Alkenes

Alkenes are unsaturated hydrocarbons containing at least one C=C double bond. The general formula for acyclic alkenes with one double bond is $\mathrm{C}_n\mathrm{H}_{2n}$. The C=C bond consists of one $\sigma$ bond (formed by $sp^2$--$sp^2$ head-on overlap) and one $\pi$ bond (formed by sideways overlap of unhybridised $p$ orbitals). The $\pi$ bond is weaker than the $\sigma$ bond ($\approx 268\,\mathrm{kJ/mol}$ vs $\approx 347\,\mathrm{kJ/mol}$ for a C--C $\sigma$ bond) and is the primary site of reactivity.

Structure and Properties

Each $sp^2$ hybridised carbon has trigonal planar geometry with bond angles of approximately $120^\circ$. The C=C bond length ($134\,\mathrm{pm}$) is shorter than the C--C single bond ($154\,\mathrm{pm}$) because the $\pi$ bond draws the atoms closer together.

The $\pi$ bond restricts rotation about the C=C axis, which has two critical consequences: E/Z isomerism (discussed in the introduction) and the existence of a substantial barrier to rotation ($\approx 270\,\mathrm{kJ/mol}$ for ethene).

Physical Properties

Alkenes are non-polar, insoluble in water, and have boiling points slightly lower than the corresponding alkanes (the $\pi$ electrons do not significantly enhance London forces). Boiling points increase with chain length and decrease with branching, as with alkanes.

Electrophilic Addition

The $\pi$ bond is a region of high electron density above and below the plane of the molecule. It acts as a nucleophile, attacking electrophilic species. This is the defining reaction class of alkenes: electrophilic addition.

General Mechanism

1. The $\pi$ electrons attack an electrophile, breaking the $\pi$ bond and forming a new $\sigma$ bond between one alkene carbon and the electrophile. The other carbon becomes a carbocation.
2. A nucleophile attacks the carbocation, forming the final product.

The rate-determining step is step 1 (formation of the carbocation). The stability of the carbocation intermediate determines the regiochemistry of the reaction.

Addition of Hydrogen Halides (HX)

Reaction with HBr:

$$\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 + \mathrm{HBr} \to \mathrm{CH}_3\mathrm{CHBrCH}_3$$

Mechanism:

Step 1 (rate-determining): The $\pi$ bond attacks the $\delta^+$ hydrogen of HBr. The H--Br bond breaks heterolytically, producing a carbocation and $\mathrm{Br}^-$.

$$\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 + \mathrm{H}\mathrm{-}\mathrm{Br} \to \mathrm{CH}_3\overset◆LB◆+◆RB◆◆LB◆\mathrm{C}◆RB◆\mathrm{HCH}_3 + \mathrm{Br}^-$$

The secondary carbocation ($\mathrm{CH}_3\overset◆LB◆+◆RB◆◆LB◆\mathrm{C}◆RB◆\mathrm{HCH}_3$) is formed in preference to the primary carbocation ($\mathrm{CH}_3\mathrm{CH}_2\overset◆LB◆+◆RB◆◆LB◆\mathrm{C}◆RB◆\mathrm{H}_2$) because it is more stable. The methyl group donates electron density through the inductive effect and hyperconjugation (the adjacent C--H $\sigma$ bonds overlap with the empty $p$ orbital of the carbocation).

Step 2 (fast): $\mathrm{Br}^-$ attacks the carbocation from either face:

$$\mathrm{CH}_3\overset◆LB◆+◆RB◆◆LB◆\mathrm{C}◆RB◆\mathrm{HCH}_3 + \mathrm{Br}^- \to \mathrm{CH}_3\mathrm{CHBrCH}_3$$

Markovnikov's Rule

When HX adds to an unsymmetrical alkene, the hydrogen atom attaches to the carbon of the double bond that already has the greater number of hydrogen atoms. Equivalently: the electrophile adds to the less substituted carbon, and the nucleophile adds to the more substituted carbon.

Markovnikov's rule is a consequence of carbocation stability. The more substituted carbocation (tertiary $>$ secondary $>$ primary $>$ methyl) is lower in energy and therefore forms preferentially. The Hammond postulate tells us that the transition state leading to a more stable intermediate is itself lower in energy.

Carbocation stability order: $(\mathrm{CH}_3)_3\mathrm{C}^+ \gt (\mathrm{CH}_3)_2\mathrm{CH}^+ \gt \mathrm{CH}_3\mathrm{CH}_2^+ \gt \mathrm{CH}_3^+$

Worked Example. Predict the major product when HCl adds to 2-methylpropene.

$$(\mathrm{CH}_3)_2\mathrm{C}=\mathrm{CH}_2 + \mathrm{HCl} \to (\mathrm{CH}_3)_3\mathrm{CCl}$$

The hydrogen adds to the less substituted carbon (the $=\mathrm{CH}_2$ carbon, which already has more hydrogens), producing a tertiary carbocation intermediate that is then attacked by $\mathrm{Cl}^-$. The product is 2-chloro-2-methylpropane (tert-butyl chloride).

