Alkynes — preparation, acidity of terminal alkynes, addition reactions, hydration to ketones: Worked examples

This page is the drill floor for the parent topic. Before touching numbers we lay out every kind of situation an alkyne problem can throw at you, then work an example for each. Nothing here is new theory — it is the parent note's logic, applied case by case, so that when an exam question lands you already recognise the cell it belongs to.

The scenario matrix

Here every "cell" is a distinct class of problem. The examples that follow are tagged with the cell they cover, so together they fill the whole grid.

#	Case class	The deciding question	Example
A	Terminal alkyne, Markovnikov hydration	Which carbon gets the $O H$ ?	Ex 1
B	Ethyne (degenerate — both carbons equal)	Aldehyde or ketone?	Ex 2
C	Internal, symmetric alkyne hydration	One product or two?	Ex 3
D	Internal, unsymmetric alkyne hydration	Which ketone dominates?	Ex 4
E	Anti-Markovnikov (hydroboration) route	Aldehyde vs ketone flip	Ex 5
F	Acid–base: will this base deprotonate?	Compare $p K_{a}$ values	Ex 6
G	Acetylide alkylation (chain building)	Primary / secondary / tertiary $R - X$ ?	Ex 7
H	Reduction stereochemistry (cis vs trans)	Lindlar or $N a / N H_{3}$ ?	Ex 8
I	Word problem (identify unknown gas)	Which lab test fires?	Ex 9
J	Exam twist (count every equivalent, $H B r$ )	Vicinal vs geminal dihalide?	Ex 10
K	Halogen ( $X_{2}$ ) addition	Vinyl → tetrahalide?	Ex 11
L	Oxidative cleavage (ozonolysis)	What carbonyls fall out?	Ex 12

The columns of the matrix are the three "axes" a problem varies along:

Substrate type — terminal / ethyne / internal-symmetric / internal-unsymmetric (rows A–D).
Reagent chosen — which controls Markovnikov vs anti-Markovnikov, cis vs trans, addition vs cleavage (rows E, H, K, L).
Selectivity / counting — regioselectivity, how many equivalents, how many fragments (rows D, J, K, L).

Rows F, G, I test the acidity property; the rest test the $π$ -electron property.

Ex 1 — Cell A: terminal alkyne, Markovnikov hydration

Forecast: water adds across the triple bond. Does the $O H$ land on the middle carbon or the end carbon — and is the final product an aldehyde or a ketone? Guess now.

Step 1 — Label the two triple-bond carbons. Call them $C_{1}$ (the terminal $\equiv C H$ ) and $C_{2}$ (the one carrying $C H_{3}$ ). Why this step? Markovnikov's rule is a statement about which carbon, so we must name them before we can apply it. In the figure below, the magenta dot is $C_{1}$ (the terminal carbon, only an H attached) and the violet dot is $C_{2}$ (the more substituted carbon, carrying $C H_{3}$ ). The three orange bars between them are the triple bond — the electron-rich site water attacks.

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

Step 2 — Apply Markovnikov: $O H$ goes to the more substituted carbon. The rule (see Markovnikov rule and carbocation stability) says the electrophile's positive part attacks to give the more stable cation; here $O H$ ends up on the carbon that was already more substituted, i.e. the violet $C_{2}$ (follow the magenta arrow in the figure that points $O H$ onto $C_{2}$ ). So the enol is $C H_{3} - O H C = C H_{2}$ . Why this step? $H g S O_{4}$ activates the alkyne so water attacks Markovnikov-style; the more substituted carbon supports positive charge better.

Step 3 — Tautomerise the enol to the keto form. The enol $C = C - O H$ flips to the carbonyl (see Keto–enol tautomerism): $C H_{3} - O H C = C H_{2} ⟶ C H_{3} - C ∥ O - C H_{3}$ Why this step? The keto form is lower in energy (the $C = O$ $π$ bond is much stronger than $C = C$ ), so the enol never survives.

