Preparation of Alkynes

Q: Why must the alkyl halide in acetylide alkylation be primary?

Alkylation is an SN2 reaction that needs back-side attack on the carbon bearing the halogen. Secondary and especially tertiary halides are too crowded, and because the acetylide is also a strong base it switches to E2 elimination, giving an alkene and regenerating the alkyne.

Q: How do you make an alkyne from a ketone?

Treat the ketone with PCl5 to replace the carbonyl oxygen with two chlorine atoms on the same carbon, giving a geminal dichloride, then eliminate two molecules of HCl with two equivalents of sodamide. Pentan-2-one becomes pent-2-yne this way.

Q: Which method of preparing alkynes lengthens the carbon chain?

Only alkylation of an acetylide. Every other preparation removes atoms from an existing skeleton, so the product has the same number of carbons as the starting material. If a synthesis question increases the carbon count, use sodium acetylide with a primary alkyl halide.

ChemistryHydrocarbonsFor JEE aspirants

Preparation of alkynes comes down to two ideas: strip two small molecules from a saturated skeleton, or build a new carbon-carbon bond onto a triple bond that already exists. The first idea covers double dehydrohalogenation of vicinal and geminal dihalides with alcoholic $KOH$ or sodamide, and dehalogenation of tetrahalides with zinc. The second covers alkylation of sodium acetylide with a primary alkyl halide. Industrially, acetylene still comes from the hydrolysis of calcium carbide. This page works through every method, its mechanism and its exam traps.

SYLLABUS Preparation of alkynes is prescribed for JEE Main, JEE Advanced and NEET (NCERT Class 11, Unit 13). Kolbe electrolysis of maleate or fumarate and the trihalide-silver route go slightly beyond the NCERT text but appear in coaching material and older papers; both are marked below.

Key Reactions - Quick Reference

$Ca C_{2} + 2 H_{2} O ⟶ HC \equiv CH + Ca (OH)_{2}$
$M g_{2} C_{3} + 4 H_{2} O ⟶ C H_{3} - C \equiv CH + 2Mg (OH)_{2}$
$R - CHBr - CHBr - R 2 NaN H_{2} R - C \equiv C - R + 2NaBr + 2N H_{3}$
$R - CC l_{2} - R 2 NaN H_{2} R - C \equiv C - R$ (from a ketone $+ PC l_{5}$ )
$R - C X_{2} - C X_{2} - R 2 Zn R - C \equiv C - R + 2Zn X_{2}$
$HC \equiv CH Na/liq. N H_{3} HC \equiv C^{-} N a^{+} R - X HC \equiv C - R$
Terminal alkyne from a dihalide needs 3 equivalents of $NaN H_{2}$ , then $N H_{4} Cl$

1. The Routes at a Glance

Figure 1: The six routes to an alkyne. Every method either removes two small molecules (

2 HX

2 X_{2}

) from a saturated skeleton or builds a new bond onto an existing

C \equiv C

Notice the pattern. Every elimination route needs a substrate in which the two carbons destined to become the triple bond already carry leaving groups, and every building route needs a nucleophilic acetylide. Choose the method by asking what starting material you are given, not by memorising a list.

2. Industrial Source: Metal Carbides and Methane Cracking

From calcium carbide

Calcium carbide is a strongly basic ionic carbide. Its $C_{2}^{2 -}$ ion snatches protons from water, so hydrolysis gives acetylene directly.

$CaO + 3C 200 0^{\circ} C Ca C_{2} + CO$

$Ca C_{2} + 2 H_{2} O ⟶ HC \equiv CH + Ca (OH)_{2}$

Figure 2: Acetylene generator.

Ca C_{2} + 2 H_{2} O ⟶ C_{2} H_{2} + Ca (OH)_{2}

is still the cheapest bulk route to

HC \equiv CH

From magnesium carbide

Magnesium carbide, $M g_{2} C_{3}$ , contains the $C_{3}^{4 -}$ unit and so delivers the next homologue instead of acetylene.

$M g_{2} C_{3} + 4 H_{2} O ⟶ C H_{3} - C \equiv CH + 2Mg (OH)_{2}$

From methane by cracking

Partial oxidation and thermal cracking of natural gas gives acetylene on a large scale.

