Chemistry · Hydrocarbons

Alkynes — Preparation, Properties, Reactions

Alkynes are unsaturated hydrocarbons of general formula $\ce{C_nH_{2n-2}}$ that carry at least one carbon-carbon triple bond. Following NCERT Class 11 Chemistry §9.4, this note builds the topic from the sp hybridised triple bond outward — through nomenclature, the two standard preparations, electrophilic addition, the Kucherov hydration that splits acetaldehyde from ketones, and the acidic terminal hydrogen. Each of these threads has surfaced in NEET, so the structure here mirrors how the examiner samples the chapter.

What Alkynes Are

Like alkenes, alkynes are unsaturated hydrocarbons, but they carry at least one triple bond between two carbon atoms. Because each triple bond consumes two more hydrogen atoms than a double bond, the number of hydrogen atoms in an alkyne is smaller still than in the corresponding alkene or alkane, and the general formula contracts to $\ce{C_nH_{2n-2}}$.

The first stable member of the series is ethyne, $\ce{HC#CH}$, popularly known as acetylene. Acetylene is used for arc welding in the form of the oxyacetylene flame, produced by mixing acetylene with oxygen. Beyond the welding torch, alkynes are starting materials for a very large number of organic compounds, which is why the class repays close study.

Structure of the Triple Bond

Ethyne is the simplest molecule of the series, and its bonding sets the template for every alkyne. Each carbon atom is sp hybridised, producing two sp orbitals oriented at 180° to one another. One sp orbital of each carbon overlaps head-on with an sp orbital of the neighbouring carbon to form the carbon-carbon sigma ($\sigma$) bond; the second sp orbital overlaps with the 1s orbital of a hydrogen atom to form a C–H sigma bond. The $\ce{H-C-C}$ bond angle is therefore 180°, and the molecule is linear.

Each carbon still holds two unhybridised 2p orbitals, mutually perpendicular and both perpendicular to the C–C axis. These overlap sideways with the matching 2p orbitals of the other carbon to form two pi ($\pi$) bonds. The ethyne molecule thus contains one C–C sigma bond, two C–H sigma bonds, and two C–C pi bonds — three shared electron pairs binding the two carbons. The resulting electron cloud is cylindrically symmetrical about the internuclear axis.

Figure 1 H H C C σ π (vertical 2p) π (horizontal 2p) ∠ H–C–C = 180° (linear)
Figure 1 — The sp framework of ethyne: a linear backbone (180°) with one C–C sigma bond plus two mutually perpendicular pi bonds formed by sideways overlap of unhybridised 2p orbitals.

The three-electron-pair binding has measurable consequences. The C≡C bond enthalpy of 823 kJ mol⁻¹ exceeds that of the C=C double bond (681 kJ mol⁻¹) and the C–C single bond, while the bond length contracts to 120 pm, shorter than both C=C and C–C. The table below collects these comparisons.

Carbon-carbon bondHybridisation at CBond lengthBond enthalpy
C–C (single)sp3longest of the threelowest of the three
C=C (double)sp2intermediate681 kJ mol⁻¹
C≡C (triple)sp120 pm (shortest)823 kJ mol⁻¹ (highest)

Nomenclature and Isomerism

In the common system, alkynes are named as derivatives of acetylene. In the IUPAC system, they are named as derivatives of the corresponding alkane by replacing the suffix -ane with -yne, and the position of the triple bond is indicated by the locant of the first triply bonded carbon.

nStructureCommon nameIUPAC name
2$\ce{HC#CH}$AcetyleneEthyne
3$\ce{CH3-C#CH}$MethylacetylenePropyne
4$\ce{CH3CH2-C#CH}$EthylacetyleneBut-1-yne
4$\ce{CH3-C#C-CH3}$DimethylacetyleneBut-2-yne

Ethyne and propyne each have only one structure, but butyne has two: but-1-yne and but-2-yne, which differ only in the position of the triple bond and are therefore position isomers. With $\ce{C5H8}$ a richer set appears — pent-1-yne, pent-2-yne and 3-methylbut-1-yne. Pent-1-yne and pent-2-yne are position isomers of each other, whereas either of them paired with 3-methylbut-1-yne is a case of chain isomerism.

