Alkynes — preparation, acidity of terminal alkynes, addition reactions, hydration to ketones: Visual walkthrough

We start from nothing. First we agree on what the drawings mean, then we add water one piece at a time.

Step 1 — What a triple bond actually looks like

WHAT. A triple bond written $- C \equiv C -$ is not "three identical sticks". It is one strong bond that lies straight along the line joining the two carbons (called a == $σ$ bond==, "sigma", the head-on bond) plus two weaker bonds made of electron clouds sitting above/below and front/back of that line (each called a == $π$ bond==, "pi", the sideways bond).

WHY it matters. Those two $π$ clouds are loose, exposed, electron-rich regions. Anything that is electron-poor (a "+" seeking partner, which chemists call an electrophile = "electron-lover") will be attracted straight to them. That attraction is the entire engine of this reaction.

PICTURE. Look at the figure. The gray line is the $σ$ bond. The blue lobes are the two $π$ clouds. Notice the two carbons and everything attached to them lie in a perfectly straight line — that is the == $s p$ == geometry ( $18 0^{\circ}$ ), which you met in Hybridisation and s-character.

Figure — Alkynes — preparation, acidity of terminal alkynes, addition reactions, hydration to ketones

Step 2 — The reagents and why each one is present

WHAT. The recipe is water $H_{2} O$ , sulfuric acid $H_{2} S O_{4}$ , and mercuric sulfate $H g S O_{4}$ . That is three ingredients. Each has one job.

WHY each.

$H g S O_{4}$ supplies $H g^{2 +}$ , a hungry electrophile that grabs the $π$ cloud first and activates the alkyne. This is the key player — the reaction is called oxymercuration. We use mercury and not just acid because plain acid on an alkyne is sluggish; $H g^{2 +}$ is the real activator.
$H_{2} O$ is the piece that actually gets added across the triple bond — the oxygen ends up in the product.
$H_{2} S O_{4}$ supplies $H^{+}$ — but not to attack the alkyne first. Its $H^{+}$ is needed later, to protonate and to help remove the mercury at the end.

PICTURE. The figure separates the three reagents and labels the job of each with an arrow pointing at where it acts.

Step 3 — Mercury attacks first: the mercurinium ion (not a vinyl cation)

WHAT. Take a terminal alkyne, $R - C \equiv C H$ ( $R$ = any carbon chain, e.g. $C H_{3}$ ). The $H g^{2 +}$ electrophile bonds to both triple-bond carbons at once, forming a three-membered bridged ring called a mercurinium ion (mercury bridging the two carbons, carrying the "+" charge).

WHY a bridge and not a free "+" carbon. A free vinyl cation (a bare "+" on a doubly-bonded carbon) is prohibitively unstable — it essentially never forms under these conditions. Nature avoids it by having mercury bridge the two carbons: the positive charge is shared over three atoms in a ring, which is far more stable. So there is no free vinyl cation here — that is the whole point of using $H g^{2 +}$ .

PICTURE. The bridged three-membered ring is drawn: $H g^{+}$ sits above the old triple bond, bonded to both carbons. A red-crossed alternative shows the (rejected) free vinyl cation.

R - both C bonded to Hg C \equiv C H + H g^{2 +} ⟶ mercurinium ion (bridged, +) R - C - C ╱ ╲ H g + H

Reading it: the $H g^{2 +}$ approaches the electron-rich triple bond; the two slashes $╱ ╲$ show mercury bonding to both carbons simultaneously; the "+" now lives on the bridged ring, spread out and stable — not stuck on one lonely carbon.

Step 4 — Water attacks the bridge (Markovnikov regiochemistry)

WHAT. The mercurinium ring is electron-poor. Water donates a lone pair ( $: O H_{2}$ ) and opens the ring. It attacks the carbon that carries the more substituted (the $R$ ) side, because that carbon holds more of the positive character — this is the Markovnikov rule and carbocation stability pattern.

WHY that carbon. In the strained bridge, the "+" leans toward the more substituted carbon (that carbon can better support positive charge). The nucleophile goes to the most electron-poor carbon, so the oxygen ends up on the $R$ -carbon — this single fact decides ketone-vs-aldehyde later. Mercury stays on the other (terminal) carbon for now, then is removed, giving the enol.

PICTURE. A curved arrow runs from an oxygen lone pair to the $R$ -carbon of the ring; the ring opens; loss of $H^{+}$ and then removal of mercury (demercuration) gives $R - C (O H) = C H_{2}$ , the enol.

mercurinium [R - + H g -bridge- C H] : O H_{2}, - H^{+} R - O on R -carbon Hg on end carbon C (O H) = C H (H g) - H g R - enol C (O H) = C H_{2}

Term by term: $: O H_{2}$ is water offering a lone pair (recall the " $:$ " means a lone pair); it opens the ring at the $R$ -carbon so the $- O H$ lands there; mercury departs (demercuration); the result $R - C (O H) = C H_{2}$ is the enol.

