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Ethyne reacts with sodium to release hydrogen gas, just as an acid would. Ethene and ethane do not. And but-1-yne gives a white precipitate with ammoniacal silver nitrate, while but-2-yne, with the same formula, gives nothing. This post explains the acidity behind both facts, puts numbers on it and shows how to use it to identify an unknown.
Where does this appear in the syllabus?
The rationalised NCERT Class 11 chapter on Hydrocarbons (Reprint 2026-27) covers the "acidic character of alkyne". The JEE (Main) 2026 syllabus lists "Alkynes - Acidic character", and the JEE (Advanced) 2026 brochure lists "Metal acetylides". The silver and copper tests are not in the NCERT text; they appear in the NIOS Senior Secondary chemistry lesson on hydrocarbons. Check the syllabus for your exam year.
Why is a terminal alkyne hydrogen acidic?
It comes down to hybridisation. In ethyne the hydrogen is attached to an sp carbon, in ethene to sp², and in ethane to sp³. NCERT explains that the sp orbital, with the highest s character (50%), is the most electronegative, so it pulls the C–H electron pair closest and lets the hydrogen leave as H⁺ more easily.
OpenStax Organic Chemistry puts numbers on it:
Ethyne, HC≡CH: pKa 25 (50% s character in the anion's orbital)
Ethene, H₂C=CH₂: pKa 44 (33%)
Methane, CH₄: pKa 60 (25%)
A 19-unit gap in pKa between ethyne and ethene means a factor of 10¹⁹ in Ka.
NCERT gives two orders to remember: HC≡CH > H₂C=CH₂ > CH₃–CH₃, and HC≡CH > CH₃–C≡CH >> CH₃–C≡C–CH₃. The last one has no hydrogen on a triple-bonded carbon at all. As NCERT notes, only the hydrogens attached to triply bonded carbons are acidic, not every hydrogen in an alkyne.
Which bases can remove the hydrogen?
An acid-base reaction lies on the side of the weaker acid. OpenStax gives pKa 35 for ammonia and 15.7 for water.
Worked example 1: sodamide. HC≡CH + NH₂⁻ ⇌ HC≡C⁻ + NH₃. The equilibrium constant is K = Ka(ethyne)/Ka(NH₃) = 10⁻²⁵/10⁻³⁵ = 10¹⁰. The reaction goes essentially to completion. This is why OpenStax says amide ion is often used to deprotonate terminal alkynes, and why NCERT shows ethyne and propyne reacting with NaNH₂.
Worked example 2: hydroxide. HC≡CH + OH⁻ ⇌ HC≡C⁻ + H₂O. Now K = 10⁻²⁵/10⁻¹⁵·⁷ = 10⁻⁹·³ ≈ 5 × 10⁻¹⁰. Essentially no acetylide forms, so aqueous NaOH is not a suitable base.
Ethyne has two acidic hydrogens. NCERT shows sodium giving monosodium ethynide, HC≡C⁻Na⁺, and then disodium ethynide, Na⁺⁻C≡C⁻Na⁺, with hydrogen gas released each time. Propyne has only one.
How do silver and copper tests identify terminal alkynes?
NIOS lists two tests. Passed through ammoniacal silver nitrate, ethyne gives a white precipitate of silver acetylide; through ammoniacal cuprous chloride, a red precipitate of copper(I) acetylide. In NIOS's comparison table, alkanes and alkenes give no change in either test.
Worked example 3: an unknown C₄H₆. A hydrocarbon C₄H₆ decolourises bromine in CCl₄ and gives a white precipitate with ammoniacal silver nitrate. Which isomer is it?
C₄H₆ has two degrees of unsaturation, so it could be a diene, an alkyne or a cyclic alkene.
Bromine decolourisation confirms a C=C or C≡C, which does not narrow it far.
The silver test needs C≡C–H. Among the alkynes, but-1-yne (HC≡C–CH₂CH₃) has it; but-2-yne (CH₃C≡CCH₃) does not. Buta-1,3-diene has no triple bond.
The compound is but-1-yne.
Metal acetylide precipitates should be treated as hazardous; follow your teacher's instructions for handling and disposal.
Which mistakes cost marks?
Saying all alkyne hydrogens are acidic. Only C≡C–H hydrogens are.
Using NaOH to make an acetylide. Water is a stronger acid than ethyne.
Expecting but-2-yne to give the silver test.
Forgetting ethyne has two acidic hydrogens and can form a disodium salt.
A practice routine
Day 1: Write the pKa list and the s-character list side by side and explain the link in two lines.
Day 2: Calculate K for ethyne with NH₂⁻, OH⁻ and H⁻, using pKa values from a table.
Day 3: Three "identify the C₄H₆ or C₅H₈" questions using bromine, Baeyer's and silver tests.
Day 4: Revise the rest of alkyne chemistry in our map of hydrocarbon reactions.
If you want to check a chain of reasoning like Example 3, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Tonight, write the products when propyne reacts with sodamide, and explain why propyne can form only a monosodium salt.
