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

Intuition How to use this page

Work top to bottom. Each problem is one level harder than the last. Cover the solution, try it yourself, then open the collapsible [!recall]- to check. After each level a [!mistake] callout shows the trap students actually fall into — read it even if you got the answer right.

A few symbols recur below — here they are once, in plain words:

$R$ = a placeholder for "any carbon group" (like $C H_{3} -$ or $C H_{3} C H_{2} -$ ); we write $R$ so a rule works for the whole family, not one molecule.
$s p, s p^{2}, s p^{3}$ = the three ways a carbon "mixes" its one $s$ orbital with its $p$ orbitals. $s p$ mixes 1 $s$ + 1 $p$ (so it is $50%$ $s$ ), $s p^{2}$ mixes 1 $s$ + 2 $p$ ( $33%$ $s$ ), $s p^{3}$ mixes 1 $s$ + 3 $p$ ( $25%$ $s$ ). Triple-bond carbons are $s p$ . See Hybridisation and s-character.
Degree of unsaturation = how many H atoms are "missing" versus a fully saturated chain; each ring or double bond removes $2$ H, a triple bond removes $4$ H.

Parent: Alkynes topic note.

Level 1 — Recognition

(Can you name, classify, and read a structure at sight?)

L1.1 Molecular formula

An open-chain alkyne has $7$ carbon atoms. How many hydrogen atoms does it have? Give the molecular formula.

Recall Solution

The general formula for an open-chain alkyne is $== C_{n} H_{2 n - 2} ==$ . Here $n$ is just the number of carbons. A saturated chain (all single bonds) is $C_{n} H_{2 n + 2}$ ; every ring or double bond you add removes $2$ H (a degree of unsaturation), and a triple bond is two of those at once, so it removes four H. Put $n = 7$ : $H = 2 n - 2 = 2 (7) - 2 = 12$ Answer: $C_{7} H_{12}$ .

L1.2 Terminal or internal?

Classify each and say which one(s) have an acidic hydrogen: (a) $C H_{3} - C \equiv C H$ (b) $C H_{3} - C \equiv C - C H_{3}$ (c) $H C \equiv C H$

Recall Solution

A terminal alkyne has the triple bond at a chain end, so one of the triple-bond ( $s p$ ) carbons carries a hydrogen ( $\equiv C - H$ ). That H is the acidic one. ( $s p$ = the carbon mixed its orbitals as 1 $s$ + 1 $p$ , giving $50%$ s-character — the reason this H is acidic, unpacked in L3.1.)

(a) propyne — terminal, has acidic $\equiv C - H$ . ✔
(b) 2-butyne — internal, both $s p$ carbons carry $C H_{3}$ , no acidic H. ✗
(c) ethyne — terminal on both ends, has acidic H. ✔ Answer: (a) and (c) have acidic H; (b) does not.

Level 2 — Application

(Apply one rule to predict a product.)

L2.1 Markovnikov hydration

Predict the organic product of: $C H_{3} - C H_{2} - C \equiv C H H_{2} O H_{2} S O_{4}, H g S O_{4} ?$

Recall Solution

Step 1 — add water Markovnikov. OH goes to the more substituted triple-bond carbon (here the internal one, bearing the ethyl group $C H_{3} C H_{2} -$ ; recall $R$ just means "a carbon group"). Why that carbon? The acid/mercury first makes $H^{+}$ (or $H g^{2 +}$ ) add to the terminal $C H$ ; this leaves a positive charge on the internal carbon, which is the more stable cation because the neighbouring carbon group donates electron density to soothe the charge (more substituted = more stable — see Markovnikov rule and carbocation stability). Water then attacks that positive carbon, so OH ends up there. The enol formed is $C H_{3} C H_{2} - O H C = C H_{2}$ Step 2 — tautomerise. The enol ( $C = C - O H$ ) flips to the far more stable keto form via Keto–enol tautomerism (the $C = O$ bond, $\approx 360$ kJ/mol, beats the $C = C$ , $\approx 270$ kJ/mol). $C H_{3} C H_{2} - C ∥ O - C H_{3}$ Answer: butan-2-one (a methyl ketone), $C_{4} H_{8} O$ .

L2.2 Which base?

You want to convert $C H_{3} C \equiv C H$ into its acetylide $C H_{3} C \equiv C^{-}$ . You have $N a O H$ and $N a N H_{2}$ . Which works, and why — using $p K_{a}$ values ( $p K_{a}$ : propyne $\approx 25$ , water $= 15.7$ , ammonia $= 38$ )?

Recall Solution

A base deprotonates an acid only if the base's conjugate acid is weaker (higher $p K_{a}$ ) than the acid attacked. Equilibrium favours forming the weaker acid.

$N a O H$ : conjugate acid is water, $p K_{a} = 15.7 < 25$ . Water is the stronger acid, so equilibrium lies left — no acetylide. ✗
$N a N H_{2}$ : conjugate acid is ammonia, $p K_{a} = 38 > 25$ . Ammonia is the weaker acid, so the proton transfers to $N H_{2}^{-}$ — acetylide forms. ✔ Answer: $N a N H_{2}$ . Rule of thumb: you need a base whose conjugate-acid $p K_{a}$ exceeds $25$ .

