Chapter 9 · Hydrocarbons

9.2 Alkenes

Alkenes, $\mathrm{C_{n}H_{2n}}$, carry a C=C double bond: one σ bond and one exposed π bond. That bond locks the molecule flat, creates cis and trans isomers, and makes alkenes add reagents — in the direction set by Markovnikov's rule, reversed for HBr by peroxides. Preparation, addition, oxidation, ozonolysis and polymerisation.

Ethene is the most-produced organic chemical in the world — more than $200$ million tonnes a year, nearly all of it turned into polythene, PVC, antifreeze and polyester. It is also a plant hormone: a ripe banana gives off a trace of ethene that makes the fruit next to it ripen too. Both uses depend on the same feature, the carbon–carbon double bond. Its second bond, the π bond, is weaker than the first and its electron cloud sits exposed above and below the molecule, where electron-seeking reagents can reach it. Alkenes therefore add other molecules across the double bond, and when one alkene adds to another, again and again, the result is a polymer.

This section explains the structure of the double bond and the geometrical (cis–trans) isomerism that its rigidity creates; four ways of making alkenes (partial reduction of alkynes, and three eliminations); the addition of hydrogen, halogens, hydrogen halides, sulphuric acid and water, with Markovnikov's rule and its explanation through carbocations; the peroxide effect, which reverses the direction of HBr addition by switching to a free-radical mechanism; and oxidation, ozonolysis and polymerisation.

The Double Bond, Naming and Geometrical Isomerism

📘 Definition 9.6 — Alkenes and the C=C bond
An alkene is an unsaturated hydrocarbon containing a carbon–carbon double bond; with one double bond and no ring the general formula is $\mathrm{C_{n}H_{2n}}$ (ethene $\mathrm{C_{2}H_{4}}$, propene $\mathrm{C_{3}H_{6}}$, …). Alkenes are also called olefins ("oil-forming"), because ethene gives an oily liquid with chlorine. The two doubly bonded carbons are sp² hybridised: each forms three σ bonds at about $120^{\circ}$ in one plane, and the unhybridised p orbitals overlap sideways to form a π bond above and below that plane (Section 4.3). So the C=C bond is one σ + one π bond, $134\ \mathrm{pm}$ long (C–C: $154\ \mathrm{pm}$), and the two carbons together with the four atoms attached to them all lie in one plane. The π bond is weaker than the σ bond: with the mean bond enthalpies of Section 5.3, $614-348\approx266\ \mathrm{kJ\,mol^{-1}}$. Its loosely held electrons make alkenes a source of electrons — they react with electrophiles (Chapter 8).

Alkenes are named with the ending -ene: the main chain must contain the double bond, and it is numbered to give the double bond the lowest locant, e.g. $\mathrm{CH_{2}{=}CHCH_{2}CH_{3}}$ but-1-ene, $\mathrm{CH_{3}CH{=}CHCH_{3}}$ but-2-ene, $\mathrm{(CH_{3})_{2}C{=}CH_{2}}$ 2-methylpropene (Chapter 8 gives the full rules). Alkenes from $\mathrm{C_{4}H_{8}}$ onwards show structural isomerism — chain isomers (but-1-ene and 2-methylpropene) and position isomers (but-1-ene and but-2-ene). They also show a kind of isomerism that alkanes cannot.

Turning one carbon of a double bond relative to the other would twist the two p orbitals apart and break the π bond, which costs about $266\ \mathrm{kJ\,mol^{-1}}$ — twenty times the barrier in ethane. So at room temperature there is no rotation about a C=C bond, and groups fixed on the same side of it stay there.

