Acidic Nature of Terminal Alkynes

What Makes a Terminal Alkyne

An alkyne carries a carbon–carbon triple bond, $\ce{-C#C-}$. The two carbons of that triple bond are sp-hybridised: each forms one $\sigma$ bond by an sp hybrid orbital and contributes two unhybridised p orbitals to the two $\pi$ bonds. Whether the molecule is acidic depends entirely on what sits on those sp carbons.

A terminal alkyne has at least one hydrogen bonded directly to a triply bonded carbon, written $\ce{-C#C-H}$. Ethyne, $\ce{HC#CH}$, has two such hydrogens; propyne, $\ce{CH3-C#CH}$, has one. An internal alkyne such as but-2-yne, $\ce{CH3-C#C-CH3}$, has no $\ce{#C-H}$ — both sp carbons carry alkyl groups. As NCERT §9.4.4 stresses, only the hydrogen on the triply bonded carbon is acidic; the alkyl C–H hydrogens elsewhere in the molecule are not.

That distinction — terminal versus internal — is the hinge of this entire topic. Every reaction that follows, from the liberation of dihydrogen with sodium to the coloured precipitates with silver and copper salts, is a direct consequence of whether or not the molecule carries that single acidic $\ce{#C-H}$. Get the classification right and the chemistry follows automatically; mistake an internal alkyne for a terminal one and a whole cluster of NEET answers flips.

Why the Terminal C–H Is Acidic

Sodium metal and sodamide ($\ce{NaNH2}$) are strong bases. They react with ethyne to give a sodium acetylide and liberate dihydrogen — reactions that simply do not happen with ethene or ethane. That observation alone establishes that ethyne is acidic relative to its alkene and alkane counterparts. The structural reason lies in hybridisation.

In ethyne the hydrogen is attached to an sp-hybridised carbon; in ethene it is attached to an sp² carbon, and in ethane to an sp³ carbon. An sp hybrid orbital has the maximum percentage of s-character (50%). Because an s orbital is spherical and lies closer to the nucleus than a p orbital, more s-character means the bonding electrons are held nearer the carbon nucleus — so the sp carbon is the most electronegative of the three.

That high electronegativity pulls the shared electron pair of the C–H bond towards the carbon, polarising the bond and leaving the hydrogen with a more positive partial charge. The proton can therefore be detached more easily. NCERT puts it directly: in ethyne, "hydrogen atoms can be liberated as protons more easily as compared to ethene and ethane."

There is a second, complementary way to see the same effect — through the anion that is left behind. When the proton departs, the bonding electron pair stays on carbon as a lone pair in the acetylide ion, $\ce{HC#C^-}$. That lone pair sits in the sp hybrid orbital. Because the sp orbital holds electrons close to the nucleus, it stabilises the negative charge better than an sp² or sp³ orbital would. A more stable conjugate base means the parent acid gives up its proton more readily. Both viewpoints — the polarised C–H bond and the stabilised acetylide anion — converge on the same conclusion: the sp $\ce{#C-H}$ is the most acidic C–H of the three hydrocarbon families.

It is worth being precise about what "electronegativity" means here. The intrinsic electronegativity of the carbon atom does not change; what changes is the electronegativity of the hybrid orbital it uses for the C–H bond. An sp orbital is effectively more electron-attracting than an sp² orbital, which in turn beats an sp³ orbital, simply because of how much s-character each contains. This orbital-electronegativity idea is the engine behind every acidity comparison in this topic, and behind the bond-length and bond-strength trends seen across ethane, ethene and ethyne.

The s-Character Acidity Ladder

The acidity of a C–H bond rises with the s-character of the carbon holding it. NIOS tabulates the percentages explicitly, and they fall straight out of the hybridisation: an sp orbital is half s and half p (50%), an sp² is one-third s (33.3%), and an sp³ is one-quarter s (25%). The figure below stacks these as a ladder.

This yields the two acidity orders NCERT highlights. Across hybridisation types the order is $\ce{HC#CH > H2C=CH2 > CH3-CH3}$. Among alkynes themselves, a terminal alkyne with two acidic hydrogens beats one with a single acidic hydrogen, and an internal alkyne with none sits far below both: $\ce{HC#CH > CH3-C#CH >> CH3-C#C-CH3}$.

