Properties of Alkynes

Q: Why are terminal alkynes acidic?

The hydrogen sits on an sp hybridised carbon with fifty percent s character. When it leaves, the lone pair occupies that sp orbital, which holds electrons close to the nucleus and stabilises the acetylide ion. That stability makes the parent C-H acidic, with pKa about 25 against 44 for an alkene and 50 for an alkane.

Q: Why do alkynes undergo electrophilic addition more slowly than alkenes?

Three reasons combine: the sp carbons are more electronegative and hold the pi electrons tightly, those electrons are more delocalised in the cylindrical cloud, and the vinyl carbocation formed after the electrophile adds is far less stable than an alkyl carbocation because its positive charge sits on an sp carbon.

Q: Why does acetylene burn with a sooty luminous flame?

Its percentage of carbon by mass is very high, about 92 percent, so in a limited air supply combustion is incomplete and unburnt carbon particles glow in the flame. In pure oxygen combustion is complete and the oxy-acetylene flame reaches roughly 3500 degrees Celsius, hot enough for welding and cutting steel.

ChemistryHydrocarbonsFor NEET aspirants

The properties of alkynes follow from one structural fact: the triply bonded carbons are $s p$ hybridised. That gives terminal alkynes an acidic hydrogen ( $p K_{a} \approx 25$ ) that forms metal acetylides, and at the same time makes the $π$ electrons less available, so alkynes undergo electrophilic addition more slowly than alkenes. Two molecules of a reagent can add across the triple bond, hydration gives a ketone through an enol, and partial reduction gives either the $c i s$ or the $t r an s$ alkene depending on the reagent. This page covers every physical and chemical property of alkynes with mechanisms, solved examples and the traps JEE and NEET set.

SYLLABUS Properties of alkynes are prescribed for JEE Main, JEE Advanced and NEET (NCERT Class 11, Unit 13). Hydroboration of alkynes with

Si a_{2} BH

and the detailed Birch mechanism go beyond the NCERT text; both are badged below.

Key Reactions - Quick Reference

Acidity: $s p^{3} (50) < s p^{2} (44) < s p (25)$ on the $p K_{a}$ scale, so acidity rises $s p^{3} < s p^{2} < s p$
Test: $R - C \equiv CH AgN O_{3} /N H_{4} OH$ white ppt; $C u_{2} C l_{2} /N H_{4} OH$ red ppt
Halogenation: $R - C \equiv C - R B r_{2}$ trans-dibromide $B r_{2}$ tetrabromide (anti in both steps)
Hydrohalogenation: $HC \equiv CH HBr C H_{2} = CHBr HBr C H_{3} CHB r_{2}$ (Markovnikov twice)
Hydration: $R - C \equiv CH H_{2} O/ H_{2} S O_{4} /HgS O_{4} R - CO - C H_{3}$ (via enol)
Reduction: $H_{2}$ /Lindlar $\to c i s$ alkene; $Na/liq. N H_{3} \to t r an s$ alkene
Ozonolysis: $R - C \equiv C - R^{'} O_{3} H_{2} O RCOOH + R^{'} COOH$
Reactivity orders: $C l_{2} > B r_{2} > I_{2}$ and $HI > HBr > HCl$

1. Physical Properties

Acetylene is a colourless gas with an ethereal smell when pure; boiling point $- 84^{\circ} C$ . Commercial acetylene smells of garlic because of $P H_{3}$ and $H_{2} S$ impurities from the carbide.
The first three members are gases, the next eight are liquids, and higher members are solids.
Liquefied acetylene is explosive. It is stored dissolved in acetone soaked into porous material inside steel cylinders.
It burns with a luminous smoky flame because of its high carbon content, which is why it was once used for lighting. In pure oxygen the oxy-acetylene flame reaches about $3500^{\circ} C$ and is used for welding.
Alkynes are slightly polar and slightly more soluble in water than alkanes and alkenes, but still largely insoluble; they dissolve freely in organic solvents.
Boiling points are a little higher than those of the corresponding alkenes because the linear molecules pack more closely.
Bond length order: $C \equiv C < C = C < C - C$ , because $s$ character rises from $25%$ through $33.3%$ to $50%$ and pulls the bonding electrons closer to the nuclei.

