Animated Solution for Chemistry - Organic Chemistry: Choose the correct statement regarding the formation of carbocations A and B.

Imagine you are observing a microscopic dance between an alkene, specifically but-1-ene (${\text{CH}}_3{\text{CH}}_2\text{CH}={\text{CH}}_2$), and a molecule of hydrogen bromide ($\text{HBr}$). The double bond in the alkene is a region rich in electron density—a perfect target for an electrophile. The ${\text{H}}^{+}$ ion from $\text{HBr}$ approaches this double bond, initiating the first step of an electrophilic addition reaction.

When the $\pi$-bond breaks to grab the proton, the hydrogen atom can attach to either of the two carbons that originally shared the double bond. This choice leads to a fork in the road, creating two distinct possible intermediates: Carbocation A and Carbocation B.

The Power of Hyperconjugation

Let's analyze these two intermediates.

Carbocation A is formed when the proton attaches to the inner carbon, leaving the positive charge on the terminal carbon. This is a primary ($1^{\circ }$) carbocation: ${\text{CH}}_3-{\text{CH}}_2-{\text{CH}}_2-\stackrel{+}{C}{\text{H}}_2$. To determine its stability, we look at the adjacent carbon (the $\alpha$-carbon). It has exactly $2\alpha$-hydrogens available to stabilize the empty p-orbital through hyperconjugation.

Carbocation B, on the other hand, is formed when the proton attaches to the terminal carbon, placing the positive charge on the inner carbon. This creates a secondary ($2^{\circ }$) carbocation: ${\text{CH}}_3-{\text{CH}}_2-\stackrel{+}{C}\text{H}-{\text{CH}}_3$. If we count the $\alpha$-hydrogens here, we find $2$ from the left ${\text{CH}}_2$ group and $3$ from the right ${\text{CH}}_3$ group, giving a total of $5\alpha$-hydrogens.

Because Carbocation B has significantly more $\alpha$-hydrogens ($5$ vs $2$), it experiences much greater stabilization through hyperconjugation. Additionally, it benefits from a stronger +I (inductive) effect from the two surrounding alkyl groups. Therefore, thermodynamically, Carbocation B is much more stable than Carbocation A.

Why Stable Means Faster

Now, we must bridge the gap between thermodynamics (stability) and kinetics (rate of formation). To do this, we rely on Hammond's Postulate, which suggests that the transition state of an endothermic step (like carbocation formation) closely resembles the intermediate it produces.

If we visualize the energy profile diagram, the path leading to the more stable Carbocation B has a lower energy peak. This peak represents the activation energy ($E_a$). Because Carbocation B is lower in energy, its corresponding transition state is also lower in energy compared to that of Carbocation A.

Mathematically, $E_{a,B}<E_{a,A}$. A lower activation energy means that a larger fraction of molecules possess enough kinetic energy to overcome the barrier at any given temperature. Consequently, the reaction proceeds much faster along this pathway.

By combining our understanding of hyperconjugation and reaction kinetics, the conclusion becomes crystal clear. Carbocation B is not only the more stable intermediate, but because of its lower activation energy, it is also formed at a relatively faster rate. This elegant interplay between stability and speed is the fundamental reason why Markovnikov's rule works in organic chemistry.