Organic:entropy

The elimination reaction in the second example can be either E₂ or E₁. If it is the former, then the following scenario would apply:

Scenario 1: The role of Entropy for Reactions of different molecularity

The fundemantal equations for understanding rates of reactions are

1. ΔG^‡ = ΔH^‡ - TΔS^‡
2. ΔG^‡ = -RT Ln k

In equation 2, k is the rate constant for the reaction. From this, you can see that the rate of a reaction depends on the entropy of activation for that reaction, ΔS^‡.

For a unimolecular reaction of the type:

A ⇒ B + C

there are more degrees of freedom on the rhs than the lhs (the two molecules B and C can translate and rotate with respect to each other freely). ΔS^‡ for such a reaction is therefore +ve. Plugged into equation 1, this reduces ΔG^‡.

For an alternative bimolecular reaction invoking a base of this type;

A + Base ⇒ BaseH + D

there are more or less an equal number of degrees of freedom on both sides of this equation, and hence ΔS^‡ for such a reaction is therefore close to zero. ΔG^‡ in equation 1 is therefore not reduced. If ΔH^‡ is the same for both reactions, then the unimolecular form will always be much faster than the bimolecular form (by a factor of 10⁵ or so).

If the reaction takes the form

A-OTs + Base ⇒ BaseH⁺ + D + TsO^-

the degrees of freedom are again increased on the rhs, with again a +ve entropy of reaction. In these circumstances, the bimolecular reaction will not be at an entropic disadvantage compared to the unimolecular variation.

Scenario 2: Neighbouring Group Participation in Reactions of similar molecularity

If the elimination reaction proceeds by initial loss of the Tosyl group (E₁) then the rate of the reaction will not depend on the concentration of base (or indeed the basicity of BH₄), and entropic differentiation between the two reactions cannot be sustained. The question then becomes whether the expulsion of the Tosyl group is assisted in any way by another group present in the same molecule. It is at this point that the alignment of the nitrogen lone pair with respect to the cleaving C-OTs bond is crucial. If the two diastereoisomers differ in this alignment, then the unimolecular rate constant k will differ for the two reactions. A good antiperiplanar alignment will increase the rate constant k (reduce ΔG^‡ by reducing ΔH^‡) with respect to a less efficient alignment.