Animated Solution for Chemistry - Organic Chemistry: Reaction of one molecule of $\text{HBr}$ with one molecule of 1, 3-butadiene at $40^{\circ }\text{C}$ gives predominantly

Stability: Kinetic vs. Thermodynamic Control

Imagine you are standing at a crossroads. One path is a quick, easy downhill stroll that leads to a small, pleasant valley. The other path requires a grueling hike over a steep mountain pass, but it eventually drops you into a massive, lush, and incredibly stable paradise. Which path do you choose?

In the world of organic chemistry, molecules face this exact dilemma every day. This is the essence of Kinetic vs. Thermodynamic Control, and it perfectly explains the reaction between 1,3-butadiene and hydrogen bromide ($\text{HBr}$).

Creating the Allylic Carbocation

When 1,3-butadiene (${\text{CH}}_2=\text{CH}-\text{CH}={\text{CH}}_2$) encounters $\text{HBr}$, the first step is an electrophilic attack. The proton (${\text{H}}^{+}$) from $\text{HBr}$ seeks out the electron-rich double bond. It specifically attaches to the terminal carbon. Why? Because doing so creates a secondary allylic carbocation:

This intermediate is highly stable because the positive charge isn't isolated; it is delocalized through resonance. The pi electrons from the adjacent double bond can shift over, moving the positive charge to the primary carbon:

Now, the remaining bromide ion (${\text{Br}}^{-}$) has a choice. It can attack the secondary carbon (forming the 1,2-adduct) or the primary carbon (forming the 1,4-adduct).

If the bromide attacks the secondary carbon, it forms 3-bromo-1-butene. This is the 1,2-addition product. Because the positive charge is more concentrated on the secondary carbon, this attack happens very quickly. It has a low activation energy ($E_a$). We call this the Kinetic Product—it wins the race of speed.

If the bromide attacks the primary carbon, it forms 1-bromo-2-butene. This is the 1,4-addition product. This pathway has a higher activation energy barrier. However, look at the final product: the double bond is now internal (disubstituted) rather than terminal (monosubstituted). According to Zaitsev's rule, a more substituted alkene is more stable. We call this the Thermodynamic Product—it wins the prize for ultimate stability.

Temperature

So, which product actually forms? It depends entirely on the temperature.

At low temperatures (like $-80^{\circ }\text{C}$), the molecules are sluggish. They don't have enough thermal energy to climb the high activation energy barrier required for the 1,4-addition. They take the easy path. The reaction is irreversible, and the kinetic product (1,2-adduct) dominates.

But at high temperatures (like $40^{\circ }\text{C}$), the molecules are energized. They have plenty of energy to cross both barriers. More importantly, the reaction becomes reversible. Even if the 1,2-adduct forms quickly, it can revert back to the carbocation and eventually fall into the deeper energy well of the 1,4-adduct. Over time, the system settles into the most stable state. The reaction is under thermodynamic control, and the 1,4-adduct dominates.

In our specific problem, the reaction occurs at $40^{\circ }\text{C}$. This is considered a high temperature for this reaction. Therefore, the reaction is under thermodynamically controlled conditions.

The major product will be the 1,4-adduct, which is 1-bromo-2-butene.

This perfectly matches option (c): 1-bromo-2-butene under thermodynamically controlled conditions.