Animated Solution for Chemistry - Organic Chemistry: The major product of the following reaction is

Imagine you are looking at a complex molecular architecture. We have a decalin ring system—two fused cyclohexane rings. At one of the bridgehead carbons, there is a methyl group ($-{\text{CH}}_3$). Right next door, at the adjacent carbon, an exocyclic methylene group ($={\text{CH}}_2$) juts out. This is our reactant, and we are introducing it to hydrochloric acid ($\text{HCl}$).

This setup is a classic playground for an electrophilic addition reaction. The double bond is electron-rich, and the $\text{HCl}$ molecule provides a hungry electrophile in the form of a proton (${\text{H}}^{+}$).

The first step is the attack of the $\pi$ electrons from the double bond onto the proton. But which carbon gets the proton? This is where Markovnikov's Rule comes into play. The rule dictates that the electrophile will add to the carbon with the greater number of hydrogen atoms.

In our exocyclic double bond, the terminal carbon has two hydrogens, while the ring carbon has none. Therefore, the proton attaches to the terminal ${\text{CH}}_2$ group, transforming it into a methyl group ($-{\text{CH}}_3$).

As the proton bonds to the terminal carbon, the $\pi$ bond breaks, leaving the ring carbon electron-deficient. This creates a carbocation intermediate.

Now, we must always pause and evaluate the stability of any carbocation we form. Look closely at the positively charged carbon. It is bonded to three other carbons: the bridgehead carbon, the adjacent ring carbon, and the newly formed methyl group. This makes it a tertiary ($3^{\circ }$) carbocation.

Tertiary carbocations are highly stable due to the inductive effect and hyperconjugation from the surrounding alkyl groups. Because it is already in a highly stable state, there is absolutely no driving force for a hydride or alkyl shift. The molecule will not undergo any complex ring expansions or rearrangements.

With a stable tertiary carbocation waiting, the second half of our reagent comes into action. The chloride ion (${\text{Cl}}^{-}$), which was left behind when the proton was taken, acts as a nucleophile.

It swoops in and attacks the positively charged carbon, forming a strong new carbon-chlorine bond. The result is that both the new methyl group and the chlorine atom are attached to the exact same carbon on the decalin ring.

This gives us our final major product: 2-chloro-1,2-dimethyldecalin. When we compare this structure to our given options, it perfectly matches option (d).