The major product of the following reaction is:

Step 1: First reaction - KOH alcoholic

Vinyl bromide + alcoholic KOH → Elimination

Starting: $\ce{CH3CH2C(Br)=CH2}$ (vinyl bromide)

Problem: Vinyl halides are very unreactive toward normal elimination.

But with strong base and heat, some elimination can occur.

$\ce{CH3CH2C(Br)=CH2 ->[KOH(alc)] CH3CH2C≡CH + KBr + H2O}$

This forms 1-butyne (terminal alkyne)

Step 2: Second reaction - NaNH₂ in liquid NH₃

Sodamide ($\ce{NaNH2}$) is a very strong base.

With terminal alkyne:

$\ce{CH3CH2C≡CH + NaNH2 -> CH3CH2C≡C-Na+ + NH3}$

Forms sodium acetylide (deprotonation)

But if no electrophile is added, the product after workup is still the alkyne.

Final product: $\ce{CH3CH2C≡CH}$ (1-butyne)

Answer: (d)

Key Reactions:

Vinyl halide elimination:

• Very difficult (requires strong base, heat)
• Forms alkynes
• Vinyl C-X bond is strong (partial double bond character)

NaNH₂ uses:

1. Deprotonation of terminal alkynes (pKa ~25)
2. Elimination from alkyl dihalides to form alkynes
3. Strong base for difficult eliminations

Alkyne acidity:

• Terminal alkynes are weakly acidic (pKa ~25)
• Can be deprotonated by strong bases (NaNH₂, n-BuLi)
• Forms nucleophilic acetylide anions
• Used in alkylation reactions