Animated Solution for Chemistry - Organic Chemistry: In the following reaction, the major product is -

KCP vs TCP

Welcome to a classic organic chemistry problem that tests your understanding of reaction mechanisms, resonance, and the delicate balance between kinetic and thermodynamic control. Let's dive into the fascinating world of conjugated dienes!

Our reactant is $2\text{-methylbuta-1,3-diene}$, commonly known as isoprene. It's a conjugated diene, meaning it has two double bonds separated by a single bond. This conjugation allows the $\pi$ electrons to delocalize, which profoundly impacts how the molecule reacts.

We are treating this diene with $1$ equivalent of $\text{HBr}$. This is an electrophilic addition reaction. Because we only have one equivalent of the reagent, it will only add across one of the double bonds, leaving the other intact. The big question is: where does it add, and why?

Protonation and Carbocation Stability

The first step is the attack of the electrophile, the ${\text{H}}^{\oplus }$ ion from $\text{HBr}$. The rule of thumb in electrophilic addition is to protonate the double bond in a way that forms the most stable carbocation intermediate.

If we protonate the left double bond (${\text{C}}_1={\text{C}}_2$) at the terminal carbon (${\text{C}}_1$), we generate a tertiary allylic carbocation at ${\text{C}}_2$. This intermediate is exceptionally stable because it benefits from both the inductive effect of the methyl group and resonance stabilization from the adjacent double bond.

The Allylic Carbocation

Because the carbocation is allylic, the $\pi$ electrons from the neighboring double bond can shift over to stabilize the positive charge. This delocalization creates a resonance structure where the positive charge is now located on the primary terminal carbon (${\text{C}}_4$).

Thus, our true intermediate is a resonance hybrid. The positive charge is shared between the tertiary carbon and the primary carbon. This means the incoming nucleophile has two potential targets!

Now, the bromide ion (${\text{Br}}^{\ominus }$) enters the scene. It can attack either of the positively charged carbons:

1. Attack at the tertiary carbon: This yields the $1,2\text{-addition}$ product, where the $\text{H}$ and $\text{Br}$ have added to adjacent carbons. 2. Attack at the primary carbon: This yields the $1,4\text{-addition}$ product, where the $\text{H}$ and $\text{Br}$ have added to the ends of the conjugated system, and a new double bond has formed in the middle.

Which product dominates? This is where temperature becomes the deciding factor.

At low temperatures (e.g., $-80^{\circ }\text{C}$), the reaction is under kinetic control. The bromide ion attacks the tertiary carbon faster because it bears a larger partial positive charge in the resonance hybrid. The $1,2\text{-addition}$ product forms more rapidly and is the major product, known as the Kinetically Controlled Product (KCP).

However, at room temperature or higher, the reaction is under thermodynamic control. The addition is reversible, allowing the system to equilibrate and favor the most stable product. According to Zaitsev's rule, the stability of an alkene increases with the number of alkyl substituents on the double bond. The $1,4\text{-addition}$ product has a trisubstituted double bond, whereas the $1,2\text{-addition}$ product only has a monosubstituted double bond. Therefore, the $1,4\text{-addition}$ product is more stable and becomes the major product, known as the Thermodynamically Controlled Product (TCP).

Since the problem implies standard conditions (room temperature), we must select the TCP. The $1,4\text{-addition}$ product is $1\text{-bromo-3-methylbut-2-ene}$, which perfectly matches Option (D).