$6C H_{4} + 2 O_{2} 150 0^{\circ} C cracking 2HC \equiv CH + 2C O_{2} + 10 H_{2}$

The quench must be extremely fast. Acetylene is thermodynamically unstable with respect to its elements, so if the hot gas is allowed to cool slowly it decomposes to carbon and hydrogen.

3. Kolbe's Electrolytic Method

BEYOND NCERT Not in the NCERT text, but a standard coaching topic and a favourite of older JEE and state-board papers.

Electrolysis of a concentrated aqueous solution of the sodium or potassium salt of maleic acid or fumaric acid gives acetylene at the anode. The carboxylate ions migrate to the anode, lose an electron each, and the resulting diradical expels two molecules of carbon dioxide.

$^{-} OOC - CH = CH - CO O^{-} - 2 e^{-} \cdot OOC - CH = CH - COO \cdot ⟶ HC \equiv CH + 2C O_{2}$

At the cathode, water is reduced, so hydrogen and alkali are produced there: $2 H_{2} O + 2 e^{-} ⟶ H_{2} + 2O H^{-}$

Overall: acetylene

+ 2 C O_{2}

at the anode,

H_{2} +

alkali at the cathode. Compare this with the Kolbe electrolysis of a saturated carboxylate salt, which gives an alkane instead.

4. Dehydrohalogenation of Vicinal Dihalides

A $1, 2$ -dihalide (vicinal dihalide) loses two molecules of hydrogen halide when treated with alcoholic $KOH$ or with sodamide, and the intermediate is a vinyl halide.

$R - CHBr - CHBr - R alc. KOH R - CBr = CH - R NaN H_{2} R - C \equiv C - R$

Figure 3: Two successive E2 eliminations convert a vicinal dihalide to an alkyne through a vinyl halide. Step 2 is the slow step, so

NaN H_{2}

is preferred over alcoholic

KOH

Why the second step is harder

In a vinyl halide the halogen sits on an $s p^{2}$ carbon. Its lone pair overlaps with the $π$ system, giving the $C - X$ bond partial double bond character, so it is shorter and stronger than an ordinary $C - X$ bond.
The hydrogen being removed is also on an $s p^{2}$ carbon and is less accessible.
Hence alcoholic $KOH$ often stops at the vinyl halide, while $NaN H_{2}$ in liquid ammonia (or in an inert solvent at $110$ - $160^{\circ} C$ ) pushes the elimination through.
Both eliminations are E2 and prefer an anti-periplanar arrangement of the $H$ and the $X$ .

Figure 4: Stoichiometry trap. Two equivalents of

NaN H_{2}

suffice for an internal alkyne; a terminal alkyne needs three plus an

N H_{4} Cl

work-up.

Solved Example 1

How many equivalents of

NaN H_{2}

are needed to convert

C H_{3} C H_{2} - CHBr - C H_{2} Br

into but-

1

-yne, and what reagent is used at the end?

Solution:

Three equivalents. Two are consumed removing the two molecules of $HBr$ . The product, but- $1$ -yne, is a terminal alkyne whose $\equiv C - H$ is acidic enough to be deprotonated by the third equivalent, so the mixture actually contains sodium but- $1$ -ynide. Adding $N H_{4} Cl$ protonates it and liberates but- $1$ -yne.

$C H_{3} C H_{2} CHBrC H_{2} Br 3 NaN H_{2} C H_{3} C H_{2} - C \equiv C^{-} N a^{+} N H_{4} Cl C H_{3} C H_{2} - C \equiv CH$

5. Dehydrohalogenation of Geminal Dihalides

A $1, 1$ -dihalide (geminal dihalide) behaves the same way with alcoholic $KOH$ or with $NaN H_{2}$ in $N H_{3}$ , again through a vinyl halide.

$R - C H_{2} - CHC l_{2} 2 NaN H_{2} R - C \equiv CH$

Geminal dihalides are easy to make: treat an aldehyde or ketone with $PC l_{5}$ , which swaps the carbonyl oxygen for two chlorines on the same carbon. That makes this route the standard way of turning a carbonyl compound into an alkyne.

Figure 6: Alkene plus

B r_{2}

gives the vic-dihalide; ketone plus

PC l_{5}

gives the gem-dihalide. Both feed the same double-dehydrohalogenation step.