2. Acidity of Terminal Alkynes

A hydrogen attached to an $s p$ carbon is appreciably acidic. This is the single most examined property of alkynes, because it separates terminal alkynes from every other hydrocarbon.

Figure 1:

p K_{a}

against hybridisation. Acidity order

s p^{3} < s p^{2} < s p

follows

s

character, because an

s p

carbanion holds its lone pair closest to the nucleus.

Why $s$ character controls acidity

An $s$ orbital is closer to the nucleus than a $p$ orbital of the same shell. The more $s$ character a hybrid orbital has, the closer its electrons sit to the positively charged nucleus, and the better it can hold a negative charge. When $HC \equiv C - H$ loses a proton, the lone pair is left in an $s p$ orbital with $50%$ $s$ character, so the acetylide ion is comparatively stable and the parent acid is comparatively strong.

Electronegativity of hybridised carbon:

s p > s p^{2} > s p^{3}

.
Acidity of the attached C-H:

HC \equiv CH > C H_{2} = C H_{2} > C H_{3} - C H_{3}

.
Basicity of the conjugate carbanions (the exact reverse):

C H_{3} C H_{2}^{-} > C H_{2} = C H^{-} > HC \equiv C^{-}

Placed on the wider scale of second-period hydrides:

Acid	$H_{2} O$	$ROH$	$HC \equiv CH$	$N H_{3}$	$C H_{2} = C H_{2}$	$C H_{3} C H_{3}$
$p K_{a}$	$15.7$	$16$ - $17$	$25$	$38$	$44$	$50$

Read that table carefully. Acetylene is more acidic than ammonia, which is why

NaN H_{2}

can deprotonate it completely. But it is less acidic than water and alcohols, so

NaOH

cannot do the job, and adding water to a solution of an acetylide destroys it instantly.

Formation of metal acetylides

$HC \equiv CH + NaN H_{2} liq. N H_{3} HC \equiv C^{-} N a^{+} + N H_{3}$

$C H_{3} - C \equiv CH + Na liq. N H_{3} C H_{3} - C \equiv C^{-} N a^{+} + \frac{1}{2} H_{2}$

Passing acetylene into a solution of sodium in liquid ammonia until the blue colour disappears also gives sodium acetylide. These salts are the nucleophiles used to build longer chains.

The laboratory test for a terminal alkyne

Figure 2: The acetylide test. Only

R - C \equiv C - H

precipitates: white with

AgN O_{3} /N H_{4} OH

, red with

C u_{2} C l_{2} /N H_{4} OH

$HC \equiv CH + 2 [Ag (N H_{3})_{2}]^{+} ⟶ Ag - C \equiv C - Ag ↓ + 2N H_{4}^{+} + 2N H_{3}$

$HC \equiv CH + C u_{2} C l_{2} N H_{4} OH Cu - C \equiv C - Cu ↓ + 2N H_{4} Cl + 2 H_{2} O$

Dry silver and copper acetylides are explosive on impact. In practice the precipitate is destroyed with dilute acid straight after the test, which also regenerates the alkyne, so the test is non-destructive.

Solved Example 1

How would you distinguish between but-

1

-yne and but-

2

-yne in the laboratory?

Solution:

Pass each gas into ammoniacal silver nitrate. But- $1$ -yne is terminal, so its $\equiv \mathrm{C-H}$ is replaced by silver and a white precipitate of $C H_{3} C H_{2} - C \equiv C - Ag$ forms. But- $2$ -yne has no hydrogen on an $s p$ carbon, so no precipitate appears. Ammoniacal cuprous chloride works equally well, giving a red precipitate with but- $1$ -yne only.

Solved Example 2

Acetylene hydrogens are acidic because (A) acetylene contains the least number of hydrogen atoms (B) acetylene has only one hydrogen on each carbon (C) acetylene has the formula

C_{n} H_{2 n - 2}

(D) the

σ

electron density of the C-H bond lies nearer a carbon with

50%

s

character.

Solution:

Answer: (D). Acidity has nothing to do with how many hydrogens a molecule has; it depends on how stable the anion left behind is. The $s p$ carbon with $50%$ $s$ character is strongly electronegative, pulls the $σ$ pair towards itself, and stabilises the acetylide ion.

