Alkynes

B.Sc. 1st Year Chemistry — detailed study notes

Structure, nomenclature, preparation, acidity, addition reactions, mechanisms, oxidation, reduction, polymerization and examination preparation.

1. Introduction to Alkynes

Alkynes are unsaturated hydrocarbons containing at least one carbon-carbon triple bond.

The simplest alkyne is ethyne, commonly called acetylene.

Ethyne

HC≡CH

For an open-chain hydrocarbon containing one triple bond, the general molecular formula is:

C_nH_2n-2

Examples:

Ethyne — C₂H₂
Propyne — C₃H₄
But-1-yne — C₄H₆
But-2-yne — C₄H₆

Memory tip: Alkane has single bond, alkene has double bond and alkyne has triple bond. The ending -yne indicates a triple bond.

2. Structure of Alkynes

The two carbon atoms involved in the triple bond are sp hybridized.

A carbon-carbon triple bond contains:

C≡C = 1 σ bond + 2 π bonds

Formation of the triple bond

In an alkyne, each triple-bond carbon uses two sp hybrid orbitals. One sp orbital forms a sigma bond with the other carbon, while the second sp orbital forms a sigma bond with hydrogen or another substituent.

The two remaining unhybridized p-orbitals on each carbon overlap sideways in two different planes. These overlaps produce two π bonds.

Shape of an alkyne

Because of sp hybridization, the atoms around the triple bond are arranged approximately in a straight line.

Bond angle around C≡C ≈ 180°

Why are alkynes reactive?

The two π bonds contain electrons that can interact with electrophiles. Therefore, alkynes commonly undergo addition reactions.

Remember: Alkene = 1 π bond. Alkyne = 2 π bonds. Therefore, an alkyne can undergo addition reactions more than once.

3. Nomenclature of Alkynes

Basic rules

Select the longest carbon chain containing the triple bond.
Number the chain from the end nearest to the triple bond.
Replace "-ane" with "-yne".
Give the triple bond the lowest possible locant.
Name substituents in the usual manner.

Examples

HC≡CH → Ethyne

CH₃C≡CH → Prop-1-yne

CH≡C-CH₂-CH₃ → But-1-yne

CH₃-C≡C-CH₃ → But-2-yne

Common mistake: The chain must contain the triple bond. Do not choose a longer chain that leaves the triple bond outside the parent chain.

4. Isomerism in Alkynes

Chain isomerism

Different carbon skeletons can produce different structural isomers.

Position isomerism

The position of the triple bond can change.

CH≡C-CH₂-CH₃ → But-1-yne

CH₃-C≡C-CH₃ → But-2-yne

These compounds have the same molecular formula, C₄H₆, but differ in the position of the triple bond.

Memory tip: Move the triple bond along the carbon chain and you can get position isomers.

5. Preparation of Alkynes

5.1 Double dehydrohalogenation of vicinal dihalides

Vicinal dihalides can be converted into alkynes by removing two molecules of hydrogen halide.

R-CHX-CHX-R' → R-C≡C-R'

Strong base, usually excess NaNH₂

The reaction occurs in two elimination steps. The first elimination produces an alkene containing a halogen. A second elimination removes another molecule of HX and produces the triple bond.

Mechanism idea

Step 1: First elimination

The strong base removes a β-hydrogen. At the same time, the halide leaves and a C=C bond is formed.

Step 2: Formation of a vinyl halide

After the first elimination, one halogen remains attached to a carbon involved in the double bond.

Step 3: Second elimination

A stronger base removes another hydrogen and the remaining halide leaves.

Step 4: Formation of C≡C

The electrons form another π bond, converting the double bond into a triple bond.

Memory: Dihalide → remove HX twice → alkyne.

5.2 From vicinal tetrahalides

Appropriate tetrahalogen derivatives can also undergo dehalogenation to form alkynes.

The important idea is that removal of halogen atoms allows the carbon atoms to form additional bonding between them.

5.3 Preparation of acetylene from calcium carbide

Ethyne can be prepared by treating calcium carbide with water.

CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂

This is an important classical method of preparing acetylene.