Solved Example 2

Starting from pentan-

2

-one, outline a two-step synthesis of pent-

2

-yne.

Solution:

$C H_{3} - CO - C H_{2} C H_{2} C H_{3} PC l_{5} C H_{3} - CC l_{2} - C H_{2} C H_{2} C H_{3}$ (geminal dichloride)
$C H_{3} - CC l_{2} - C H_{2} C H_{2} C H_{3} 2 NaN H_{2} C H_{3} - C \equiv C - C H_{2} C H_{3}$

Only two equivalents of base are needed, because pent- $2$ -yne is an internal alkyne with no acidic hydrogen.

6. Dehalogenation of Tetrahalides and Trihalides

A $1, 1, 2, 2$ -tetrahaloalkane loses two molecules of halogen when refluxed with zinc dust, exactly as a vicinal dihalide loses one to give an alkene.

$R - C X_{2} - C X_{2} - R + 2Zn ⟶ R - C \equiv C - R + 2Zn X_{2}$

BEYOND NCERT Two molecules of a haloform can be coupled by finely divided silver to give acetylene:

CH X_{3} + 6Ag + X_{3} CH Δ HC \equiv CH + 6AgX

This is an Ullmann-type coupling; silver removes all six halogens.

7. Alkylation of Acetylene and Terminal Alkynes

This is the only route in the list that lengthens the carbon chain, which makes it the one that synthesis questions are built around. A terminal alkyne is first deprotonated to its acetylide, then that acetylide displaces halide from an alkyl halide.

$HC \equiv CH + Na liq. N H_{3} HC \equiv C^{-} N a^{+} R - X HC \equiv C - R$

$R - C \equiv CH + Na liq. N H_{3} R - C \equiv C^{-} N a^{+} R^{'} - X R - C \equiv C - R^{'}$

For example, propyne can be extended to pent- $2$ -yne in one step:

$C H_{3} - C \equiv CH (i) Na/liq. N H_{3} (ii) C H_{3} C H_{2} I C H_{3} - C \equiv C - C H_{2} C H_{3}$

Figure 5:

HC \equiv C^{-}

is both a strong nucleophile and a strong base. With

1^{\circ}

R - X

it substitutes; with

3^{\circ}

R - X

it eliminates.

The one rule that decides every alkylation question.

R - X

and

R^{'} - X

must be methyl or primary halides. Secondary and tertiary halides give mainly alkenes, because the acetylide is a strong base and prefers E2 elimination over

S_{N} 2

substitution at a hindered carbon.

The Grignard variant

A Grignard reagent can replace sodium metal. It deprotonates the terminal alkyne to give an alkynyl magnesium halide, which then alkylates in the usual way.

$HC \equiv CH + C H_{3} MgBr ⟶ HC \equiv C - MgBr + C H_{4}$

$HC \equiv C - MgBr + C H_{3} Br ⟶ HC \equiv C - C H_{3} + MgB r_{2}$

The methane released in the first step is the basis of the Zerewitinoff determination of active hydrogen: measure the volume of

C H_{4}

and you know how many acidic hydrogens the sample had.

Solved Example 3

Synthesise dec-

3

-yne from acetylene, using any alkyl halides you need.

Solution:

Dec- $3$ -yne is $C H_{3} C H_{2} - C \equiv C - C H_{2} C H_{2} C H_{2} C H_{2} C H_{2} C H_{3}$ , so an ethyl group and a hexyl group must be added to acetylene, one at a time.

$HC \equiv CH NaN H_{2} /liq. N H_{3} HC \equiv C^{-} N a^{+} C H_{3} (C H_{2})_{5} Br C H_{3} (C H_{2})_{5} - C \equiv CH$ (oct- $1$ -yne)
$C H_{3} (C H_{2})_{5} - C \equiv CH NaN H_{2} C H_{3} C H_{2} Br C H_{3} (C H_{2})_{5} - C \equiv C - C H_{2} C H_{3}$ (dec- $3$ -yne)

Both halides are primary, so both steps are clean $S_{N} 2$ alkylations. The order can be reversed without changing the product.