3. Electrophilic Addition: Why Alkynes React More Slowly

A triple bond has four $π$ electrons against an alkene's two, so students expect alkynes to be more reactive. They are not. Alkynes undergo electrophilic addition less readily than alkenes, for three reasons.

The $s p$ carbons are more electronegative and hold the $π$ electrons more tightly.
The $π$ electrons of an alkyne are more delocalised over the cylindrical cloud and so are less available to an approaching electrophile.
The vinyl carbocation formed after the electrophile adds is far less stable than the alkyl carbocation formed from an alkene.

Figure 3: Alkene gives a secondary carbocation; alkyne gives a vinyl cation with the charge on an

s p

carbon. The higher barrier is why

C \equiv C

reacts more slowly than

C = C

The third reason is the decisive one. Rate is controlled by the energy of the transition state leading to the intermediate, and a vinyl cation with its positive charge on an

s p

carbon sits much higher in energy than a secondary or tertiary alkyl cation.

4. Addition of Halogens

Halogens add across the triple bond in two separate steps. One equivalent gives the $t r an s$ -dihalide, and a second equivalent gives the tetrahalide.

$R - C \equiv C - R B r_{2} (1 eq) R - CBr = CBr - R B r_{2} (1 eq) R - CB r_{2} - CB r_{2} - R$

Addition is anti in both steps, through a cyclic halonium ion, which is why the first product is the $t r an s$ isomer.
Order of reactivity of the halogen: $C l_{2} > B r_{2} > I_{2}$ .
Decolourising a red-brown solution of bromine in carbon tetrachloride is the standard test for unsaturation, and alkynes decolourise it just as alkenes do, only more slowly.
Alkynes are less reactive than alkenes towards $B r_{2}$ because the bromonium ion formed from an alkyne contains a full double bond inside a three-membered ring, which is badly strained and therefore less stable.

5. Addition of Hydrogen Halides

Hydrogen halides add across the triple bond with Markovnikov orientation in both steps. The first addition gives a vinyl halide, and the second gives a geminal dihalide.

$HC \equiv CH HBr C H_{2} = CHBr HBr C H_{3} - CHB r_{2}$

The second addition obeys Markovnikov's rule because the intermediate $C H_{3} - C + H - Br$ is stabilised by the lone pair on bromine, whereas the alternative primary cation is not. Order of reactivity: $HI > HBr > HCl$ .

Because of the

- I

effect of the halogen already present, the vinyl halide is electron poor, so the second addition is much slower than the first and than the corresponding addition to ethylene. This lets you stop the reaction at the vinyl halide stage with one equivalent.

Industrially this is how vinyl chloride, the monomer of PVC, was once made:

$HC \equiv CH + HCl H g^{2 +} C H_{2} = CHCl$

Solved Example 3

Predict the product when but-

1

-yne reacts with two equivalents of

HCl

Solution:

Markovnikov addition puts the chlorine on the more substituted carbon each time, that is on $C_{2}$ .

$C H_{3} C H_{2} - C \equiv CH HCl C H_{3} C H_{2} - CCl = C H_{2} HCl C H_{3} C H_{2} - CC l_{2} - C H_{3}$

The product is $2, 2$ -dichlorobutane, a geminal dihalide with both chlorines on the same carbon.

6. Addition of Water: Hydration

Alkynes add water across the triple bond in dilute sulphuric acid at about $60^{\circ} C$ with $HgS O_{4}$ as catalyst. The first-formed product is an enol, which immediately tautomerises to the far more stable carbonyl compound.

Figure 4: Hydration of

C H_{3} - C \equiv CH

. The enol

C H_{3} - C (OH) = C H_{2}

tautomerises to

C H_{3} - CO - C H_{3}

; the keto form is far more stable.

Acetylene is the exception and gives an aldehyde, because both carbons are equivalent and unsubstituted:

$HC \equiv CH + H_{2} O HgS O_{4} H_{2} S O_{4} [C H_{2} = CH - OH] ⟶ C H_{3} CHO$

Every homologue gives a ketone:

$C H_{3} - C \equiv CH + H_{2} O HgS O_{4} H_{2} S O_{4} C H_{3} - CO - C H_{3}$

Keto-enol tautomerism. Tautomers are constitutional isomers that interconvert rapidly by the movement of a hydrogen and a

π

bond. The keto form is normally more stable than the enol, because a

C = O

π

bond is much stronger than a

C = C

π

bond.