Easy memory: Calcium carbide + water → acetylene + calcium hydroxide.

6. Acidity of Terminal Alkynes

One important difference between alkynes and alkenes is that terminal alkynes contain a hydrogen directly attached to the sp-hybridized carbon.

R-C≡C-H

This hydrogen is more acidic than the hydrogen attached to an alkene or alkane carbon.

Why is terminal alkyne hydrogen acidic?

When the hydrogen is removed, a carbanion is formed. The negative charge is located on an sp-hybridized carbon.

An sp orbital contains 50% s-character. The greater s-character holds the electrons closer to the nucleus and helps stabilize the negative charge.

s-character: sp = 50% > sp² = 33% > sp³ = 25%

Therefore, terminal alkynes are more acidic than alkenes and alkanes.

Reaction with sodium metal

2R-C≡C-H + 2Na → 2R-C≡C⁻Na⁺ + H₂

Hydrogen gas is liberated.

Formation of metal acetylides

Strong bases such as NaNH₂ can remove the terminal hydrogen and form an acetylide ion.

R-C≡C-H + NaNH₂ → R-C≡C⁻Na⁺ + NH₃

Memory tip: Terminal alkyne has C≡C-H. That H is acidic because the resulting negative charge is stabilized on an sp carbon.

Important: Internal alkynes do not have a hydrogen directly attached to the triple-bond carbon, so they do not show this terminal alkyne acidity in the same way.

7. Chemical Reactions of Alkynes

Alkynes mainly undergo addition reactions because the two π bonds can interact with reagents.

Important reactions include:

Hydrogenation
Halogenation
Addition of hydrogen halides
Hydration
Hydroboration-oxidation
Oxidation
Ozonolysis
Polymerization
Reactions of terminal alkynes

Main idea: Alkyne has two π bonds, so addition can occur once or twice. Always check whether the reaction stops at an alkene or continues to an alkane.

8. Hydrogenation of Alkynes

Complete hydrogenation

An alkyne can react with hydrogen in the presence of a metal catalyst to give an alkane.

R-C≡C-R' + 2H₂ → R-CH₂-CH₂-R'

H₂, Ni / Pt / Pd

Two molecules of hydrogen are required because the alkyne contains two π bonds.

Partial hydrogenation

If the reaction is stopped after addition of one molecule of hydrogen, an alkene can be obtained.

Using Lindlar catalyst

Alkyne + H₂ → cis-alkene

Lindlar catalyst

Lindlar catalyst is a poisoned palladium catalyst and is used when the desired product is a cis-alkene.

Reduction with sodium in liquid ammonia

Alkyne + Na / NH₃(l) → trans-alkene

This reaction gives the trans-alkene.

Reagent	Main product
H₂, Ni/Pt/Pd	Alkane
H₂, Lindlar catalyst	Cis-alkene
Na/NH₃(l)	Trans-alkene

Best memory:
Lindlar → cis
Na/NH₃ → trans
Excess H₂ → alkane

9. Halogenation of Alkynes

Halogens such as Br₂ and Cl₂ can add across the triple bond.

Addition of one molecule of bromine

R-C≡C-R' + Br₂ → R-CBr=CBr-R'

A dibromoalkene is produced.

Further addition

R-CBr=CBr-R' + Br₂ → R-CBr₂-CBr₂-R'

With excess bromine, the reaction can continue to give a tetrabromoalkane.

Mechanism

Step 1: The electron-rich π system attacks Br₂.

Step 2: Br₂ becomes polarized.

Step 3: A bromonium-type intermediate is formed.

Step 4: Br⁻ attacks and the first addition product forms.

Step 5: If another Br₂ molecule is available, the remaining double bond can react again.

Remember: Alkyne + 1 Br₂ → dibromoalkene. Alkyne + 2 Br₂ → tetrabromoalkane.

10. Addition of Hydrogen Halides

Alkynes react with HCl, HBr or HI to form haloalkenes and, with further addition, geminal dihalides.

First addition

R-C≡CH + HBr → R-C(Br)=CH₂

With a suitable unsymmetrical alkyne, addition normally follows Markovnikov orientation under ionic conditions.