📘 Definition 9.7 — Geometrical (cis–trans) isomerism
Geometrical isomers are stereoisomers that have the same structural formula but a different arrangement of groups about a double bond (or a ring), and that cannot be interconverted at room temperature because rotation is restricted. In the cis isomer two similar groups lie on the same side of the double bond; in the trans isomer they lie on opposite sides.
Condition: an alkene $\mathrm{abC{=}Cxy}$ shows geometrical isomerism only if each doubly bonded carbon carries two different groups ($\mathrm{a\ne b}$ and $\mathrm{x\ne y}$). If either carbon has two identical groups — as in propene $\mathrm{CH_{3}CH{=}CH_{2}}$, but-1-ene or 2-methylpropene — swapping them gives the same molecule and there are no cis–trans isomers.
📐 Rule 9.5 — Physical properties of cis and trans isomers
  • Dipole moment: in the trans isomer of $\mathrm{abC{=}Cab}$ the two bond (or group) dipoles point in opposite directions and cancel, so $\mu\approx0$; in the cis isomer they add, so the cis isomer is more polar. trans-1,2-Dichloroethene: $\mu=0$; cis: $\mu=1.90\ \mathrm{D}$.
  • Boiling point: the more polar cis isomer has dipole–dipole attractions as well as dispersion forces, so it usually boils higher: 1,2-dichloroethene, cis $60.3\ ^{\circ}\mathrm{C}$, trans $47.5\ ^{\circ}\mathrm{C}$; but-2-ene, cis $3.7\ ^{\circ}\mathrm{C}$, trans $0.9\ ^{\circ}\mathrm{C}$.
  • Melting point: the more symmetrical trans isomer packs better in the crystal and usually melts higher: but-2-ene, trans $-106\ ^{\circ}\mathrm{C}$, cis $-139\ ^{\circ}\mathrm{C}$.
  • Stability: the trans isomer is usually slightly more stable, because the bulky groups are further apart.
✏️ Example 1 — Which alkenes show geometrical isomerism?
(a) Decide which of these show cis–trans isomerism: propene, but-1-ene, but-2-ene, 2-methylbut-2-ene, pent-2-ene, 1,1-dichloroethene, 1,2-dichloroethene, $\mathrm{CH_{3}CH{=}CHCl}$. (b) How many alkenes have the formula $\mathrm{C_{4}H_{8}}$, counting geometrical isomers separately? (c) Which isomer of 1,2-dichloroethene has the higher boiling point, and why?
Method Look at each doubly bonded carbon in turn and list its two groups; both pairs must be different.
(a) Propene $\mathrm{CH_{3}CH{=}CH_{2}}$: C1 has H, H — no. But-1-ene: C1 has H, H — no. But-2-ene $\mathrm{CH_{3}CH{=}CHCH_{3}}$: (H, $\mathrm{CH_{3}}$) and (H, $\mathrm{CH_{3}}$) — yes. 2-Methylbut-2-ene $\mathrm{(CH_{3})_{2}C{=}CHCH_{3}}$: C2 has two $\mathrm{CH_{3}}$ — no. Pent-2-ene $\mathrm{CH_{3}CH{=}CHCH_{2}CH_{3}}$: (H, $\mathrm{CH_{3}}$) and (H, $\mathrm{C_{2}H_{5}}$) — yes. 1,1-Dichloroethene $\mathrm{CH_{2}{=}CCl_{2}}$ — no. 1,2-Dichloroethene $\mathrm{ClCH{=}CHCl}$ — yes. 1-Chloroprop-1-ene $\mathrm{CH_{3}CH{=}CHCl}$: (H, $\mathrm{CH_{3}}$) and (H, Cl) — yes.
(b) But-1-ene, cis-but-2-ene, trans-but-2-ene and 2-methylpropene: 4 alkenes (3 structural isomers, one of which has two geometrical forms).
(c) cis ($60.3\ ^{\circ}\mathrm{C}$ against $47.5\ ^{\circ}\mathrm{C}$): its two C–Cl dipoles add to give $\mu=1.90\ \mathrm{D}$, so dipole–dipole attractions add to the dispersion forces; in the trans isomer the dipoles cancel.
✏️ Example 2 — The alkenes of formula C₅H₁₀
Draw and name every open-chain alkene of formula $\mathrm{C_{5}H_{10}}$. How many structural isomers are there, and how many alkenes in total when geometrical isomers are counted? Which of them gives 2-methylbutane on hydrogenation?
Plan Take each $\mathrm{C_{5}}$ skeleton (pentane, 2-methylbutane, 2,2-dimethylpropane) and put the double bond in every distinct position; a carbon may not exceed four bonds.
Unbranched $\mathrm{CH_{2}{=}CHCH_{2}CH_{2}CH_{3}}$ pent-1-ene; $\mathrm{CH_{3}CH{=}CHCH_{2}CH_{3}}$ pent-2-ene (cis and trans).
Branched $\mathrm{CH_{2}{=}C(CH_{3})CH_{2}CH_{3}}$ 2-methylbut-1-ene; $\mathrm{(CH_{3})_{2}C{=}CHCH_{3}}$ 2-methylbut-2-ene; $\mathrm{(CH_{3})_{2}CHCH{=}CH_{2}}$ 3-methylbut-1-ene. The 2,2-dimethylpropane skeleton has no C–C bond between two carbons that could both take another bond (the central carbon already has four), so it gives no alkene.
Count 5 structural isomers; 6 alkenes including cis- and trans-pent-2-ene. The three branched ones give 2-methylbutane (Example 3 of Section 9.1).
Interactive 1 — Cis–Trans Builder

Choose the two groups on each carbon of the double bond, then flip between the two arrangements with the cis/trans switch (or press ▶ Play to cycle through examples). The lab tests the condition of Definition 9.7, draws the bond (or group) dipoles as arrows and their resultant, and — for compounds whose data are known — shows the measured boiling points and dipole moments of both isomers.

compound—
cis–trans isomers?—
dipole moment—
boiling point—

Preparing Alkenes

Alkenes are made industrially by cracking petroleum fractions (Section 9.1). In the laboratory they are made either by partly reducing a triple bond, or by an elimination: two atoms or groups on neighbouring carbon atoms are removed, and a new π bond forms between those carbons.

