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Author: Subject: Why does NaCNBH3+GAA+formaldehyde not do pictet Spengler?
zephler1
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[*] posted on 1-5-2014 at 07:59
Why does NaCNBH3+GAA+formaldehyde not do pictet Spengler?


The mechanism shows that tryptamine would do the pictet Spengler ring closure due to a proton being present, however, Rhodiums page has a bunch of procedures that say the authors used GAA, NaCNBH3 & formaldehyde with tryptamine to afford the dimethylated species. Yes NaCNBH3 is mild enough to leave the formaldehyde alone, but umm win out the competing reaction of ring closure? BTW, the above reaction was reported to yield material with a MP of 47-50c
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Tsjerk
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[*] posted on 1-5-2014 at 09:26


pictet Spengler ring closure is acid catalyzed, the acetic acid reacts with the cyanoborohydride to di-acetate-cyanoborohydride. There is no free acid present.

You can also use NaBH(CN)3, instead of reacting NaBH3CN with acetic acid.
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Methyl.Magic
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[*] posted on 2-5-2014 at 03:33


you can di-alkylate using MeI (or other alkylating agent) and CsCO3. works well
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zephler1
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[*] posted on 2-5-2014 at 08:30


thanks for the info. I also wonder why N-alkylation of the indole N would not occur? Sterics with an itty bitty methyl wouldn't be much of an issue I think?
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Dr.Bob
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[*] posted on 2-5-2014 at 10:17


Because indole N's are not very nucleophilic or very basic, due to resonance effects with the ring. They are more like like deactivated anilines, rather than like a alkyl amine. But it is possible to react them, usually requires either long times, higher temps, stronger bases or more reactive reagents.

But you can do a reductive amination on an aminoindole at the ring -NH2 without reacting at the indole NH in most cases. But if you deprotonate it first with NaH, then it will react with most alkylating reagents. But most indole will not react quickly with only K2CO3, I did one alkylation with it that took 2 weeks to go to completion.
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Nicodem
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3-5-2014 at 09:30
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[*] posted on 4-5-2014 at 13:25


This is a little off topic but is GAA glacial acetic acid?



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