Alkenes

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

Structure, nomenclature, preparation, reactions, reaction mechanisms, oxidation, ozonolysis, polymerization and examination preparation.

1. Introduction to Alkenes

Alkenes are unsaturated hydrocarbons containing at least one carbon-carbon double bond.

The carbon atoms joined by the double bond are generally sp² hybridized. The double bond consists of one sigma (σ) bond and one pi (π) bond.

General formula of an open-chain alkene with one double bond: C_nH_2n

Examples:

Ethene — C₂H₄
Propene — C₃H₆
Butene — C₄H₈
Pentene — C₅H₁₀

Memory tip: Alkane = single bond, alkene = double bond, alkyne = triple bond. The "ene" ending reminds you of a double bond.

2. Structure of the Carbon-Carbon Double Bond

In ethene, each carbon atom is attached to three other atoms through sigma bonds and uses one unhybridized p-orbital to form the pi bond.

C = C contains one σ bond + one π bond

Formation of the double bond

Each carbon atom in ethene is sp² hybridized. Three sp² orbitals lie approximately in one plane, while the remaining unhybridized p-orbital is perpendicular to that plane.

The two carbon atoms first form a sigma bond by head-on overlap. Their unhybridized p-orbitals then overlap sideways to form the pi bond.

Why is the π bond important?

The pi bond is weaker and more exposed than the sigma bond. Therefore, many reactions of alkenes occur by attacking or breaking the pi bond.

Remember: The sigma bond forms the strong framework. The pi bond is the part that usually reacts first.

Restricted rotation

Rotation around a C=C bond is restricted because rotation would require breaking the sideways overlap of the p-orbitals.

This restricted rotation is responsible for geometrical isomerism in suitable alkenes.

3. Nomenclature of Alkenes

Basic rules

Select the longest carbon chain containing the double bond.
Number the chain from the end nearest to the double bond.
Replace the "-ane" ending of the alkane by "-ene".
Give the position of the double bond the lowest possible number.
Name and locate substituents in the usual way.

Examples

Example 1

CH₂=CH₂ → Ethene

Example 2

CH₃CH=CH₂ → Propene

Example 3

CH₃CH=CHCH₃ → But-2-ene

Common mistake: Do not number an alkene from the end that gives a substituent the lower number. The double bond gets priority when deciding the direction of numbering.

4. Isomerism in Alkenes

Structural isomerism

Alkenes can show chain isomerism and position isomerism.

For example, C₄H₈ can form but-1-ene and but-2-ene due to different positions of the double bond.

Geometrical isomerism

Geometrical isomerism occurs because rotation around the carbon-carbon double bond is restricted.

For geometrical isomerism to occur, each carbon atom of the double bond should generally have two different groups attached.

Cis and trans form

If similar groups are on the same side of the double bond, the compound may be described as cis. If they are on opposite sides, it may be described as trans.

Memory tip: Cis = same side. Trans = opposite side.

5. Preparation of Alkenes

5.1 Dehydration of alcohols

One of the most important laboratory methods of preparing alkenes is dehydration of alcohols.

General reaction

Alcohol → Alkene + H₂O

Concentrated H₂SO₄ and heat

Example:

CH₃CH₂OH → CH₂=CH₂ + H₂O

Conc. H₂SO₄, heat

Mechanism idea

The exact mechanism depends on the type of alcohol. For secondary and tertiary alcohols, dehydration commonly occurs through carbocation formation under acidic conditions.

Step 1: Protonation of the -OH group

The oxygen of the alcohol accepts H⁺. This converts the poor leaving group -OH into water, which is a much better leaving group.

Step 2: Removal of water

In an appropriate secondary or tertiary alcohol, water leaves and a carbocation is formed.

Step 3: Removal of β-hydrogen

A base removes a hydrogen from the carbon next to the positively charged carbon. The electrons from the C-H bond form the C=C bond.

Step 4: Alkene formation

The alkene is produced and the acid catalyst can be regenerated.

Remember the dehydration idea: Remove H from one carbon and OH from the neighboring carbon. The two carbons then form C=C.

5.2 Dehydrohalogenation of alkyl halides

Alkyl halides can give alkenes when heated with alcoholic KOH or another suitable strong base.

CH₃CH₂Br → CH₂=CH₂ + HBr

Alcoholic KOH, heat

In the actual reaction, KOH provides the basic conditions and the hydrogen halide is removed from the molecule.

Mechanism: β-elimination

The base approaches a hydrogen attached to the β-carbon.
The base removes that hydrogen.
The C-H bond electrons move toward the adjacent carbon.
The leaving group leaves.
A C=C double bond is formed.