Worked Example

List the chain isomers of the alkyne $\ce{C6H10}$ and identify the relationship between the straight-chain members.

The continuous-chain hexynes are hex-1-yne $\ce{HC#C-CH2-CH2-CH2-CH3}$, hex-2-yne $\ce{CH3-C#C-CH2-CH2-CH3}$ and hex-3-yne $\ce{CH3-CH2-C#C-CH2-CH3}$. These three differ only in the position of the triple bond, so they are position isomers of one another. Branched skeletons such as 3-methylpent-1-yne, 4-methylpent-1-yne, 4-methylpent-2-yne and 3,3-dimethylbut-1-yne are chain isomers relative to the hexynes.

Preparation of Alkynes

Two routes dominate the NEET syllabus: a bulk industrial method that delivers acetylene itself, and a laboratory dehydrohalogenation that builds the triple bond from a dihalide. A third, building higher alkynes from acetylides, follows from the acidic character discussed later.

From calcium carbide

On an industrial scale, ethyne is prepared by treating calcium carbide with water. The carbide is itself made from quick lime and coke, and the quick lime comes from heating limestone — a clean limestone-to-acetylene chain.

$$\ce{CaCO3 ->[\Delta] CaO + CO2}$$

$$\ce{CaO + 3C ->[\Delta] CaC2 + CO}$$

$$\ce{CaC2 + 2H2O -> Ca(OH)2 + C2H2 ^}$$

From vicinal and geminal dihalides

Vicinal dihalides (halogens on adjacent carbons) are treated with alcoholic potassium hydroxide and undergo dehydrohalogenation. One molecule of hydrogen halide is eliminated to form an alkenyl (vinylic) halide, which on further treatment with sodamide loses a second molecule of hydrogen halide to give the alkyne. Geminal dihalides reach the same triple bond by the analogous double elimination.

$$\ce{CH3-CHBr-CH2Br ->[alc.\ KOH][-HBr] CH3-CH=CHBr ->[NaNH2][-HBr] CH3-C#CH}$$

MethodReagents / conditionsWhat it gives
From calcium carbide$\ce{CaC2}$ + water, room temperatureEthyne (acetylene) only; industrial scale
From vicinal dihalidealc. KOH (1st HX), then $\ce{NaNH2}$ (2nd HX)Any terminal/internal alkyne via double dehydrohalogenation
From geminal dihalidealc. KOH then $\ce{NaNH2}$ (double elimination)Alkyne with triple bond at the original C–X carbon
From lower acetylide$\ce{NaNH2}$, then primary alkyl halideHigher alkyne by C–C chain extension

Physical Properties

The physical trends of alkynes parallel those of alkanes and alkenes. The first three members are gases, the next eight are liquids, and the higher homologues are solids. All alkynes are colourless; ethyne carries a characteristic (garlic-like) odour while the other members are essentially odourless. The molecules are weakly polar, lighter than water and immiscible with it, but soluble in organic solvents such as ethers, carbon tetrachloride and benzene. Melting point, boiling point and density all rise as molar mass increases.

Acidic Character of Terminal Alkynes

Strong bases such as sodium metal and sodamide ($\ce{NaNH2}$) react with ethyne to form sodium acetylide with liberation of dihydrogen. Ethene and ethane show no such reaction, marking ethyne as acidic by comparison. The cause is hybridisation: the hydrogen atoms of ethyne sit on sp carbon (50% s character), the most electronegative of the carbon hybrids, which pulls the C–H bonding pair tightly toward carbon and lets the hydrogen leave as a proton more readily than from $\ce{sp^2}$ ethene or $\ce{sp^3}$ ethane.