Addition of Sulphuric Acid

Alkenes react with concentrated sulphuric acid at room temperature to form alkyl hydrogensulphates:

$$\mathrm{CH}_2=\mathrm{CH}_2 + \mathrm{H}_2\mathrm{SO}_4 \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{OSO}_3\mathrm{H}$$

This follows Markovnikov's rule. The product can be hydrolysed to an alcohol by heating with water:

$$\mathrm{CH}_3\mathrm{CH}_2\mathrm{OSO}_3\mathrm{H} + \mathrm{H}_2\mathrm{O} \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} + \mathrm{H}_2\mathrm{SO}_4$$

This two-step process was historically important for industrial alcohol production (the "indirect hydration" of alkenes) but has been superseded by direct catalytic hydration over a phosphoric acid catalyst at $300^\circ\mathrm{C}$ and high pressure.

Addition of Halogens

Alkenes decolourise bromine water (orange to colourless) and chlorine water. This is used as the standard test for unsaturation.

Addition of Br$_2$:

$$\mathrm{CH}_2=\mathrm{CH}_2 + \mathrm{Br}_2 \to \mathrm{CH}_2\mathrm{BrCH}_2\mathrm{Br}$$

Mechanism: The $\pi$ electrons attack one bromine atom, causing heterolytic fission of the Br--Br bond. A bromonium ion intermediate (three-membered ring) forms rather than a simple carbocation. The second $\mathrm{Br}^-$ then attacks the bromonium ion from the opposite side, giving anti addition (stereospecific trans addition).

The bromonium ion mechanism explains the observed stereospecificity. For cyclic alkenes, this means the two bromine atoms end up on opposite faces of the ring (anti addition), which would not be the case with a free carbocation intermediate.

Addition Polymerisation

Alkenes with a C=C double bond can undergo addition polymerisation, in which the double bond opens and monomer units link together to form a long-chain polymer.

General Equation

$$n\,\mathrm{CH}_2=\mathrm{CHR} \to \mathrm{--}(\mathrm{CH}_2\mathrm{CHR})_n\mathrm{--}$$

The repeating unit is enclosed in parentheses with the subscript $n$. The double bond is consumed in the polymerisation.

Key Polymers

Monomer	Polymer	Uses
Ethene ($\mathrm{CH}_2=\mathrm{CH}_2$)	Poly(ethene) (PE)	Plastic bags, bottles
Propene ($\mathrm{CH}_2=\mathrm{CHCH}_3$)	Poly(propene) (PP)	Ropes, containers
Chloroethene ($\mathrm{CH}_2=\mathrm{CHCl}$)	Poly(chloroethene) (PVC)	Pipes, window frames
Tetrafluoroethene ($\mathrm{CF}_2=\mathrm{CF}_2$)	PTFE	Non-stick coatings
Phenylethene (styrene)	Poly(phenylethene) (polystyrene)	Packaging, insulation

Properties and Structure

Low-density poly(ethene) (LDPE): Produced at high pressure ($1000$--$3000\,\mathrm{atm}$) with an oxygen initiator. The free radical mechanism produces a branched polymer with lower crystallinity, lower density ($0.91$--$0.94\,\mathrm{g/cm}^3$), lower melting point, and greater flexibility.

High-density poly(ethene) (HDPE): Produced at lower pressure using a Ziegler-Natta catalyst (titanium(IV) chloride with triethylaluminium). The ionic mechanism produces a largely unbranched polymer with higher crystallinity, higher density ($0.94$--$0.97\,\mathrm{g/cm}^3$), higher melting point, and greater strength.

Disposal of Polymers

The persistence of synthetic polymers in the environment is a major problem. Disposal methods:

• Landfill: Slow decomposition (decades to centuries). Leaching of additives into groundwater.
• Incineration: Recovers energy but produces $\mathrm{CO}_2$ and, for halogenated polymers (PVC), toxic hydrogen halides and dioxins.
• Recycling: Mechanical recycling (melting and reforming) degrades polymer quality over cycles. Chemical recycling (depolymerisation back to monomers) is more promising but energy-intensive.
• Biodegradable polymers: Polylactic acid (PLA, from corn starch) and polyhydroxyalkanoates (PHAs, produced by bacteria) degrade under appropriate conditions. Their use is growing but limited by cost and performance relative to conventional polymers.

Test for Unsaturation

Bromine water is the standard test for C=C double bonds:

• Add bromine water (orange/brown) to the unknown compound.
• Shake or stir.
• If the solution decolourises (turns colourless), a C=C double bond is present.

The reaction is the addition of $\mathrm{Br}_2$ across the double bond, as described above. Alkanes do not react with bromine water at room temperature (they require UV light for radical substitution).