Second addition

R-C(Br)=CH₂ + HBr → R-CBr₂-CH₃

The final product can contain both halogen atoms on the same carbon. Such a compound is called a geminal dihalide.

Mechanism idea

Step 1: Protonation

The π electrons of the alkyne interact with H⁺. The proton adds in the direction that gives the more favorable carbocation-type intermediate.

Step 2: Halide attack

Br⁻ or another halide ion attacks the positively charged center.

Step 3: Haloalkene formation

The first addition gives a haloalkene.

Step 4: Second addition

If excess HX is present, the remaining double bond reacts again and a geminal dihalide is formed.

Memory: Alkyne + HX once → haloalkene. Alkyne + HX twice → geminal dihalide.

11. Hydration of Alkynes

Hydration means addition of water across the triple bond. An acid and Hg²⁺ catalyst are commonly used.

Alkyne + H₂O → Enol → Carbonyl compound

H₂SO₄, HgSO₄

Important idea: enol formation

The first product is an enol. It contains both a C=C bond and an -OH group.

The enol usually changes into a more stable carbonyl compound through keto-enol tautomerism.

Example: Ethyne

HC≡CH + H₂O → CH₂=CHOH → CH₃CHO

H₂SO₄, HgSO₄

The enol CH₂=CHOH changes into acetaldehyde, CH₃CHO.

Mechanism

Step 1: Activation of the alkyne

The mercury(II) ion coordinates with the π system and makes the triple bond more reactive toward water.

Step 2: Water adds

Water attacks the activated system and, after proton transfers, an enol-type product is obtained.

Step 3: Enol formation

The intermediate contains a C=C bond and an -OH group.

Step 4: Tautomerization

Proton movement and movement of the π bond convert the enol into the more stable carbonyl compound.

Very important memory:
Alkyne hydration does not usually stop at an enol. Think: Alkyne → enol → carbonyl compound.

12. Hydroboration-Oxidation

Hydroboration-oxidation is another important method of converting an alkyne into a carbonyl compound.

Alkyne → Enol → Aldehyde or ketone

1. BH₃ or suitable borane
2. H₂O₂ / OH⁻

For terminal alkynes, hydroboration-oxidation generally gives an aldehyde, whereas acid-catalyzed hydration commonly gives a methyl ketone.

Mechanism idea

Step 1: Hydroboration

The B-H bond adds across the triple bond. Boron attaches to one carbon while hydrogen attaches to the other.

Step 2: Oxidation

Hydrogen peroxide in basic medium converts the carbon-boron bond into a carbon-oxygen bond.

Step 3: Enol formation

The immediate oxygen-containing product rearranges to an enol-type structure.

Step 4: Tautomerization

The enol changes into the more stable carbonyl compound.

Memory tip: Terminal alkyne:
Hg²⁺ hydration → usually methyl ketone
Hydroboration-oxidation → aldehyde

13. Oxidation of Alkynes

Strong oxidation can break the carbon-carbon triple bond.

Depending on the structure of the alkyne and the conditions, carboxylic acids or carbon dioxide can be formed.

Oxidative cleavage

R-C≡C-R' → R-COOH + R'-COOH

For a terminal alkyne, the terminal carbon can ultimately be oxidized to carbon dioxide under strong oxidative conditions.

How to think about oxidation: Locate the triple bond first. Then imagine breaking the C≡C bond and determine what each carbon becomes under the given oxidative conditions.

14. Ozonolysis of Alkynes

Ozone can oxidatively cleave an alkyne.

R-C≡C-R' + O₃ → oxidative cleavage products

Unlike simple alkene ozonolysis, alkyne cleavage under oxidative conditions generally leads to carboxylic acids when hydrogen is not the only substituent involved.

How to solve cleavage questions

Find the C≡C bond.
Cut the bond between the two triple-bond carbons.
Consider the oxidation state of each carbon.
Write the appropriate carboxylic acid or CO₂ product.

Memory tip: Alkene ozonolysis often gives aldehydes/ketones under reductive work-up, while strong oxidative cleavage of alkynes commonly gives carboxylic acids and, for a terminal carbon, ultimately CO₂.