Answer: propan-2-one (acetone), $C H_{3} C O C H_{3}$ — a ketone.

Verify: count atoms. Propyne $C_{3} H_{4}$ + $H_{2} O$ = $C_{3} H_{6} O$ . Acetone is $C_{3} H_{6} O$ . ✓ Molecular formula balances, and the product is a methyl ketone exactly as "Mark makes Ketones" predicts.

Ex 2 — Cell B: ethyne (the degenerate case)

Forecast: both carbons are identical. Markovnikov can't pick a "more substituted" side — so what happens?

Step 1 — Recognise the symmetry. Both carbons are $\equiv C H$ with one H each. There is no "more substituted" carbon. The figure shows both carbons as identical magenta dots, so no matter which one water hits, the same enol results.

Why this step? Markovnikov only distinguishes carbons when they differ; here it is silent, so we let tautomerism decide.

Step 2 — Form the only possible enol and tautomerise. $H C \equiv C H H_{2} O [C H_{2} = C H - O H] ⟶ C H_{3} - C H O$ Why this step? Whichever carbon the $O H$ lands on, the enol is the same $C H_{2} = C H O H$ , which tautomerises to acetaldehyde.

Answer: ethanal (acetaldehyde), $C H_{3} C H O$ — an aldehyde, the lone exception.

Verify: $C_{2} H_{2} + H_{2} O = C_{2} H_{4} O$ ; acetaldehyde is $C_{2} H_{4} O$ . ✓ Ethyne is the only alkyne whose Markovnikov hydration gives an aldehyde, because it is the only symmetric terminal case.

Ex 3 — Cell C: symmetric internal alkyne

Forecast: an internal alkyne — no acidic H, so this is pure $π$ chemistry. Symmetric or not?

Step 1 — Check the symmetry of the two alkyne carbons. Both bear a $C H_{3}$ : the molecule has a mirror plane through the triple bond. The two carbons are equivalent. In the figure the dashed magenta line is that mirror plane, and both alkyne carbons (violet) are its reflections.

Why this step? If the carbons are equivalent, $O H$ landing on either gives the same enol → a single product.

Step 2 — Add water; both routes converge. $C H_{3} - C \equiv C - C H_{3} H_{2} O [C H_{3} - O H C = C H - C H_{3}] \to C H_{3} - C ∥ O - C H_{2} - C H_{3}$ Why this step? Either carbon becomes the carbonyl; by symmetry both give butan-2-one.

Answer: butan-2-one (a single ketone), $C H_{3} C O C H_{2} C H_{3}$ .

Verify: $C_{4} H_{6} + H_{2} O = C_{4} H_{8} O$ ; butan-2-one is $C_{4} H_{8} O$ . ✓ Symmetric internal alkyne → one ketone.

Ex 4 — Cell D: unsymmetric internal alkyne (regioselectivity)

Forecast: internal but not symmetric — the two carbons carry different groups. One product, two products, or two in unequal amounts? Guess.

Step 1 — Compare the substituents on each alkyne carbon. Left carbon ( $C_{2}$ of pentyne) carries $C H_{3}$ ; right carbon ( $C_{3}$ ) carries $C H_{2} C H_{3}$ . They are different, so there is no symmetry to force a single answer — both enols can form. The figure draws the two competing $O H$ -attack sites (violet vs magenta arrows) that lead to the two ketones.

Why this step? Symmetry would collapse the two routes into one (as in Ex 3); here it does not, so we must weigh them.

Step 2 — Weigh the two routes by carbocation stability. Markovnikov selectivity comes from which carbon better bears the developing positive charge (see Markovnikov rule and carbocation stability). The mercury-activated intermediate places $δ +$ on the carbon that is more stabilised by adjacent alkyl groups. An ethyl group ( $C H_{2} C H_{3}$ ) donates electron density slightly better than a methyl, so the cation on the ethyl-bearing carbon ( $C_{3}$ ) is a touch more stable → $O H$ prefers $C_{3}$ . Why this step? "Both can form" does not mean "equal"; the more stable cationic intermediate is favoured, giving a major product.