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

$$\ce{HC#C^{-}Na^{+} + Na -> Na^{+}\,^{-}C#C^{-}Na^{+} + 1/2 H2 ^}$$

$$\ce{CH3-C#C-H + NaNH2 -> CH3-C#C^{-}Na^{+} + NH3}$$

Only the hydrogen on the triply bonded carbon is acidic, not every hydrogen of the molecule. This is why a terminal alkyne reacts but an internal one such as but-2-yne does not, giving the two acidity orders below.

$$\ce{HC#CH > H2C=CH2 > CH3-CH3}$$

$$\ce{HC#CH > CH3-C#CH >> CH3-C#C-CH3}$$

Go Deeper

The sp-acidity argument and the acetylide alkylation route get a dedicated walkthrough in Acidic Terminal Alkynes.

Addition Reactions

The triple bond lets an alkyne add two molecules of a reagent — dihydrogen, halogen or hydrogen halide. Most of these are electrophilic additions proceeding through a vinylic cation, and for unsymmetrical alkynes the orientation follows Markovnikov's rule: the electrophile's positive part adds to the carbon already bearing more hydrogen.

Addition of dihydrogen

Over a platinum, palladium or nickel catalyst, an alkyne first adds one molecule of $\ce{H2}$ to the alkene and then a second to the alkane.

$$\ce{HC#CH ->[H2][Pt/Pd/Ni] H2C=CH2 ->[H2] CH3-CH3}$$

$$\ce{CH3-C#CH ->[H2][Pt/Pd/Ni] CH3-CH=CH2 ->[H2] CH3-CH2-CH3}$$

Addition of halogens

Two molecules of halogen add successively to give first a dihalo-alkene and then a tetrahalo-alkane. The decolourisation of the reddish-orange bromine-in-$\ce{CCl4}$ solution is the standard test for unsaturation.

$$\ce{HC#CH ->[Br2] CHBr=CHBr ->[Br2] CHBr2-CHBr2}$$

Addition of hydrogen halides

Two molecules of a hydrogen halide (HCl, HBr, HI) add to give a gem-dihalide, in which both halogens land on the same carbon. The second addition obeys Markovnikov's rule.

$$\ce{HC#CH ->[HBr] CH2=CHBr ->[HBr] CH3-CHBr2}$$

Reagent (2 equivalents)Final product classDiagnostic point
$\ce{H2}$ / Pt, Pd or NiAlkaneGoes via the alkene intermediate
$\ce{X2}$ ($\ce{Br2}$ in $\ce{CCl4}$)TetrahaloalkaneDecolourises bromine water — unsaturation test
$\ce{HX}$ (HCl, HBr, HI)gem-DihalideMarkovnikov orientation; both X on one carbon
$\ce{H2O}$ / $\ce{HgSO4}$, dil. $\ce{H2SO4}$Carbonyl compoundKucherov reaction (see below)

Hydration — The Kucherov Reaction

Alkynes are immiscible with water and do not react with it directly. However, on warming with mercuric sulphate and dilute sulphuric acid at 333 K, one molecule of water adds across the triple bond. The first-formed product is an unstable enol that immediately tautomerises to a carbonyl compound. This is the Kucherov reaction, and the identity of the carbonyl is the most heavily examined point in the whole subtopic.

With acetylene the addition can only place the OH on a terminal carbon, so the enol $\ce{CH2=CH-OH}$ rearranges to acetaldehyde. With every higher alkyne, Markovnikov addition puts the OH on the more substituted carbon, and tautomerisation yields a ketone.

$$\ce{HC#CH + H2O ->[HgSO4,\ dil.\ H2SO4][333\ K] [CH2=CH-OH] -> CH3-CHO}$$

$$\ce{CH3-C#CH + H2O ->[HgSO4,\ dil.\ H2SO4][333\ K] [CH3-C(OH)=CH2] -> CH3-CO-CH3}$$

Figure 2 Alkyne R–C≡C–H H₂O HgSO₄ / 333K Enol R–C(OH)=CH₂ tautomerise Carbonyl acetaldehyde (R=H) or ketone
Figure 2 — Kucherov hydration: water adds across the triple bond to a transient enol, which tautomerises to a carbonyl. R = H (acetylene) gives an aldehyde; any other R gives a ketone.
NEET Trap

Acetylene gives an aldehyde — every other alkyne gives a ketone

The single most repeated error in Kucherov questions is writing a ketone for acetylene. Because ethyne is symmetrical with no carbon bearing two carbon substituents, its only carbonyl is the aldehyde $\ce{CH3CHO}$ (acetaldehyde). For propyne and all higher alkynes, Markovnikov addition forces the OH onto the internal carbon, so tautomerisation delivers a ketone.

