Animated Solution for Chemistry - Organic Chemistry: The major product $Y$ in the following reaction is

Let's begin by carefully analyzing our starting material. We are given 2-chloro-3-methylbutane, which is a secondary alkyl halide. The reaction conditions specify the use of sodium ethoxide ($\text{EtONa}$) and heat ($\Delta$).

What does this combination scream to you? Yes, a strong, bulky base coupled with heat is the classic recipe for an E2 elimination reaction. The heat provides the necessary activation energy to favor the entropically driven elimination pathway over a simple $S_{N}2$ substitution.

E2 Elimination

In an E2 elimination, the base needs to abstract a proton from a $\beta$-carbon. The carbon directly attached to the halogen is our $\alpha$-carbon. The carbons adjacent to it are the $\beta$-carbons.

If we look at our substrate, we have two distinct $\beta$-carbons, meaning we have two different types of $\beta$-hydrogens available for elimination.

If the base abstracts the ${\beta }_1$ hydrogen (from the $\text{CH}$ group), we form a double bond in the middle of the chain. If it abstracts the ${\beta }_2$ hydrogen (from the ${\text{CH}}_3$ group), the double bond forms at the terminal end.

According to Saytzeff's rule, the more substituted alkene is the major product because it is thermodynamically more stable due to greater hyperconjugation. Therefore, the base will preferentially attack the ${\beta }_1$ hydrogen. The ethoxide ion grabs the proton, the $\text{C-H}$ bond electrons swing down to form a $\pi$ bond, and the chloride ion is kicked out. This synchronized dance gives us our major intermediate $X$, which is 2-methylbut-2-ene.

Electrophilic Addition

Now we have our intermediate $X$. The second part of the reaction treats this alkene with hydrogen bromide ($\text{HBr}$). This is a classic electrophilic addition reaction. The electron-rich $\pi$ bond of the alkene acts as a nucleophile and attacks the electrophilic hydrogen of $\text{HBr}$.

When the $\pi$ bond attacks the proton, we have a choice. Which carbon gets the hydrogen, and which carbon becomes the positive carbocation? Markovnikov's rule dictates that the electrophile adds in a way that generates the most stable carbocation intermediate.

If the hydrogen attaches to the less substituted carbon on the right, we form a tertiary ($3^{\circ }$) carbocation on the left. A tertiary carbocation is highly stabilized by hyperconjugation and the inductive effect ($+I$) of three alkyl groups. This is vastly more stable than the alternative secondary ($2^{\circ }$) carbocation.

Now that we have our stable tertiary carbocation, the final step is fast and easy. The bromide ion (${\text{Br}}^{-}$), which is a good nucleophile, sees that positive charge and attacks it directly.

This forms our final carbon-bromine bond, yielding the final product $Y$: 2-bromo-2-methylbutane. By carefully following the rules of elimination and addition, we've successfully navigated this two-step synthesis. The correct option is (b).