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

Internal alkynes such as 2-butyne give a white precipitate with ammoniacal silver nitrate. ::: False. The silver test needs an acidic $\equiv C-H$ to replace with silver; 2-butyne ($C{H}_{3}C\equiv CC{H}_3$) has no H on the triple-bond carbons, so nothing precipitates. An alkyne is always more acidic than an alkene of the same carbon count. ::: True in the sense that matters: the terminal $sp$ C–H ($p{K}_a\approx 25$) is more acidic than a vinyl $s{p}^2$ C–H ($p{K}_a\approx 44$), because $50\%$ $s$-character holds the conjugate-base lone pair tighter than $33\%$ does. Sodium hydroxide is a strong enough base to convert ethyne quantitatively into sodium acetylide. ::: False. Water ($p{K}_{a}15.7$) is far more acidic than ethyne ($p{K}_{a}25$), so $O{H}^{-}$ prefers to stay $O{H}^{-}$; the equilibrium sits on the alkyne side. You need $NaN{H}_2$ ($N{H}_3$, $p{K}_{a}38$). Markovnikov hydration of any terminal alkyne gives a ketone. ::: False for ethyne only. Every higher terminal alkyne gives a methyl ketone, but ethyne's two carbons are equivalent, so its enol can only tautomerise to acetaldehyde (an aldehyde). Hydroboration–oxidation and $H_{2}O/H_{2}S{O}_4/HgS{O}_4$ give the same product from 1-butyne. ::: False. Mercuric-acid hydration is Markovnikov → butan-2-one (ketone); hydroboration is anti-Markovnikov → butanal (aldehyde). Same "add water," opposite regiochemistry. See Aldehydes and Ketones — carbonyl chemistry. The enol is the final, isolable product of alkyne hydration. ::: False. The enol ($C=C-OH$) is a fleeting intermediate; it tautomerises to the keto form because the new $C=O$ $\pi$ bond ($\approx 360$ kJ/mol) is far stronger than the $C=C$ it replaces ($\approx 270$). See Keto–enol tautomerism. Lindlar's catalyst and $Na/\text{liq. }N{H}_3$ both stop reduction at the alkene stage, so they are interchangeable. ::: False. Both stop at the alkene, but stereochemistry differs: Lindlar delivers two H's to the same face → cis; dissolving-metal reduction goes through a trans radical anion → trans.

"The acetylide ion does $S_{N}2$ well on tert-butyl bromide to build a bulky alkyne." ::: Error: substrate. Acetylide is a strong base and nucleophile; on a tertiary halide it forces elimination, not $S_{N}2$. Alkylation of acetylides works only on primary (unhindered) R–X. See SN2 reactions. "Both HX eliminations from a vic-dihalide use alcoholic KOH." ::: Error: second base. The first HX leaves with alc. KOH giving a vinyl halide, but the second elimination from an $s{p}^2$ (vinyl) C–X is hard and needs the much stronger $NaN{H}_2$. "In $C{H}_3-C\equiv CH+HBr$, Br adds to the terminal $CH$ because it is less hindered." ::: Error: it's charge, not sterics. $H^{+}$ adds first to give the more stable (more substituted) vinyl cation; $B{r}^{-}$ then attaches to that cationic carbon, putting Br on the more substituted carbon — Markovnikov. See Markovnikov rule and carbocation stability. "Ethyne's acidity comes from resonance stabilising the carbanion." ::: Error: wrong cause. There is no resonance to delocalise the lone pair here. The stabilisation is purely hybridisation: the $sp$ orbital's $50\%$ $s$-character holds the lone pair close to the nucleus. "$Ca{C}_2+H_{2}O$ gives ethene." ::: Error: product. Carbide $C_2^{2-}$ is the conjugate base of ethyne; water protonates it twice to give ethyne ($C_2{H}_2$), not ethene. "Anti-Markovnikov hydration puts OH on the more substituted carbon." ::: Error: definition flipped. Anti-Markovnikov (hydroboration) puts OH on the less substituted (terminal) carbon — that's exactly why terminal alkynes then give aldehydes.

Why does more s-character make a carbanion more stable, not less? ::: Because a higher fraction of $s$ character means the orbital is rounder and centred closer to the positive nucleus, so the negative lone pair feels a stronger pull and is held tighter — lower energy = more stable. Why does $NaN{H}_2$ work as a base for terminal alkynes while $NaOH$ does not? ::: $NaN{H}_2$'s conjugate acid $N{H}_3$ ($p{K}_{a}38$) is weaker than the alkyne ($p{K}_{a}25$), so deprotonation is favourable; $NaOH$'s conjugate acid water ($p{K}_{a}15.7$) is stronger, so the equilibrium runs backward. Why is a mercury salt ($HgS{O}_4$) needed for Markovnikov hydration? ::: $H{g}^{2+}$ is the electrophile that activates the triple bond (forms a mercurinium bridge) so water can add Markovnikov under mild conditions; plain dilute acid alone hydrates alkynes far too sluggishly. Why does the enol always collapse to the keto form and not the reverse? ::: The keto form trades a weaker $C=C$ + $O-H$ arrangement for a much stronger $C=O$ bond, releasing energy; equilibrium therefore lies overwhelmingly on the keto side. Why can a triple bond add two equivalents of HX while a double bond adds only one? ::: An alkyne has two $\pi$ bonds; the first HX consumes one $\pi$ bond (making a vinyl halide/alkene), and a second HX can then consume the remaining $\pi$ bond. Why does dehydrohalogenation to make an alkyne need heat and a strong base, unlike simple alkene formation? ::: Two consecutive eliminations are required, and the second (from an $s{p}^2$ vinyl halide) has a high barrier — strong base ($NaN{H}_2$) plus heat is needed to push both steps. Why is the $C\equiv C$ bond (120 pm) shorter than $C=C$ (134 pm)? ::: Three bonding interactions (one $\sigma$ + two $\pi$) pull the nuclei closer, and the $sp$ carbons hold their bonding electrons in tighter, higher-s-character orbitals, both shrinking the internuclear distance.

What happens if you try the silver/copper acetylide test on propyne vs 2-butyne? ::: Propyne (terminal, has $\equiv C-H$) gives a precipitate; 2-butyne (internal, no acidic H) gives none — this is precisely how the test tells terminal from internal alkynes. Ethyne is the lone exception in Markovnikov hydration — why, exactly? ::: Its two triple-bond carbons are identical, so "more substituted vs less substituted" is meaningless; the only possible enol tautomerises to acetaldehyde, the sole aldehyde among terminal-alkyne hydrations. Does an internal alkyne like 2-pentyne give a single clean ketone on mercuric-acid hydration? ::: Not necessarily — with an unsymmetrical internal alkyne water can add to either carbon, giving a mixture of two ketones; only symmetric internal alkynes give one product. Can acetylide alkylation build a ring or a quaternary centre directly? ::: No — it needs a primary R–X, and pushing it toward a hindered (secondary/tertiary) centre triggers elimination instead of the desired $S_{N}2$ coupling. What is the product if you fully hydrogenate 2-butyne with excess $H_2/Pt$ (no poison)? ::: The alkane butane — an unpoisoned catalyst does not stop at the alkene; you need Lindlar or $Na/N{H}_3$ to halt at the alkene stage. Is a terminal alkyne's $\equiv C-H$ acidic enough to react with water and change the pH noticeably? ::: No — at $p{K}_a\approx 25$ it is roughly $10^9$ times less acidic than water, so it behaves as a non-acid toward water; you must use a base whose conjugate acid is weaker than the alkyne.