📘 Definition 9.8 — Partial reduction and β-elimination
  • Partial reduction of alkynes. With Lindlar's catalyst — palladium on charcoal (or calcium carbonate) partly deactivated ("poisoned") with sulphur compounds or quinoline — hydrogen adds only once, and both H atoms add to the same side, giving the cis alkene. Sodium in liquid ammonia gives the trans alkene. $$\mathrm{HC{\equiv}CH+H_{2}\xrightarrow{\text{Lindlar}}CH_{2}{=}CH_{2}}$$ $$\begin{aligned}&\mathrm{CH_{3}C{\equiv}CCH_{3}+H_{2}}\\&\quad\xrightarrow{\text{Lindlar}}\textit{cis}\text{-}\mathrm{CH_{3}CH{=}CHCH_{3}}\end{aligned}$$
  • Dehydrohalogenation of alkyl halides: heating with alcoholic potassium hydroxide removes HX — the halogen from the α-carbon and an H atom from the neighbouring β-carbon (a β-elimination): $$\begin{aligned}&\mathrm{CH_{3}CH_{2}Br+KOH(alc)}\\&\quad\xrightarrow{\Delta}\mathrm{CH_{2}{=}CH_{2}+KBr+H_{2}O}\end{aligned}$$ For a given alkyl group the ease of reaction is iodide $\gt$ bromide $\gt$ chloride, and for a given halogen tertiary $\gt$ secondary $\gt$ primary.
  • Dehalogenation of vicinal dihalides (two halogens on adjacent carbons) with zinc dust: $$\begin{aligned}&\mathrm{BrCH_{2}CH_{2}Br+Zn}\\&\quad\mathrm{\rightarrow CH_{2}{=}CH_{2}+ZnBr_{2}}\end{aligned}$$
  • Acidic dehydration of alcohols: heating with concentrated sulphuric acid removes $\mathrm{H_{2}O}$ (OH from the α-carbon, H from the β-carbon): $$\begin{aligned}&\mathrm{CH_{3}CH_{2}OH}\\&\quad\xrightarrow[443\,\text{K}]{\text{conc. }\mathrm{H_{2}SO_{4}}}\mathrm{CH_{2}{=}CH_{2}+H_{2}O}\end{aligned}$$
⚠️ Common mistakes
Alcoholic KOH eliminates HX to give an alkene; aqueous KOH substitutes OH for X to give an alcohol (Class 12). Zinc with a vicinal dihalide gives an alkene; zinc and dilute HCl with a monohalide give an alkane (Section 9.1). A molecule with no H on the β-carbon, such as $\mathrm{CH_{3}Br}$ or $\mathrm{(CH_{3})_{3}CCH_{2}Br}$, cannot undergo β-elimination.
✏️ Example 3 — Making alkenes
Give the organic product(s) of: (a) 2-bromopropane heated with alcoholic KOH; (b) but-2-yne with $\mathrm{H_{2}}$ over Lindlar's catalyst, and with Na in liquid $\mathrm{NH_{3}}$; (c) 1,2-dibromopropane with zinc dust; (d) propan-2-ol heated with concentrated $\mathrm{H_{2}SO_{4}}$; (e) 2-bromobutane heated with alcoholic KOH. (f) Calculate the mass of ethene obtainable from $23.0\ \mathrm{g}$ of ethanol if the yield is $80.0\%$. ($M$: ethanol $46.07$, ethene $28.05\ \mathrm{g\,mol^{-1}}$)
(a) HBr is removed from C2 and a neighbouring $\mathrm{CH_{3}}$: $\mathrm{CH_{3}CHBrCH_{3}\rightarrow CH_{3}CH{=}CH_{2}}$, propene (both β-carbons are equivalent).
(b) Lindlar: cis-but-2-ene. Na/liquid $\mathrm{NH_{3}}$: trans-but-2-ene. Same structure, different geometry.
(c) $\mathrm{CH_{3}CHBrCH_{2}Br+Zn\rightarrow CH_{3}CH{=}CH_{2}+ZnBr_{2}}$: propene.
(d) $\mathrm{(CH_{3})_{2}CHOH\rightarrow CH_{3}CH{=}CH_{2}+H_{2}O}$: propene.
(e) $\mathrm{CH_{3}CH_{2}CHBrCH_{3}}$ has two different β-carbons (C1 and C3), so a mixture forms: but-1-ene (H removed from C1) and but-2-ene (H from C3, as cis and trans) — in practice mostly but-2-ene.
(f) $n(\mathrm{C_{2}H_{5}OH})=23.0/46.07=0.4992\ \mathrm{mol}$; $1:1$, so theoretical ethene $0.4992\times28.05=14.0\ \mathrm{g}$; at $80.0\%$, $11.2\ \mathrm{g}$.

Addition Reactions: Markovnikov's Rule and the Peroxide Effect

In an addition reaction the π bond breaks and two new σ bonds form, one to each carbon of the former double bond; nothing is lost. Hydrogen, the halogens, the hydrogen halides, sulphuric acid and water all add to alkenes.

📐 Rule 9.6 — Markovnikov's rule
When an unsymmetrical reagent HX ($\mathrm{HCl}$, $\mathrm{HBr}$, $\mathrm{HI}$, $\mathrm{H_{2}SO_{4}}$, $\mathrm{H_{2}O}$) adds to an unsymmetrical alkene, the negative part of the reagent (X) goes to the doubly bonded carbon that carries fewer hydrogen atoms — equivalently, the H goes to the carbon that already has more H atoms (Vladimir Markovnikov, 1869): $$\mathrm{CH_{3}CH{=}CH_{2}+HBr\rightarrow CH_{3}CHBrCH_{3}}$$ 2-bromopropane is the main product; 1-bromopropane is formed only in traces. The rule holds because the reaction goes through the more stable carbocation.

Mechanism of electrophilic addition of HBr. HBr supplies an electrophile, $\mathrm{H^{+}}$. In step 1 the π electrons of the alkene attack it; this is the slow step and it leaves a positive charge on the other carbon. With propene there are two possibilities: adding $\mathrm{H^{+}}$ to C1 gives the secondary carbocation $\mathrm{CH_{3}\overset{+}{C}HCH_{3}}$, while adding it to C2 gives the primary carbocation $\mathrm{CH_{3}CH_{2}\overset{+}{C}H_{2}}$. A secondary carbocation is more stable than a primary one (it is stabilised by more alkyl groups through their inductive effect and hyperconjugation — Chapter 8), so it forms faster. In step 2 the bromide ion attacks the carbocation.