Stereochemistry of Alkenes

E/Z Isomerism

Restricted rotation about the C=C bond means that substituents on either side of the double bond have fixed positions. When each carbon of the double bond carries two different substituents, two distinct stereoisomers exist: the $E$ (entgegen, opposite) and $Z$ (zusammen, together) isomers.

The Cahn-Ingold-Prelog (CIP) priority rules assign priority to substituents on each carbon:

1. Higher atomic number = higher priority.
2. If the same atom, compare the next atoms in the chain.
3. Double bonds are treated as if each doubly-bonded atom is duplicated.

$Z$ isomer: The higher-priority substituents on each carbon are on the same side of the double bond.

$E$ isomer: The higher-priority substituents on each carbon are on opposite sides.

Worked Example. Assign $E$ or $Z$ to the following structure: $\mathrm{CH}_3\mathrm{CH}=\mathrm{C}(\mathrm{CH}_3)\mathrm{COOH}$.

Left carbon of the double bond: substituents are $\mathrm{CH}_3$ and $\mathrm{H}$. Priority: $\mathrm{CH}_3 \gt \mathrm{H}$ (atomic number of $\mathrm{C} \gt \mathrm{H}$).

Right carbon: substituents are $\mathrm{CH}_3$ and $\mathrm{COOH}$. For $\mathrm{COOH}$: the atoms directly bonded are $\mathrm{C}$, $\mathrm{O}$, $\mathrm{O}$; for $\mathrm{CH}_3$: the atoms directly bonded are $\mathrm{H}$, $\mathrm{H}$, $\mathrm{H}$. Since $\mathrm{O} \gt \mathrm{H}$, $\mathrm{COOH}$ has higher priority than $\mathrm{CH}_3$.

If the higher-priority groups ($\mathrm{CH}_3$ on the left, $\mathrm{COOH}$ on the right) are on the same side: $Z$. If on opposite sides: $E$.

Physical Differences Between E and Z Isomers

$E$ and $Z$ isomers have different physical properties:

• Different boiling points (differences in dipole moment and packing).
• Different melting points.
• Different $R_f$ values in chromatography.
• Different NMR spectra (different coupling constants between vinyl protons).

The coupling constant $J$ between the two vinyl protons is typically larger for trans ($E$) isomers ($J \approx 12$--$18\,\mathrm{Hz}$) than for cis ($Z$) isomers ($J \approx 6$--$12\,\mathrm{Hz}$).

Hydrogenation of Alkenes

Alkenes react with hydrogen gas in the presence of a metal catalyst (Ni, Pd, or Pt) to form alkanes:

$$\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 + \mathrm{H}_2 \xrightarrow{\mathrm{Ni}} \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_3$$

Conditions: Room temperature and pressure with a nickel catalyst (or $\mathrm{H}_2$ with Pd/C at room temperature).

The reaction is stereospecific: hydrogen adds from the same face of the double bond (syn addition), producing the alkane. For cyclic alkenes, both hydrogens end up on the same face.

The enthalpy of hydrogenation provides a quantitative measure of alkene stability:

Alkene	$\Delta H_\mathrm{hydrogenation}$ ($\mathrm{kJ/mol}$)
Ethene	$-137$
(Z)-but-2-ene	$-120$
(E)-but-2-ene	$-116$
2-methylpropene (tetrasubstituted)	$-119$

More substituted alkenes are more stable and have less negative (less exothermic) hydrogenation enthalpies. (E)-but-2-ene is more stable than (Z)-but-2-ene because the two methyl groups are on opposite sides, minimising steric repulsion.

Addition of Water (Hydration)

Direct Hydration

Alkenes react with steam in the presence of a phosphoric acid catalyst ($\mathrm{H}_3\mathrm{PO}_4$) at $300^\circ\mathrm{C}$ and $60\,\mathrm{atm}$ to form alcohols:

$$\mathrm{CH}_2=\mathrm{CH}_2 + \mathrm{H}_2\mathrm{O} \xrightarrow{\mathrm{H}_3\mathrm{PO}_4,\,300^\circ\mathrm{C}} \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}$$

This follows Markovnikov's rule. The reaction is reversible, so yield is limited by the equilibrium position. Industrial conditions use high pressure to shift equilibrium towards the product (fewer moles of gas on the product side).

Hydroboration-Oxidation (Anti-Markovnikov Hydration)

Hydroboration-oxidation gives the anti-Markovnikov alcohol:

$$\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 \xrightarrow{\mathrm{BH}_3\cdot\mathrm{THF}} \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH}$$

The borane adds in a syn fashion, and the oxidation step replaces boron with $-\mathrm{OH}$. The overall result is that the $-\mathrm{OH}$ ends up on the less substituted carbon (anti-Markovnikov), the opposite of direct hydration or acid-catalysed hydration.

