Animated Solution for Chemistry - Organic Chemistry: The major product formed in the following reaction is $C{H}_{3}CH=CHCH(C{H}_3{)}_2\stackrel{HBr}{\to }$

Imagine you are observing a molecular battlefield. We start with our reactant, 4-methylpent-2-ene, a seemingly peaceful alkene. But the moment we introduce hydrogen bromide ($HBr$), the peace is shattered.

The $\pi$ bond of the alkene is an electron-rich region, essentially a cloud of negative charge. It acts as a magnet for anything seeking electrons.

The $HBr$ molecule dissociates, releasing a hungry $H^{+}$ ion (an electrophile). This proton aggressively attacks the $\pi$ bond, breaking it open.

Now, the molecule faces a critical decision. When the $\pi$ bond breaks and the proton attaches to one of the carbons, the other carbon is left electron-deficient, forming a carbocation.

According to Markovnikov's Rule, the proton will attach itself to the carbon that results in the most stable carbocation possible.

If it attaches to the third carbon, we get a $2^{\circ }$ carbocation:

This $2^{\circ }$ carbocation is reasonably stable due to hyperconjugation from adjacent alpha hydrogens. But in the world of organic chemistry, molecules are always striving for maximum stability.

I know this intermediate looks fine, but let's take a breath and look closer. There is a catch here!

Right next to our positively charged carbon is a tertiary carbon holding a hydrogen atom. What if this hydrogen, along with its bonding electrons, simply migrated over?

This is the beautiful phenomenon known as a $1,2$-hydride shift.

By shifting the hydride ion ($H^{-}$) to the adjacent carbon, the positive charge moves to the tertiary carbon.

Suddenly, we have a $3^{\circ }$ carbocation! This is a massive upgrade in stability because the positive charge is now stabilized by even more hyperconjugation from the surrounding methyl groups. The molecule has found its thermodynamic sweet spot.

Now that our highly stable $3^{\circ }$ carbocation is ready and waiting, the final act of the reaction unfolds.

The bromide ion ($B{r}^{-}$), which has been waiting patiently in the solution, sees this concentrated positive charge. Acting as a nucleophile, it swoops in and attacks the carbocation.

The bromine atom forms a strong covalent bond with the tertiary carbon, neutralizing the charge and completing the reaction.

We have successfully navigated through electrophilic addition and carbocation rearrangement to find our major product. The correct option is indeed (d).