Whether hydration gives an aldehyde or a ketone depends on the substrate: a terminal alkyne bears an sp C–H, an internal alkyne buries the triple bond mid-chain.
1. A triple bond is two π bonds, so electrophiles can add twice.
The first addition leaves a still-nucleophilic double bond; equivalents of reagent decide whether you stop there or add again.
2. Hydrohalogenation adds HX with Markovnikov regiochemistry: one equivalent gives a vinyl halide, two give a geminal dihalide.
One equivalent of HX adds Markovnikov (halogen to the more substituted carbon) and stops at a vinyl halide.
Markovnikov addition of one HBr → a vinyl bromide.
A second equivalent adds again Markovnikov, so both halogens land on the same carbon — a geminal dihalide.
Two equivalents drive a second Markovnikov addition to the gem-dibromide.
3. Acid-catalyzed hydration gives a Markovnikov enol that tautomerizes to a ketone.
With H2O, H2SO4, HgSO4, water adds Markovnikov (OH on the more substituted carbon) to give an enol.
Markovnikov hydration of propyne → acetone (via an enol).
The enol immediately tautomerizes to the stable carbonyl, and since OH sat on the more substituted carbon it is a ketone (a methyl ketone from a terminal alkyne).
4. Hydroboration–oxidation delivers water anti-Markovnikov, turning a terminal alkyne into an aldehyde.
A bulky dialkylborane (R2BH) then H2O2, NaOH puts boron — and OH — on the terminal carbon, giving an anti-Markovnikov enol.
Anti-Markovnikov hydration of propyne → propanal, an aldehyde.
This enol tautomerizes with the oxygen on the terminal carbon, so it becomes an aldehyde — the standard route from a terminal alkyne.
5. Reduction reagents let you choose the cis alkene, the trans alkene, or the alkane.
The reagent sets both how far reduction goes and the geometry:
- H2, Lindlar catalyst — syn addition stops at the cis (Z) alkene.
- Na (or Li) in liquid NH3 — dissolving-metal reduction gives the trans (E) alkene.
- H2, Pd/C — reduces all the way to the alkane.
6. Summary
1 eq HX → vinyl halide · 2 eq HX → gem-dihalide · H2O/H2SO4/HgSO4 → ketone · R2BH then H2O2/NaOH → aldehyde · H2/Lindlar → cis · Na/NH3 → trans · H2/Pd-C → alkane.
Worked example
- HBr adds with Markovnikov orientation: H to the terminal carbon, Br to the internal (more-substituted) carbon → 2-bromopropene.
- The second HBr again adds Markovnikov, placing the second Br on the same carbon (now stabilised by the first Br).
Answer. 2,2-dibromopropane (a geminal dihalide).
How each reagent works — the arrow pushing
Electron flow only. Follow the arrows; the structures do the talking.
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2,2-Dibromopropane, a geminal dihalide. Both additions are Markovnikov, so both bromines go to the same (more substituted, internal) carbon. One equivalent would stop at the vinyl bromide 2-bromopropene.
Both routes make an enol that tautomerizes to a carbonyl; the difference is where the OH lands. Hg-catalyzed hydration is Markovnikov (OH on the more substituted carbon → ketone). Hydroboration puts boron, and then OH, on the terminal carbon (anti-Markovnikov) → aldehyde.
Use Na (or Li) in liquid NH3 — dissolving-metal reduction gives the trans (E) alkene. For the cis (Z) isomer use H2 with Lindlar catalyst (syn addition). Ordinary H2/Pd-C would overshoot to butane.
A triple bond contains two π bonds. The first addition consumes one π bond and leaves a double bond behind, which still has a π bond available to react with a second equivalent. An alkene starts with only one π bond, so it is saturated after a single addition.
Draw this on the whiteboard
Open the OChem Board whiteboard — benzene rings, curved arrows, wedge/dash bonds and a clickable periodic table built in. No account needed.