Level 3 — Analysis

(Compare, rank, or explain WHY.)

L3.1 Rank the acidity

Rank these by acidity (most acidic first) and justify with hybridisation: $C H_{3} - C H_{3}, C H_{2} = C H_{2}, H C \equiv C H$

Recall Solution

Acidity = stability of the conjugate base (the carbanion after losing $H^{+}$ ). The lone pair sits in the carbon's hybrid orbital. s-character = the fraction of that orbital that is $s$ -type; $s$ orbitals hug the nucleus, so more s-character pulls the lone pair closer to the nucleus, holding it tighter → more stable anion → stronger acid. See Hybridisation and s-character. $50% s s p > 33% s s p^{2} > 25% s s p^{3}$ Answer (most → least acidic): $H C \equiv C H (s p, p K_{a} \approx 25) > C H_{2} = C H_{2} (s p^{2}, \approx 44) > C H_{3} C H_{3} (s p^{3}, \approx 50)$

L3.2 Same molecule, two reagents

Compound $C H_{3} - C H_{2} - C \equiv C H$ is treated separately with: (a) $H_{2} O / H_{2} S O_{4} / H g S O_{4}$ (b) $B H_{3}$ then $H_{2} O_{2} / O H^{-}$ . Give both products and explain why they differ.

Recall Solution

Both add water across $C \equiv C$ , but with opposite regiochemistry, so they give an enol on different carbons. (a) Markovnikov (mercury-catalysed): $H^{+}$ adds first, leaving the positive charge on the more substituted internal carbon (more stable cation), so OH → internal carbon → enol $C H_{3} C H_{2} C (O H) = C H_{2}$ → tautomerises to the ketone butan-2-one $C H_{3} C H_{2} C O C H_{3}$ . (b) Anti-Markovnikov (hydroboration–oxidation): here the boron of $B H_{3}$ is the atom that adds to carbon, and boron is electron-poor (electrophilic) yet bulky. Why the terminal carbon? Two reasons pointing the same way: (i) sterics — the small $- C H_{2}$ terminal carbon is far less crowded, so the bulky boron slots onto it easily; (ii) electronics — boron adds so as to leave any partial positive charge on the more substituted internal carbon (the more stabilised position), which means boron sits on the terminal carbon. In the oxidation step, OH replaces boron at the same carbon it was on — the terminal one. So OH ends up terminal → enol $C H_{3} C H_{2} C H = C H (O H)$ → tautomerises to the aldehyde butanal $C H_{3} C H_{2} C H_{2} C H O$ . Answer: (a) butan-2-one (ketone); (b) butanal (aldehyde). Mark makes Ketones, Boron makes Aldehydes.

Level 4 — Synthesis

(Design a route; combine several steps.)

L4.1 Build a bigger alkyne

Starting from ethyne ( $H C \equiv C H$ ) and any alkyl halides, synthesise pent-1-yne ( $C H_{3} C H_{2} C H_{2} - C \equiv C H$ ).

Recall Solution

Pent-1-yne is a terminal alkyne; we must attach a propyl group ( $C H_{3} C H_{2} C H_{2} -$ ) to one carbon of ethyne while keeping the other $\equiv C - H$ . Step 1 — make the acetylide. Deprotonate one end of ethyne: $H C \equiv C H N a N H_{2} H C \equiv C^{-} N a^{+}$ Step 2 — alkylate by $S_{N} 2$ . The acetylide is a strong nucleophile; use a primary halide (1-bromopropane) so $S_{N} 2$ wins over elimination. See SN2 reactions. $H C \equiv C^{-} + C H_{3} C H_{2} C H_{2} - B r ⟶ H C \equiv C - C H_{2} C H_{2} C H_{3} + B r^{-}$ Answer: pent-1-yne. Reagents in order: (1) $N a N H_{2}$ ; (2) $C H_{3} C H_{2} C H_{2} B r$ .

L4.2 From an alkane skeleton to a ketone

Convert 1,2-dibromopropane ( $C H_{3} - C H B r - C H_{2} B r$ ) into acetone ( $C H_{3} C O C H_{3}$ ).

Recall Solution

Plan: dibromide → alkyne → hydration. Step 1 — double dehydrohalogenation. Alc. KOH removes the first HX (easy); $N a N H_{2}$ removes the second (hard — a vinylic/ $s p^{2}$ C–Br needs a very strong base). This builds the triple bond: $C H_{3} C H B r C H_{2} B r KOH (alc.) C H_{3} C B r = C H_{2} N a N H_{2} C H_{3} - C \equiv C H$ Step 2 — Markovnikov hydration. Propyne + water (Hg-catalysed) → enol on the internal carbon → acetone: $C H_{3} C \equiv C H H_{2} O H_{2} S O_{4}, H g S O_{4} [C H_{3} C (O H) = C H_{2}] \to C H_{3} - C ∥ O - C H_{3}$ Answer: acetone (propan-2-one).

Level 5 — Mastery

(Multi-step reasoning, exceptions, and quantitative comparison.)