Memory trick: In elimination, think: β-H leaves + leaving group leaves → C=C forms.

5.3 Dehalogenation of vicinal dihalides

Vicinal dihalides can be converted into alkenes by treatment with zinc.

R-CHX-CHX-R' + Zn → R-CH=CH-R' + ZnX₂

Both halogen atoms are removed from neighboring carbon atoms, allowing a double bond to form between those carbons.

6. Physical Properties of Alkenes

Lower alkenes are gases at room temperature.
Higher members may be liquids or solids.
They are generally insoluble or only slightly soluble in water.
They are soluble in many organic solvents.
Boiling point generally increases with molecular mass.
Branching can affect boiling point.

Alkenes are less polar than water and therefore do not mix well with water.

7. Chemical Reactions of Alkenes

The main chemical behavior of alkenes comes from the carbon-carbon double bond, especially the π bond.

Most common reactions are addition reactions. The double bond is converted into two new single bonds as new atoms or groups become attached to the two carbon atoms.

7.1 Hydrogenation

Addition of hydrogen

CH₂=CH₂ + H₂ → CH₃CH₃

Ni, Pt or Pd catalyst

Hydrogen is added across the double bond and an alkane is formed.

Mechanism idea

On the metal surface, hydrogen molecules interact with the catalyst and become activated. The alkene also interacts with the catalyst surface. Hydrogen atoms are then transferred to the two carbon atoms of the double bond.

Memory tip: Hydrogenation = add H₂ = reduce the double bond.

7.2 Halogenation

Addition of bromine

CH₂=CH₂ + Br₂ → CH₂Br-CH₂Br

Bromine adds across the double bond to form a vicinal dibromide.

Mechanism of bromination

Step 1: Attack of the alkene on Br₂

The electron-rich π bond approaches the bromine molecule. This polarizes Br-Br because the electron cloud is pushed away from the approaching alkene.

Step 2: Formation of bromonium ion

Instead of simply producing a free carbocation, the bromine atom becomes temporarily attached to both carbon atoms, forming a three-membered cyclic bromonium ion.

Step 3: Attack by Br⁻

The bromide ion attacks from the opposite side and opens the bromonium ion.

Step 4: Dibromide formation

The final product contains bromine atoms attached to the two carbon atoms.

Remember bromination: Alkene attacks Br₂ → bromonium ion → Br⁻ attacks → dibromide.

7.3 Addition of hydrogen halides

Hydrogen halides such as HCl, HBr and HI add across the carbon-carbon double bond.

CH₃CH=CH₂ + HBr → CH₃CHBrCH₃

In the absence of special radical conditions, this generally follows Markovnikov orientation.

8. Markovnikov's Rule

Markovnikov's rule is used to predict the major product when an unsymmetrical alkene reacts with an unsymmetrical reagent such as HBr.

In simple terms, when HX adds to an unsymmetrical alkene, hydrogen generally goes to the double-bond carbon that already has more hydrogen atoms, while X goes to the carbon with fewer hydrogen atoms.

Example: Propene + HBr

CH₃CH=CH₂ + HBr → CH₃CHBrCH₃

Why does this happen?

The important point is not just memorizing which carbon gets hydrogen. The rule comes from carbocation stability.

Step 1: Protonation

The π electrons of the alkene attack H⁺. Depending on which carbon receives H, two possible carbocations could theoretically form.

Step 2: Compare carbocations

One pathway produces a secondary carbocation, while the other would produce a primary carbocation.

Step 3: More stable carbocation forms

The secondary carbocation is more stable than the primary carbocation, so the pathway producing the secondary carbocation is favored.

Step 4: Nucleophilic attack

Br⁻ attacks the carbocation and forms the final alkyl bromide.

Best memory method: Do not memorize Markovnikov only as "H goes to the carbon having more H." Instead remember: choose the pathway that gives the more stable carbocation.

Carbocation stability

3° carbocation > 2° carbocation > 1° carbocation

Alkyl groups help stabilize the positive charge through electron-releasing effects and hyperconjugation.

9. Peroxide Effect

The peroxide effect is mainly observed with HBr. In the presence of peroxides, HBr can add to an alkene by a free radical mechanism and give the anti-Markovnikov product.

CH₃CH=CH₂ + HBr → CH₃CH₂CH₂Br

Peroxide / radical conditions

Why only HBr?

The radical chain steps involving HBr have favorable enough energetics to sustain the chain process. Comparable reactions with HCl and HI do not proceed effectively by the same peroxide mechanism.

