Acidity and acetylide formation

Hybridisation sets the acidity. A carbon–hydrogen bond's acidity is governed by how tightly the carbon holds its bonding electrons — the more tightly, the more stable the carbanion left behind once the proton departs, and the more acidic the C–H bond. Hybrid orbitals differ in $s$-character: $sp$ is $50\%$ $s$, $sp^2$ is $33\%$ $s$, and $sp^3$ is $25\%$ $s$. Because an $s$ orbital keeps electron density closer to the nucleus than a $p$ orbital does, more $s$-character makes an orbital pull electrons in harder — the carbon behaves as if it were more electronegative. In a terminal alkyne, the carbon bonded to the acidic hydrogen is $sp$ hybridised, so it grips the C–H bonding pair more strongly than the $sp^2$ carbon of an alkene or the $sp^3$ carbon of an alkane would. This gives the acidity order

where: the ordering reflects decreasing $s$-character of the carbon orbital forming the C–H bond ($50\%$, $33\%$, $25\%$ respectively), and hence decreasing stability of the conjugate base formed on deprotonation. Terminal alkynes remain only weakly acidic overall — far weaker than water or alcohols — so only strong bases achieve the deprotonation.

Forming the acetylide. Sodium metal or sodium amide strips the terminal proton, generating a sodium acetylide and hydrogen gas or ammonia as the by-product:

The acetylide carbanion is a strong nucleophile: treated with a primary alkyl halide it displaces the halide, forming a new C–C bond and building a longer or branched alkyne — a standard route from a short alkyne to a bigger one.

The diagnostic precipitate test. Ammoniacal silver nitrate and ammoniacal cuprous chloride both react with a terminal C–H to give an insoluble metal acetylide:

Silver acetylide precipitates white; copper(I) acetylide precipitates red/reddish-brown. Internal alkynes, which have alkyl groups on both triple-bond carbons and so no terminal C–H, give no reaction at all — which is exactly what makes this test diagnostic.

The reasoning chain behind the acidity order is short and worth stating explicitly, since it is exactly what an exam answer must reproduce. Higher $s$-character means the orbital's electron density is, on average, closer to the nucleus, because $s$ orbitals are spherically concentrated around the nucleus while $p$ orbitals have a node there. A carbon atom using an $sp$ hybrid orbital to bond to hydrogen therefore behaves as an effectively more electronegative carbon than one using an $sp^2$ or $sp^3$ hybrid, even though all three are the same element. A more electronegative carbon polarises the C–H bond more strongly, pulling electron density toward itself and away from hydrogen, and — crucially — it stabilises the negative charge left on carbon once the proton is removed. Since acidity is set by how favourable it is for a species to lose a proton and leave behind a stable conjugate base, the $sp$ carbanion (the acetylide anion) is the most stable of the three, the $sp^3$ carbanion the least, and the acidity order follows directly: $sp > sp^2 > sp^3$. Note what this argument does not invoke — it is not about the triple bond being "electron-withdrawing" in some vague inductive sense; the mechanistic reason expected at this level is specifically the s-character/hybridisation argument above.

Problem: Two gaseous hydrocarbons, X and Y, both have molecular formula $\text{C}_4\text{H}_6$ and are known to be alkynes. X gives a white precipitate with ammoniacal $\text{AgNO}_3$; Y gives no visible reaction. Identify X and Y, and explain why propyne's terminal C–H ($\text{CH}_3-\text{C}\equiv\text{CH}$) is more acidic than a C–H of propane.

Step 1. $\text{C}_4\text{H}_6$ matches the alkyne general formula $\text{C}_n\text{H}_{2n-2}$ for $n=4$, so both X and Y are butynes: but-1-yne, $\text{CH}_3\text{CH}_2-\text{C}\equiv\text{CH}$, or but-2-yne, $\text{CH}_3-\text{C}\equiv\text{C}-\text{CH}_3$.

Step 2. But-1-yne has its triple bond at the chain end, so one triple-bond carbon carries a hydrogen directly — it is a terminal alkyne. But-2-yne has the triple bond between C2 and C3, with alkyl groups on both triple-bond carbons and no terminal C–H — it is an internal alkyne.

Step 3. Only a terminal alkyne can lose its terminal proton to $Ag(NH_3)_2^+$ and form the white silver acetylide precipitate; an internal alkyne has no such proton to lose, so it gives no reaction.

Step 4. Therefore X is but-1-yne (the terminal alkyne, giving the white precipitate) and Y is but-2-yne (the internal alkyne, unreactive in this test).

Step 5. For propyne versus propane: propyne's terminal C–H sits on an $sp$-hybridised carbon ($50\%$ s-character), while every C–H in propane sits on an $sp^3$-hybridised carbon ($25\%$ s-character). The $sp$ carbon holds its bonding electrons closer to the nucleus, behaves as more electronegative, and stabilises the acetylide anion formed on deprotonation better than an $sp^3$ carbon stabilises an ordinary alkyl carbanion. Answer: X = but-1-yne, Y = but-2-yne; propyne's terminal C–H is more acidic than propane's C–H because it is attached to an sp carbon of higher s-character, which better stabilises the conjugate-base carbanion.

• Calling terminal alkynes "strongly acidic" — they are only weakly acidic, much weaker than water and far weaker than carboxylic acids; deprotonating them needs a strong base like Na or $\text{NaNH}_2$, not a mild one.
• Assuming any alkyne reacts with ammoniacal $\text{AgNO}_3$/$\text{Cu}_2\text{Cl}_2$ — only terminal alkynes (a C–H directly on a triple-bond carbon) give the precipitate; internal alkynes such as but-2-yne do not react.
• Swapping the precipitate colours — silver acetylide is white, copper(I) acetylide is red/reddish-brown; mixing these up is a common recall slip.
• Explaining the acidity as "the triple bond withdraws electrons" instead of naming the real mechanism — the higher s-character of the sp orbital, and the resulting closer-to-nucleus electron density and greater effective electronegativity, is the explanation the syllabus expects.

Ask the tutor: if the s-character argument still feels abstract, ask the AI tutor to walk it back through the hybridisation orbital diagrams you already drew for sp, sp2 and sp3 carbon in Chemical Bonding.