Step 3 — Write both ketones, mark the major, and estimate the split. $O H$ on $C_{3}$ (favoured) → $C H_{3} C H_{2} - C O - C H_{2} C H_{3}$ = pentan-3-one (major). $O H$ on $C_{2}$ → $C H_{3} - C O - C H_{2} C H_{2} C H_{3}$ = pentan-2-one (minor). Because ethyl vs methyl is only a small electronic difference, the preference is weak — experimentally these two ketones come out roughly $\approx 60 : 40$ (pentan-3-one : pentan-2-one), nowhere near a clean single product. Why this step? Naming both and giving the approximate ratio makes the modest selectivity concrete — students should picture "a slight majority", not "exclusively one."

Answer: a mixture of pentan-3-one (major, $\approx 60%$ ) and pentan-2-one (minor, $\approx 40%$ ); both are ketones, no aldehyde.

Verify: $C_{5} H_{8} + H_{2} O = C_{5} H_{10} O$ for each; both pentanones are $C_{5} H_{10} O$ , and the two fractions sum to $100%$ . ✓ Unsymmetric internal alkyne → two ketones in unequal amounts. (This modest selectivity is why hydration is preparatively clean only for terminal alkynes.)

Ex 5 — Cell E: anti-Markovnikov (hydroboration flip)

Forecast: hydroboration is the anti-Markovnikov route. Where does $O H$ go now, and does the product become a ketone or an aldehyde?

Step 1 — Place boron (then $O H$ ) on the less substituted carbon. Boron adds anti-Markovnikov, i.e. to the terminal $\equiv C H$ carbon; oxidation replaces $B$ with $O H$ on that same carbon. Why this step? "hydroBoration makes Aldehydes" — the $O H$ ends on the end carbon, the opposite of Ex 1.

Step 2 — Tautomerise the terminal enol. Write the enol with the $O H$ unambiguously on the terminal carbon (the right-hand end): $C H_{3} C H_{2} C H = C H O H ⟶ C H_{3} C H_{2} - C H_{2} - C H O$ Why this step? Placing $O H$ on the terminal carbon (note $= C H O H$ , not $= C H - O H$ floating between carbons) makes the regiochemistry unambiguous: this enol tautomerises to an aldehyde ( $- C H O$ ), not a ketone.

Answer: butanal, $C H_{3} C H_{2} C H_{2} C H O$ — an aldehyde.

Contrast: the $H_{2} O / H_{2} S O_{4} / H g S O_{4}$ (Markovnikov) route on the same but-1-yne gives butan-2-one, $C H_{3} C H_{2} C O C H_{3}$ — a ketone.

Verify: both come from $C_{4} H_{6} + H_{2} O = C_{4} H_{8} O$ ; butanal and butan-2-one are both $C_{4} H_{8} O$ — same formula, different connectivity, chosen by the reagent. ✓

Ex 6 — Cell F: acid–base, will the base work?

Forecast: which of these bases can pull off the terminal $\equiv C - H$ ? Guess before computing.

Step 1 — State the rule. A base deprotonates an acid only if the base's own conjugate acid has a higher $p K_{a}$ than the acid being attacked. Higher $p K_{a}$ = weaker acid = the proton would rather sit on the alkyne product. The figure lines up all four $p K_{a}$ values on a number line so you can see which base's conjugate acid sits above 25.

Why this step? Acid–base reactions run downhill toward the weaker acid; we compare $p K_{a}$ values (rooted in Hybridisation and s-character for why $25$ ).

Step 2 — Compare each base's conjugate-acid $p K_{a}$ against $25$ .

$N a O H$ : conjugate acid $H_{2} O$ , $p K_{a} = 15.7 < 25$ → fails (water is the stronger acid, equilibrium sits left).
$N a N H_{2}$ : conjugate acid $N H_{3}$ , $p K_{a} = 38 > 25$ → works (equilibrium sits right).
$N a H C O_{3}$ : conjugate acid $H_{2} C O_{3}$ , $p K_{a} = 6.3 < 25$ → fails badly.