The second order rewards careful reading. Ethyne has two acidic $\ce{#C-H}$ hydrogens, propyne has one (the methyl group has replaced the other), and but-2-yne has none. So as we move from ethyne to propyne to but-2-yne we strip away acidic hydrogens one at a time, and the acidity collapses. The double inequality sign before but-2-yne in NCERT's notation is deliberate: the drop from a terminal to an internal alkyne is not a small gradation but a qualitative loss — the very feature the acetylide tests are built to detect.

Reaction with Active Metals and Bases

The acidic hydrogen is removable by strong bases, generating acetylide (ethynide) anions. The base must be strong because, in absolute terms, a terminal alkyne is still a very feeble acid; the bases that do the job — sodium metal, sodamide, and the heavy-metal ammine complexes — are far stronger than anything that would touch an alkane or alkene C–H. With sodium metal, ethyne first forms monosodium ethynide, releasing half a mole of dihydrogen per acidic hydrogen:

$$\ce{HC#CH + Na -> HC#CNa + 1/2 H2}$$

Because both hydrogens of ethyne are acidic, excess sodium converts the monosodium salt further into disodium ethynide:

$$\ce{HC#CNa + Na -> NaC#CNa + 1/2 H2}$$

Sodamide acts the same way, abstracting the proton to leave the acetylide and ammonia. For propyne only the $\ce{#C-H}$ hydrogen is touched, giving sodium propynide:

$$\ce{CH3-C#CH + NaNH2 -> CH3-C#CNa + NH3}$$

The acetylide anion is also a useful carbon nucleophile: alkylating sodium acetylide with a primary alkyl halide builds a longer-chain alkyne, the basis of the classic NEET 2016 acetylide-alkylation problem in the snapshot below. Treating sodium acetylide with ethyl bromide, for instance, extends ethyne into but-1-yne, and a second round of deprotonation and alkylation extends it again. The chemistry only works because the starting material is a terminal alkyne whose acidic hydrogen can first be removed to make the nucleophile — yet another place where the acidity of the $\ce{#C-H}$ does real synthetic work, not just a qualitative test.

One caution on the metal-acetylide salts: dry silver and copper acetylides are sensitive and can decompose explosively, so in the laboratory they are destroyed with dilute acid as soon as the test result is read. For NEET the examinable points are the formation, the colours, and the structural requirement of a terminal $\ce{#C-H}$, all of which flow directly from the acidity of that hydrogen.

Acetylide Precipitation Tests

The same acidic hydrogen lets terminal alkynes form insoluble heavy-metal acetylides, which is the qualitative test most asked in exams. Passing the gas through an ammoniacal silver nitrate solution gives a white precipitate of silver acetylide; passing it through ammoniacal cuprous chloride gives a red precipitate of copper acetylide. For ethyne:

$$\ce{HC#CH + 2[Ag(NH3)2]+ -> AgC#CAg v + 2NH4+ + 2NH3}$$

$$\ce{HC#CH + 2[Cu(NH3)2]+ -> CuC#CCu v + 2NH4+ + 2NH3}$$

Terminal versus Internal Alkynes

The acetylide and metal tests are decisive precisely because internal alkynes fail them. But-1-yne, $\ce{CH3CH2-C#CH}$, has a terminal $\ce{#C-H}$ and so gives the white and red precipitates and reacts with sodamide. But-2-yne, $\ce{CH3-C#C-CH3}$, has its triple bond buried between two methyl groups, no $\ce{#C-H}$ at all, and so gives a negative result on every acetylide-forming test even though both are C₄H₆ alkynes.

This is the practical payoff of the whole topic. Bromine water and Baeyer's reagent confirm that a molecule is unsaturated, but they respond identically to alkenes and to either kind of alkyne, so they cannot pin down which alkyne is present. The acetylide tests add the missing discrimination: a positive silver or copper precipitate is, in effect, a direct chemical readout of an acidic terminal hydrogen. Run in sequence, the four reagents in the table above can sort an unknown gas into alkane, alkene, terminal alkyne, or internal alkyne — a classic identification chain that NEET likes to compress into a single statement-based question.

A useful mental check: count the hydrogens directly bonded to the triple-bonded carbons. If that count is zero, the molecule is internal and every acetylide test reads negative; if it is one or two, the molecule is terminal and the tests read positive. Substituents elsewhere — methyl, ethyl, phenyl — never make a difference, because their hydrogens sit on sp³ or sp² carbons and carry no measurable acidity.

Worked Reasoning