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Author: Subject: BF3 + NaBH4 -> BH3 to reduce 3,4 mdns

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[*] posted on 17-9-2014 at 08:57
BF3 + NaBH4 -> BH3 to reduce 3,4 mdns

Since BBr3 is know to cleave methoxy groups, I assume it can also cleave methylenedioxygroups, which leads me to believe that using the BF3 + NaBH4 combo to produce in situ BH3, with the goal of using the BH3 to reduce 3,4-methylenedioxynitrostyrene to 3,4-methylenedioxyphenylethylamine, to not be feasible, as the methylene dioxy bridge would be cleaved resulting in the phenolic product.

Can anyone comment on this speculation? Something that gives me a glimmer of hope that it will work is the Synth. Comm. 15(9), 843-847, 1985 paper, where they reduce a nitro propene that has a 3,4 diethoxynitropropene to the respective alkyl amine. I know there is a difference between the methylenedioxy group and an ethoxy group however; this is where my question comes from.

[Edited on 17-9-2014 by zephler1]
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[*] posted on 17-9-2014 at 11:29

There is a small different between ethoxy groups and methoxy groups but that is not what will determine if the reaction works the way you want it to.

The main difference is that of boron tribromide and boron trifluoride. See the attached mechanism.

The reason this works is that bromide is a good leaving group/nucleophile. Fluorine is not as good of a leaving group/nucleophile. Fluorine will most likely leave the alkoxy groups untouched.

Attachment: demethylation mechanism.tif (98kB)
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