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JEE Mains Chemistry · Hydrocarbons

Alkynes: Preparation, Acidity, Reduction and Addition

Alkynes are made by removing two HX from a dihalide; a terminal alkyne's C–H is acidic enough to give an acetylide, which builds longer chains; H₂ with Lindlar's catalyst gives the cis alkene and Na in liquid NH₃ the trans; water adds with Hg²⁺ to give a ketone.

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Why this matters

Twenty PYQs, seventeen of them multiple choice, and five from 2026. Six are about making alkynes and the acidity of the terminal H: sodium or sodamide, the gas released and the chain built from an acetylide. Eight are about reducing alkynes to cis or trans alkenes, or turning one alkene isomer into the other. Six follow additions to the triple bond: bromine, water with Hg²⁺, ozone, and cyclisation to an arene.

Concept 1 of 3: Making alkynes and using the acidic terminal H

Two eliminations of HX from a vicinal or geminal dihalide give a triple bond; the first needs alcoholic KOH, the second a stronger base, NaNH₂. The H on a triple-bonded carbon sits on an sp carbon with 50% s-character, so the C–H bond gives up H⁺ to Na or NaNH₂. The acetylide ion formed is a strong nucleophile: with a primary alkyl halide it forms a new C–C bond, which lengthens the chain.

Definition

  • Preparation: R−CHBr−CH2Br→alc. KOHR−CBr=CH2→NaNH2R−C≡CH\mathrm{R{-}CHBr{-}CH_2Br \xrightarrow{alc.\ KOH} R{-}CBr{=}CH_2 \xrightarrow{NaNH_2} R{-}C{\equiv}CH}. Alcoholic KOH alone stops at the vinyl halide.
  • Acidity: HC≡CH>H2C=CH2>H3C−CH3\mathrm{HC{\equiv}CH > H_2C{=}CH_2 > H_3C{-}CH_3}. Only a terminal alkyne (R−C≡C−H\mathrm{R{-}C{\equiv}C{-}H}) reacts; but-2-yne has no acidic H.
  • With Na: 2 mol alkyne give 1 mol H₂. With NaNH₂: 1 mol alkyne gives 1 mol NH₃. Moles of gas × 22.4 L gives the volume at STP.
  • Chain building: R−C≡C−Na++R′−X→R−C≡C−R′+NaX\mathrm{R{-}C{\equiv}C^-Na^+ + R'{-}X \rightarrow R{-}C{\equiv}C{-}R' + NaX}, best with a primary R'X.
  • Terminal alkynes give a white precipitate with ammoniacal AgNO₃ and a red one with ammoniacal Cu₂Cl₂.

Acidic terminal H

2R−C≡CH+2Na→2R−C≡C−Na++H2R−C≡CH+NaNH2→R−C≡C−Na++NH3\mathrm{2R{-}C{\equiv}CH + 2Na \rightarrow 2R{-}C{\equiv}C^-Na^+ + H_2}\qquad \mathrm{R{-}C{\equiv}CH + NaNH_2 \rightarrow R{-}C{\equiv}C^-Na^+ + NH_3}

Worked example

2.7 g of but-1-yne (M = 54) is treated separately with excess Na and with excess NaNH₂. What volume of gas at STP does each give?
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The same idea in a real exam question:

JEE Mains · 2025 · 22 Jan 2025 · Q36Moderate

Example 1 · Hydrocarbons · Alkynes: Preparation, Acidity, Reduction and Addition

Given below are two statements : Statement I : One mole of propyne reacts with excess of sodium to liberate half a mole of H2H_{2} gas. Statement II : Four g of propyne reacts with NaNH2NaNH_{2} to liberate NH3NH_{3} gas which occupies 224 mL at STP. In the light of the above statements, choose the most appropriate answer from the options given below:

Na gives half a mole of H₂ per acidic H

Two acidic hydrogens make one H₂ molecule, so one mole of a terminal alkyne gives half a mole of H₂. NaNH₂ gives one NH₃ for every H removed.

Convert moles to millilitres carefully

One mole of gas is 22,400 mL at STP. 0.1 mol is 2240 mL, not 224 mL; a slip of ten is a common wrong option.

Only a terminal alkyne has the acidic H

An internal alkyne such as but-2-yne has no H on a triple-bonded carbon, so it gives no acetylide, no H₂ with Na and no silver precipitate.
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Concept 2 of 3: Reducing alkynes to cis or trans alkenes

The catalyst decides the geometry. On a metal surface both H atoms arrive from the same side, so a partly poisoned catalyst that stops at the alkene gives the cis isomer. Sodium in liquid ammonia adds one electron and one proton at a time, and the intermediate settles with its groups apart, so the trans isomer forms. With an active catalyst and excess H₂, the reduction runs on to the alkane.

Definition

  • Lindlar's catalyst: Pd on CaCO₃ (or BaSO₄) poisoned with lead acetate or quinoline. H₂ adds syn: cis-alkene.
  • Na in liquid NH₃ (Birch-type): trans-alkene.
  • Excess H₂ with Pt, Pd or Ni: alkane.
  • cis isomers are polar and boil higher (cis-but-2-ene 277 K, trans 274 K); trans isomers are non-polar or less polar and pack better, so they melt higher.
  • To turn a trans alkene into the cis one: Br₂, then alcoholic KOH and NaNH₂ (back to the alkyne), then H₂ with Lindlar's catalyst.
ReagentHow H addsProduct from pent-2-yneDipole of product
H₂, Lindlar's catalystSyn, stops at the alkenecis-Pent-2-eneNon-zero
Na in liquid NH₃Anti, stepwisetrans-Pent-2-eneClose to zero
Excess H₂, Pt or NiSyn, twicePentaneClose to zero
Lindlar gives cis, sodium in ammonia gives trans, and an unpoisoned catalyst with excess H₂ goes to the alkane.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 31 January 2023 · Q35Moderate