Hydroboration-oxidation: the opposite regiochemistry

BEYOND NCERT Disiamylborane hydration of alkynes is a JEE Advanced topic, not part of the NCERT Class 11 text.

Hydroboration places boron on the less substituted carbon, so after oxidation the hydroxyl ends up on the terminal carbon and the enol tautomerises to an aldehyde rather than a ketone. A bulky borane such as $Si a_{2} BH$ is used so that only one addition occurs.

Figure 5: Same substrate, opposite regiochemistry.

H g^{2 +}

hydration gives

R - CO - C H_{3}

; hydroboration-oxidation gives

R - C H_{2} - CHO

$R - C \equiv CH (i) Si a_{2} BH (ii) H_{2} O_{2} /O H^{-} R - C H_{2} - CHO$

For an internal alkyne the same reagents give a ketone. A symmetrical internal alkyne gives one pure ketone; an unsymmetrical one gives a mixture, because boron can attach at either carbon.

Solved Example 4

2-Pentyne is treated with (a)

H_{2} O/ H_{2} S O_{4} /HgS O_{4}

and (b)

Si a_{2} BH

followed by

H_{2} O_{2} /O H^{-}

. What is formed in each case, and why?

Solution:

Pent- $2$ -yne, $C H_{3} - C \equiv C - C H_{2} C H_{3}$ , is an unsymmetrical internal alkyne, so both routes give a mixture of two ketones.

Attack at $C_{2}$ gives pentan- $2$ -one, $C H_{3} - CO - C H_{2} C H_{2} C H_{3}$ ; attack at $C_{3}$ gives pentan- $3$ -one, $C H_{3} C H_{2} - CO - C H_{2} C H_{3}$ . Neither carbon is clearly favoured, so neither method is synthetically useful here.

Contrast but- $2$ -yne, which is symmetrical: either orientation gives the same compound, butan- $2$ -one, as a single pure product.

7. Reduction of Alkynes

An alkyne can be reduced all the way to the alkane, or stopped cleanly at either alkene stereoisomer. Which one you get is decided entirely by the reagent.

Reagent	Product	Stereochemistry
$H_{2}$ (excess), $Pt$ , $Pd$ or $Ni$	Alkane	Both $π$ bonds reduced
$H_{2}$ , Lindlar catalyst ( $Pd/CaC O_{3}$ or $Pd/BaS O_{4}$ , quinoline)	$c i s$ alkene	syn addition
$H_{2}$ , P-2 catalyst ( $N i_{2} B$ , nickel boride)	$c i s$ alkene	syn addition
$Na$ or $Li$ in liquid $N H_{3}$	$t r an s$ alkene	anti addition
$B H_{3}$ -THF, then $C H_{3} COOH$	$c i s$ alkene	syn addition

Figure 6: Choose your alkene.

H_{2}

with Lindlar catalyst gives the

c i s

alkene by syn addition;

Na

in liquid

N H_{3}

gives the

t r an s

alkene by anti addition.

How Lindlar catalyst works

The alkyne and hydrogen are both adsorbed on the metal surface, so both hydrogen atoms are delivered to the same face of the triple bond: a syn addition giving the $c i s$ alkene. Quinoline occupies part of the metal surface and prevents the alkene from re-adsorbing, so reduction stops at the alkene. That is why quinoline is called a catalyst poison and the palladium is called a deactivated or poisoned catalyst.

How Birch reduction works

Sodium dissolved in liquid ammonia releases solvated electrons. One electron adds to the triple bond to give a radical anion, which takes a proton from ammonia; a second electron and a second proton follow. The intermediate vinyl anion prefers the arrangement in which the two bulky $R$ groups are as far apart as possible, so the product is the $t r an s$ alkene.

