
whats the question?
Billy_Bathgate
Community Veteran, Chemistry Guru
Since your starting with M1T, youd want to start at step 4 below.
OXANDROLONE (Anavar)
Step 1: Starting with 1-test base, the 3-one is protected as the ethylene ketal as in [above is stirred in boiling benzene with ethylene glycol and catalytic acid. A Dean-Stark trap should be used to remove water as it is formed. This protects the 3-one as an ethylene ketal.].
Step 2: The 17-OH is oxidized to a 17-one with chomium VI reagent.
step 3: The 17-methyl is attached with methyl magnesium iodide and the crude reaction mixture is treated with water and acid to quench the reaction and remove the ketal protecting group. The product is 17-methyl-1-test.
Step 4: the 17-methy-1-test is treated with ozone in methanol and then NaOH is added. This allows the purification of the intermediate by recrystallization of the sodium salt. The result is that the double bond is cleaved to give a carboxyolic acid (salt) on one side and an aldehyde on the other. The salt is dissolved in water and acidified to pH=4. NaBH4 is added to reduce the intermediate aldehyde. The resulting alcohol will spontaniously close with the carboxylic acid to give the desired lactone ring. The product is oxandrolone
OXANDROLONE (Anavar)
Step 1: Starting with 1-test base, the 3-one is protected as the ethylene ketal as in [above is stirred in boiling benzene with ethylene glycol and catalytic acid. A Dean-Stark trap should be used to remove water as it is formed. This protects the 3-one as an ethylene ketal.].
Step 2: The 17-OH is oxidized to a 17-one with chomium VI reagent.
step 3: The 17-methyl is attached with methyl magnesium iodide and the crude reaction mixture is treated with water and acid to quench the reaction and remove the ketal protecting group. The product is 17-methyl-1-test.
Step 4: the 17-methy-1-test is treated with ozone in methanol and then NaOH is added. This allows the purification of the intermediate by recrystallization of the sodium salt. The result is that the double bond is cleaved to give a carboxyolic acid (salt) on one side and an aldehyde on the other. The salt is dissolved in water and acidified to pH=4. NaBH4 is added to reduce the intermediate aldehyde. The resulting alcohol will spontaniously close with the carboxylic acid to give the desired lactone ring. The product is oxandrolone
Billy_Bathgate
Community Veteran, Chemistry Guru
Well, Im not PHD in chem...but from what I do know...
I think its possible. The permange should be strong enough....
In the last step, the var should appear in your solution. A plain coffe filter should extract it. I dont know how pure it will be though. You could do a simple recrystalilzation with EtOH though
Its worth more looking into...
I think its possible. The permange should be strong enough....
In the last step, the var should appear in your solution. A plain coffe filter should extract it. I dont know how pure it will be though. You could do a simple recrystalilzation with EtOH though
Its worth more looking into...
Billy_Bathgate
Community Veteran, Chemistry Guru
EtOH = Ethanol
carboxil and aldehyde are functional groups, nothing that you add, its something that forms
kinda hard to explain without jsut drawing it out, but your ripping out a double bond which leaves the ring open, this is what forms on each end. the last step recloses this ring.
1:1 molar should do it for all of them, maybe a 1.5:1 to have some excess
carboxil and aldehyde are functional groups, nothing that you add, its something that forms
kinda hard to explain without jsut drawing it out, but your ripping out a double bond which leaves the ring open, this is what forms on each end. the last step recloses this ring.
1:1 molar should do it for all of them, maybe a 1.5:1 to have some excess
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