Hello,

 

This is the first reaction in the synthesis of C60.

 

As you can see from the attached picture the starting molecule is benzene with a chlorine attached and a bromine para to it. After a grignard reaction with Mg and MeCHO(Acetaldehyde) you get a chlorine attached and a methyl and alcohol para to it. Can someone please confirm that my thinking below is correct on how I think the initial carbonyl is reduced to an alcohol.

 

1. The bromine which was kicked of by the Grignard reagent now carries a negative charge and attacks the carbon double bonded to the oxygen. The 2 elections from one of the double bonds goes onto the oxygen

 

2. The oxygen now attacks the adjacent hydrogen and is protonated. The bond which was attached to it collapses and kicks of the bromine?

 

I might be completely wrong!

 

 

Many thanks.

 

Nick.

The mechanism isn't fully agreed upon I don't think, but you're more or less correct. I would add that the bromine isn't kicked out by the Grignard reagent, though. It forms part of the Grignard reagent following oxidative insertion of the magnesium into the C-Br bond. The end result is the same in that it still gives you what is essentially a source of carbanions. As well, the protonation step generally occurs during quenching of the reaction with acid.

Thank you for your answer it really helped. I'm glad I found this forum.