A Clear Guide to Alkene Chemistry
For an 11th Grader
June 10, 2025
Part 1: How to Make Alkenes
(Synthesis)
Introduction: What is an Alkene?
Alkenes are a class of hydrocarbons that contain at least one carbon-carbon double
bond (C – C). This double bond is the ”functional group” of an alkene – it’s where most
of the chemical action happens!
A double bond consists of one strong sigma (σ) bond and one weaker, more exposed
pi (π) bond. The electrons in the pi bond are located above and below the plane of the
atoms, making them easily accessible to other reacting molecules. This electron-rich pi
bond is the key to understanding almost all alkene reactions.
Preparation via Elimination Reactions
Most methods for making alkenes involve elimination reactions. The name says it all:
you eliminate (remove) two atoms or groups from adjacent carbon atoms in a molecule
to form a double bond.
H
Y
Elimination
C
C + XY
X
C
C
H
We will focus on two main starting materials: alcohols and haloalkanes.
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From Alcohols (Dehydration)
Dehydration literally means ”removal of water.” We can force an alcohol to lose a
water molecule (an -H from one carbon and an -OH from the adjacent carbon) to form
an alkene.
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• Reagents: This is done by heating the alcohol with a strong acid catalyst, typically
concentrated sulfuric acid (H2 SO4 ) or phosphoric acid (H3 PO4 ).
• Ease of Dehydration: Not all alcohols dehydrate equally well. The order of
reactivity is:
Tertiary (3◦ ) Alcohol > Secondary (2◦ ) Alcohol > Primary (1◦ ) Alcohol
This is because the reaction often proceeds through a carbocation intermediate
(a carbon atom with a positive charge). Tertiary carbocations are the most stable,
so they form most easily.
Mechanism for 2◦ and 3◦ Alcohols (E1 Mechanism)
This mechanism happens in steps. Let’s look at the dehydration of propan-2-ol:
conc. H SO
2
4
−→
CH3 CH – CH2 + H2 O
CH3 CH(OH)CH3 −−−−−−
∆
1. Step 1: Protonation of the Alcohol. The acid catalyst donates a proton (H+ )
to the lone pair on the oxygen atom. This turns the -OH group, which is a poor
leaving group, into an – OH2 + group, which is an excellent leaving group (it’s a water
+
OH
OH2
molecule waiting to happen!). CH3
CH
CH3 + H+
CH3
CH
CH3
2. Step 2: Formation of the Carbocation. The C-O bond breaks, and the stable
water molecule leaves, taking the bonding electrons with it. This leaves behind a
carbon with a +
positive charge – a carbocation. This is the slow, rate-determining
OH2
+
Slow
CH3
CH
CH3 + H2 O
step. CH3
CH
CH3
3. Step 3: Deprotonation to form the Alkene. A weak base (like a water molecule
or the HSO4 – ion) removes a proton from a carbon atom adjacent to the carbocation. The electrons from that C-H bond swing over to form the new pi bond, creat+
ing the alkene. The acid catalyst (H+ ) is regenerated. H2 2, O + H
CH2
CH
Question 1: Why is a tertiary alcohol easier to dehydrate than a primary alcohol?
(Hint: Think about the stability of the intermediate formed in Step 2).
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From Haloalkanes (Dehydrohalogenation)
Dehydrohalogenation means ”removal of a hydrogen and a halogen.” We remove a
hydrogen atom (H) from one carbon and a halogen atom (X = Cl, Br, I) from an adjacent
carbon.
2
CH3
• Reagents: This requires a strong base dissolved in ethanol. A common choice
is potassium hydroxide in ethanol (KOH/EtOH), which forms potassium ethoxide
(CH3 CH2 OK), a very strong base.
• Ease of Reaction: The reactivity order is the same as for alcohols:
Tertiary (3◦ ) Haloalkane > Secondary (2◦ ) > Primary (1◦ )
Mechanism (E2 Mechanism)
This mechanism is concerted, meaning it all happens in one single, coordinated step!
Let’s look at the reaction of 2-bromopropane with the strong ethoxide ion (CH3 CH2 O – ).
Br
7, EtO + H
CH2
C
EtOH + CH2
CH
CH3 + Br
CH3
single step explained: The strong base (EtO – ) attacks and removes a proton. As the
C-H bond breaks, its electrons swing down to form the C – C double bond. At the exact
same time, the C-Br bond breaks and the bromine leaves as a bromide ion (Br – ).
Question 2: In dehydrohalogenation, what is the role of the strong base like potassium
ethoxide?
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[-¿] (c1)
Part 2: What Alkenes Do
(Reactions)
Introduction: Electrophilic Addition
The pi (π) bond of an alkene is a region of high electron density. This makes it a
nucleophile (a ”nucleus-lover”). Therefore, alkenes react with electrophiles (”electronlovers”), which are electron-deficient species.
The most common reaction of alkenes is electrophilic addition. In this reaction,
the pi bond breaks, and two new sigma bonds are formed as a new molecule adds across
C + X
Y
the double bond. C
X
C
C
Y
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Addition of Hydrogen Halides (Hydrohalogenation)
A hydrogen halide (H – X, like HBr or HCl) adds across the double bond. When the
alkene is unsymmetrical (like propene), this reaction follows a specific rule.
