A chemist wants to deprotonate a terminal alkyne to generate a nucleophilic alkynide anion for a carbon-carbon bond forming reaction. Which reagent is appropriate?
ANaOH in aqueous solution (conjugate acid pKa ≈ 16)
BNaHCO₃ in aqueous solution (conjugate acid pKa ≈ 10)
CNaNH₂ in THF (conjugate acid pKa ≈ 38)
DAcetic acid (pKa ≈ 5)
To deprotonate a terminal alkyne (pKa ≈ 25), the base must have a conjugate acid with a pKa *higher* than 25 — the base's conjugate acid must be weaker than the terminal alkyne's conjugate acid. NaNH₂ has a conjugate acid (NH₃) with pKa ≈ 38, making it sufficiently basic. NaOH (conjugate acid pKa ≈ 16) and NaHCO₃ (pKa ≈ 10) are far too weak. Acetic acid is an acid, not a base. This is a classic misconception: terminal alkynes feel 'relatively acidic' compared to alkanes, but pKa 25 still requires strong bases like NaNH₂ or n-BuLi — not aqueous hydroxide.
Question 2 Multiple Choice
Why is the C–H bond on a terminal alkyne (pKa ≈ 25) more acidic than a vinyl C–H bond on an alkene (pKa ≈ 44)?
AThe triple bond withdraws electron density through resonance, weakening the C–H bond inductively
BThe sp carbon of the alkyne has 50% s-character, stabilizing the negative charge on the alkynide anion better than the 33% s-character of an sp² carbon
CThe linear geometry of the alkyne reduces steric hindrance around the acidic proton
DThe two pi bonds in the triple bond donate electron density to the sigma framework, polarizing the C–H bond
The key is hybridization and s-character. An sp orbital has 50% s-character; sp² has 33%; sp³ has 25%. S orbitals hold electrons closer to the nucleus at lower energy. The greater the s-character of the orbital holding the lone pair on the carbanion (alkynide anion), the better it stabilizes the negative charge. Therefore, sp (50%) > sp² (33%) > sp³ (25%) in carbanion stability, giving the acidity order: terminal alkyne > vinyl C–H > alkane C–H. This is a hybridization effect, not resonance.
Question 3 True / False
Treating an internal alkyne with sodium metal dissolved in liquid ammonia (dissolving metal reduction) gives the trans-alkene as the major product.
TTrue
FFalse
Answer: True
True. Dissolving metal reduction proceeds via a radical anion mechanism: an electron from Na is added to the alkyne to give a vinyl radical anion, which is protonated by NH₃ to give a vinyl radical; then a second electron gives a vinyl carbanion, which is protonated again. The most stable vinyl radical and carbanion intermediates have the two substituents *trans* (anti) to minimize steric strain. This mechanism delivers hydrogens to *opposite* faces, producing the *trans* (E) alkene exclusively. In contrast, Lindlar's catalyst delivers both hydrogens from the same catalyst surface, giving the *cis* (Z) alkene.
Question 4 True / False
Terminal alkynes are strongly acidic and can be readily deprotonated by common bases like aqueous sodium hydroxide solution.
TTrue
FFalse
Answer: False
False. Terminal alkynes have pKa ≈ 25 — more acidic than alkenes (~44) or alkanes (~50), but very weakly acidic by ordinary standards. Water has pKa 16, meaning NaOH (conjugate acid pKa ≈ 16) is not strong enough to deprotonate a terminal alkyne. Strong bases like NaNH₂ (conjugate acid pKa ~38) or n-BuLi (~50) are required. The relative acidity of terminal alkynes is important for synthetic utility but should not be overstated — an alkyne proton is roughly 10 billion times less acidic than an acetic acid proton.
Question 5 Short Answer
Explain, using hybridization and s-character, why terminal alkynes are more acidic than vinyl C–H bonds, which are in turn more acidic than alkane C–H bonds.
Think about your answer, then reveal below.
Model answer: Acidity reflects the stability of the conjugate base (carbanion). In each case, the negative charge resides in a hybrid orbital on carbon: sp for alkynide (50% s-character), sp² for vinyl carbanion (33%), sp³ for alkyl carbanion (25%). S-orbital electrons experience greater nuclear attraction and are held at lower energy. The greater the s-character, the more the negative charge is stabilized, and the more acidic the corresponding C–H bond. This gives the order: terminal alkyne (pKa ~25) > vinyl C–H (pKa ~44) > alkane C–H (pKa ~50), mirroring s-character: 50% > 33% > 25%.
This is purely a hybridization argument — not resonance or inductive effects. The same reasoning explains why sp-hybridized carbanions are *more* stable (higher s-character stabilizes negative charge) while sp-hybridized carbocations would be *less* stable (pulling positive charge toward the nucleus destabilizes it). This is one of the cleaner structure-reactivity relationships in organic chemistry: the s-character numbers map directly and predictably to the observed acidity trend.