Hello,
I have a reaction question about allyl bromide with an aldehyde using an indium intermediate. I have been
for a while.
What is the product from:
3-bromopropene + o-chlorobenzaldehyde ---Indium, H2O----> Product
Thanks
I'll start with some background.
The reason you are using Indium & water together is due to the avoidance of performing the experiment in an inert atmosphere. Back in the day people used Indium with the aprotic solvent DMF for indium-mediated allyl reactions. With the use of water (instead of DMF), no inert atmosphere or other dry solvents were required (which is why "probably" you are performing the experiment in an open atmosphere).
Indium is in fact a good choice for this experiment because it has a lower ionization potential and energy than most metals used in organic chemistry experiments such as magnesium, zinc, etc. It is also unaffected by the boiling point of water during the experiment, since indium's boiling point is around 2000 degrees C.
With the use of indium, you form a transient organoindium intermediate which drives the reaction further.
As an end product, alkene will still be present and the carbonyl of the benzaldehyde will turn into an alcohol. Keep in mind that alkene carbons do not attack the carbonyl carbon. In 3-bromopropene, carbon #3 is attached to Br. In the product, that carbon #3
will make a new bond with the carbonyl carbon of the benzaldehyde, and you will end up with an alcohol.
So the product will have both an alcohol and an alkene. Br goes somewhere, but I don't know where.
I'm
assuming indium makes a bond with Br to make a InBr-propene...I got this assumption from the fact that Mg makes a bond with a halide in a Grignard reaction to make a MgX-R to make the R group act as a nucleophile and MgX is released, where X is a halide.
Hope that helped a little.