📐 Rule 9.7 — The peroxide effect (anti-Markovnikov addition of HBr)
In the presence of an organic peroxide such as benzoyl peroxide, $\mathrm{(C_{6}H_{5}CO)_{2}O_{2}}$, HBr adds to an unsymmetrical alkene against Markovnikov's rule — the Br goes to the carbon with more hydrogen atoms (M. S. Kharasch and F. R. Mayo, 1933): $$\begin{aligned}&\mathrm{CH_{3}CH{=}CH_{2}+HBr}\\&\quad\xrightarrow{\text{peroxide}}\mathrm{CH_{3}CH_{2}CH_{2}Br}\end{aligned}$$ The peroxide switches the mechanism to a free-radical chain: $$\mathrm{(C_{6}H_{5}COO)_{2}\xrightarrow{\Delta}2C_{6}H_{5}COO^{\bullet}}$$ $$\mathrm{C_{6}H_{5}COO^{\bullet}\rightarrow C_{6}H_{5}^{\bullet}+CO_{2}}$$ $$\mathrm{C_{6}H_{5}^{\bullet}+HBr\rightarrow C_{6}H_{6}+Br^{\bullet}}$$ $$\mathrm{Br^{\bullet}+CH_{3}CH{=}CH_{2}\rightarrow CH_{3}\dot{C}HCH_{2}Br}$$ $$\begin{aligned}&\mathrm{CH_{3}\dot{C}HCH_{2}Br+HBr}\\&\quad\mathrm{\rightarrow CH_{3}CH_{2}CH_{2}Br+Br^{\bullet}}\end{aligned}$$ Now the attacking species is $\mathrm{Br^{\bullet}}$, and it adds to C1 because that gives the more stable secondary radical (adding to C2 would give a primary radical). The H is delivered last. The effect is seen only with HBr: the H–Cl bond ($431\ \mathrm{kJ\,mol^{-1}}$) is too strong to be broken by the radical, and iodine atoms formed from HI combine with each other rather than add to the double bond.
⚠️ Same reagent, opposite product — what decides?
In both mechanisms the reagent adds so as to form the more stable intermediate; what changes is which part of HBr arrives first. Ionic (no peroxide): $\mathrm{H^{+}}$ arrives first → it goes to the $\mathrm{CH_{2}}$ end → the Br ends up on the more substituted carbon. Radical (peroxide): $\mathrm{Br^{\bullet}}$ arrives first → it goes to the $\mathrm{CH_{2}}$ end → the Br ends up on the less substituted carbon. HCl and HI follow Markovnikov's rule with or without peroxide, and a symmetrical alkene such as but-2-ene gives the same product either way.
✏️ Example 4 — Predicting addition products
Name the main product of each reaction: (a) propene + HBr; (b) propene + HBr in the presence of benzoyl peroxide; (c) propene + HCl in the presence of benzoyl peroxide; (d) 2-methylpropene + HBr, with and without peroxide; (e) but-2-ene + HBr, with and without peroxide; (f) but-1-ene + water, $\mathrm{H^{+}}$; (g) 2-methylbut-2-ene + HI.
(a) Markovnikov: $\mathrm{CH_{3}CHBrCH_{3}}$, 2-bromopropane.
(b) Peroxide effect: $\mathrm{CH_{3}CH_{2}CH_{2}Br}$, 1-bromopropane.
(c) No peroxide effect with HCl: $\mathrm{CH_{3}CHClCH_{3}}$, 2-chloropropane.
(d) Without peroxide: via the tertiary cation $\mathrm{(CH_{3})_{2}\overset{+}{C}CH_{3}}$, giving $\mathrm{(CH_{3})_{3}CBr}$, 2-bromo-2-methylpropane. With peroxide: $\mathrm{Br^{\bullet}}$ adds to the $\mathrm{CH_{2}}$ to give the tertiary radical, giving $\mathrm{(CH_{3})_{2}CHCH_{2}Br}$, 1-bromo-2-methylpropane.
(e) Both carbons are equivalent: 2-bromobutane, $\mathrm{CH_{3}CHBrCH_{2}CH_{3}}$, in both cases.
(f) $\mathrm{CH_{3}CH_{2}CH(OH)CH_{3}}$, butan-2-ol.
(g) $\mathrm{(CH_{3})_{2}C{=}CHCH_{3}}$: H goes to C3 (it has one H; C2 has none), giving the tertiary carbocation at C2; product $\mathrm{(CH_{3})_{2}CICH_{2}CH_{3}}$, 2-iodo-2-methylbutane.
✏️ Example 5 — Writing the mechanism
(a) Write the mechanism for the addition of HBr to 2-methylpropene and explain why 1-bromo-2-methylpropane is hardly formed. (b) Write the propagation steps for the addition of HBr to 2-methylpropene in the presence of a peroxide. (c) An alkene $\mathrm{C_{4}H_{8}}$ gives the same product with HBr whether or not a peroxide is present, and it shows cis–trans isomerism. Identify it.
(a) step 1 $\mathrm{(CH_{3})_{2}C{=}CH_{2}+H^{+}\rightarrow (CH_{3})_{2}\overset{+}{C}{-}CH_{3}}$ (H to the $\mathrm{CH_{2}}$): a tertiary carbocation, stabilised by three methyl groups. The alternative, $\mathrm{(CH_{3})_{2}CH\overset{+}{C}H_{2}}$, is primary — much less stable, so it hardly forms.
(a) step 2 $\mathrm{(CH_{3})_{3}C^{+}+Br^{-}\rightarrow (CH_{3})_{3}CBr}$, 2-bromo-2-methylpropane.
(b) $\mathrm{Br^{\bullet}+(CH_{3})_{2}C{=}CH_{2}\rightarrow (CH_{3})_{2}\dot{C}CH_{2}Br}$ (tertiary radical); $\mathrm{(CH_{3})_{2}\dot{C}CH_{2}Br+HBr\rightarrow (CH_{3})_{2}CHCH_{2}Br+Br^{\bullet}}$.
(c) A symmetrical alkene, so that both carbons are equivalent: but-2-ene (cis or trans), which gives 2-bromobutane either way.
Interactive 2 — Markovnikov or Anti-Markovnikov?

