Chemistry · Hydrocarbons · NEET
Almost always a KETONE. Water adds by Markovnikov rule using dilute H2SO4 with HgSO4 (this is called the Kucherov reaction). The -OH first goes on the more substituted carbon and you get an unstable enol, which then rearranges (tautomerises) to a carbonyl. For any alkyne except acetylene, this carbonyl is a ketone. The ONLY exception is acetylene (HC#CH) itself, which gives ethanal (CH3CHO), an aldehyde. So NEET rule: acetylene -> aldehyde; every other alkyne -> ketone.
It is about HOW the two hydrogens are delivered. Lindlar's catalyst is a metal surface. The alkyne lies flat on the metal and both H atoms are added to the SAME side (syn addition). Same side = the two big groups end up on the same side = cis (Z) alkene. Na in liquid NH3 works by a different path: it adds one electron, then H, then another electron, then H, through a radical/anion. Here the more stable trans arrangement forms, so the two big groups end up on OPPOSITE sides = trans (E) alkene. Both stop cleanly at the alkene; they do NOT go all the way to the alkane.
It is palladium on calcium carbonate (or charcoal) that has been PARTLY poisoned (deactivated) with quinoline or a sulphur compound. NCERT calls it 'partially deactivated palladised charcoal'. The poison weakens the catalyst just enough so that it adds only ONE molecule of H2 (alkyne -> alkene) and then stops. Without the poison, plain Pd or Pt would keep going and reduce the alkene all the way to the alkane.
If the question wants the CIS (Z) alkene, choose H2 with Lindlar's catalyst (Pd/C + quinoline). If it wants the TRANS (E) alkene, choose Na (or Li) in liquid NH3. If it wants the ALKANE (full reduction), choose H2 with plain Pt/Pd/Ni. Do not mix these up: this exact choice was asked in NEET 2019 and again in ReNEET 2026.
It does NOT stay. The first product is an enol (a C=C carbon carrying an -OH). Enols are unstable and immediately rearrange to the keto form by moving one H and shifting the double bond. This is called keto-enol tautomerism. That is why the FINAL product is a carbonyl compound (ketone or aldehyde), not an alcohol. Many students wrongly pick the enol as the answer; it is only an intermediate.
Yes. Alkynes have pi bonds, so they also do electrophilic addition and follow Markovnikov rule for unsymmetrical alkynes. In hydration, the -OH (from water) ends up on the more substituted carbon, and after tautomerism that carbon becomes the C=O carbon of the ketone. So propyne (CH3-C#CH) gives propanone (acetone), not propanal.
Predict the correct intermediate (A) and product (B): H3C-C#CH --(H2O, H2SO4, HgSO4)--> (A) --> (B)
The most suitable reagent for the conversion: CH3-C#C-CH3 -> cis-2-butene
Ph-C#C-Me with Na/liq. NH3 -> L; with H2, Lindlar's catalyst -> K. Which statement is correct?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Syn addition. Both hydrogen atoms are delivered to the same side of the triple bond from the metal surface, which is why the product is the cis (Z) alkene.
Acetylene (HC#CH) is symmetrical with H on both carbons, so hydration gives ethanal, an aldehyde. In propyne one carbon already carries a CH3 group, so after Markovnikov hydration and tautomerism the carbonyl sits on the internal carbon, giving a ketone (propanone).
No. It stops at the alkene stage and gives the trans (E) alkene. To go fully to the alkane you would use H2 with plain Pt, Pd or Ni catalyst.
It is the acid-catalysed hydration of an alkyne using dilute H2SO4 with HgSO4 as catalyst. It follows Markovnikov rule and gives a carbonyl compound (a ketone, or ethanal from acetylene) through an enol intermediate.
Hydrocarbons is a steady scorer, and the Lindlar-versus-Na/NH3 (cis vs trans) idea plus alkyne hydration have appeared in NEET 2017, 2019 and ReNEET 2026. Learning this one page reliably locks in about 1 question (4 marks) most years.