15. Polymerization of Alkynes

Alkynes can undergo polymerization under suitable conditions. Acetylene is an important example.

Acetylene can polymerize to form larger unsaturated products, and under appropriate catalytic conditions aromatic products can also be obtained.

Trimerization of acetylene

3HC≡CH → C₆H₆

Suitable catalyst and elevated temperature

Three molecules of acetylene can combine to form benzene. This is an important example of cyclization and trimerization.

Memory: Three acetylene molecules → benzene.

16. How to Remember Alkyne Reaction Mechanisms

1. Always identify the π bonds

An alkyne contains two π bonds. Most electrophilic addition reactions begin with these electrons.

2. Ask what type of reagent is present

Reagent	Think of
H₂	Reduction / hydrogenation
Br₂	Halogen addition
HBr / HCl	Hydrohalogenation
H₂O, Hg²⁺	Hydration → enol → carbonyl
BH₃, H₂O₂/OH⁻	Hydroboration → aldehyde/ketone
Strong oxidant	Triple-bond cleavage

3. Remember the two-stage addition idea

Because an alkyne has two π bonds, many addition reactions can happen twice.

Alkyne → alkene → alkane

Whether the reaction stops at the alkene depends on the reagent and reaction conditions.

4. Terminal vs internal alkyne

Terminal: R-C≡C-H
Internal: R-C≡C-R'

If you see C≡C-H, immediately think: acidic hydrogen and acetylide formation.

5. Hydration memory

Two hydration routes:

Hg²⁺ route: Alkyne → enol → usually ketone

Hydroboration route: Alkyne → enol → terminal alkyne usually gives aldehyde

6. Reduction memory

Lindlar = cis
Na/NH₃ = trans
Excess H₂ = alkane

17. Alkene vs Alkyne

Property	Alkene	Alkyne
Characteristic bond	C=C	C≡C
Number of π bonds	1	2
Hybridization	sp²	sp
Approximate bond angle	120°	180°
General formula	C_nH_2n	C_nH_2n-2
Terminal acidic hydrogen	No	Yes, in terminal alkynes
Common addition reactions	Yes	Yes

Easy comparison: Alkene = double bond, sp², one π bond. Alkyne = triple bond, sp, two π bonds.

18. Common Mistakes

Writing C_nH_2n as the general formula of an open-chain alkyne.
Forgetting that an alkyne contains two π bonds.
Confusing sp and sp² hybridization.
Calling every alkyne hydrogen acidic. Only terminal alkynes contain the characteristic C≡C-H hydrogen.
Forgetting that hydration first gives an enol.
Forgetting tautomerization of the enol to the carbonyl compound.
Confusing Lindlar catalyst with Na/NH₃.
Writing cis-alkene for Na/NH₃ reduction.
Assuming every addition reaction stops after one addition.
Treating hydroboration-oxidation as identical to Hg²⁺-catalyzed hydration.
Forgetting the second elimination when preparing an alkyne from a vicinal dihalide.
Writing only the final product in a mechanism question.

Exam warning: In mechanism questions, show electron movement or clearly explain which bond is broken, which bond is formed, and why the intermediate/product is favored.

19. Worked Problems

Problem 1: Molecular formula

Find the molecular formula of an open-chain alkyne containing 5 carbon atoms.

General formula = C_nH_2n-2

n = 5

Formula = C₅H₈

Therefore, the molecular formula is C₅H₈.

Problem 2: Identify terminal alkyne

Which one is a terminal alkyne?

A: CH₃-C≡C-CH₃

B: CH₃-C≡CH

Compound B is a terminal alkyne because the triple bond is at the end and one triple-bond carbon is attached directly to H.

Problem 3: Complete hydrogenation

What is formed when propyne reacts with excess hydrogen?

CH₃-C≡CH + 2H₂ → CH₃-CH₂-CH₃

The triple bond is completely reduced to a single bond. Therefore, propane is formed.

Problem 4: Partial reduction

What is the major product when an alkyne is treated with H₂/Lindlar catalyst?