Solved Example 4

Why does

HC \equiv C^{-} N a^{+}

fail to give

HC \equiv C - C (C H_{3})_{3}

when treated with

2

-bromo-

2

-methylpropane?

Solution:

$S_{N} 2$ attack at a tertiary carbon is blocked by the three methyl groups: the backside of the $C - Br$ bond is completely shielded. The acetylide therefore behaves as a base and removes a $β$ -hydrogen in an E2 reaction, so the products are acetylene (regenerated) and $2$ -methylpropene.

$HC \equiv C^{-} N a^{+} + (C H_{3})_{3} C - Br ⟶ HC \equiv CH + C H_{2} = C (C H_{3})_{2} + NaBr$

8. Isomerisation: Moving the Triple Bond

Heating an internal alkyne with $NaN H_{2}$ in an inert solvent walks the triple bond to the end of the chain, because the terminal acetylide that forms is the thermodynamic sink.

$C H_{3} C H_{2} - C \equiv C - C H_{3} NaN H_{2}, Δ C H_{3} C H_{2} C H_{2} - C \equiv C^{-} N a^{+} H_{2} O C H_{3} C H_{2} C H_{2} - C \equiv CH$

The reverse walk, terminal to internal, is achieved with alcoholic $KOH$ , since there the internal alkyne is the more stable product and no acetylide can form.

9. Summary Table of Methods

Starting material	Reagent	Product	Key point
$Ca C_{2}$	$H_{2} O$	Acetylene	Industrial, cheapest route
$M g_{2} C_{3}$	$H_{2} O$	Propyne	$C_{3}^{4 -}$ gives the $C_{3}$ alkyne
Maleate or fumarate salt	Electrolysis	Acetylene (at anode)	$2 C O_{2}$ lost
vic-dihalide	alc. $KOH$ , then $NaN H_{2}$	Alkyne	Via vinyl halide, E2 twice
gem-dihalide (from ketone $+ PC l_{5}$ )	$2 NaN H_{2}$	Alkyne	Carbonyl to alkyne in two steps
Tetrahaloalkane	$Zn$ dust	Alkyne	Loses $2 X_{2}$
Terminal alkyne $+ 1^{\circ} R - X$	$NaN H_{2}$ , then $R - X$	Higher alkyne	Only chain-lengthening method
Internal alkyne	$NaN H_{2}$ , $Δ$	Terminal alkyne	Triple bond walks to the end

Solved Example 5

Give three isomeric dibromides that could each be used to prepare

3, 3

-dimethylbut-

1

-yne.

Solution:

The target is $(C H_{3})_{3} C - C \equiv CH$ . Any dibromide whose two bromines sit on the two carbons that become the triple bond will work.

$(C H_{3})_{3} C - C H_{2} - CHB r_{2}$ (geminal, on the terminal carbon)
$(C H_{3})_{3} C - CHBr - C H_{2} Br$ (vicinal)
$(C H_{3})_{3} C - CB r_{2} - C H_{3}$ (geminal, on the internal carbon)

All three give the same alkyne on treatment with excess sodamide, because in each case two molecules of $HBr$ leave from the same pair of carbons. Since the product is terminal, use three equivalents and finish with $N H_{4} Cl$ .

Solved Example 6

Convert but-

1

-ene into but-

2

-yne.

Solution:

$C H_{3} C H_{2} CH = C H_{2} B r_{2} /CC l_{4} C H_{3} C H_{2} - CHBr - C H_{2} Br$
$3 NaN H_{2} N H_{4} Cl C H_{3} C H_{2} - C \equiv CH$ (but- $1$ -yne)
$alc. KOH, Δ C H_{3} - C \equiv C - C H_{3}$ (but- $2$ -yne)

The last step is the terminal-to-internal isomerisation. Using $NaN H_{2}$ there would be a mistake: it would drive the equilibrium the other way, back to the terminal acetylide.

