# Why is the yield of substitution reactions decreased at higher temperatures?

This is according to my text :

Increasing temperature favours elimination over substitution. This is because energy of activation for elimination is higher than that of substitution since higher degree o charges in bonding is involved in elimination reaction. [sic!]

Well I understand why more elimination occurs at high temperatures (I guess this is due to $\Delta G$ being more negative due to multiple product formation as well), but why do substitution reactions have their yields decreased as temperature goes up?

This is according to the master of organic chemistry; (Even though it says that there is a relative increase, the graph shows a decrease in yield of the substituted products): Aren't they supposed to have their output increased as well, as the activation energy barrier has already passed and rate increases with increase in temperature, or am I missing something fundamental here?

• – Mockingbird Oct 16 '17 at 14:02

In short, this is because both elimination and substitution occur parallely and are competing; amount of elimination product + substitution product = constant. If the elimination rate increases more than the increase in substitution rate, then the ratio of elimination product to substitution product will increase. This is what happens on heating, though the rate of both elimination and substitution increase with temperature.

Broadly, there are two types of reactions: those that are kinetically controlled, and those that are thermodynamically controlled. Most elimination-substitution reactions are exothermic and are kinetically controlled. That means that the proportion of products (elimination or substitution) depends only on the rate of elimination and substitution, and not on the thermodynamic favour for the reaction.

Given that the reaction proceeds to completion, the ratio of elimination product to substitution product is equal to the ratio of elimination rate to substitution rate. Suppose elimination and substitution rates are equal, then the products form in a 50:50 proportion.

The ratio of elimination rate to substitution rate is

$$\frac{e^{\frac{-\Delta G_{e}^{*}}{RT}}}{e^{\frac{-\Delta G_{s}^{*}}{RT}}} = e^{\frac{\Delta G_{s}^{*}-\Delta G_{e}^{*}}{RT}}$$

where $\frac{-\Delta G_{e}^{*}}{RT}$ and $\frac{-\Delta G_{s}^{*}}{RT}$ are the energy barriers for elimination and substitution respectively.

Since the barrier for elimination is more than the barrier for substitution, the above ratio increases with increase in temperature, and hence more elimination product is formed at higher temperatures.

• What do you mean by 'constant'? Are you saying that the net yield of products don't change irrespective of the temperature? I do not understand. – user73157 Sep 30 '16 at 13:30
• @user73157 , The yield does change with the temperature, but only slightly. Suppose you start with 1 mole of reactant & run the reaction at 80 degrees C, suppose the yield is 90% & elim:subst ratio is 50:50. We'll get 0.45 mol elim & 0.45 mole subst product. Suppose we run the reaction at 100 degrees C, the yield may be 85%, but elim:subst ratio may be 80:20. We'll get 0.68 mol elim & 0.17 mol subst product. The yield does change with temperature, but this effect is (usually) much smaller than the difference in amount of products due to difference in their activation barriers. – Ananth Kamath Oct 2 '16 at 5:21
• Thank you for the explanation! I have now understood it. – user73157 Oct 2 '16 at 17:55