Sciencemadness Discussion Board

FC-alkylation Positioning

Maui3 - 26-12-2024 at 08:36

I am gonna do a FC-alkylation soon with ethyl bromide, 1,3-benzodioxole (1,2-methylenedioxybenzene) and AlCl3 as catalyst. I am a bit confused as to what position on the benzodioxole that the ethyl-group is introduced. Also, I read somewhere that for the FC-alkylation of toluene the positioning of the ethyl group can depend on the temperature of the reaction.

It might be my bad searching-skills, but I also couldn't find a lot of info on this - so I'd appreciate if somebody could help clarify it for me.

Thanks.

Keras - 26-12-2024 at 08:51

FC is a nucleophilic reaction, it forms a carbanion which will attack on the most electron rich carbon of the aromatic cycle. With 1,3-benzodioxole there is no obvious such point, because oxygens somehow cancel each other influence – each free carbon is both favoured and unfavoured. Because of steric hindrance, I would say that attack at ortho- positions w/r to any of the oxygens is unlikely, so that leaves you with the para/meta positions. Also, for reasons of steric hindrance, it is unlikely that you have more than one alkylation.

My tuppence, though.

DraconicAcid - 26-12-2024 at 10:05

FC is electrophilic addition, and will form a carbocation that will add to the most electron-rich carbon of the aromatic ring.

Maui3 - 27-12-2024 at 03:11

Thank you for the clarification! But this still depends on temperature, right?

Keras - 27-12-2024 at 03:32

Not that I am aware of. But you may be right, increasing the temperature might lead to a higher level of multiple alkylation.

Dr.Bob - 27-12-2024 at 08:08

Lower temp can favor the monoreaction in FC as well as pushing the reaction mostly to the lowest transition state product. I have had to cool some reactions to below 0C to get primarily one product over another possible one, but then the reaction took 2 days. What fun that was.

Keras - 28-12-2024 at 00:13

Although the primary means of reducing multiple alkylation is to use a large excess of the aromatic compound, in order to ensure that the chances that the alkyl moiety has to encounter an already alkylated molecule are low.

Dr.Bob - 28-12-2024 at 08:11

In our case, the aromatic was the rare/expensive part, so we had to go the other way, which makes everything tougher. But your point is often the easiest way. I have done that as well, but not much, sadly, as much easier.