Terminal alkynes are the exception. A terminal alkyne is not reduced by

Na/liq. N H_{3}

at all, because its acidic hydrogen is removed first and the resulting acetylide anion is unreactive towards further electron transfer:

C H_{3} - C \equiv CH + Na \to C H_{3} - C \equiv C^{-} N a^{+} + \frac{1}{2} H_{2}

Solved Example 5

But-

2

-yne is treated as follows. Identify A, B, C and D.
(i)

H_{2}

/Lindlar

\to

A; then

B r_{2}

\to

B
(ii)

Na/liq. N H_{3}

\to

C; then

B r_{2}

\to

Solution:

A is $c i s$ -but- $2$ -ene: Lindlar delivers both hydrogens to the same face (syn addition).
C is $t r an s$ -but- $2$ -ene: $Na/liq. N H_{3}$ adds the hydrogens to opposite faces (anti addition).
Bromine adds anti across a double bond, through a bromonium ion. Applying anti addition to the $c i s$ alkene A gives B $=$ the pair of enantiomers ( $d$ and $ℓ$ ) of $2, 3$ -dibromobutane, that is a racemic mixture.
Applying the same anti addition to the $t r an s$ alkene C gives D $=$ meso- $2, 3$ -dibromobutane, which is optically inactive because of its internal mirror plane.

The lesson: a stereospecific reaction turns different stereoisomers of the starting material into different stereoisomers of the product.

8. Oxidation of Alkynes

Ozonolysis

Ozone cleaves the triple bond completely and the fragments are isolated as carboxylic acids.

$R - C \equiv C - R^{'} (i) O_{3} (ii) H_{2} O RCOOH + R^{'} COOH$

$C H_{3} - C \equiv C - C H_{3} (i) O_{3} (ii) H_{2} O 2 C H_{3} COOH$

Acetylene is again exceptional: it gives glyoxal ( $OHC - CHO$ ) as well as formic acid.

Oxidation with potassium permanganate

Conditions	Substrate	Product
Cold, neutral or slightly alkaline $KMn O_{4}$	$R - C \equiv C - R^{'}$	$α$ -diketone $R - CO - CO - R^{'}$
Hot or acidic $KMn O_{4}$	$R - C \equiv C - R^{'}$	$RCOOH + R^{'} COOH$
Hot or acidic $KMn O_{4}$	Terminal $R - C \equiv CH$	$RCOOH + C O_{2} + H_{2} O$

Note the diagnostic detail: a terminal carbon is oxidised all the way to carbon dioxide, so seeing $C O_{2}$ among the products tells you the triple bond was at the end of the chain.

Solved Example 6

A hydrocarbon

C_{16} H_{26}

on ozonolysis followed by hydrolysis gives only

C H_{3} (C H_{2})_{4} COOH

and succinic acid

HOOC - C H_{2} C H_{2} - COOH

. Identify the hydrocarbon.

Solution:

First find the degrees of unsaturation: $DoU = \frac{2 ( 16 ) + 2 - 26}{2} = \frac{8}{2} = 4$ , which is consistent with two triple bonds.

Succinic acid is a four-carbon diacid, so it came from a $- C H_{2} C H_{2} -$ unit flanked by two triple bonds. Each $C H_{3} (C H_{2})_{4} COOH$ came from a hexyl chain capping one end.

$C H_{3} (C H_{2})_{4} - C \equiv C - C H_{2} C H_{2} - C \equiv C - (C H_{2})_{4} C H_{3}$

Counting carbons: $5 + 2 + 2 + 2 + 5 = 16$ , and the hydrogen count matches $C_{16} H_{26}$ .

9. Polymerisation

To benzene. Passing acetylene through a red-hot tube (iron, copper or nickel) cyclically trimerises it: $3 HC \equiv CH red hot tube C_{6} H_{6}$ .
To mesitylene. Propyne trimerises the same way to $1, 3, 5$ -trimethylbenzene (mesitylene).
To vinyl acetylene. Passing acetylene into cuprous chloride in ammonium chloride dimerises it: $2 HC \equiv CH C u_{2} C l_{2} N H_{4} Cl C H_{2} = CH - C \equiv CH$ , but- $3$ -en- $1$ -yne. Adding $HCl$ to this gives chloroprene, the monomer of neoprene rubber.

10. Isomerisation

Heating an alkyne with $NaN H_{2}$ in an inert solvent walks the triple bond towards the end of the chain, because the terminal acetylide that forms is a thermodynamic sink. Alcoholic $KOH$ walks it the other way, inward, where the internal alkyne is the more stable product.