Answer: only $N a N H_{2}$ deprotonates the terminal alkyne.

Verify: the "works" test is 38 > 25; the two "fails" are 15.7 < 25 and 6.3 < 25. ✓ Confirms the parent note's " $N a N H_{2}$ not $N a O H$ " rule quantitatively.

Ex 7 — Cell G: acetylide alkylation (chain building, all three halide classes)

Forecast: we need to graft a $C H_{3} C H_{2} -$ group onto ethyne. Which reagents, and how does the answer change as the halide goes primary → secondary → tertiary?

Step 1 — Deprotonate ethyne to the acetylide. $H C \equiv C H N a N H_{2} H C \equiv C^{-} N a^{+}$ Why this step? Ex 6 showed $N a N H_{2}$ is the base that actually works; the acetylide is the nucleophile we need. The figure shows this acetylide attacking each of the three halide classes and the branching outcomes.

Step 2 — Primary halide: clean $S_{N} 2$ . $H C \equiv C^{-} N a^{+} + C H_{3} C H_{2} B r S_{N} 2 H C \equiv C - C H_{2} C H_{3}$ Why this step? The acetylide is a strong nucleophile that does a clean backside $S_{N} 2$ (see SN2 reactions) on an unhindered primary carbon. This gives our target 1-butyne.

Step 3 — Secondary halide: $S_{N} 2$ vs $E 2$ competition. With 2-bromopropane the carbon is more crowded, so backside attack is slower and the acetylide is also a strong base. The two pathways compete:

$S_{N} 2$ → $H C \equiv C - C H (C H_{3})_{2}$ (some substitution product),
$E 2$ → propene $C H_{3} C H = C H_{2}$ + regenerated ethyne (elimination). For a strong, bulky base–nucleophile like acetylide on a secondary substrate, elimination is significant; substitution yields are only moderate. Why this step? Secondary is the borderline case — you must name both competing outcomes, not just one.

Step 4 — Tertiary halide: elimination only. $H C \equiv C^{-} + (C H_{3})_{3} C - B r E 2 (C H_{3})_{2} C = C H_{2} + H C \equiv C H$ The $3^{\circ}$ carbon is far too hindered for $S_{N} 2$ ; the acetylide acts purely as a base → isobutylene, no alkyne coupling. Why this step? It shows the failure mode and why (steric bulk blocks $S_{N} 2$ , so only E2 remains).

Answer: primary halide → 1-butyne cleanly (good $S_{N} 2$ , high yield); secondary halide → a mixture of substitution ( $H C \equiv C - C H (C H_{3})_{2}$ ) and elimination (propene + ethyne), so only moderate yield of the coupled alkyne; tertiary halide → only elimination (isobutylene + ethyne), the alkylation fails entirely. Rule of thumb: use primary halides for acetylide alkylation.

Verify: carbon count for the target: $C_{2}$ (ethyne) $+ C_{2}$ (ethyl) $= C_{4}$ ; 1-butyne is $C_{4} H_{6}$ . ✓

Ex 8 — Cell H: reduction stereochemistry

Forecast: which reducing method delivers both H's to the same face, and which to opposite faces?

Step 1 — cis: Lindlar catalyst. $C H_{3} C \equiv C C H_{3} H_{2}, Lindlar P d / C a C O_{3} c i s C H_{3} C H = C H C H_{3}$ Why this step? Poisoned Pd delivers both hydrogens from the same face ("Lindlar = Left-together = cis"). In the figure, the cis product has both $C H_{3}$ groups on the same side of the double bond.

Step 2 — trans: dissolving metal. $C H_{3} C \equiv C C H_{3} N a / liq. N H_{3} t r an s C H_{3} C H = C H C H_{3}$ Why this step? The radical-anion intermediate adopts the lower-energy trans geometry ("Na/NH₃ = Not-together = trans") — the two $C H_{3}$ groups end on opposite sides.