Oxidation of Alkenes

Reaction with Acidified Potassium Manganate(VII)

Cold, dilute, acidified $\mathrm{KMnO}_4$ oxidises the C=C bond to form a vicinal diol (glycol):

$$\mathrm{CH}_2=\mathrm{CH}_2 + [\mathrm{O}] + \mathrm{H}_2\mathrm{O} \to \mathrm{CH}_2\mathrm{OHCH}_2\mathrm{OH}$$

The purple $\mathrm{MnO}_4^-$ is reduced to colourless $\mathrm{Mn}^{2+}$, providing a second test for unsaturation alongside bromine water.

Oxidative Cleavage with Hot Concentrated $\mathrm{KMnO}_4$

Hot concentrated $\mathrm{KMnO}_4$ cleaves the double bond, converting each carbon to its highest oxidation state:

Alkene type	Product(s)
$\mathrm{RCH}=\mathrm{CH}_2$	$\mathrm{RCOOH} + \mathrm{CO}_2$
$\mathrm{RCH}=\mathrm{CHR}'$	$\mathrm{RCOOH} + \mathrm{R}'\mathrm{COOH}$
$\mathrm{R}_2\mathrm{C}=\mathrm{CR}'_2$	$\mathrm{R}_2\mathrm{C}=\mathrm{O} + \mathrm{R}'_2\mathrm{C}=\mathrm{O}$

This reaction can be used to determine the position of the double bond in an unknown alkene by identifying the carbonyl products.

Ozonolysis

Ozone ($\mathrm{O}_3$) cleaves the C=C bond at low temperature ($-78^\circ\mathrm{C}$) to form ozonides, which are reduced by dimethyl sulphide or zinc to give carbonyl compounds:

$$\mathrm{RCH}=\mathrm{CHR}' \xrightarrow{\mathrm{O}_3} \xrightarrow{\mathrm{Zn}} \mathrm{RCHO} + \mathrm{R}'\mathrm{CHO}$$

Ozonolysis is more useful than permanganate cleavage because it gives aldehydes (not further oxidised to carboxylic acids), providing more information about the original alkene structure.

Industrial Importance of Alkenes

Ethene

Ethene is the most important organic chemical in industry:

• Poly(ethene) -- the world's most widely used plastic.
• Ethanol -- direct hydration with steam.
• Ethanol (indirect) -- absorption in $\mathrm{H}_2\mathrm{SO}_4$ followed by hydrolysis.
• Ethane-1,2-diol (antifreeze) -- oxidation of ethene with oxygen and a silver catalyst.
• Halogenoethanes -- addition of HCl, HBr, etc.

Propene

• Poly(propene) -- high-strength plastic for ropes and containers.
• Propan-2-ol -- hydration of propene.
• Propanal / propanone -- hydration followed by oxidation.

Common Pitfalls

1. Violating Markovnikov's rule. The hydrogen of HX always adds to the carbon with more hydrogens. If you find yourself attaching H to the more substituted carbon, check again. This rule is a direct consequence of carbocation stability.

2. Confusing the bromine water test with radical bromination. Bromine water reacts with alkenes via electrophilic addition at room temperature (no UV needed). Bromine reacts with alkanes only under UV light via radical substitution. These are fundamentally different mechanisms.

3. Drawing the polymer repeating unit incorrectly. The repeating unit must show the opened double bond. The monomer $\mathrm{CH}_2=\mathrm{CHCl}$ becomes the repeating unit $-\mathrm{CH}_2\mathrm{CHCl}-$, not $-\mathrm{CH}_2=\mathrm{CHCl}-$.

4. Assuming all polymerisations produce the same type of poly(ethene). LDPE (radical, branched) and HDPE (ionic, linear) have very different properties. The mechanism determines the polymer structure.

5. Forgetting the stereochemistry of halogen addition. Bromine adds anti (trans) across the double bond via a bromonium ion intermediate, not syn (cis). For cyclic alkenes, this has consequences for the stereochemistry of the product.

Practice Problems

Problem 1

Predict the major organic product(s) when 2-methylbut-2-ene reacts with HBr. Show the mechanism and explain the regiochemistry.

Solution:

$$(\mathrm{CH}_3)_2\mathrm{C}=\mathrm{CHCH}_3 + \mathrm{HBr} \to (\mathrm{CH}_3)_2\mathrm{CBrCH}_2\mathrm{CH}_3$$

The $\pi$ bond attacks the $\delta^+$ hydrogen of HBr. Following Markovnikov's rule, the hydrogen adds to the less substituted carbon of the double bond. There are two carbons in the double bond:

• $\mathrm{C}_1$: $(\mathrm{CH}_3)_2\mathrm{C}=$ -- bonded to two methyl groups (0 hydrogens on this carbon directly from the double bond? No, each of the two methyls is a substituent). The carbon has zero hydrogens directly attached.
• $\mathrm{C}_2$: $=\mathrm{CHCH}_3$ -- bonded to one hydrogen and one methyl group.