Free-radical mechanism

Step 1: Initiation

The peroxide bond breaks to generate radicals.

RO-OR → 2RO•

Step 2: Formation of bromine radical

The radical reacts with HBr and produces a bromine radical.

Step 3: Propagation

The bromine radical adds to the alkene in such a way that the more stable carbon radical is formed.

That carbon radical then removes hydrogen from HBr, producing the product and regenerating Br•.

Step 4: Chain continues

The newly formed Br• can react with another alkene, so the process continues as a chain reaction.

Memory tip: Normal HBr addition → Markovnikov. HBr + peroxide → Anti-Markovnikov.

The peroxide effect is associated with HBr, not all hydrogen halides.

Common mistake: Do not write peroxide effect for HCl or HI in ordinary undergraduate reaction questions.

10. Oxidation of Alkenes

10.1 Mild oxidation

Cold, dilute alkaline KMnO₄ can oxidize an alkene to a vicinal diol.

Alkene + [O] → Vicinal diol

Cold, dilute alkaline KMnO₄

For ethene:

CH₂=CH₂ → HO-CH₂-CH₂-OH

The purple color of permanganate may disappear during the reaction. This is one reason KMnO₄ is useful in detecting unsaturation.

10.2 Strong oxidation

Strong oxidative conditions can cleave the C=C bond. Depending on the groups attached to the double-bond carbons, aldehydes, ketones, or carboxylic acids may be formed.

Remember: Mild oxidation adds OH groups. Strong oxidation breaks the double bond.

11. Ozonolysis

Ozonolysis is an important reaction used to locate the position of a double bond in an alkene.

Ozone first reacts with the double bond to form an ozonide, which is then decomposed during work-up to produce carbonyl compounds.

Alkene + O₃ → Ozonide → Carbonyl compounds

Reductive work-up such as Zn/H₂O

Example: Propene

CH₃CH=CH₂ → CH₃CHO + HCHO

How to solve ozonolysis questions

Find the C=C bond.
Imagine cutting the bond between the two double-bond carbons.
Turn each double-bond carbon into a carbonyl carbon.
Keep the groups originally attached to each carbon.
Write the resulting aldehyde or ketone products.

Best memory trick: For ozonolysis, mentally "cut the double bond." Each half becomes a C=O-containing product.

Using ozonolysis to identify an unknown alkene

If the products of ozonolysis are known, the original alkene can often be reconstructed by joining the two carbonyl carbons with a double bond.

12. Polymerization of Alkenes

Many alkenes undergo addition polymerization. The π bond opens and many alkene molecules join together to form a long-chain polymer.

Example: Ethene

n CH₂=CH₂ → [-CH₂-CH₂-]_n

The monomer is ethene and the polymer is polyethylene.

Another example: Propene

n CH₂=CHCH₃ → [-CH₂-CH(CH₃)-]_n

The monomer is propene and the polymer is polypropylene.

Memory tip: Addition polymerization = open the C=C and connect many monomers together.

13. Stability of Alkenes

Alkene stability generally increases as the number of alkyl groups attached to the double-bond carbons increases.

Tetrasubstituted > Trisubstituted > Disubstituted > Monosubstituted

Alkyl groups stabilize the double bond through electron releasing effects and hyperconjugation.

Hydrogenation heat as an indication of stability

More stable alkenes generally release less heat during hydrogenation because they begin at a lower energy level.

Remember: More alkyl substitution usually means a more stable alkene.

14. How to Remember Alkene Reaction Mechanisms

Tip 1: Find the π bond first

In most alkene reactions, begin by asking: What is attacking the π bond?

Tip 2: Separate ionic and radical reactions

Ionic mechanism	Radical mechanism
Involves ions	Involves radicals
Often involves carbocations	Often uses single-electron movement
Normal HX addition	Peroxide effect with HBr

Tip 3: For Markovnikov addition, ask one question

Which direction gives the more stable carbocation?

Once you answer that, the major product usually becomes much easier to determine.

Tip 4: For elimination, look for β-hydrogen

If a leaving group is present, check the neighboring carbon. If a suitable β-hydrogen can be removed, a C=C bond can form.

Tip 5: For ozonolysis, cut C=C

Do not try to memorize every product separately. Cut the double bond and convert both alkene carbons into carbonyl carbons.

Tip 6: For bromination, remember bromonium ion

Alkene → bromonium ion → Br⁻ attack.