Alkyne → cis-alkene

Problem 5: Hydration of ethyne

What is obtained when ethyne is hydrated in the presence of HgSO₄/H₂SO₄?

HC≡CH + H₂O → CH₂=CHOH → CH₃CHO

The final product is acetaldehyde.

Problem 6: Calcium carbide method

Write the preparation of acetylene from calcium carbide.

CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂

20. TU Examination Preparation

2-mark questions

Define alkyne with an example.
Write the general formula of alkynes.
What is a terminal alkyne?
Why are terminal alkynes acidic?
What is Lindlar catalyst?
What is hydroboration?
Write the preparation of acetylene from calcium carbide.
What is acetylide ion?

5-mark questions

Explain the structure and hybridization of alkynes.
Explain the preparation of alkynes from vicinal dihalides.
Explain the acidic character of terminal alkynes.
Explain hydrogenation of alkynes.
Explain hydration of alkynes.
Explain hydroboration-oxidation of alkynes.

10-mark questions

Discuss the preparation and chemical reactions of alkynes with suitable mechanisms.
Explain the addition reactions of alkynes in detail.
Discuss the acidic character of terminal alkynes and reactions of acetylides.
Explain hydration and hydroboration-oxidation of alkynes with mechanisms.
Describe the reduction of alkynes and explain the formation of cis- and trans-alkenes.

How to write a strong 10-mark reaction answer

Start with the reaction or a short definition.
Mention the reagent and conditions.
Write the reaction in normal structural form.
Explain the first mechanistic step.
Show the important intermediate.
Explain the second step and product formation.
Explain why the major product is formed if orientation is involved.
Add one example.
Finish with an important limitation or memory point when appropriate.

For mechanism marks: Do not write a mechanism as a story only. Show the reaction structure and then explain the electron movement in words. This makes the answer easier to follow and revise.

21. Quick Revision

Alkynes contain a carbon-carbon triple bond.
General formula of an open-chain monoalkyne: C_nH_2n-2.
Triple-bond carbons are sp hybridized.
C≡C contains one σ bond and two π bonds.
The bond angle around a triple bond is approximately 180°.
Alkynes commonly undergo addition reactions.
Vicinal dihalides can give alkynes by double elimination.
Calcium carbide reacts with water to produce acetylene.
Terminal alkynes contain an acidic C≡C-H hydrogen.
Terminal alkynes can form acetylide ions.
Complete hydrogenation gives an alkane.
Lindlar catalyst gives a cis-alkene.
Na/NH₃(l) gives a trans-alkene.
Halogen addition can occur twice.
HX can add twice to produce geminal dihalides.
Hg²⁺-catalyzed hydration gives an enol first, followed by tautomerization.
Hydroboration-oxidation of a terminal alkyne generally gives an aldehyde.
Strong oxidation can cleave the triple bond.
Three acetylene molecules can form benzene by trimerization.

One-minute memory map:

C≡C → 2 π bonds
Terminal C≡C-H → acidic H
H₂ / Ni → alkane
Lindlar → cis-alkene
Na / NH₃ → trans-alkene
Br₂ → dibromoalkene → tetrabromoalkane
HX → haloalkene → geminal dihalide
H₂O / Hg²⁺ → enol → ketone
BH₃, H₂O₂/OH⁻ → enol → aldehyde for terminal alkyne
Strong oxidation → cleavage

22. Quick Check

1. What is the hybridization of the carbon atoms in an alkyne?

sp hybridization.

2. How many π bonds are present in a C≡C bond?

Two π bonds.

3. Which type of alkyne has an acidic hydrogen?

A terminal alkyne containing C≡C-H.

4. What does Lindlar catalyst produce from an alkyne?

A cis-alkene by partial hydrogenation.

5. What does Na/NH₃(l) give?

A trans-alkene.

6. What is the first product of alkyne hydration?

An enol, which usually tautomerizes to a carbonyl compound.

7. What is the important difference between terminal and internal alkynes?

Terminal alkynes have C≡C-H; internal alkynes have carbon groups on both triple-bond carbons.