L5.1 The ethyne exception, justified

Two students hydrate (a) ethyne and (b) propyne under $H_{2} O / H_{2} S O_{4} / H g S O_{4}$ . One gets an aldehyde, the other a ketone. Say which is which and explain why the same reagent gives different classes of product.

Recall Solution

The reagent is Markovnikov in both cases — OH prefers the more substituted carbon. The difference is symmetry. (a) Ethyne $H C \equiv C H$ : both carbons are identical, so "more substituted" is meaningless. The enol $C H_{2} = C H - O H$ can only tautomerise to $C H_{3} C H O$ — acetaldehyde, an aldehyde. This is the unique exception. (b) Propyne $C H_{3} C \equiv C H$ : the internal carbon bears $C H_{3}$ , so OH lands there. Enol $C H_{3} C (O H) = C H_{2}$ → acetone, a ketone. Answer: ethyne → acetaldehyde (aldehyde); propyne → acetone (ketone). All higher terminal alkynes give methyl ketones; only ethyne gives an aldehyde by this method.

L5.2 Distinguish three isomers in the lab

You have three colourless liquids of formula $C_{4} H_{6}$ : but-1-yne ( $C H_{3} C H_{2} C \equiv C H$ ), but-2-yne ( $C H_{3} C \equiv C C H_{3}$ ), and 1,3-butadiene ( $C H_{2} = C H - C H = C H_{2}$ ). Using ammoniacal silver nitrate $[A g (N H_{3})_{2}]^{+}$ , which give a precipitate and which do not? Write the equation for the positive test.

Recall Solution

The Tollens-type reagent $[A g (N H_{3})_{2}]^{+}$ reacts only with an acidic terminal $\equiv C - H$ , replacing it by silver to give an insoluble silver acetylide.

But-1-yne — terminal, has acidic $\equiv C - H$ : white/grey precipitate. ✔ $C H_{3} C H_{2} C \equiv C H + [A g (N H_{3})_{2}]^{+} \to C H_{3} C H_{2} C \equiv C A g ↓ + N H_{4}^{+} + N H_{3}$
But-2-yne — internal, no acidic H: no precipitate. ✗
1,3-Butadiene — an alkene (see Alkenes — electrophilic addition), no $\equiv C - H$ : no precipitate. ✗ Answer: only but-1-yne gives a precipitate; the test cleanly picks out the terminal alkyne.

L5.3 Quantitative acidity check

Ethyne has $p K_{a} \approx 25$ ; water has $p K_{a} = 15.7$ . For the reaction $H C \equiv C H + O H^{-} ⇌ H C \equiv C^{-} + H_{2} O$ estimate the equilibrium constant $K$ and state whether $N a O H$ is a practical base here.

Recall Solution

For a proton-transfer equilibrium, the working rule is $lo g_{10} K = p K_{a} (acid consumed) - p K_{a} (acid formed) .$ Read the forward reaction: the acid consumed is ethyne ( $p K_{a} = 25$ ), and the acid formed is water ( $p K_{a} = 15.7$ ). So $lo g_{10} K = p K_{a} (ethyne) - p K_{a} (H_{2} O) = 25 - 15.7 = 9.3 ?$ Careful with direction. The formula subtracts the product acid's $p K_{a}$ from the reactant acid's $p K_{a}$ only when you want $K$ for making the weaker acid. Doing it cleanly by definition: $K = \frac{K _{a} ( ethyne )}{K _{a} ( H _{2} O )} = 1 0^{- p K_{a} (ethyne)} /1 0^{- p K_{a} (H_{2} O)} = 1 0^{p K_{a} (H_{2} O) - p K_{a} (ethyne)}$ . $lo g_{10} K = p K_{a} (H_{2} O) - p K_{a} (ethyne) = 15.7 - 25 = - 9.3$ $K = 1 0^{- 9.3} \approx 5 \times 1 0^{- 10}$ This is astronomically small — the equilibrium sits essentially entirely on the left. $N a O H$ is not a practical base for making acetylide. Contrast $N a N H_{2}$ (conjugate acid $N H_{3}$ , $p K_{a} = 38$ ): $lo g_{10} K = 38 - 25 = + 13$ , i.e. $K \approx 1 0^{13}$ , effectively complete.

Wrap-up recall

Recall One-line takeaways per level

L1 — alkyne formula ::: $C_{n} H_{2 n - 2}$ (triple bond = 2 degrees unsaturation, $- 4$ H)
L2 — Markovnikov hydration of a higher terminal alkyne ::: methyl ketone
L3 — anti-Markovnikov (hydroboration) ::: aldehyde
L4 — acetylide alkylation ::: $N a N H_{2}$ then primary R–X ( $S_{N} 2$ )
L5 — ethyne is the lone hydration exception ::: gives acetaldehyde
L5 — why NaOH fails on ethyne ::: $K \approx 1 0^{- 9.3}$ , equilibrium far left

Connections

Markovnikov rule and carbocation stability
Keto–enol tautomerism
Hybridisation and s-character
SN2 reactions
Aldehydes and Ketones — carbonyl chemistry
Alkenes — electrophilic addition