Common Mistakes to Avoid

Watch out

Using two equivalents of $NaN H_{2}$ for a terminal alkyne. The third equivalent is eaten by the acidic $\equiv C - H$ ; without it, and without the $N H_{4} Cl$ work-up, you isolate the sodium acetylide, not the alkyne.
Alkylating with a secondary or tertiary halide. The acetylide eliminates instead of substituting, and you get an alkene plus the original alkyne back.
Expecting alcoholic $KOH$ alone to finish the job. It normally stops at the vinyl halide; sodamide is needed for the second elimination.
Confusing $Ca C_{2}$ with $M g_{2} C_{3}$ . Calcium carbide gives acetylene; magnesium carbide gives propyne.
Writing $Na$ metal with a terminal alkyne and expecting reduction. Sodium in liquid ammonia reduces internal alkynes to trans alkenes, but with a terminal alkyne it simply forms the acetylide salt, so no reduction occurs.
Forgetting that dehalogenation needs a tetrahalide, not a dihalide. Zinc plus a vicinal dihalide gives an alkene; you need four halogens on two carbons to reach an alkyne.
Choosing $NaN H_{2}$ when you want the internal isomer. Sodamide walks the triple bond to the chain end; alcoholic $KOH$ walks it inward.

Frequently Asked Questions

How is acetylene prepared in the laboratory?

Water is dripped onto lumps of calcium carbide in a flask and the acetylene that bubbles off is collected over water: $Ca C_{2} + 2 H_{2} O \to C_{2} H_{2} + Ca (OH)_{2}$ . The gas smells of garlic because of phosphine and hydrogen sulphide impurities from the technical carbide, and it is purified by passing it through acidified copper sulphate solution.

Why are three equivalents of sodamide used to make a terminal alkyne?

Two equivalents remove the two molecules of hydrogen halide from the dihalide. The alkyne formed is terminal, and its $\equiv C - H$ has $p K_{a} \approx 25$ , acidic enough to be deprotonated by the third equivalent of the strongly basic amide ion. Adding $N H_{4} Cl$ at the end returns the neutral alkyne.

Why must the alkyl halide in acetylide alkylation be primary?

Alkylation is an $S_{N} 2$ reaction requiring back-side attack on the carbon bearing the halogen. Secondary and especially tertiary halides are too crowded for that approach, and since the acetylide is a strong base it switches to E2 elimination instead, giving an alkene and regenerating the alkyne.

What is the intermediate in the dehydrohalogenation of a vicinal dihalide?

A vinyl halide, that is an alkene carrying the halogen on one of the doubly bonded carbons. Removing the first molecule of $HX$ is easy, but the $C - X$ bond in the vinyl halide is strengthened by resonance with the $π$ system, so the second elimination needs sodamide rather than alcoholic potassium hydroxide.

How do you make an alkyne from a ketone?

Treat the ketone with $PC l_{5}$ to replace the carbonyl oxygen with two chlorine atoms on the same carbon, giving a geminal dichloride, then eliminate two molecules of $HCl$ with two equivalents of sodamide. Pentan- $2$ -one becomes pent- $2$ -yne this way.

Which method lengthens the carbon chain?

Only alkylation of an acetylide. Every other preparation simply removes atoms from an existing skeleton, so the product has the same number of carbons as the starting material. In multi-step synthesis questions, spotting that the carbon count has increased is the signal to use $HC \equiv C^{-} N a^{+}$ plus a primary halide.

What happens when an internal alkyne is heated with sodamide?

The triple bond migrates to the end of the chain. Sodamide removes a propargylic proton, the resulting anion is reprotonated at the other end, and the sequence repeats until the terminal acetylide forms. Because that salt cannot revert, the equilibrium is pulled all the way to the terminal alkyne, which is released on aqueous work-up.

Why does calcium carbide give acetylene while magnesium carbide gives propyne?

The anion decides the product. Calcium carbide contains the two-carbon acetylide ion $C_{2}^{2 -}$ , so protonation by water gives $HC \equiv CH$ . Magnesium carbide $M g_{2} C_{3}$ contains the three-carbon allylide ion $C_{3}^{4 -}$ , and protonating that gives $C H_{3} - C \equiv CH$ .

Can zinc dust convert a vicinal dihalide into an alkyne?

No. Zinc removes one molecule of halogen from two adjacent carbons, so a vicinal dihalide gives an alkene. To reach an alkyne you need a tetrahaloalkane with two halogens on each of the two carbons, from which zinc removes two molecules of $X_{2}$ .

Previous year questions on Preparation of Alkynes

5 questions from past papers, each with a step-by-step solution.

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