Question

Question: Explain the apparent paradox. Although the addition of one equivalent of HX to an alkyne is more exothermic than the addition of HX to an alkene, an alkene reacts faster with HX.

Short answer

Answer

• Reaction rate (which is determined by${\mathrm{E}}_{\mathrm{a}}$) and enthalpy $\left(∆\mathrm{H}°\right)$are not related.
• Reactions that are more exothermic are not necessarily faster. Since the addition of HX to an alkene forms a more stable carbocation in an endothermic, rate-determining step, this carbocation is formed faster by the Hammond postulate.

Step-by-step solution

Step-by-Step Solution Step 1: Hammond postulate

• The transition state is not a normal covalent molecule as some bonds are partially broken, whereas others are only partially formed.

• In any individual step in a reaction, the geometry and structure of the transition state resemble either the reactants or the products or the intermediate, depending on which of these is closer in free energy to the transition state.

• Stable carbocation gives a faster reaction, and less stable carbocation gives a slow reaction.

 Step 2: Reaction mechanism

a. The sp- hybridized carbocation is less stable, and the reaction is slow.

Reaction a

b. The ${\mathrm{sp}}^2$hybridized carbocation is more stable, and the reaction is a faster one.

Reaction b