Animated Solution for Chemistry - Haloalkanes & Haloarenes: The major product (A) formed in the reaction given below is

Substitution vs. Elimination

In organic chemistry, whenever an alkyl halide meets a strong base that is also a good nucleophile, a fierce battle ensues. The two primary combatants are the $S_{N}2$ (Bimolecular Nucleophilic Substitution) pathway and the $E2$ (Bimolecular Elimination) pathway.

To predict the winner, we must act as molecular detectives, carefully analyzing both the structure of the substrate and the nature of the reagent. Let's dive into the specifics of this fascinating reaction.

Our substrate is 1-bromo-2-(4-methoxyphenyl)butane. At first glance, the leaving group (the bromide ion, ${\text{Br}}^{\ominus }$) is attached to a primary ($1^{\circ }$) carbon.

Normally, primary alkyl halides are highly unhindered, making them prime targets for a backside attack by a nucleophile. This structural feature strongly screams, "$S_{N}2$ substitution!"

Our reagent is sodium methoxide (${\text{CH}}_3\text{ONa}$) dissolved in methanol. The methoxide ion (${\text{CH}}_3{\text{O}}^{\ominus }$) is small, highly reactive, and carries a full negative charge. It is both a powerful nucleophile and a strong base. Because it is a strong base, it is fully capable of ripping a proton off the molecule to trigger an $E2$ elimination.

If we only looked at the primary nature of the halide, we might incorrectly assume that the methoxide ion would simply attack the alpha-carbon, kick out the bromine, and form an ether (Option A or D).

However, we must look at the beta-carbon—the carbon directly adjacent to the one holding the bromine. This beta-carbon is not just any ordinary carbon; it is directly attached to a benzene ring! This makes it a benzylic carbon.

Protons attached to a benzylic carbon are unusually acidic. Why? Because the transition state formed during their removal is stabilized by the adjacent aromatic ring. The strong methoxide base "sees" this highly acidic proton and finds it much easier to abstract it rather than navigating to the alpha-carbon for a substitution.

Once the methoxide ion targets the benzylic beta-hydrogen, the $E2$ mechanism kicks into high gear. This is a concerted process, meaning everything happens simultaneously in one smooth, elegant motion:

1. The oxygen of the methoxide ion donates its electron pair to the beta-hydrogen, beginning to form an $\text{O-H}$ bond. 2. As the $\text{C-H}$ bond breaks, the electrons from that bond swing down toward the alpha-carbon. 3. These electrons form a new $\pi$-bond (a carbon-carbon double bond). 4. The formation of the double bond forces the leaving group, the bromide ion (${\text{Br}}^{\ominus }$), to depart with its bonding electrons.

Conjugation

The ultimate reason $E2$ dominates over $S_{N}2$ in this specific molecule lies in the stability of the final product.

The newly formed carbon-carbon double bond is located directly adjacent to the benzene ring. This creates an extended conjugated $\pi$-system. The electrons in the double bond can delocalize and resonate with the $\pi$-electrons of the aromatic ring.

In chemistry, conjugation equals massive thermodynamic stability. Nature always favors the path of least resistance leading to the most stable outcome. The immense stability gained from forming a conjugated alkene drastically lowers the activation energy for the $E2$ pathway, allowing it to completely outcompete the $S_{N}2$ pathway.

Therefore, the major product is the conjugated alkene, 2-(4-methoxyphenyl)but-1-ene, which perfectly matches Option (b). The ether formed via substitution will only be a minor byproduct.