Answer: (a) $H_{2}$ /Lindlar → cis-but-2-ene; (b) $N a$ /liq. $N H_{3}$ → trans-but-2-ene.

Verify: both products are $C_{4} H_{8}$ (one $π$ bond added one $H_{2}$ to $C_{4} H_{6}$ ). ✓ Same formula, opposite stereochemistry set by the reagent.

Ex 9 — Cell I: word problem, identify the gas

Forecast: what structural feature does the silver test detect? Guess which cylinder reacts.

Step 1 — Recall what the test detects. The Tollens-type reagent forms an insoluble silver acetylide only with a terminal $\equiv C - H$ (an acidic H). Why this step? The reaction is an acid–base + precipitation; no acidic H → no salt → no precipitate.

Step 2 — Match to structures.

1-butyne $C H_{3} C H_{2} C \equiv C H$ has a terminal $\equiv C - H$ → white precipitate $C H_{3} C H_{2} C \equiv C A g ↓$ .
2-butyne $C H_{3} C \equiv C C H_{3}$ is internal, no acidic H → no reaction.

Answer: $Y$ (1-butyne) gives the precipitate; $X$ (2-butyne) does not. The reacting cylinder is the terminal alkyne.

Verify (reasoning check): exactly one of {terminal, internal} has an acidic H → exactly one precipitate. The count of terminal-alkyne reactants here is $1$ . ✓

Ex 10 — Cell J: exam twist, count every equivalent of $H B r$

Forecast: two $π$ bonds, two possible additions. Where does each $B r$ end up — on the same carbon (geminal) or on neighbours (vicinal)? Guess.

Step 1 — First $H B r$ (Markovnikov). $C H_{3} - C \equiv C H H B r C H_{3} - C B r = C H_{2}$ $H^{+}$ adds to the terminal carbon to give the more stable (secondary vinyl) cation; $B r^{-}$ lands on the central carbon. Why this step? Markovnikov puts $B r$ on the more substituted carbon — this is the 1 equivalent answer: 2-bromopropene. The figure tracks both $B r$ atoms onto the central carbon across the two steps.

Step 2 — Second $H B r$ (Markovnikov again). $C H_{3} - C B r = C H_{2} H B r C H_{3} - C B r_{2} - C H_{3}$ The existing $B r$ directs the new $H$ / $B r$ so the second $B r$ also lands on the central carbon. Why this step? The cation is again most stable on the carbon already bearing $B r$ (halogen lone-pair stabilisation), so both $B r$ end up on the same (central) carbon → a geminal dihalide.

Answer: 1 equivalent → 2-bromopropene $C H_{3} C B r = C H_{2}$ ; 2 equivalents → 2,2-dibromopropane $C H_{3} C B r_{2} C H_{3}$ (geminal — both Br on the same carbon, never vicinal).

Verify: atom count for the final product: propyne $C_{3} H_{4} + 2 H B r \to C_{3} H_{6} B r_{2}$ ; 2,2-dibromopropane is $C_{3} H_{6} B r_{2}$ . ✓ Two Markovnikov additions → geminal dihalide, the hallmark pattern.

Ex 11 — Cell K: halogen ( $X_{2}$ ) addition, one vs two equivalents

Forecast: halogen adds across $π$ bonds like $H B r$ but with two halogens each time. What does one equivalent leave, and what does the second do?

Step 1 — First $B r_{2}$ gives a vinyl dihalide (trans). $H C \equiv C H B r_{2} (E)-1,2-dibromoethene B r C H = C H B r$ Why this step? $B r_{2}$ adds across one $π$ bond through a bromonium-type intermediate; anti addition puts the two $B r$ on opposite sides → the trans (E) vinyl dihalide. One $π$ bond remains.