The hydrogen adds to $\mathrm{C}_2$ (which has more hydrogens), producing a tertiary carbocation at $\mathrm{C}_1$: $(\mathrm{CH}_3)_2\overset◆LB◆+◆RB◆◆LB◆\mathrm{C}◆RB◆\mathrm{CH}_2\mathrm{CH}_3$. The bromide ion then attacks this tertiary carbocation to give 2-bromo-2-methylbutane.

Problem 2

A sample of poly(propene) has an average chain length of $n = 5000$. Calculate the molecular mass of one average chain. The molar mass of the propene monomer ($\mathrm{C}_3\mathrm{H}_6$) is $42.08\,\mathrm{g/mol}$.

Solution:

$$M_{\mathrm{polymer}} = n \times M_{\mathrm{monomer}} = 5000 \times 42.08 = 210,400\,\mathrm{g/mol}$$

Each repeating unit has the same molar mass as the monomer (the double bond opens but no atoms are lost or gained in addition polymerisation).

Problem 3

Predict the products of the following reactions and state the mechanism:

(a) But-2-ene with $\mathrm{Br}_2$ in $\mathrm{CCl}_4$ at room temperature.

(b) 2-methylpropene with $\mathrm{HBr}$ in the presence of peroxides.

Solution:

(a) Anti addition of bromine via the bromonium ion mechanism. But-2-ene exists as two stereoisomers (E and Z).

For (E)-but-2-ene: the product is the racemic mixture $(2R,3R)$-2,3-dibromobutane and $(2S,3S)$-2,3-dibromobutane (the meso compound $(2R,3S)$ is NOT formed from the E isomer because the bromonium ion intermediate is opened by anti attack).

For (Z)-but-2-ene: the product is meso-$(2R,3S)$-2,3-dibromobutane (the two bromine atoms add to the same face and then anti opening gives the meso form).

(b) In the presence of peroxides, the reaction follows the anti-Markovnikov pathway (Kharasch effect). The peroxide generates a bromine radical, which adds to the less substituted carbon of the double bond:

$$(\mathrm{CH}_3)_2\mathrm{C}=\mathrm{CH}_2 + \mathrm{Br}\cdot \to (\mathrm{CH}_3)_2\mathrm{CBrCH}_2\cdot$$$$(\mathrm{CH}_3)_2\mathrm{CBrCH}_2\cdot + \mathrm{HBr} \to (\mathrm{CH}_3)_2\mathrm{CBrCH}_3 + \mathrm{Br}\cdot$$

Product: 1-bromo-2-methylpropane (anti-Markovnikov). Without peroxides, the product would be 2-bromo-2-methylpropane (Markovnikov).

Problem 4

Explain how ozonolysis can be used to determine the position of the double bond in an unknown alkene. An unknown alkene $\mathrm{C}_6\mathrm{H}_{12}$ produces ethanal and butanal on ozonolysis followed by reduction. Identify the alkene.

Solution:

Ozonolysis cleaves the C=C bond and replaces each carbon of the double bond with a carbonyl group. The products tell us what was on each side of the double bond.

The products are ethanal ($\mathrm{CH}_3\mathrm{CHO}$) and butanal ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CHO}$). The original double bond was between the carbon that became ethanal and the carbon that became butanal:

$$\mathrm{CH}_3\mathrm{CH}=\mathrm{CHCH}_2\mathrm{CH}_2\mathrm{CH}_3$$

The alkene is hex-2-ene (which exists as E and Z isomers).

Check: $\mathrm{C}_6\mathrm{H}_{12}$ has one degree of unsaturation (one C=C bond). Hex-2-ene: $\mathrm{C}_6\mathrm{H}_{12}$. Correct.

Worked Examples: Electrophilic Addition in Detail

Example 1: Addition of HBr to 3-methylpent-2-ene

Reaction: $(\mathrm{CH}_3)_2\mathrm{C}=\mathrm{CHCH}_2\mathrm{CH}_3 + \mathrm{HBr}$

Step 1 (rate-determining): The $\pi$ electrons attack $\delta^+\mathrm{H}$ of HBr. Markovnikov addition: H adds to the less substituted carbon.

The less substituted carbon is $\mathrm{CH}$ (bonded to one methyl and one $\mathrm{CH}_2\mathrm{CH}_3$ group). The more substituted carbon is $\mathrm{C}(\mathrm{CH}_3)_2$ (bonded to two methyl groups). H adds to the $\mathrm{CH}$ carbon.

Carbocation formed: $(\mathrm{CH}_3)_2\overset◆LB◆+◆RB◆◆LB◆\mathrm{C}◆RB◆\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_3$ (tertiary carbocation).

Step 2 (fast): $\mathrm{Br}^-$ attacks the tertiary carbocation.

