Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: bunsenburning on April 10, 2012, 04:22:58 PM
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Hey guys, I have no idea how to approach this synthesis problem! Thanks!
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Hey guys, I have no idea how to approach this synthesis problem! Thanks!
Funny, neither do I have any ideas. Have you got any at all?
p.s. check the forum rules!
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Funny, neither do I have any ideas. Have you got any at all?
p.s. check the forum rules!
Honestly, no Ideas. We have never encountered a problem in class where the aromaticity of rings disappeared, so I dont know where to start! It says to do Robinson Annulation, but you can't do that without a ketone, right? And you cant oxidize the OH to make one because its attached to an aromatic ring! AHHH
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Funny, neither do I have any ideas. Have you got any at all?
p.s. check the forum rules!
Honestly, no Ideas. We have never encountered a problem in class where the aromaticity of rings disappeared, so I dont know where to start! It says to do Robinson Annulation, but you can't do that without a ketone, right? And you cant oxidize the OH to make one because its attached to an aromatic ring! AHHH
Try making the aromaicity disappear and see what you get!
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WHat I cant figure out is which rings the starting material should be. at first I thought it was the bottom two, but the methyl group is really confusing me.
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Suppose you discard your initial assumption about which rings go where and see if that helps.
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Well, the only other option is the middle two rings, and I was thinking the methyl group in B is the one that is coming out at the top of the molecule, but that puts the Bromine on the left where there is nothing on the final molecule.
Honestly I have no idea where to start.
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If your starting material will form the B and C rings of the steroid, then the bromine can be on either side of the methyl group. Therefore, the carbon that bears the bromine atom can either be one of the carbon atoms at the fusion of the C and the D ring (the 5 membered ring) or at the 12 O'clock position of the C ring. Do you see how to add the A ring?
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I am not sure about how and when to form the 5-membered ring. It seems to me that one might need to protect the ketone elsewhere on the molecule, but then I am stuck. This is not my original problem, but I am curious.
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I am usually decent on these types of problems, but I really don't have a clue here. It would be easier to start with air, fire, and water to do this synthesis than that starting material. Maybe there is a synthesis from this starting material, but I don't see it, particularly the bromine.
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This looks like one of those "sell product B, buy product A with proceeds" problems. I can't find any synthesis that starts with a naphthalene derivative and ends up with a steroid.
There is a morphine synthesis that might give you a start: Synthesis of (-)-Morphine; Taber, Douglass F. et al From Journal of the American Chemical Society, 124(42), 12416-12417; 2002. It will get you at least three rings of the steroid using a starting material not too different from yours.
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Here is a retero synthesis open to comments etc.
I thought protect the naphthalene to get cpd 5, partially hydrog the naphthalene ring giving 4 after conversion of the Br to OH. Oxidise the OH to the ketone 3. Use a three carbon synthon (need to come up with a useful one, perhaps someone can suggest something) containing an aldehyde in an aldol process followed by attack of the other end at the ketone. Elinimate all the water possible to get compound 2. Then hydrog everything down remove the R in RO do the Robinson and you get compound 1.
Sounds easy no?
Don't forget this is a suggestion only from my one active brain cell, so don't be too hard on me if you think it's all B.S.
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Unfortunately your compound 5 isn't the same as the starting material compound B
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Even if we put aside fledarmus's point, the first step looks like a problem, 4 to 5.
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Unfortunately your compound 5 isn't the same as the starting material compound B
I apologise for my error, I said the brain cell wasn't working well!
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Did 75% of my Ph.D. on steroid partial synthesis and do know the old literature pretty well but that one is tricky.
Somehow in starting material B the position of the methyl group is illegitimate. Bromo and methyl have to be interchanged.
For the six-membered ring I would follow disco´s way. 2-Step reduction of the naphthol to the corrsponding tetralone, then Robinson Annulation using but-4-en-2-one will give the A, B and C rings (all six-membered). But for the five-membered ring I have no clue so far.
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Did 75% of my Ph.D. on steroid partial synthesis and do know the old literature pretty well but that one is tricky.
Somehow in starting material B the position of the methyl group is illegitimate. Bromo and methyl have to be interchanged.
For the six-membered ring I would follow disco´s way. 2-Step reduction of the naphthol to the corrsponding tetralone, then Robinson Annulation using but-4-en-2-one will give the A, B and C rings (all six-membered). But for the five-membered ring I have no clue so far.
Would it be worth using the tetralone to construct Dane's diene? Then use the interchanged bromine and methyl in my scheme to make the A ring? See the following http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2852147/
by the way my diene should have a methyl ether.
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Diels-Alder approach is of course very elegant. Woodward used it in his syntheses of cortisone and cholesterol.
But, as mentioned in the post from bunsenburning Robinson Annulation has to be used!!
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I think the burningbunsen problem may have an error. I just don't see how the bromine is being used.
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I think the burningbunsen problem may have an error. I just don't see how the bromine is being used.
I think you may be correct.
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I think the burningbunsen problem may have an error. I just don't see how the bromine is being used.
I was on the same track.
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Went through SciFinder and Reaxys. Starting naphthalene has never been described.
Looked also through ~20 syntheses of androstenedione A.
@ bunsenburning: From where do you have this?