Pick an alkene and a reagent, then step through the mechanism with the slider (or ▶ Play). Step 1 shows the two intermediates that the first-arriving species ($\mathrm{H^{+}}$ in the ionic route, $\mathrm{Br^{\bullet}}$ in the radical route) could make, with their class (1°, 2°, 3°); the more stable one is highlighted. Step 2 completes the addition and names the product.

mechanism—
first species adds to—
intermediate—
product—

Oxidation, Ozonolysis and Polymerisation

📘 Definition 9.9 — Oxidation of alkenes by permanganate
  • Baeyer's reagent is a cold, dilute ($\approx1\%$), alkaline aqueous solution of potassium permanganate. It adds two OH groups across the double bond, giving a vicinal glycol (a 1,2-diol), and its purple colour disappears as brown $\mathrm{MnO_{2}}$ forms — a second test for unsaturation: $$\begin{aligned}&\mathrm{CH_{2}{=}CH_{2}+H_{2}O+[O]}\\&\quad\mathrm{\rightarrow HOCH_{2}CH_{2}OH}\end{aligned}$$ (ethane-1,2-diol); propene gives propane-1,2-diol.
  • Acidified potassium permanganate or potassium dichromate is a stronger oxidant: it breaks the C=C bond. Each doubly bonded carbon becomes a C=O group and is then oxidised as far as it can be: a $\mathrm{{=}CR_{2}}$ end gives a ketone $\mathrm{R_{2}C{=}O}$; a $\mathrm{{=}CHR}$ end gives a carboxylic acid $\mathrm{RCOOH}$; a $\mathrm{{=}CH_{2}}$ end gives $\mathrm{CO_{2}+H_{2}O}$: $$\begin{aligned}&\mathrm{(CH_{3})_{2}C{=}CH_{2}}\\&\quad\xrightarrow{\mathrm{KMnO_{4}/H^{+}}}\mathrm{(CH_{3})_{2}C{=}O+CO_{2}+H_{2}O}\end{aligned}$$ $$\begin{aligned}&\mathrm{CH_{3}CH{=}CHCH_{3}}\\&\quad\xrightarrow{\mathrm{KMnO_{4}/H^{+}}}\mathrm{2CH_{3}COOH}\end{aligned}$$
📐 Rule 9.8 — Ozonolysis
Ozonolysis is the reaction of an alkene with ozone, followed by treatment of the ozonide formed with zinc dust and water. The C=C bond is cut and each doubly bonded carbon becomes a C=O group, giving aldehydes and/or ketones (zinc removes the hydrogen peroxide formed, so the aldehydes are not oxidised further): $$\mathrm{R_{2}C{=}CR'_{2}\xrightarrow[\text{2. Zn}/H_{2}O]{\text{1. }O_{3}}R_{2}C{=}O+O{=}CR'_{2}}$$ For example, propene gives ethanal + methanal; 2-methylpropene gives propanone + methanal. Because the carbonyl carbons in the products were the doubly bonded carbons, the position of the double bond in an unknown alkene can be found by joining the carbonyl carbons of the products back together.
📘 Definition 9.10 — Addition polymerisation
A polymer is a very large molecule built from many small repeating units, the monomers. In addition polymerisation alkene molecules add to one another, each π bond becoming a σ bond to the next unit, so that no atoms are lost. Ethene heated under high pressure with a catalyst gives polythene: $$\mathrm{n\,CH_{2}{=}CH_{2}\xrightarrow[\text{high }T,\ p]{\text{catalyst}}{-}\!\left(CH_{2}{-}CH_{2}\right)_{n}\!{-}}$$ Propene gives polypropene, $\mathrm{{-}(CH_{2}{-}CH(CH_{3}))_{n}{-}}$ — used for ropes, crates and bottles; polythene is used for carry-bags, films and pipes.
✏️ Example 6 — Reading an alkene from its ozonolysis products
(a) Give the ozonolysis products of but-1-ene and of 2-methylbut-2-ene. (b) An alkene $\mathrm{C_{6}H_{12}}$ gives only propanone on ozonolysis. Identify it. (c) An alkene gives a mixture of ethanal and propanal. Identify it and say whether it shows geometrical isomerism. (d) What does acidified permanganate give with 2-methylbut-2-ene? (e) An alkene $\mathrm{C_{5}H_{10}}$ gives propanone and an aldehyde on ozonolysis. Identify the alkene and the aldehyde.
(a) $\mathrm{CH_{3}CH_{2}CH{=}CH_{2}\rightarrow CH_{3}CH_{2}CHO+HCHO}$: propanal and methanal. $\mathrm{(CH_{3})_{2}C{=}CHCH_{3}\rightarrow (CH_{3})_{2}CO+CH_{3}CHO}$: propanone and ethanal.
(b) Two identical halves $\mathrm{(CH_{3})_{2}C{=}}$: join the carbonyl carbons, $\mathrm{(CH_{3})_{2}C{=}C(CH_{3})_{2}}$, 2,3-dimethylbut-2-ene. Check: $\mathrm{C_{6}H_{12}}$ ✓.
(c) $\mathrm{CH_{3}CH{=}}$ + $\mathrm{{=}CHCH_{2}CH_{3}}$: $\mathrm{CH_{3}CH{=}CHCH_{2}CH_{3}}$, pent-2-ene. Each doubly bonded carbon has H and an alkyl group, so yes, cis and trans (ozonolysis gives the same products from both).
(d) The $\mathrm{{=}C(CH_{3})_{2}}$ end gives propanone; the $\mathrm{{=}CHCH_{3}}$ end gives ethanoic acid (the aldehyde is oxidised further): $\mathrm{(CH_{3})_{2}CO+CH_{3}COOH}$.
(e) Propanone supplies $\mathrm{(CH_{3})_{2}C{=}}$ (3 C); the other half has $5-3=2$ carbons: $\mathrm{{=}CHCH_{3}}$, from ethanal. Alkene: 2-methylbut-2-ene; aldehyde: ethanal.
🚀 Beyond the board syllabus (enrichment only) — bromine adds through a bromonium ion

Enrichment only. JEE Main, NEET and JEE Advanced list the addition of halogens to alkenes, but none lists the stereochemistry of the addition — JEE Advanced excludes it explicitly. Read the anti addition below for interest; exam questions do not depend on it.

