Test: Alkynes & Aromatic hydrocarbon (October 3) - NEET MCQ

The correct answer is option C

• In ethyne (C₂H₂), each carbon is sp-hybridized.
• The C≡C σ-bond is formed by the overlap of two sp orbitals (one from each carbon) — that’s sp–sp overlap.
• The two π-bonds arise from the side-on overlap of unhybridized p orbitals (p–p).
• Each C–H σ-bond is formed by sp–s overlap with the hydrogen 1s orbital.

Because vinyl bromide is less reactive and Alc. KoH is not that strong a base; we need a strong base like NaNH₂ .

• Acid–base reaction: NaH contains H⁻, which pulls off the acidic hydrogen from the terminal alkyne, forming the acetylide anion and liberating H₂ gas.

• Visible effervescence: The H₂ gas bubbles out immediately, giving a clear fizz.

• Why not others?

  • Na metal also deprotonates but is more reactive and can give side reactions (and is less commonly used for clean H₂ evolution with alkynes).

  • NaNH₂ deprotonates to make the acetylide but produces NH₃ (ammonia), which dissolves rather than effervescing noticeably.

  • CH₃MgBr deprotonates to give CH₄ (methane), not H₂, so no hydrogen gas effervescence occurs.

• Statement I (C₆H₅MgBr + propyne → 1-phenyl propene): Incorrect. Grignard reagents with terminal alkynes undergo acid-base reactions, not direct alkene formation. The product of reacting C6H5MgBr with propyne is benzene and a magnesium acetylide, not 1-phenyl propene directly.
• Statement II (C₆H₅^- is a strong nucleophile): Correct. The phenyl anion (C₆H₅^- ) is a strong nucleophile due to its carbanionic nature.