Example 2 · Hydrocarbons · Alkynes: Preparation, Acidity, Reduction and Addition

Choose the correct set of reagents for the following conversion. trans (Ph−CH=CH−CH3)→cis(Ph−CH=CH−CH3)\left( Ph - CH = CH -CH_{3} \right)\rightarrow cis\left( Ph - CH = CH -CH_{3} \right)

Aqueous KOH substitutes, alcoholic KOH eliminates

To return a dibromide to the alkyne you need alcoholic KOH (then NaNH₂). Aqueous KOH replaces Br by OH and gives a diol, which cannot become an alkyne.

The cis isomer is the more polar one

In the cis isomer the two C–R bond dipoles add; in the trans isomer they cancel. So cis boils higher and dissolves better in polar solvents, while trans melts higher.
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Concept 3 of 3: Adding water, halogens and ozone to alkynes

A triple bond can add two molecules of a reagent. Water adds once, with Hg²⁺ and dilute acid as catalysts, to give an enol, which at once becomes the carbonyl form. By Markovnikov's rule the OH goes to the inner carbon, so every alkyne except ethyne gives a ketone, and a terminal alkyne gives a methyl ketone. Bromine adds twice, to a tetrabromide. Three ethyne molecules can also join into a benzene ring in a red-hot iron tube.

Definition

  • Hydration: HgSO₄, dil. H₂SO₄, 333 K. Ethyne → ethanal; R–C≡CH → R–CO–CH₃ (a methyl ketone: iodoform positive, Tollens negative).
  • Halogen: R−C≡CH+2Br2→R−CBr2−CHBr2\mathrm{R{-}C{\equiv}CH + 2Br_2 \rightarrow R{-}CBr_2{-}CHBr_2}.
  • HX: two moles add, both Markovnikov, giving a geminal dihalide.
  • Ozonolysis: C≡C is cut into two carboxylic acids (a terminal ≡CH gives HCOOH, which reduces Tollens' reagent).
  • Cyclic polymerisation: 3HC≡CH→red-hot Fe, 873 KC6H6\mathrm{3HC{\equiv}CH \xrightarrow{red\text{-}hot\ Fe,\ 873\,K} C_6H_6}; propyne gives 1,3,5-trimethylbenzene (mesitylene).

Hydration of a terminal alkyne

R−C≡CH+H2O→Hg2+, H2SO4[R−C(OH)=CH2]→R−CO−CH3\mathrm{R{-}C{\equiv}CH + H_2O \xrightarrow{Hg^{2+},\ H_2SO_4} [R{-}C(OH){=}CH_2] \rightarrow R{-}CO{-}CH_3}

Worked example

Pent-1-yne is warmed with HgSO₄ and dilute H₂SO₄. Name the product and say whether it gives the iodoform test.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 2 Apr 2025 · Q27Moderate

Example 3 · Hydrocarbons · Alkynes: Preparation, Acidity, Reduction and Addition

An optically active alkyl halide C4H9Br[A]C_{4}H_{9}Br\lbrack A\rbrack reacts with hot KOH dissolved in ethanol and forms alkene [B] as major product which reacts with bromine to give dibromide [C]. The compound [C] is converted into a gas [D] upon reacting with alcoholic NaNH2NaNH_{2}. During hydration 18 gram of water is added to 1 mole of gas [D] on warming with mercuric sulphate and dilute acid at 333 K to form compound [E]. The IUPAC name of compound [E] is :

An enol is not the final product

Hydration first gives an enol, but the enol turns into the aldehyde or ketone at once. Options that show a vinyl alcohol as the product are wrong.

A terminal alkyne gives a methyl ketone, not an aldehyde

Markovnikov addition puts OH on C-2, so R–C≡CH gives R–CO–CH₃. Only ethyne gives an aldehyde.
Drill 5 more on adding water, halogens and ozone to alkynes

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Making alkynes and using the acidic terminal H

    Acidic terminal H

    2R−C≡CH+2Na→2R−C≡C−Na++H2R−C≡CH+NaNH2→R−C≡C−Na++NH3\mathrm{2R{-}C{\equiv}CH + 2Na \rightarrow 2R{-}C{\equiv}C^-Na^+ + H_2}\qquad \mathrm{R{-}C{\equiv}CH + NaNH_2 \rightarrow R{-}C{\equiv}C^-Na^+ + NH_3}
  • Adding water, halogens and ozone to alkynes

    Hydration of a terminal alkyne

    R−C≡CH+H2O→Hg2+, H2SO4[R−C(OH)=CH2]→R−CO−CH3\mathrm{R{-}C{\equiv}CH + H_2O \xrightarrow{Hg^{2+},\ H_2SO_4} [R{-}C(OH){=}CH_2] \rightarrow R{-}CO{-}CH_3}

Reference tables (1)

Reducing alkynes to cis or trans alkenes3 rows
ReagentHow H addsProduct from pent-2-yneDipole of product
H₂, Lindlar's catalystSyn, stops at the alkenecis-Pent-2-eneNon-zero
Na in liquid NH₃Anti, stepwisetrans-Pent-2-eneClose to zero
Excess H₂, Pt or NiSyn, twicePentaneClose to zero
Lindlar gives cis, sodium in ammonia gives trans, and an unpoisoned catalyst with excess H₂ goes to the alkane.

Watch out for (7)

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