As a $\mathrm{Br_{2}}$ molecule approaches the π cloud it is polarised, $\mathrm{Br^{\delta+}{-}Br^{\delta-}}$, and the alkene attacks $\mathrm{Br^{\delta+}}$. Instead of an open carbocation, a three-membered cyclic bromonium ion forms, with Br bridging both carbons. The $\mathrm{Br^{-}}$ left behind can then attack only from the opposite face, so the two Br atoms end up on opposite sides: anti addition. Cyclopentene, for example, gives only trans-1,2-dibromocyclopentane, and cis-but-2-ene gives a different stereoisomer of 2,3-dibromobutane from trans-but-2-ene. Evidence that the two Br atoms arrive in separate steps: if NaCl is present, some $\mathrm{BrCH_{2}CH_{2}Cl}$ forms, but never $\mathrm{ClCH_{2}CH_{2}Cl}$ — the first atom added always comes from $\mathrm{Br_{2}}$.

Exercises

I
Structure, naming and isomerism. Atomic masses: C $12.01$, H $1.008$.
  • 1 Write the molecular formula and molar mass of the open-chain alkene with six carbon atoms and one double bond.
  • 2 Describe the C=C bond of ethene: hybridisation of carbon, the bonds present, bond length, and why the molecule is planar.
  • 3 Name: (a) $\mathrm{CH_{2}{=}C(CH_{3})CH_{2}CH_{3}}$; (b) $\mathrm{CH_{3}CH{=}C(CH_{3})_{2}}$; (c) $\mathrm{CH_{2}{=}CHCH(CH_{3})_{2}}$; (d) $\mathrm{CH_{3}CH_{2}CH{=}CHCH_{2}CH_{3}}$.
  • 4 Which show geometrical isomerism: hex-3-ene, 2-methylbut-1-ene, 3-methylpent-2-ene, 2,3-dimethylbut-2-ene, $\mathrm{BrCH{=}CHCl}$?
  • 5 Draw the cis and trans isomers of pent-2-ene.
  • 6 Explain why trans-but-2-ene has no dipole moment while cis-but-2-ene has a small one.
  • 7 Of cis- and trans-1,2-dichloroethene, which has the higher boiling point and which the higher melting point? Explain each.
  • 8 How many open-chain alkenes have the formula $\mathrm{C_{6}H_{12}}$ — structural isomers only, and counting cis–trans isomers separately?
1 $\mathrm{C_{6}H_{12}}$ ($\mathrm{C_{n}H_{2n}}$): $6(12.01)+12(1.008)=84.16\ \mathrm{g\,mol^{-1}}$.
2 Both carbons sp²; one σ bond (sp²–sp² overlap) and one π bond (sideways p–p overlap); $134\ \mathrm{pm}$. The p orbitals can overlap only if they are parallel, which holds the two $\mathrm{CH_{2}}$ planes together in one plane.
3 (a) 2-methylbut-1-ene; (b) 2-methylbut-2-ene; (c) 3-methylbut-1-ene; (d) hex-3-ene.
4 Hex-3-ene (H, $\mathrm{C_{2}H_{5}}$ on each C): yes. 2-Methylbut-1-ene ($\mathrm{{=}CH_{2}}$): no. 3-Methylpent-2-ene $\mathrm{CH_{3}CH{=}C(CH_{3})CH_{2}CH_{3}}$: C2 (H, $\mathrm{CH_{3}}$), C3 ($\mathrm{CH_{3}}$, $\mathrm{C_{2}H_{5}}$): yes. 2,3-Dimethylbut-2-ene: no. $\mathrm{BrCH{=}CHCl}$: yes.
5 cis: $\mathrm{CH_{3}}$ and $\mathrm{C_{2}H_{5}}$ on the same side of C=C, both H on the other side; trans: $\mathrm{CH_{3}}$ and $\mathrm{C_{2}H_{5}}$ on opposite sides.
6 Each C–$\mathrm{CH_{3}}$ bond has a small dipole. In the trans isomer the two are equal and opposite (the molecule has a centre of symmetry), so they cancel; in the cis isomer they point partly the same way and add.
7 cis boils higher ($60.3$ vs $47.5\ ^{\circ}\mathrm{C}$) — it is polar ($1.90\ \mathrm{D}$), so dipole–dipole forces add to dispersion forces. trans melts higher ($-50$ vs $-80\ ^{\circ}\mathrm{C}$) — it is more symmetrical and packs better in the solid.
8 13 structural isomers; 4 of them (hex-2-ene, hex-3-ene, 4-methylpent-2-ene, 3-methylpent-2-ene) have cis and trans forms, so 17 alkenes in all (not counting optical isomers).
II
Preparation of alkenes. $M$: bromoethane $108.96$, ethene $28.05\ \mathrm{g\,mol^{-1}}$.
  • 1 Write the equation for the reduction of ethyne with hydrogen over Lindlar's catalyst. What is the catalyst?
  • 2 Pent-2-yne is reduced (a) with $\mathrm{H_{2}}$/Lindlar's catalyst, (b) with Na in liquid ammonia. Name each product, including its geometry.
  • 3 Name the product when 1-chlorobutane is heated with alcoholic KOH, and write the equation.
  • 4 Name the product of 2-chloro-2-methylpropane with alcoholic KOH. Why does it react more easily than 1-chlorobutane?
  • 5 Name the product of 2,3-dibromobutane with zinc dust.
  • 6 Name the alkene formed when 2-methylpropan-2-ol is heated with concentrated sulphuric acid, and write the equation.
  • 7 Explain why bromomethane cannot give an alkene with alcoholic KOH.
  • 8 What mass of ethene can be made from $10.9\ \mathrm{g}$ of bromoethane, assuming complete conversion?
1 $\mathrm{HC{\equiv}CH+H_{2}\xrightarrow{\text{Lindlar}}CH_{2}{=}CH_{2}}$. Palladium on charcoal or calcium carbonate, partly deactivated with sulphur compounds or quinoline, so that it stops at the alkene.
2 (a) cis-pent-2-ene; (b) trans-pent-2-ene.
3 $\mathrm{CH_{3}CH_{2}CH_{2}CH_{2}Cl+KOH(alc)\rightarrow CH_{3}CH_{2}CH{=}CH_{2}+KCl+H_{2}O}$: but-1-ene.
