On this page
Alkene plus water gives an alcohol. So students expect alkyne plus water to give an alcohol too, and write ethenol or propan-2-ol. The real products are carbonyl compounds, and only one alkyne gives an aldehyde. This post explains the two steps behind that, works through product prediction for six alkynes and a yield calculation, and lists the slips examiners look for.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus both list, under Hydrocarbons, "Alkynes - Acidic character, addition of hydrogen, halogens, water and hydrogen halides". The JEE (Advanced) 2026 brochure lists "Acid catalysed hydration". The NCERT Class 11 chapter Hydrocarbons (2026-27 reprint) says that one molecule of water adds to alkynes "on warming with mercuric sulphate and dilute sulphuric acid at 333 K to form carbonyl compounds". The Class 12 chapter Aldehydes, Ketones and Carboxylic Acids repeats it as a preparation: ethyne gives acetaldehyde (ethanal), and "all other alkynes give ketones". Check your year's syllabus.
The reaction has a long history. Kucherov (spelt Kutscheroff in the German journal) reported the mercury-catalysed method in 1881, in a short paper titled "on a new method of direct addition of water (hydration) to hydrocarbons of the acetylene series" (Kutscheroff, Berichte, 1881).
Why does the product end up as a carbonyl?
Two steps, as OpenStax Organic Chemistry sets them out.
Markovnikov addition. Catalysed by Hg²⁺, water adds across one π bond. The OH goes to the more substituted carbon, the H to the less substituted one. This gives an enol, an alcohol with OH on a C=C carbon.
Keto-enol tautomerism. The enol is not the isolated product. It rearranges at once: a hydrogen moves from oxygen to carbon and the C=C–OH becomes CH–C=O. The keto form is far more stable, so that is what you collect.
Ethyne has two identical carbons with one H each. Its enol, CH₂=CH–OH, becomes CH₃–CHO, ethanal. For any other alkyne, the OH lands on a carbon that also carries a carbon group, so tautomerism gives a C=O inside the chain: a ketone.
Predicting the product: six alkynes
Ethyne, HC≡CH: ethanal, CH₃CHO.
Propyne, CH₃C≡CH: OH goes to C2 by Markovnikov's rule. Propanone, CH₃COCH₃.
But-1-yne, CH₃CH₂C≡CH: butan-2-one. A terminal alkyne always gives a methyl ketone, as OpenStax notes.
But-2-yne, CH₃C≡CCH₃: symmetrical, so either carbon gives the same enol. Butan-2-one again.
Pent-2-yne, CH₃C≡CCH₂CH₃: unsymmetrical and internal. The two carbons are about equally substituted, so you get a mixture of pentan-2-one and pentan-3-one. OpenStax says this is why the method is most useful for terminal alkynes.
Phenylethyne, C₆H₅C≡CH: acetophenone, C₆H₅COCH₃.
Working backwards. Which alkyne gives 3-methylbutan-2-one, CH₃COCH(CH₃)₂, cleanly? Put the triple bond where the C=O is, at the end of the chain: 3-methylbut-1-yne, HC≡C–CH(CH₃)₂. Hydration gives the methyl ketone and nothing else.
Worked example: how much product?
13.0 g of ethyne is hydrated completely. What mass of ethanal forms? (Sample quantities.)
Molar masses: C₂H₂ = 26.04 g/mol; CH₃CHO = 44.05 g/mol.
Moles of ethyne = 13.0 ÷ 26.04 ≈ 0.499 mol. The ratio is 1 : 1.
Mass of ethanal = 0.499 × 44.05 ≈ 22.0 g.
If 10.0 g of propyne (40.07 g/mol) gives propanone (58.08 g/mol) in 75% yield: 10.0 ÷ 40.07 ≈ 0.250 mol, theoretical 14.5 g, actual 0.75 × 14.5 ≈ 10.9 g.
Which mistakes cost marks?
Stopping at the enol. Ethenol is an intermediate, not the answer.
Writing propanal from propyne. That would need anti-Markovnikov addition.
Mixing up alkene and alkyne hydration. NCERT's alkene hydration, with a few drops of concentrated sulphuric acid, gives alcohols. Alkynes need Hg²⁺ and give carbonyls.
Leaving out the catalyst. NCERT notes alkynes do not react with water alone.
Giving one product for an unsymmetrical internal alkyne.
The acidity of terminal alkynes is a separate topic that questions often pair with this one; see our post on terminal alkyne acidity. If you can follow the mechanism but keep picking the wrong carbon for the OH, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning.
A practice routine
Day 1: Write the enol and the keto form for ethyne, propyne and but-1-yne, with arrows showing where the hydrogen moves.
Day 2: Ten alkynes, terminal and internal: predict each product, marking mixtures.
Day 3: Five backwards questions: from a ketone to the alkyne that gives it cleanly.
Day 4: A conversion chain: calcium carbide to ethyne to ethanal, then on to ethanoic acid, using our aldehydes and ketones reactions guide for the last step.
Tonight, predict the hydration product of hex-1-yne and name it.
