The major product formed in the following reaction is

\(\mathrm{H}_{3} \mathrm{C}-\mathrm{C} \equiv\mathrm{C}-\mathrm{CH}_{3}+\mathrm{Na} \xrightarrow{\text { Liquid } \mathrm{NH}_{3}}\)

CONCEPT:

Reduction of Alkynes using Sodium in Liquid Ammonia

• When an alkyne is treated with sodium (Na) in liquid ammonia (NH₃), it undergoes reduction to form a trans-alkene.
• This reaction is also known as the Birch reduction.
• The reduction occurs via a radical anion intermediate, leading to the formation of a trans-alkene as the major product.

EXPLANATION:

• In the given reaction:

  H₃C-C≡C-CH₃ + Na → (Liquid NH₃)

  • The alkyne (H₃C-C≡C-CH₃) is reduced by sodium in liquid ammonia.
  • During the Birch reduction, sodium donates an electron to the alkyne, forming a radical anion.
  • Protonation of this radical anion by ammonia results in the formation of a trans-alkene.
• The major product formed is:

  H₃C-CH=CH-CH₃

Other Options:

• Option 1: H₃C-CH₂-C=CH^- Na⁺
  • This is not the major product as it represents an intermediate rather than the final reduced product.
• Option 2: H₃C-C≡C-CH₂^- Na⁺
  • This is also not the major product as it indicates an incomplete reduction where the alkyne is still present.

Therefore, the correct answer is option 4: H₃C-CH=CH-CH₃.