$C H_{3} C H_{2} - C \equiv C - C H_{3} NaN H_{2}, Δ H_{2} O C H_{3} C H_{2} C H_{2} - C \equiv CH$

11. Reaction Map and Comparison

Figure 7: Reaction map for

R - C \equiv C - R^{'}

. Addition, reduction and cleavage, all from one functional group.

Property	Alkane	Alkene	Alkyne
Hybridisation	$s p^{3}$	$s p^{2}$	$s p$
$p K_{a}$ of C-H	$50$	$44$	$25$
Rate of electrophilic addition	-	Faster	Slower
Rate of catalytic hydrogenation	-	Slower	Faster
Heat of combustion (per mole, $C_{2}$ )	$- 373$ kcal	$- 337$ kcal	$- 317$ kcal
Density ( $g c m^{- 3}$ , $C_{3}$ )	$0.50$	$0.52$	$0.67$
Reaction with ammoniacal $AgN O_{3}$	No	No	Yes, if terminal

Solved Example 7

An alkyne

X

of formula

C_{4} H_{6}

gives a white precipitate with ammoniacal

AgN O_{3}

and, on hydration with

H_{2} S O_{4} /HgS O_{4}

, gives a compound

Y

that answers the iodoform test. Identify

X

and

Y

Solution:

The precipitate proves $X$ is a terminal alkyne, so among the $C_{4} H_{6}$ alkynes it must be but- $1$ -yne, $C H_{3} C H_{2} - C \equiv CH$ (but- $2$ -yne is internal and would give no precipitate).

Markovnikov hydration puts the oxygen on $C_{2}$ : $Y = C H_{3} C H_{2} - CO - C H_{3}$ , butan- $2$ -one. It contains a $C H_{3} - CO -$ group, so it gives a yellow precipitate of iodoform with $I_{2} /NaOH$ , confirming the structure.

Solved Example 8

Explain why alkynes are less reactive than alkenes towards addition of

B r_{2}

Solution:

Both react through a cyclic bromonium ion. The ion formed from an alkyne still contains a full $C = C$ inside the three-membered ring, so it is far more strained than the ion from an alkene, which contains only a single bond in the ring.

In addition, the ring carbons derived from an alkyne have more $s$ character than those derived from an alkene, which makes them poorer at tolerating the positive charge. Both effects raise the energy of the intermediate, so the alkyne reacts more slowly.

Common Mistakes to Avoid

Watch out

Assuming alkynes are more reactive than alkenes because they have more $π$ electrons. Towards electrophiles they are less reactive, because the vinyl cation intermediate is unstable. Towards catalytic hydrogenation they are more reactive.
Writing an alcohol as the product of hydration. The enol is never isolated; it always tautomerises to a ketone (or, for acetylene only, an aldehyde).
Expecting an aldehyde from mercuric-catalysed hydration of a terminal alkyne. Markovnikov orientation puts the oxygen on the inner carbon, giving a methyl ketone. Only $Si a_{2} BH$ then $H_{2} O_{2} /O H^{-}$ gives the aldehyde.
Swapping the reduction reagents. Lindlar gives $c i s$ , sodium in liquid ammonia gives $t r an s$ . Remembering "Lindlar = liquid-free = same side" is one way to keep them apart.
Using $Na/liq. N H_{3}$ on a terminal alkyne to make a $t r an s$ alkene. It does not reduce; it only forms the acetylide salt.
Claiming that acetylides form with $NaOH$ . Acetylene ( $p K_{a} 25$ ) is a weaker acid than water ( $15.7$ ), so hydroxide cannot deprotonate it. $NaN H_{2}$ can, because ammonia has $p K_{a} 38$ .
Forgetting $C O_{2}$ when a terminal alkyne is oxidised. The $\equiv CH$ carbon goes all the way to carbon dioxide, which is itself a useful structural clue.
Drawing a $c i s$ or $t r an s$ product from addition to the triple bond and stopping there. With two equivalents of reagent the stereochemistry of the first step is erased in the second.

Frequently Asked Questions

Why are terminal alkynes acidic?

The hydrogen sits on an $s p$ hybridised carbon with $50%$ $s$ character. When it leaves, the lone pair occupies that $s p$ orbital, which holds electrons close to the nucleus and stabilises the acetylide ion. That stability is what makes the parent C-H acidic, with $p K_{a} \approx 25$ against $44$ for an alkene and $50$ for an alkane.

