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NaBH4 reduction of harmaline to THH

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150 mg harmaline (214.268 g/mol) 1 eq. (0.7 mmol)
80 mg NaBH4 (37.83 g/mol) 3 eq. (2.1 mmol)
20 mL MeOH

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150mg of purified harmalas (harmaline enriched)

[image1]

dissolved in 20mL MeOH, dissolved and set up with mag stirring

[image2]

added 80mg of NaBH4, which evolved vigorous amounts of H2, and an almost instantaneous colour change to a pale yellow, almost colourless solution.

[image3]

TLC (40:10:1) MeOH:EtOAc:10%aqNH3 Left side is starting material harmalas (bottom spot is harmaline, top is harmine) and the right side is post reaction.

After 5 min indicated complete reduction of the harmaline... The spot was colourless compared to the yellow spot of the starting harmalas...

When viewed under 365nm UV light, THH is visible as a weakly fluorescent dark spot with a rf slightly less than that of harmaline.

[image 4]

Very clean reaction, goes 100% in under a minute. I perfer this to zinc just because its so easy and fast and you don't need to deal with zinc salts or use ammonia.
 

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this is a sweet writup mind. I've got a ton of NaBH4 I havent touched in ages. Maybe I'll try it myself :3
 
arcologist said:
Have you tried doing the same with zinc?


This has already been covered:


NaBH4 has a few advantages:

-Need much less material
-400mg of NaBH4 is plenty to reduce 1g of harmaline, where zinc powder requires many grams of metal to reduce.
-Zinc metal reduction requires a extremely finely powdered zinc metal to be effective. This can be just as hard to obtain as sodium borohydride, unless you live near a pottery shop.


- Is faster.
-Reaction appears to run to completion in seconds.

- Work up is simple and non-messy.
-The THH can be recovered by simply letting the methanol evaporate, quenching any remaining borohydride with a little acetic acid or HCl (just a few dilute drops) and then crashing out the THH with sodium carbonate. Borates are very soluble in water and will stay in solution. No need to worry about a giant gelatinous mass of zinc acetate.
 
I haven't had much luck with zinc and would like to give NaBH4 a try.

Will this still work if you dissolve the harmaline in vinegar and work on harmaline acetate instead of dissolving the freebase in methanol? Just like you do it with zinc?

Is it possible to recover the excess NaBH4 and re-use it in subsequent reductions?

And are there heavy metal contamination concerns with this as is the case with zinc?
 
Jagube said:
I haven't had much luck with zinc and would like to give NaBH4 a try.

Will this still work if you dissolve the harmaline in vinegar and work on harmaline acetate instead of dissolving the freebase in methanol? Just like you do it with zinc?

Is it possible to recover the excess NaBH4 and re-use it in subsequent reductions?

And are there heavy metal contamination concerns with this as is the case with zinc?

NaBH4 reacts in aqueous media, especially acidic water. Vinegar is what I normally use to quench excess NaBH4. Not really possible to recover excess NaBH4, but you don't need to use excess, you could get away with 1.1eq , just .1 eq excess to ensure complete reaction but it's not wasting much at all. With zinc you must uses many times excess 5-50eq ive seen.

You could probably use ethanol instead, but NaBH4 reacts slower in it.
 
Great idea Jagube to try.

Mindlusion, does the methanol (or conversely ethanol) have to be dry?
 
Thanks.

I'm not sure how the 1.1eq excess translates to mg figures, but 1000mg harmaline to 400mg NaBH4 seems cost-effective.

And I still don't see what role the methanol plays, considering that vinegar, too, dissolves harmaline. Is it only preferable because it's faster?

Is 99.85% methanol good enough?

I'm assuming this reduction would work as well on a harmine/harmaline mix, reducing the harmaline and leaving the harmine intact?
 
Jagube said:
...I'm assuming this reduction would work as well on a harmine/harmaline mix, reducing the harmaline and leaving the harmine intact?
The top spot in Mindlusions TLC plate (see photo in that post) is harmine, so it works on a mix and leaves harmine unchanged.
 
1.1eq is Molar equivalent, you can calculate this for yourself. NaBH4 actually has 4 hydride equivalents, but if your trying to get away with wet solvents, you'll most certainly need excess. So one equivalent of NaBH4 is technically 4x excess, but practically speaking it's better to treat NaBH4 as one equivalent, considering only the first hydride that reacts. Even still like this its cost effective.

methanol should be dry, but as ethanol is less reactive you might get away with 96% ethanol, I would dry it to be safe. Drying methanol with MgSO4 works ok, not for ethanol though.
 
Mindlusion said:
-The THH can be recovered by simply letting the methanol evaporate, quenching any remaining borohydride with a little acetic acid or HCl (just a few dilute drops) and then crashing out the THH with sodium carbonate. Borates are very soluble in water and will stay in solution. No need to worry about a giant gelatinous mass of zinc acetate.

I may be able to get some NaBH4 and would like to give it a try, so could you elaborate a bit on how to recover the THH in the end? What exactly do you mean by "quenching"?
 
Dasein said:
Mindlusion said:
-The THH can be recovered by simply letting the methanol evaporate, quenching any remaining borohydride with a little acetic acid or HCl (just a few dilute drops) and then crashing out the THH with sodium carbonate. Borates are very soluble in water and will stay in solution. No need to worry about a giant gelatinous mass of zinc acetate.

I may be able to get some NaBH4 and would like to give it a try, so could you elaborate a bit on how to recover the THH in the end? What exactly do you mean by "quenching"?

Quench is to neutralize leftover borohydride, use dilute acetic acid in water. Also I should mention, On small scale the borates can be dissolved or decanted out, but the borates are not that soluble in water so beware, depending on how much is used.
 
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