4 2-Methylpropene. Ease of dehydrohalogenation is 3° $\gt$ 2° $\gt$ 1°, and this halide is tertiary.
5 $\mathrm{CH_{3}CHBrCHBrCH_{3}+Zn\rightarrow CH_{3}CH{=}CHCH_{3}+ZnBr_{2}}$: but-2-ene.
6 $\mathrm{(CH_{3})_{3}COH\xrightarrow{\text{conc. }H_{2}SO_{4},\ \Delta}(CH_{3})_{2}C{=}CH_{2}+H_{2}O}$: 2-methylpropene.
7 β-Elimination needs an H on a carbon next to the one carrying the halogen; $\mathrm{CH_{3}Br}$ has only one carbon.
8 $n=10.9/108.96=0.1000\ \mathrm{mol}$; ethene $0.1000\times28.05=2.81\ \mathrm{g}$.
III
Addition reactions. $M$: ethene $28.05$, $\mathrm{Br_{2}}$ $159.80\ \mathrm{g\,mol^{-1}}$.
  • 1 What is seen when ethene is bubbled into bromine in $\mathrm{CCl_{4}}$? Write the equation and name the product.
  • 2 State Markovnikov's rule and use it to predict the product of propene with HI.
  • 3 Name the product of propene with HBr in the presence of benzoyl peroxide, and name the effect.
  • 4 Name the main product of 2-methylbut-1-ene with HCl.
  • 5 Name the product of acid-catalysed hydration of but-1-ene.
  • 6 Why does a peroxide not change the direction of addition of HCl?
  • 7 Arrange in order of increasing stability: $\mathrm{(CH_{3})_{3}C^{+}}$, $\mathrm{CH_{3}^{+}}$, $\mathrm{(CH_{3})_{2}CH^{+}}$, $\mathrm{CH_{3}CH_{2}^{+}}$. How does this order explain Markovnikov's rule?
  • 8 What mass of bromine is decolourised by $2.80\ \mathrm{g}$ of ethene?
1 The reddish-orange colour disappears. $\mathrm{CH_{2}{=}CH_{2}+Br_{2}\rightarrow BrCH_{2}CH_{2}Br}$: 1,2-dibromoethane.
2 The negative part of the reagent adds to the doubly bonded carbon with fewer H atoms. $\mathrm{CH_{3}CH{=}CH_{2}+HI\rightarrow CH_{3}CHICH_{3}}$: 2-iodopropane.
3 1-Bromopropane; the peroxide (Kharasch) effect, or anti-Markovnikov addition.
4 $\mathrm{CH_{2}{=}C(CH_{3})CH_{2}CH_{3}+HCl\rightarrow (CH_{3})_{2}CClCH_{2}CH_{3}}$: 2-chloro-2-methylbutane (via the tertiary carbocation).
5 Butan-2-ol, $\mathrm{CH_{3}CH(OH)CH_{2}CH_{3}}$.
6 The radical chain needs the step $\mathrm{R^{\bullet}+H{-}X\rightarrow R{-}H+X^{\bullet}}$. The H–Cl bond ($431\ \mathrm{kJ\,mol^{-1}}$) is too strong for this step to be fast, so the chain does not run and the ionic (Markovnikov) addition takes place.
7 $\mathrm{CH_{3}^{+}\lt CH_{3}CH_{2}^{+}\lt (CH_{3})_{2}CH^{+}\lt (CH_{3})_{3}C^{+}}$ (more alkyl groups, more stabilisation). $\mathrm{H^{+}}$ adds to whichever carbon gives the more substituted (more stable) carbocation; the negative ion then joins that carbon — the one with fewer H atoms.
8 $n=2.80/28.05=0.0998\ \mathrm{mol}$; $1:1$; $0.0998\times159.80=16.0\ \mathrm{g}$ of $\mathrm{Br_{2}}$.
IV
Oxidation, ozonolysis and polymerisation.
  • 1 What is Baeyer's reagent? Give the product of its reaction with propene and the observation.
  • 2 Give the products of acidified $\mathrm{KMnO_{4}}$ with (a) but-2-ene, (b) propene.
  • 3 Give the ozonolysis products of pent-1-ene.
  • 4 Give the ozonolysis products of 2-methylpent-2-ene.
  • 5 An alkene gives two moles of propanal per mole on ozonolysis. Identify it.
  • 6 An alkene gives methanal and butanone on ozonolysis. Identify it.
  • 7 Explain why no open-chain alkene $\mathrm{C_{5}H_{10}}$ gives only a single carbonyl compound on ozonolysis.
  • 8 Write the repeating unit of polypropene. What mass of polythene can, at most, be made from $1.00$ tonne of ethene, and why?
1 Cold, dilute, alkaline aqueous $\mathrm{KMnO_{4}}$. Propene gives propane-1,2-diol, $\mathrm{CH_{3}CH(OH)CH_{2}OH}$; the purple colour disappears and a brown precipitate ($\mathrm{MnO_{2}}$) appears.
2 (a) $\mathrm{2CH_{3}COOH}$ (ethanoic acid). (b) $\mathrm{CH_{3}COOH}$ from the $\mathrm{{=}CHCH_{3}}$ end, and $\mathrm{CO_{2}+H_{2}O}$ from the $\mathrm{{=}CH_{2}}$ end.
3 $\mathrm{CH_{3}CH_{2}CH_{2}CHO}$ (butanal) + $\mathrm{HCHO}$ (methanal).
4 $\mathrm{(CH_{3})_{2}C{=}CHCH_{2}CH_{3}\rightarrow (CH_{3})_{2}CO+CH_{3}CH_{2}CHO}$: propanone + propanal.
5 Join two $\mathrm{CH_{3}CH_{2}CH{=}}$ halves: $\mathrm{CH_{3}CH_{2}CH{=}CHCH_{2}CH_{3}}$, hex-3-ene.
6 $\mathrm{CH_{2}{=}}$ + $\mathrm{{=}C(CH_{3})CH_{2}CH_{3}}$: $\mathrm{CH_{2}{=}C(CH_{3})CH_{2}CH_{3}}$, 2-methylbut-1-ene.
7 A single product means two identical halves, so the alkene would have an even number of carbon atoms; five carbons cannot be split into two equal halves.
8 $\mathrm{{-}CH_{2}{-}CH(CH_{3}){-}}$. At most $1.00$ tonne: addition polymerisation joins the monomers without losing any atoms, so mass is conserved.
Check: ozonolysis reverse-engineer — products → alkene
join the C=O carbons—
IUPAC name—
molecular formula—
cis–trans isomers?—
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. NEET The peroxide effect is observed in the addition of (A) HCl (B) HBr (C) HI (D) all three hydrogen halides

