Friedel-Crafts formylation?

[Rhodium]

React a halobenzene grignard with ethylene oxide to get the phenethyl alcohol. This can either be oxidized to the phenylacetaldehyde and be reductively aminated to a PEA, or you could make the bromide (with PBr3) or the tosylate (with tosyl chloride) and react that with sodium azide followed by NaBH4 (the latter two PTC catalyzed) to get the PEA.

The general procedure for bromide/tosylate swap to the azide to the amine can be found at:

https://www.thevespiary.org/rhodium/Rhodium/chemistry/mda.azide.html

[bottleneck]

Psyloxy:

>Nope. The amine is too acidic for the R-MgX. But there are some less reactive organometallics that could do it. Zn I think ... I'll look it up.

Thank you.

Would it be best to make the reagent of 2-chloroethylamine or of the halobenzene?

Would it be better to use a chlorobenzene or a bromobenzene?

Rhodium:

>React a halobenzene grignard with ethylene oxide to get the phenethyl alcohol. This can either be oxidized to the phenylacetaldehyde and be reductively aminated to a PEA,

Great! I guess you can do a lot of interesting things with these types of reagents. I also like the ultrasonic bath thing very much.


Thank you.

[Rhodium]

Bromobenzenes are way better than chlorobenzenes when it comes to formation of grignard reagents.

[bottleneck]

Okay, thank you.

Are there other ways to substitute an aromatic halogen than these organometallic reactions?


Thank you.

[Rhodium]

What do you want to substitute the halogen with?

[bottleneck]

Anything that could eventually be extended to aminoethane.

I thought the halogen on the benzene would be a nice "target" for other reagents, but I can't find a lot of reactions where the halogen is substituted, except Grignard reactions, making phenols, making sulfonic acids, etc.

I am looking for a way in which one could use 1-chloro-2,5-dimethoxybenzene as a precursor.


Thanks.

[psyloxy]

Had a look in that book on organometallics again and, well, there's a lot left to
explore and no really hard facts.

It might work like this:

Br-Me-CN + Zn --> BrZn-Me-CN

BrZn-Me-CN + CuCN*2LiCl --> BrZn[Cu(CN)-Me-CN]

BrZn[Cu(CN)-Me-CN] + (MeO)₂Ph-Br --> (MeO)₂Ph-Me-CN

but you better forget about this and try to find some useful info on
higher order cuprates and organo-zinc reagents on the net yourself.
Sorry for the disappointment.

--psyloxy--

[bottleneck]

It's okay.

Thank you very much.