Making Testosterone from Androstenedione

i dont think condensation is an important step here. it dosent matter if its saturated or not, i used 1g/1000ml concentration in my oraganic chemistry lab and it worked, but the trick is in choosing the solvent. the solvent should dissolve the solid in temperature higher than room temp (thus you boil it), in order to be able to recrystallize it again in regular room temp 30 c. I dont know if water is a good solvent (most androgens dont dissolve in water), alcohol maybe a better option, im not sure though because we used water to dissolve Benzoic Acid in my lab. i think the error comes from the choice of solvent, im not sure though maybe a more knowledgeable bro can comment on that.
if you need help with the exact recrystallization procedure i have an experimental organic chemistry lab book with steps of recrystallizing many organic compound. i can copy the exact steps from the book if you wish.
 
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Thanks for the offer but I know how to re-crystallize. I used the method that is on the first post of this thread. MeOH is the solvent, but many organic synthisis reactions are carried out at lower temps. The MeOH dissolves the Andro and the NaBH4 in an ice bath. Then the acetic is added to catylize the reaction. I have the mechanism posted a few posts up. You might have synthisized Nerolin, which is simmilar in technique to what I'm trying here. I don't think this reaction is an Sn2 reaction though. Any ideas you have are appreciated though, as you can see that my first attempt was a failure.
 
i synthesized naphthalene and Benzoic Acid, which is very similar to what you doing.....keep the good work bro, you are actually encouraging me study my chemistry. if i managed to get the materials maybe i will try it on my own :D
 
I did some more research today and found that NaBH4 degrades quickly in MeOH. It is rather stable in EtOH though. Would it be a good idea to attempt this reaction in EtOH? I also ordered 15g Mn (IV) oxide in case I need to redissolve my crude product in dichloromethane and treat it with Mn. That should oxidize the over reduced portion back to TNE. Does anyone have any suggestions or ideas about this?
 
More usefull info:

"Chemoselective Reductions with Sodium Borohydride. Aldehydes vs. Ketones." D. E. Ward and C. K. Rhee, Synthetic Communications, 1988, 18, 1927-1933.

Abstract: Aldehydes can be reduced in the presence of ketones by sodium borohydride in 30% ethanol in dichloromethane at -78 C. The selectivity is generally greater than 95%.

"A General Method for the Selective Reduction of Ketones in the Presence of Enones." D. E. Ward, C. K. Rhee, and W. M. Zoghaib, Tetrahedron Letters, 1988, 29, 517-520.

Abstract: Ketones can be reduced in the presence of conjugated enones by sodium borohydride in 50% methanol in dichloromethane at -78 °C. The selectivity is generally excellent. In favorable cases the reaction can be carried out at room temperature in dichloromethane with acetic acid as catalyst.
 
sicosico said:
or try doing this : from superior muscle:


MESTEROLONE (proviron)

1-Test base is reacted with tetramethyl dilithium pentacuprate (formed in-situ from CuI and meLi). The reaction is quenched with water to afford mesterolone in a single step.

OXANDROLONE (Anavar)

Step 1: Starting with 1-test base, the 3-one is protected as the ethylene ketal as in the above synthesis of methandrostenolone.

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.
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ok im interested in these two.
 
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