Product: 3-bromo-3-methylpentane: $(\mathrm{CH}_3)_2\mathrm{CBrCH}_2\mathrm{CH}_2\mathrm{CH}_3$

Example 2: Stereochemistry of Bromine Addition to Cyclohexene

Reaction: $\mathrm{C}_6\mathrm{H}_{10}$ (cyclohexene) $+ \mathrm{Br}_2 \to$ trans-1,2-dibromocyclohexane

Mechanism:

Step 1: The $\pi$ electrons of cyclohexene attack one bromine atom of $\mathrm{Br}_2$, forming a bromonium ion intermediate. The bromonium ion is a three-membered ring with a positive charge distributed over the bromine and both carbons.

Step 2: $\mathrm{Br}^-$ attacks the bromonium ion from the opposite side (anti addition) at either carbon of the three-membered ring.

Product: The two bromine atoms end up on opposite faces of the ring (trans configuration). The product is a racemic mixture of $(1R,2R)$-1,2-dibromocyclohexane and $(1S,2S)$-1,2-dibromocyclohexane.

Example 3: Hydration of Propene to Propan-2-ol

Direct hydration:

$\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 + \mathrm{H}_2\mathrm{O} \xrightarrow{\mathrm{H}_3\mathrm{PO}_4,\,300^\circ\mathrm{C},\,60\,\mathrm{atm}} \mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{CH}_3$

Markovnikov addition: H adds to the less substituted carbon (terminal carbon with more H), giving the secondary alcohol propan-2-ol.

Acid-catalysed mechanism:

1. Protonation of the alkene by $\mathrm{H}_3\mathrm{O}^+$ to form the more stable secondary carbocation.
2. Nucleophilic attack by water on the carbocation.
3. Deprotonation to yield propan-2-ol.

Example 4: Oxidative Cleavage of an Unknown Alkene

An unknown alkene $\mathrm{C}_7\mathrm{H}_{12}$ is treated with hot, concentrated $\mathrm{KMnO}_4$. The products are pentanoic acid and ethanoic acid. Identify the alkene.

Analysis: Oxidative cleavage of the C=C bond converts each carbon to its highest oxidation state. The products tell us what was on each side of the double bond:

• Pentanoic acid ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{COOH}$) comes from the fragment $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}=$.
• Ethanoic acid ($\mathrm{CH}_3\mathrm{COOH}$) comes from the fragment $=\mathrm{CHCH}_3$.

Reconstructing: $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}=\mathrm{CHCH}_3$

Identity: Hept-2-ene ($\mathrm{C}_7\mathrm{H}_{14}$... wait, check). $\mathrm{C}_7\mathrm{H}_{12}$ has $\text{DoU} = \frac{2(7) + 2 - 12}{2} = 1$. Hept-2-ene is $\mathrm{C}_7\mathrm{H}_{14}$ with $\text{DoU} = 1$.

The formula must be $\mathrm{C}_7\mathrm{H}_{14}$, not $\mathrm{C}_7\mathrm{H}_{12}$. If the problem states $\mathrm{C}_7\mathrm{H}_{12}$, the molecule has an additional degree of unsaturation (a ring or a second double bond). In this case, the product analysis is still valid, and the structure would be hept-2-ene (the given formula may contain a typo; in an exam, use the product information to deduce the structure).

Example 5: E/Z Assignment with CIP Rules

Assign $E$ or $Z$ to 1-bromo-2-chloropropene: $\mathrm{BrCH}=\mathrm{C}(\mathrm{Cl})\mathrm{CH}_3$.

Left carbon of the double bond: substituents are $\mathrm{Br}$ and $\mathrm{H}$.

Priority: $\mathrm{Br}$ (atomic number 35) $>$ $\mathrm{H}$ (atomic number 1). Higher priority: $\mathrm{Br}$.

Right carbon: substituents are $\mathrm{Cl}$ and $\mathrm{CH}_3$.

Priority: $\mathrm{Cl}$ (atomic number 17) $>$ $\mathrm{C}$ (atomic number 6). Higher priority: $\mathrm{Cl}$.

If $\mathrm{Br}$ and $\mathrm{Cl}$ are on the same side: $Z$. If on opposite sides: $E$.

Example 6: Anti-Markovnikov Addition via Peroxide Effect

Reaction: Propene + HBr (with organic peroxides) $\to$ 1-bromopropane

Mechanism (free radical):

Initiation: Peroxide bond homolysis:

$\mathrm{ROOR} \xrightarrow{\Delta} 2\mathrm{RO}\cdot$

Propagation 1: The radical adds to the less substituted carbon (to form the more stable secondary radical):

$\mathrm{RO}\cdot + \mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 \to \mathrm{ROCH}_2\mathrm{CH}\cdot\mathrm{CH}_3$

Wait -- the $\mathrm{RO}\cdot$ radical is not the chain carrier for HBr addition. The correct mechanism involves the bromine radical:

$\mathrm{RO}\cdot + \mathrm{HBr} \to \mathrm{ROH} + \mathrm{Br}\cdot$

$\mathrm{Br}\cdot + \mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 \to \mathrm{CH}_3\mathrm{CH}\cdot\mathrm{CH}_2\mathrm{Br}$

The bromine radical adds to the less substituted carbon to form the more stable secondary radical (on C2).

