[faust]

Hello.

I have a little question about physical organic chemistry.

In an exercice, they say : CH3-CH(Br)-COOMe will do a really quick SN2 with a nucleophile because the pi star orbital from the carbonyle is an electrowithdrawing group.

But I don't really understand... The pi star is conjugated with the sigma C-Br and with the forming Nu-C bond. And how the electrowithdrawing group can influence the nucleophilic attack?

Thank you very much

[movies]

I agree with you, that argument has always bugged me too because you would think that an EWG next to the halide would make an S_N2 transition state less favorable. 

The way that I have rationalized this observation to myself has been to assume that the C bearing the Br has a partial positive charge and the C of the C=O also has a positive charge.  Since these two charges conflict, the result is that you have a higher energy starting material which is therefore more reactive.  So, the assumption would be that the transition state for the two variants (regular alkyl-Br and alpha-bromo carbonyl) would be comparable in energy, but in the latter case you start out at a higher E and therefore the activation energy is lowered.

You could also make a slightly hand-wavy argument about the orbitals: the pi* and the C-Br sigma* could be aligned and have the right symmetry for some overlap of these two vacant orbitals, thereby lowering the energy of anti-bonding MO that gets attacked.  I kinda like that explanation, but it seems a little convoluted.

[bluemonster]

Can give me the conclusion !?
In this case, does reaction follows as SN2 at Carbon bearing bromide, or nucleophilic addition at Carbon of carbonyl group (apply Felkin – Anh model) ?
Please support !
Thanks pro !