Markovnikov’s Rule
Markovnikov’s Rule states: When an unsymmetrical reagent (like H-X) adds to an
unsymmetrical alkene, the hydrogen atom adds to the carbon of the double bond that
already has more hydrogen atoms. The other part of the reagent (the X atom) adds to
the other carbon.
A simpler way to remember it: ”The rich get richer.” The carbon atom that is
already ”rich” with hydrogens gets the next hydrogen.
• Why? The rule is based on carbocation stability. The H+ adds first, creating a
carbocation. It will add in a way that forms the most stable carbocation (3◦ > 2◦ >
1◦ ).
• Example: CH3 CH – CH2 + HBr −−→ CH3 CH(Br)CH3 (Major Product)
Mechanism:
1. The electron-rich pi bond attacks the partial positive hydrogen of H-Br. The
H adds to the end carbon to form the more stable secondary (2◦ ) carbocation.
+
CH3
CH
CH3 + Br
[-¿] (c1) .. c
CH3
CH
CH2 + Hδ
Brδ
2. The bromide ion (Br – ) then attacks the positive carbocation to form the final
Br
+
product. CH3
CH
CH3 + Br
CH3
CH
CH3
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Anti-Markovnikov Addition
There is one famous exception! In the presence of peroxides (ROOR), the addition of HBr happens via a different (free-radical) mechanism and follows the AntiMarkovnikov Rule. The bromine adds to the carbon with more hydrogens.
Peroxide, ROOR
CH3 CH−CH2 + HBr −−−−−−−−−→
CH3 CH2 CH2 Br
Anti-Markovnikov Product
Note: This exception only works for HBr, not for HCl or HI.
Question 3: Predict the major product of the reaction between 1-butene (CH3 CH2 CH – CH2 )
and HCl. Will it follow Markovnikov’s rule?
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Addition of Halogens (Halogenation)
A halogen molecule (Cl2 or Br2 ) can add across the double bond.
CH3 CH−CH2 +
Propene
Br2
CCl
4
−−−→
(red-brown)
CH3 CH(Br)CH2 Br
1,2-dibromopropane (colorless)
Test for Unsaturation: This reaction is used as a chemical test for the presence of a
C – C double bond. Aqueous bromine (Br2 (aq), ”bromine water”) is red-brown. When
you shake it with an alkene, the color disappears. Alkanes do not react with bromine
water.
Question 4: If you have two unlabeled test tubes, one with hexane (an alkane) and one
with hexene (an alkene), how could you use bromine water to identify which is which?
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Addition of Water (Hydration)
This reaction adds an H and an OH across the double bond to produce an alcohol. It
follows Markovnikov’s Rule.
• Reagents: Water and a strong acid catalyst, such as dilute sulfuric acid (H2 SO4 ).
H+
CH3 CH−CH2 + H2 O −−→
Propene
CH3 CH(OH)CH3
Propan-2-ol (Major Product)
The mechanism is the reverse of the E1 dehydration mechanism shown earlier.
Question 5: What is the name of the alcohol formed when water is added to ethene
(CH2 – CH2 ) in the presence of an acid catalyst?
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Oxidation with Potassium Manganate(VII)
This is another visual test for unsaturation, known as the Baeyer’s Test.
• Reagents: Cold, dilute, alkaline potassium manganate(VII) solution (KMnO4 ).
This solution is deep purple.
• Reaction: The alkene is oxidized to a diol (a molecule with two -OH groups), and
the purple MnO4 – ion is reduced to brown manganese(IV) oxide (MnO2 ).
Cold
3CH2 −CH2 +2MnO4 − (aq)+4H2 O(l) −−→ 3HOCH2 −CH2 OH+2MnO2 (s)+2OH− (aq)
Ethane-1,2-diol
• Observation: The purple color of the KMnO4 solution disappears and a brown
precipitate of MnO2 is formed.
Question 6: What functional group is formed when an alkene reacts with cold, dilute
KMnO4 ?
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Ozonolysis
This is a very powerful reaction used by chemists to break an alkene apart at the double
bond to figure out its original structure.
• Process: It’s a two-step reaction. (i) React with ozone (O3 ). (ii) React the
intermediate product with Zinc and water (Zn/H2 O).
• The Result: The C – C double bond is cleaved, and each carbon from the double
1. O3
CR3 R4
R1 R2 C
O + R3 R4 C
bond becomes a carbonyl group (C – O). R1 R2 C
2. Zn/H2 O
can think of it as ”cutting” the double bond with scissors and putting an oxygen
atom on each new end.
– A – CH2 group becomes formaldehyde (HCHO).
– A – CHR group becomes an aldehyde (RCHO).
– A – CR2 group becomes a ketone (R2 C – O).
Question 7: An unknown alkene undergoes ozonolysis to produce only propanone
(CH3 COCH3 ). What was the structure of the original alkene?
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Polymerization
Polymers are giant molecules made by linking many small repeating units called monomers.
Alkenes can act as monomers in addition polymerization.
• Process: Under high pressure, high temperature, and with a catalyst, the pi bond
of many alkene molecules can break and link together to form a long alkane-like
chain.
• Example: Ethene monomers polymerize to form poly(ethene), also known as polythene, a common plastic.
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