Step 5 — Meet the enol (an unstable in-between)

WHAT. The molecule we just made is $R - C (O H) = C H_{2}$ . It has two features on adjacent carbons: a double bond ( $C = C$ , the "-ene") and an $- O H$ (the "-ol"). Glue those names: enol.

WHY it can't stay. The enol is a real molecule, but it is higher in energy than an isomer it can flip into. Chemistry always rolls downhill toward lower energy — so the enol will rearrange. It does not fall apart; it just shuffles one hydrogen and one double bond. That shuffle is the topic of the next step.

PICTURE. The enol is drawn with its two defining parts colour-coded: orange for the $C = C$ , blue for the $O - H$ . A dashed "downhill" energy ramp shows it sitting above the product it will become.

Step 6 — Tautomerism: the enol flips into a ketone

WHAT. The enol $C = C - O H$ converts to a $C = O$ next to a $C - H$ . In words: the $O - H$ hydrogen hops onto the neighbouring carbon, and the $C = C$ double bond slides over to become a $C = O$ double bond. The two forms are tautomers (constitutional isomers that interconvert by moving one H); this dance is Keto–enol tautomerism.

WHY the keto side wins. Count bond strengths. We destroy a $C = C$ $π$ bond (worth about $270$ kJ/mol) and an $O - H$ bond, and we build a $C = O$ $π$ bond (about $360$ kJ/mol) and a $C - H$ bond. The new $C = O$ is far stronger than the old $C = C$ , so the total energy drops. Lower energy = favoured. Hence essentially all of it ends up as the keto (carbonyl) form.

PICTURE. Three snapshots: (1) enol, (2) H mid-hop with the double bond mid-slide, (3) ketone. A little energy bar on the right shows keto sitting well below enol.

enol R - ∣ C O H = C H_{2} ⇌ keto (methyl ketone) R - C ∥ O - C H_{3}

Reading it: on the left the $O H$ sits on the $R$ -carbon and there is a $C = C$ . The half-arrows $⇌$ mean the two live in equilibrium, but the balance lies almost entirely to the right. On the right the oxygen has become a full $C = O$ (a carbonyl) on the $R$ -carbon, and the end carbon has picked up the extra H to become $C H_{3}$ . Because the carbonyl carbon has two carbon groups on it ( $R$ and $C H_{3}$ ), the product is a ketone, specifically a methyl ketone. These are the molecules of Aldehydes and Ketones — carbonyl chemistry.

Step 7 — The degenerate case: ethyne gives an aldehyde, not a ketone

WHAT. Do the whole walk again for the smallest alkyne, ethyne $H C \equiv C H$ . Now both carbons are identical — each carries only an H, there is no $R$ group.

WHY it's different. In Step 4 water opened the mercurinium ring at the more-substituted carbon. Here neither carbon is more substituted — they are twins. So it makes no difference which carbon water hits; the enol is forced to be $C H_{2} = C H - O H$ . When that tautomerises, the carbonyl carbon ends up bonded to an $R$ ... but there is no $R$ , only an $H$ . A carbonyl carbon carrying (at least one) $H$ is an aldehyde, not a ketone.

PICTURE. Ethyne's two identical carbons are shaded the same colour to stress the symmetry; the arrow leads to $C H_{3} C H O$ (acetaldehyde), labelled "aldehyde — the lone exception".

H C \equiv C H H_{2} O H_{2} S O_{4}, H g S O_{4} enol [C H_{2} = C H - O H] ⟶ acetaldehyde C H_{3} C H O

Step 8 — The messy case: unsymmetrical internal alkynes give mixtures

WHAT. Take an internal alkyne whose two triple-bond carbons carry different groups, e.g. $R - C \equiv C - R^{'}$ with $R \neq = R^{'}$ (say $C H_{3} - C \equiv C - C H_{2} C H_{3}$ ). Water can add across the triple bond in either direction.

WHY a mixture. For a terminal alkyne, one carbon was clearly "more substituted" (bears $R$ ) and the other bore only H — the choice was decisive. But when both carbons are substituted and comparably so, there is no strong preference: the mercurinium ion can be opened at either carbon, so the $O H$ (and hence the $C = O$ after tautomerism) can land on either carbon. Result: a mixture of two ketones.

PICTURE. The unsymmetrical internal alkyne branches to two enols and then two different ketones, both boxed as valid products.