Step 2 — Second $B r_{2}$ saturates the remaining double bond. $B r C H = C H B r B r_{2} B r_{2} C H - C H B r_{2}$ Why this step? The leftover $π$ bond is still nucleophilic (though deactivated by the first two Br), so a second $B r_{2}$ adds to give the tetrahalide 1,1,2,2-tetrabromoethane.

Answer: 1 equiv → (E)-1,2-dibromoethene $B r C H = C H B r$ (trans vinyl dihalide); 2 equiv → 1,1,2,2-tetrabromoethane $B r_{2} C H C H B r_{2}$ .

Verify: atom count for the tetrahalide: $C_{2} H_{2} + 2 B r_{2} = C_{2} H_{2} B r_{4}$ ; 1,1,2,2-tetrabromoethane is $C_{2} H_{2} B r_{4}$ . ✓ This is the classic bromine-water decolourisation test for unsaturation.

Ex 12 — Cell L: oxidative cleavage (ozonolysis)

Forecast: ozonolysis cuts the molecule at the multiple bond. Where does pent-2-yne break, and what does each cut end turn into?

Step 1 — Cut the molecule at the triple bond. Ozonolysis severs the $C \equiv C$ completely, splitting the chain into two pieces at that bond: $C H_{3} - C C - C H_{2} C H_{3}$ Why this step? Unlike addition (which keeps the skeleton whole), cleavage separates the carbon chain — the position of the break reports where the triple bond was. The figure shows the scissors line and the two carboxylic-acid fragments that fall out.

Step 2 — Cap each cut carbon as a carboxylic acid. Each internal alkyne carbon becomes a $- C O O H$ group on oxidative work-up: $\to C H_{3} C O O H + C H_{3} C H_{2} C O O H$ Why this step? An internal $C \equiv C$ carbon carries no H, so oxidation drives it all the way to a carboxylic acid (a terminal $\equiv C H$ carbon would instead give $C O_{2}$ ).

Answer: ethanoic acid $C H_{3} C O O H$ + propanoic acid $C H_{3} C H_{2} C O O H$ . The two different acids ( $C_{2}$ and $C_{3}$ fragments) prove the triple bond sat between carbons 2 and 3.

Verify: carbon bookkeeping — fragments carry $2 + 3 = 5$ carbons, exactly the $C_{5}$ of pent-2-yne. ✓ Different-sized acids confirm an unsymmetric internal alkyne cleaved at the 2,3-position.

Recall

Recall Cover the right — which cell, which answer?

Terminal alkyne + $H_{2} O / H g S O_{4}$ ::: methyl ketone (Markovnikov)
Ethyne + $H_{2} O / H g S O_{4}$ ::: acetaldehyde (only aldehyde)
Symmetric internal alkyne hydration ::: one ketone
Unsymmetric internal alkyne hydration ::: mixture ( $\approx$ 60:40), major = more stabilised ketone
Hydroboration–oxidation of terminal alkyne ::: aldehyde
Base that makes an acetylide ::: $N a N H_{2}$ (not $N a O H$ , $p K_{a}$ 15.7 < 25)
Acetylide alkylation needs which halide ::: primary (secondary = SN2/E2 mix, tertiary → only elimination)
Lindlar vs $N a / N H_{3}$ ::: cis vs trans alkene
Silver-acetylide test detects ::: terminal $\equiv C - H$ only
Propyne + 2 $H B r$ ::: 2,2-dibromopropane (geminal)
Ethyne + 2 $B r_{2}$ ::: 1,1,2,2-tetrabromoethane
Ozonolysis of internal alkyne ::: two carboxylic acids

Connections

Parent topic
Markovnikov rule and carbocation stability (Ex 1, 4, 5, 10)
Keto–enol tautomerism (Ex 1–5)
Aldehydes and Ketones — carbonyl chemistry (the products)
Hybridisation and s-character (Ex 6, root of $p K_{a}$ )
SN2 reactions (Ex 7 alkylation)
Alkenes — electrophilic addition (Ex 10, 11, same mechanism twice)

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