Propagation 2:

$\mathrm{CH}_3\mathrm{CH}\cdot\mathrm{CH}_2\mathrm{Br} + \mathrm{HBr} \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{Br} + \mathrm{Br}\cdot$

Product: 1-bromopropane (anti-Markovnikov).

Important: The peroxide effect only works with HBr, not HCl or HI. HCl has a bond that is too strong for radical homolysis, and HI reacts so readily via ionic addition that the radical pathway cannot compete.

Example 7: Polymer Identification from Monomer

Monomer: $\mathrm{CH}_2=\mathrm{CHCl}$ (chloroethene/vinyl chloride)

Polymer: Poly(chloroethene) / PVC

$n\,\mathrm{CH}_2=\mathrm{CHCl} \to \mathrm{--}(\mathrm{CH}_2\mathrm{CHCl})_n\mathrm{--}$

Properties: Rigid, flame-retardant (due to chlorine content), used for pipes and window frames. Addition of plasticisers makes flexible PVC for cable insulation and flooring.

Environmental concern: PVC disposal by incineration produces $\mathrm{HCl}$ gas and potentially toxic dioxins. Recycling is preferred.

Practical Techniques for Alkene Reactions

Required Practical: Testing for Unsaturaton (AQA RP 6)

Objective: To use bromine water to distinguish between an alkane and an alkene.

Procedure:

1. Add $2\,\mathrm{cm}^3$ of cyclohexane to one test tube and $2\,\mathrm{cm}^3$ of cyclohexene to another.
2. Add bromine water (orange-brown) dropwise to each tube, shaking after each addition.
3. Record the number of drops required for the bromine colour to persist.

Expected results:

• Cyclohexane: bromine colour persists (no reaction at room temperature).
• Cyclohexene: bromine colour is immediately decolourised (electrophilic addition).

Safety: Bromine water is toxic and corrosive. Wear gloves and eye protection. Work in a fume cupboard.

Preparation and Purification of an Organic Liquid (Distillation)

After an addition reaction, the product may need to be purified by distillation:

1. Set up a simple distillation apparatus with a round-bottom flask, fractionating column (or Liebig condenser for simple distillation), thermometer, condenser, and receiving flask.
2. Add anti-bumping granules to the flask.
3. Heat gently and collect the fraction that distils at the expected boiling point of the product.
4. For alkenes produced by elimination, the product can be passed through a drying tube (containing $\mathrm{CaCl}_2$) to remove water vapour before collection.

Exam-Style Questions with Full Mark Schemes

Q1 (5 marks)

Explain, with reference to the mechanism, why the reaction of propene with HBr produces 2-bromopropane as the major product.

Mark Scheme:

5 marks:

• The mechanism is electrophilic addition (1 mark).
• The $\pi$ electrons of the C=C bond attack the $\delta^+$ hydrogen of HBr (1 mark).
• The H adds to the less substituted carbon (terminal carbon), forming a secondary carbocation (1 mark).
• The secondary carbocation is more stable than the alternative primary carbocation due to inductive electron donation from the methyl group and hyperconjugation (1 mark).
• $\mathrm{Br}^-$ attacks the secondary carbocation to give 2-bromopropane (1 mark).

Q2 (6 marks)

Describe the mechanism for the reaction between cyclohexene and bromine. Explain why the product has a trans configuration.

Mark Scheme:

6 marks:

• The $\pi$ electrons of cyclohexene attack one bromine atom of $\mathrm{Br}_2$ (1 mark).
• The Br--Br bond breaks heterolytically, forming a bromonium ion intermediate (three-membered ring with Br) (1 mark).
• The bromonium ion prevents free rotation, and the second $\mathrm{Br}^-$ attacks from the opposite side of the ring (anti addition) (1 mark).
• This results in the two bromine atoms being on opposite faces (trans) of the cyclohexane ring (1 mark).
• The product is a racemic mixture of $(1R,2R)$- and $(1S,2S)$-1,2-dibromocyclohexane (1 mark).
• Diagram showing the bromonium ion intermediate and curly arrows (1 mark).

Q3 (4 marks)

State two chemical tests that could be used to distinguish between cyclohexane and cyclohexene. For each test, state the reagent, the observation with cyclohexane, and the observation with cyclohexene.

Mark Scheme:

4 marks (2 marks per test, 1 mark for reagent and conditions, 1 mark for observations):

Test 1: Bromine water. Cyclohexane: no reaction, orange colour persists. Cyclohexene: orange colour decolourised (2 marks).