How do you distinguish a terminal alkyne from an internal alkyne?

Use ammoniacal silver nitrate or ammoniacal cuprous chloride. A terminal alkyne replaces its acidic hydrogen with the metal and gives a white precipitate with silver or a red precipitate with copper. An internal alkyne has no hydrogen on an $s p$ carbon and gives no precipitate.

Why do alkynes undergo electrophilic addition more slowly than alkenes?

Three reasons combine: the $s p$ carbons are more electronegative and hold the $π$ electrons tightly, those electrons are more delocalised in the cylindrical cloud, and the vinyl carbocation formed after the electrophile adds is far less stable than an alkyl carbocation because its positive charge sits on an $s p$ carbon.

What is the product of hydration of propyne?

Acetone. Water adds with Markovnikov orientation in dilute sulphuric acid containing mercuric sulphate, so the hydroxyl lands on the middle carbon and the first product is the enol $C H_{3} - C (OH) = C H_{2}$ . That enol tautomerises immediately to the keto form $C H_{3} - CO - C H_{3}$ .

Which alkyne gives an aldehyde on hydration?

Only acetylene, and it gives acetaldehyde, because both its carbons are equivalent and unsubstituted. Every higher terminal alkyne gives a methyl ketone under the same conditions. To obtain an aldehyde from a higher terminal alkyne you must use hydroboration with $Si a_{2} BH$ followed by alkaline hydrogen peroxide.

What is the difference between Lindlar catalyst and sodium in liquid ammonia?

Both stop the reduction at the alkene, but with opposite stereochemistry. Lindlar catalyst, palladium poisoned with quinoline, adsorbs the alkyne and delivers both hydrogens to the same face, giving the $c i s$ alkene. Sodium in liquid ammonia supplies electrons and protons in separate steps through a vinyl anion, so the hydrogens end up on opposite faces and the $t r an s$ alkene is formed.

Why is acetylene stored dissolved in acetone?

Liquefied or compressed acetylene is explosive, since it is thermodynamically unstable with respect to carbon and hydrogen. Dissolving it in acetone absorbed on a porous filler inside the cylinder keeps the pressure low and prevents shock-initiated decomposition, so it can be transported safely.

What does ozonolysis of an alkyne give?

Two carboxylic acids, one from each side of the triple bond, which makes ozonolysis a standard way of locating the triple bond in an unknown. But- $2$ -yne gives two molecules of acetic acid. Acetylene is exceptional and gives glyoxal along with formic acid.

Why does acetylene burn with a sooty luminous flame?

Its percentage of carbon by mass is very high, about $92%$ , so in a limited air supply combustion is incomplete and unburnt carbon particles glow in the flame. In pure oxygen combustion is complete and the oxy-acetylene flame reaches roughly $3500^{\circ} C$ , hot enough for welding and cutting steel.

Previous year questions on Properties of Alkynes

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

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1. Physical Properties

2. Acidity of Terminal Alkynes

Why s character controls acidity

Formation of metal acetylides

The laboratory test for a terminal alkyne

3. Electrophilic Addition: Why Alkynes React More Slowly

4. Addition of Halogens

5. Addition of Hydrogen Halides

6. Addition of Water: Hydration

Hydroboration-oxidation: the opposite regiochemistry

7. Reduction of Alkynes

How Lindlar catalyst works

How Birch reduction works

8. Oxidation of Alkynes

Ozonolysis

Oxidation with potassium permanganate

9. Polymerisation

10. Isomerisation

11. Reaction Map and Comparison

Common Mistakes to Avoid

Frequently Asked Questions

Why are terminal alkynes acidic?

How do you distinguish a terminal alkyne from an internal alkyne?

Why do alkynes undergo electrophilic addition more slowly than alkenes?

What is the product of hydration of propyne?

Which alkyne gives an aldehyde on hydration?

What is the difference between Lindlar catalyst and sodium in liquid ammonia?

Why is acetylene stored dissolved in acetone?

What does ozonolysis of an alkyne give?

Why does acetylene burn with a sooty luminous flame?

Previous year questions on Properties of Alkynes

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Why $s$ character controls acidity