$\ce{HC#CH ->[H2O/HgSO4] CH3CHO}$ (aldehyde)  |  $\ce{CH3-C#CH ->[H2O/HgSO4] CH3COCH3}$ (ketone).

Polymerisation

Ethyne polymerises in two distinct ways. Under suitable conditions, linear polymerisation gives polyacetylene (polyethyne), a high-molecular-weight polyene with repeating $\ce{-(CH=CH-CH=CH)-}$ units; under special doping conditions thin films of this polymer conduct electricity and have been used as battery electrodes.

Passing ethyne through a red-hot iron tube at 873 K instead triggers cyclic polymerisation: three molecules of ethyne trimerise to one molecule of benzene. This trimerisation is the classic gateway from aliphatic to aromatic chemistry and is the preferred route in conversion problems that ask for benzene from a two-carbon feedstock.

$$\ce{3\, C2H2 ->[Fe][873\ K] C6H6}$$

NEET Trap

Only the terminal hydrogen is acidic

Acidity questions reward the candidate who remembers that the acidic hydrogen must sit on the sp (triply bonded) carbon. But-1-yne reacts with sodamide; but-2-yne does not, because both its triply bonded carbons carry alkyl groups and no C–H. Do not extend "alkynes are acidic" to internal alkynes.

$\ce{CH3CH2-C#CH}$ reacts with $\ce{NaNH2}$; $\ce{CH3-C#C-CH3}$ does not.

Quick Recap

Alkynes at a glance

  • General formula $\ce{C_nH_{2n-2}}$; carbon is sp hybridised, the molecule is linear (180°), with one sigma and two pi bonds; C≡C is 120 pm long and 823 kJ mol⁻¹ strong.
  • Prepared industrially from $\ce{CaC2 + H2O}$, and in the lab by double dehydrohalogenation of vicinal/geminal dihalides (alc. KOH then $\ce{NaNH2}$).
  • Electrophilic addition of $\ce{H2}$, $\ce{X2}$ and $\ce{HX}$ adds twice; HX follows Markovnikov to a gem-dihalide; bromine decolourisation is the unsaturation test.
  • Kucherov hydration ($\ce{HgSO4}$, dil. $\ce{H2SO4}$, 333 K): acetylene → acetaldehyde; all higher alkynes → ketones.
  • Terminal (sp) C–H is acidic — reacts with Na and $\ce{NaNH2}$; internal alkynes do not. Ethyne trimerises over hot iron (873 K) to benzene.

NEET PYQ Snapshot — Alkynes

Real NEET questions sampling alkyne hydration, acetylide alkylation and the acetylene-to-benzene route.

NEET 2017

Predict the correct intermediate (A) and product (B): $\ce{H3C-C#CH ->[H2O,\ H2SO4][HgSO4] (A) -> (B)}$.

  • (1) A: $\ce{H3C-C(OH)=CH2}$ ; B: $\ce{H3C-CO-CH3}$
  • (2) A: enol with $\ce{OSO4}$ group ; B: ketone
  • (3) A: $\ce{H3C-C(OH)=CH2}$ ; B: $\ce{H3C-C(OH)=CH2}$
  • (4) A: $\ce{H3C-C(OH)(SO4)-CH3}$ ; B: alkyne
Answer: (1)

Markovnikov hydration of propyne gives the enol $\ce{H3C-C(OH)=CH2}$ as intermediate A, which tautomerises to the ketone acetone (B). A higher alkyne yields a ketone — never an aldehyde.

