The major product (A) formed in the reaction given below is:

The given substrate is 1-bromo-2-phenylbutane, in which the bromine atom is attached to a primary carbon.

Although primary alkyl halides generally undergo $$S_N2$$ substitution with strong nucleophiles, the adjacent $$\beta$$-carbon in this molecule is highly substituted, bearing both a phenyl group and an ethyl group. This creates significant steric hindrance around the reaction center.

The reagent used is methoxide ion $$\mathrm{CH_3O^-}$$ in methanol, which acts as both a strong nucleophile and a strong base.

Because backside attack required for the $$S_N2$$ mechanism is hindered by the bulky substituents on the adjacent carbon, substitution becomes unfavorable. Instead, the methoxide ion abstracts the $$\beta$$-hydrogen, leading to an $$E2$$ elimination.

During the elimination process:

• The base removes the hydrogen atom from the $$\beta$$-carbon.
• Simultaneously, the bromide ion leaves from the $$\alpha$$-carbon.
• A double bond is formed between the $$\alpha$$- and $$\beta$$-carbons.

The resulting product is

$$\mathrm{CH_3CH_2C(Ph)=CH_2}$$

This alkene is especially stable because the newly formed double bond is conjugated with the phenyl ring, allowing resonance stabilization of the $$\pi$$-system. The additional stability of the conjugated alkene strongly favors elimination over substitution.

Hence, the major product is 2-phenylbut-1-ene,

$$\mathrm{CH_3CH_2C(Ph)=CH_2}$$

Therefore, the correct answer is Option (A).