# Rate of substitution and elimination reactions at high temperatures

From High School, I've been reading that elimination reactions are favorable at high temperatures because their entropy change is positive, so at higher temperature the Gibbs free energy will be more negative so the equilibrium will be more towards product's side.

But I've encountered a different statement in Clayden's Organic Chemistry:

This explanation is simplified because what matters is the rate of the reaction, not the stability of the products. A detailed discussion is beyond the scope of the book, but the general argument still holds.

So what is the reason behind this? Why are substitution reactions not that much faster at high temperature than elimination reactions?

• What makes you think they aren't? – Zhe Nov 6 '16 at 13:29
• When you are discussing the kinetics of a reaction, what is central to the discussion $\Delta G$ or $\Delta G^ \dagger$ ? – getafix Nov 6 '16 at 14:16
• As Far as I know- $\Delta G^ \dagger$ determines the rate of reaction . And It is used to be positive otherwise the transition state will be more stable . More It is Positive the more reaction gets slows down . – Aditya Shrivastav Nov 10 '16 at 15:20

For the purpose of answering this question, let us consider $S_N2$ and $E_2$ type reactions.
For $S_N2$ type reactions, the HOMO of the nucleophile has to be oriented with the LUMO of the electrophile. Here, a linear orientation is required. So, the entropic requirement is that the nucleophile should approach the electrophile in a certain way.
For $E2$ type reactions, the base attacks the $\alpha$-proton, which finally leads to the elimination. So, the HOMO of the base interacts with the $\sigma^*$ orbital of the $C-H$ bond, and then the $\sigma$ orbital interacts with the $\sigma^*$ orbital of the leaving group. So here the entropic requirement is the linear orientation of the base, the proton and the carbon it is attached to, and the leaving group should be antiperiplanar with respect to the proton.