NEET 2016

In the sequences $\ce{X-C#CH ->[(1)\ NaNH2/liq.\ NH3][(2)\ CH3CH2Br] X}$ and a parallel acetylide alkylation, X and Y are:

  • (1) X = 2-Butyne ; Y = 3-Hexyne
  • (2) X = 2-Butyne ; Y = 2-Hexyne
  • (3) X = 1-Butyne ; Y = 2-Hexyne
  • (4) X = 1-Butyne ; Y = 3-Hexyne
Answer: (4)

Sodamide deprotonates the terminal alkyne to the acetylide; reaction with $\ce{CH3CH2Br}$ extends the chain. Acetylene gives 1-butyne (X), and ethylacetylide alkylated again gives 3-hexyne (Y). Only terminal (sp) C–H is acidic enough for this sequence.

NEET 2025

Which one of the following reactions does NOT give benzene as the product?

  • (1) — (2) — (3) — (4) (structural options as printed in the paper)
Answer: (1)

Cyclic trimerisation of ethyne over red-hot iron at 873 K is a standard benzene route, alongside decarboxylation and aromatisation paths; the option whose reactant/conditions cannot close to a six-membered aromatic ring is the exception. (Refer to the original figure-based options when attempting.)

Concept

Which reagent distinguishes but-1-yne from but-2-yne in a single test?

  • (1) Bromine in $\ce{CCl4}$
  • (2) Ammoniacal/sodamide test (terminal C–H acidity)
  • (3) Catalytic $\ce{H2}$/Ni
  • (4) Dilute $\ce{H2SO4}$
Answer: (2)

Both alkynes decolourise bromine and both hydrogenate, so those tests fail to distinguish them. Only but-1-yne has a terminal (sp) hydrogen, so only it forms an acetylide with sodamide — the acidity test separates terminal from internal alkynes.

FAQs — Alkynes

The recurring conceptual sticking points, answered against NCERT §9.4.

Why are terminal alkynes acidic while alkenes and alkanes are not?

In ethyne the hydrogen atoms are attached to sp hybridised carbon atoms, which carry the maximum percentage of s character (50%). This makes the sp orbital the most electronegative, so it pulls the shared electron pair of the C–H bond strongly towards carbon. The hydrogen attached to the triply bonded carbon can therefore be released as a proton more easily than in ethene (sp2) or ethane (sp3). Only the hydrogen on the triply bonded carbon is acidic, not every hydrogen of the alkyne.

What is the Kucherov reaction and why does acetylene give a different product from other alkynes?

The Kucherov reaction is the addition of one molecule of water to an alkyne on warming with mercuric sulphate and dilute sulphuric acid at 333 K to give a carbonyl compound. Acetylene (ethyne) gives acetaldehyde because the only possible carbonyl on a two-carbon chain is an aldehyde. Every other alkyne follows Markovnikov addition, so the OH ends up on the more substituted carbon, and after tautomerisation the product is a ketone, not an aldehyde.

How is ethyne prepared from calcium carbide?

On an industrial scale ethyne is prepared by treating calcium carbide with water, which gives ethyne and calcium hydroxide. The calcium carbide itself is made by heating quick lime (CaO) with coke, and the quick lime is obtained by heating limestone (CaCO3). The overall route is limestone to quick lime to calcium carbide to ethyne.

How do you prepare alkynes from dihalides?

Vicinal dihalides are treated with alcoholic potassium hydroxide and undergo dehydrohalogenation. One molecule of hydrogen halide is eliminated to give an alkenyl (vinylic) halide, which on further treatment with sodamide loses a second molecule of hydrogen halide to form the alkyne. Geminal dihalides can also be used by the same double dehydrohalogenation.

Why is the C≡C bond shorter and stronger than C=C and C–C bonds?

The carbon-carbon triple bond consists of one sigma bond and two pi bonds, so three shared electron pairs hold the two carbons together. This greater bonding density shortens the bond to about 120 pm and raises its bond enthalpy to about 823 kJ per mole, which is larger than that of the C=C double bond (681 kJ per mole) and the C–C single bond.

How can acetylene be converted into benzene?

When ethyne is passed through a red hot iron tube at 873 K it undergoes cyclic polymerisation. Three molecules of ethyne combine to form one molecule of benzene. This cyclic trimerisation is the standard route for moving from aliphatic to aromatic chemistry.