One-line reaction memory map:
H₂ → alkane
Br₂ → dibromide
HX → alkyl halide
HBr + peroxide → anti-Markovnikov product
cold KMnO₄ → diol
O₃ → carbonyl products
polymerization → long chain

15. Common Mistakes

Forgetting that the double bond receives priority in numbering.
Confusing Markovnikov addition with peroxide-effect addition.
Applying the peroxide effect to HCl or HI.
Forgetting that bromination proceeds through a bromonium ion.
Writing a free carbocation as the normal intermediate in bromination.
Forgetting the β-hydrogen in elimination reactions.
Confusing dehydration with hydration.
Writing only the reagent without explaining the reaction mechanism in a long-answer question.
Forgetting that strong oxidation can cleave the C=C bond.
Memorizing ozonolysis products without understanding the double-bond cleavage method.

Exam warning: In a 5- or 10-mark mechanism question, writing only the final product usually does not show enough understanding. Write the important intermediate and explain why that intermediate forms.

16. Worked Problems

Problem 1: Molecular formula

Find the molecular formula of an open-chain alkene containing 6 carbon atoms.

General formula = C_nH_2n

n = 6

Formula = C₆H₁₂

Therefore, the molecular formula is C₆H₁₂.

Problem 2: Markovnikov product

Predict the major product of propene with HCl.

CH₃CH=CH₂ + HCl → CH₃CHClCH₃

The reaction proceeds through the more stable secondary carbocation, so the major product is 2-chloropropane.

Problem 3: Peroxide effect

Predict the major product of propene with HBr in the presence of peroxide.

CH₃CH=CH₂ + HBr → CH₃CH₂CH₂Br

Under radical peroxide conditions, HBr gives the anti-Markovnikov product.

Problem 4: Ozonolysis

Predict the products obtained when propene undergoes ozonolysis followed by reductive work-up.

CH₃CH=CH₂ → CH₃CHO + HCHO

The double bond is cut and each carbon becomes part of a carbonyl compound.

17. TU Examination Preparation

2-mark questions

Define alkene with an example.
Write the general formula of alkenes.
State Markovnikov's rule.
What is the peroxide effect?
What is ozonolysis?
Why do alkenes undergo addition reactions?

5-mark questions

Explain the preparation of alkenes by dehydration of alcohols.
Explain Markovnikov's rule with a suitable example.
Explain bromination of an alkene with mechanism.
Describe the peroxide effect.
Explain ozonolysis of alkenes.

10-mark questions

Discuss the preparation and chemical reactions of alkenes.
Explain important addition reactions of alkenes with mechanisms.
Explain Markovnikov and anti-Markovnikov addition in detail.
Describe oxidation and ozonolysis of alkenes with examples.
Discuss the mechanism of electrophilic addition to alkenes.

How to write a 10-mark alkene reaction answer

Start with a short definition or introduction.
Write the general reaction.
Write the reagent and conditions.
Explain the mechanism step by step.
Show the important intermediate where applicable.
Explain why the major product is formed.
Give one suitable example.
Add an important memory point or limitation if relevant.

18. Quick Revision

Alkenes contain a carbon-carbon double bond.
General formula of simple open-chain monoalkenes: C_nH_2n.
Double-bond carbons are sp² hybridized.
C=C contains one σ bond and one π bond.
Rotation around C=C is restricted.
Alcohol dehydration is an important preparation method.
Alkyl halides can undergo dehydrohalogenation.
Most common alkene reactions are additions.
H₂/Ni gives hydrogenation.
Br₂ gives dibromination through a bromonium ion.
HX addition normally follows Markovnikov orientation.
HBr + peroxide gives anti-Markovnikov addition.
Cold dilute KMnO₄ can give vicinal diols.
Strong oxidation can cleave the double bond.
Ozonolysis is useful for locating a C=C bond.
Alkenes can undergo addition polymerization.
Greater alkyl substitution generally increases alkene stability.

Final memory line:
Double bond = electron-rich π bond → addition is common. For every reaction, first identify what attacks the π bond, then follow where the electrons move.

19. Quick Check

1. What is the general formula of an open-chain monoalkene?

C_nH_2n

2. What intermediate is formed during bromination of an alkene?

A cyclic bromonium ion.

3. Which carbocation is generally more stable: 1° or 2°?

A 2° carbocation is generally more stable than a 1° carbocation.

4. What does ozonolysis do to a C=C bond?

It cleaves the double bond and produces carbonyl compounds after suitable work-up.

5. What special reagent gives anti-Markovnikov addition of HBr?

Peroxide under radical conditions.