Proton ejection, hydrogen bonds

When a hemi-orthoamide tetrahedral intermediate exists in the T ionic form, the amide ion is not ejected previous to protonation by the solvent, to give the secondary amine. The formation of an amide ion 24-25 is a process so high in energy, that both the protonation and the ejection processes must be synchronized 24 - 26 - 27 (28). This means that in aqueous solution, the nitrogen electron pair must first be hydrogen bonded with the solvent, so that the group can leave as a secondary amine. 

The dissociation of an acidic residue in water is essentially a downhill proton transfer between the donor and the solvent, and in the case of a large value for ApfCa, the reaction can be as fast as a barrier-less reaction [35]. To achieve the maximal rate of the reaction, the proton-donor atom must establish a hydrogen bond with the acceptor, and the reaction then takes place along this bond. As the rate of the reaction can be as fast as the hydroxyl vibration frequency, - 60-150 fs [36, 37], it is implicit that, for a fast reaction, the hydrogen bond must be established before the proton transfer event. In the case of proton transfer to water, pre-orientation of a few water molecules is required to stabilize the ejected proton and the conjugated base from which the proton was released. Accordingly, reduction in the availability of free water molecules will reduce the probability of the proton acceptor configurations, and this effect will appear as a reduction in the rate of dissociation. 

At the other extreme of elimination mechanisms is a concerted process. In an E2 reaction (here illustrated by the reaction of 2-bromobutane with sodium ethoxide) proton transfer to the base, formation of the carbon-carbon double bond, and ejection of the bromide ion occur simultaneously all bond-breaking and bond-forming steps are concerted. Because the base removes a j8-hydrogen at the same time that the C— Br bond is broken to form a halide ion, the transition state has considerable double-bond character (Figure 9.7). 

STEP 2 Take a proton away and break a bond to form a stable molecule. Water acts as a base to abstract an a-hydrogen and eject H2O as a leaving group, giving the a,/3-unsaturated carbonyl compound and regenerating the acid. 

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