Question

What happens to the More-O'Ferral-Jencks-Plot based on the following scenario? What happens to the extent of depronotation and extend of leaving-group departure at transition state?

1. Replace the base with a much stronger base.

My guess would be that it would shift to th right correct? Because of a stronger base, it'll lean more towards E1cb because deprotonation increases. But the extent of the leaving group doesn't change?

2. Replace the terminal methyl with a phenyl (instead of CH3CH2CH2Br, its now PhCH2CH2Br)

My guess would be it also shifts right, because the electron withdrawing phenyl makes it lead towards E1cb and doesn't change the leaving groups leaving ability.

Please help, thanks!

Answer 1

More-O'Ferral-Jencks-Plot :- This is 2-D representation of those reactions coordinates that involves simultaneous changes in two bonds during reaction mechanism.This plot explain any kind of alteration in the reactant and reaction condition can affect the geometry and position of the transition state of a reaction for which there are competing pathways.These two changes will causes the reaction to follow E1cb mechanism.

Horizontal axis tells us about the extent of deprotonation (C-H bond distance) and vertical axes tells us about the extent of leaving group departure(C-LG distance).

1)Strong base will abstract acidic proton easily that leads to the formation of carboanion,so it will follow E1cb mechanism.

2) Phenyl group will favour E1cb mechanism.Leaving group in this compound is bromide ion which is poor leaving group and it will favour E1cb mechanism.

Leaving group ability is the rate at which a reaction takes place.Reactant with good leaving group have fast reaction (low activation energy) and leads to the stable transition state.