C H_{3} - C \equiv C - C H_{2} C H_{3} H_{2} O H_{2} S O_{4}, H g S O_{4} ketone A C H_{3} - C ∥ O - C H_{2} C H_{2} C H_{3} + ketone B C H_{3} C H_{2} - C ∥ O - C H_{2} C H_{3}

Step 9 — The mirror route: anti-Markovnikov hydration gives an aldehyde

WHAT. If instead of $H_{2} O / H_{2} S O_{4} / H g S O_{4}$ we use hydroboration–oxidation ( $B H_{3}$ , then $H_{2} O_{2} / O H^{-}$ ), the water adds the opposite way round on a terminal alkyne.

WHY the flip. In hydroboration the boron (which becomes the $O H$ after oxidation) attaches to the less hindered, terminal carbon — the reverse of Markovnikov. So the $O H$ lands on the end carbon, the enol is $R - C H = C H - O H$ , and tautomerising puts the carbonyl on the terminal carbon → that carbon carries an H → an aldehyde $R - C H_{2} - C H O$ .

PICTURE. Two side-by-side pipelines from the same terminal alkyne: the Markovnikov route (top, green) ending in a ketone, the hydroboration route (bottom, orange) ending in an aldehyde. The only difference — which carbon gets the OH — is highlighted.

The one-picture summary

The single diagram compresses everything: $π$ cloud grabs $H g^{2 +}$ → mercurinium bridge (no free vinyl cation) → water opens it Markovnikov-style → enol → tautomerise downhill to the methyl ketone. Off the trunk sit the special cases: ethyne → aldehyde (symmetry, Step 7), unsymmetrical internal alkyne → mixture of ketones (Step 8), and the hydroboration route → aldehyde (Step 9).

Recall Feynman retelling — say it to a friend

A triple bond is a little pile of loose electrons. Splash on acid + mercury and it's the mercury that dives in first, not a hydrogen. Mercury doesn't just grab one carbon — it throws a bridge across both, forming a stable three-membered ring, so we never get a wobbly bare "+" carbon (a vinyl cation, which nature hates). Now water sneaks in with its lone pair and cracks the bridge open on the fatter, more-substituted carbon, dropping its oxygen there. Mercury leaves, and we have a wobbly molecule with a double bond and an $O H$ side by side: the enol. The enol is uncomfortable, so a hydrogen hops over and the double bond slides onto the oxygen, forming a super-strong $C = O$ . That's the ketone, much lower in energy, so nature stays there. Three twists: tiny symmetric ethyne has no fatter carbon, so it gives the aldehyde acetaldehyde; an unsymmetrical internal alkyne can't decide which carbon to favour, so it gives a mixture of two ketones; and the boron route deliberately puts the OH on the other carbon, giving an aldehyde. Everything else — a bigger terminal alkyne with mercury — gives one clean methyl ketone.

Recall Quick self-test

What attacks the alkyne first under $H g S O_{4}$ ? ::: $H g^{2 +}$ , forming a mercurinium bridge (not $H^{+}$ , no vinyl cation)
Why a bridged ion, not a vinyl cation? ::: a free vinyl cation is far too unstable; the bridge spreads the + charge
Where does water open the bridge on $R - C \equiv C H$ ? ::: at the more-substituted ( $R$ ) carbon (Markovnikov)
What is the unstable intermediate before the ketone? ::: the enol ( $C = C - O H$ )
Why does keto win over enol? ::: the $C = O$ $π$ bond (~360 kJ/mol) is stronger than $C = C$ (~270), energy drops
Product of $C H_{3} C \equiv C H$ + $H_{2} O / H_{2} S O_{4} / H g S O_{4}$ ? ::: propan-2-one (acetone), a methyl ketone
Product of $H C \equiv C H$ by the same route? ::: acetaldehyde $C H_{3} C H O$ (the exception)
Product of an unsymmetrical internal alkyne? ::: a mixture of two ketones
Product of $R - C \equiv C H$ by hydroboration–oxidation? ::: aldehyde $R - C H_{2} C H O$

Connections

Parent topic
Alkenes — electrophilic addition (same $π$ -attacks-electrophile logic, once)
Markovnikov rule and carbocation stability (decides which carbon gets the +)
Keto–enol tautomerism (the enol → ketone flip)
Aldehydes and Ketones — carbonyl chemistry (the products)
Hybridisation and s-character ( $s p$ , linear triple bond)

Under HgSO4 hydration the alkyne is first attacked by

Hg2+ forming a bridged mercurinium ion (no free vinyl cation)

Unsymmetrical internal alkyne on Markovnikov hydration gives

a mixture of two ketones