Test 2: Acidified $\mathrm{KMnO}_4$. Cyclohexane: no reaction, purple colour persists. Cyclohexene: purple colour decolourised, colourless $\mathrm{Mn}^{2+}$ formed (2 marks).

Q4 (5 marks)

An alkene $\mathrm{C}_5\mathrm{H}_{10}$ is treated with hot, concentrated $\mathrm{KMnO}_4$. The products are propanoic acid and ethanoic acid. Deduce the structure of the alkene.

Mark Scheme:

5 marks:

• Oxidative cleavage converts each carbon of the double bond to a carboxylic acid (1 mark).
• Propanoic acid ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{COOH}$) indicates the fragment $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}=$ on one side (1 mark).
• Ethanoic acid ($\mathrm{CH}_3\mathrm{COOH}$) indicates the fragment $=\mathrm{CHCH}_3$ on the other side (1 mark).
• Combining: $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}=\mathrm{CHCH}_3$ (pent-2-ene) (1 mark).
• Verification: $\mathrm{C}_5\mathrm{H}_{10}$, one degree of unsaturation, consistent with one C=C bond (1 mark).

Q5 (4 marks)

Explain why the addition polymerisation of ethene requires an initiator, whereas the addition of bromine to ethene does not.

Mark Scheme:

4 marks:

• Addition polymerisation involves free radical initiation (1 mark). The C=C bond in ethene has a high bond enthalpy and will not spontaneously open; an initiator (e.g. organic peroxide or oxygen under high pressure) generates free radicals that attack the C=C bond (1 mark).
• Bromine addition is an electrophilic addition (1 mark). The $\mathrm{Br}_2$ molecule is polarised by the $\pi$ electrons of ethene ($\mathrm{Br}^{\delta+}\text{-}\mathrm{Br}^{\delta-}$), and the electrophilic $\mathrm{Br}^{\delta+}$ is attacked directly by the nucleophilic $\pi$ bond without any initiator (1 mark).

Multi-Step Synthesis Planning with Alkenes

Alkenes serve as key starting materials in multi-step synthesis because they can be transformed into a wide range of functional groups.

Synthesis Planning Example

Target: 2-hydroxypropanoic acid (lactic acid) from propene.

Step 1: Anti-Markovnikov addition of HBr (peroxide effect):

$\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 + \mathrm{HBr} \xrightarrow{\text{peroxides}} \mathrm{BrCH}_2\mathrm{CH}_2\mathrm{CH}_3$

Step 2: Hydrolysis to propan-1-ol:

$\mathrm{BrCH}_2\mathrm{CH}_2\mathrm{CH}_3 + \mathrm{NaOH}(aq) \to \mathrm{HOCH}_2\mathrm{CH}_2\mathrm{CH}_3 + \mathrm{NaBr}$

Step 3: Oxidation to propanal (distillation conditions):

$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH} \xrightarrow{[\mathrm{O}],\,\text{distillation}} \mathrm{CH}_3\mathrm{CH}_2\mathrm{CHO}$

Step 4: Cyanohydrin formation:

$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CHO} + \mathrm{HCN} \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}(\mathrm{OH})\mathrm{CN}$

Step 5: Acid hydrolysis of the nitrile:

$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}(\mathrm{OH})\mathrm{CN} + 2\mathrm{H}_2\mathrm{O} + \mathrm{H}^+ \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}(\mathrm{OH})\mathrm{COOH} + \mathrm{NH}_4^+$

This gives 2-hydroxybutanoic acid. For lactic acid specifically, a different approach:

$\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 \xrightarrow{\mathrm{HBr}/\text{peroxides}} \mathrm{BrCH}_2\mathrm{CH}_2\mathrm{CH}_3$

Actually, lactic acid is 2-hydroxypropanoic acid ($\mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{COOH}$). A more direct route from propene:

$\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 \xrightarrow{\mathrm{H}_2\mathrm{O}/\mathrm{H}^+} \mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{CH}_3 \xrightarrow{[\mathrm{O}]} \mathrm{CH}_3\mathrm{COCH}_3$

Since direct oxidation of propan-2-ol gives propanone (ketone, not further oxidised), we need a different approach. The most efficient route is:

$\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 \xrightarrow{\mathrm{H}_2\mathrm{O}/\mathrm{H}^+} \mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{CH}_3$

Then use the cyanohydrin route on ethanal (derived from ethanol oxidation):

$\mathrm{CH}_3\mathrm{CHO} + \mathrm{HCN} \to \mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{CN} \xrightarrow{\mathrm{H}_3\mathrm{O}^+} \mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{COOH}$

This demonstrates the importance of choosing the correct disconnection strategy in retrosynthesis.

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tip

Diagnostic Test Ready to test your understanding of Alkenes? The diagnostic test contains the hardest questions within the A-Level specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Alkenes with other chemistry topics to test synthesis under exam conditions.

See Diagnostic Guide for instructions on self-marking and building a personal test matrix.