Idea The radical chain needs both propagation steps to be fast. With HCl the H–Cl bond is too strong; with HI the iodine atoms combine to $\mathrm{I_{2}}$ instead of adding to the alkene. Only HBr works.
Answer (B).

E2. JEE Main An alkene gives only ethanal on ozonolysis (followed by Zn/H₂O). The alkene is (A) but-1-ene (B) but-2-ene (C) 2-methylpropene (D) propene

Reverse Join two $\mathrm{CH_{3}CH{=}}$ units through their carbonyl carbons: $\mathrm{CH_{3}CH{=}CHCH_{3}}$.
Check others But-1-ene → propanal + methanal; 2-methylpropene → propanone + methanal; propene → ethanal + methanal.
Answer (B) (cis or trans).

E3. Board · 3 marks State Markovnikov's rule. Write the mechanism of the addition of HBr to propene and use it to explain why 2-bromopropane is the main product.

1 mark Rule: in the addition of an unsymmetrical reagent to an unsymmetrical alkene, the negative part of the reagent goes to the doubly bonded carbon carrying fewer hydrogen atoms.
1 mark Step 1: $\mathrm{HBr\rightarrow H^{+}+Br^{-}}$; the π electrons attack $\mathrm{H^{+}}$. Adding H to C1 gives the secondary carbocation $\mathrm{CH_{3}\overset{+}{C}HCH_{3}}$; adding it to C2 would give the primary $\mathrm{CH_{3}CH_{2}\overset{+}{C}H_{2}}$. The secondary cation is more stable, so it forms.
1 mark Step 2: $\mathrm{Br^{-}}$ attacks the positive carbon: $\mathrm{CH_{3}\overset{+}{C}HCH_{3}+Br^{-}\rightarrow CH_{3}CHBrCH_{3}}$ (2-bromopropane).

E4. JEE Main Numerical answer. How many of the following show geometrical isomerism? But-2-ene, pent-1-ene, pent-2-ene, 2-methylbut-2-ene, hex-3-ene, 1,2-dichloroethene, 1,1-dichloroethene, 3-methylpent-2-ene, 2-methylpropene.

Test Both doubly bonded carbons must carry two different groups. Fails: pent-1-ene ($\mathrm{{=}CH_{2}}$), 2-methylbut-2-ene (two $\mathrm{CH_{3}}$ on C2), 1,1-dichloroethene, 2-methylpropene.
Pass But-2-ene, pent-2-ene, hex-3-ene, 1,2-dichloroethene, 3-methylpent-2-ene.
Answer 5.

E5. JEE Advanced One or more options correct. For 2-methylbut-2-ene, $\mathrm{(CH_{3})_{2}C{=}CHCH_{3}}$: (A) with HBr it gives mainly 2-bromo-2-methylbutane; (B) with HBr and a peroxide it gives mainly 2-bromo-3-methylbutane; (C) on ozonolysis it gives propanone and ethanal; (D) it exists as cis and trans isomers.

(A) True: $\mathrm{H^{+}}$ adds to C3, giving the tertiary cation at C2; Br joins C2.
(B) True: $\mathrm{Br^{\bullet}}$ adds to C3 to give the tertiary radical at C2; then H joins C2, so Br is on C3: $\mathrm{(CH_{3})_{2}CHCHBrCH_{3}}$, numbered from the other end as 2-bromo-3-methylbutane.
(C) True: $\mathrm{(CH_{3})_{2}C{=}O+O{=}CHCH_{3}}$.
(D) False: C2 carries two identical $\mathrm{CH_{3}}$ groups.
Answer (A), (B), (C).

E6. Board · Case study A hydrocarbon X, $\mathrm{C_{4}H_{8}}$, decolourises bromine water and does not show geometrical isomerism. On ozonolysis it gives methanal and propanone. (i) Identify X and write its ozonolysis. (2 marks) (ii) Name the product of X with HBr. (1 mark) (iii) Name the product of X with HBr in the presence of benzoyl peroxide. (1 mark)

(i) Join $\mathrm{CH_{2}{=}}$ and $\mathrm{{=}C(CH_{3})_{2}}$: X is 2-methylpropene, $\mathrm{(CH_{3})_{2}C{=}CH_{2}}$ (no cis–trans: $\mathrm{{=}CH_{2}}$ end). $\mathrm{(CH_{3})_{2}C{=}CH_{2}\xrightarrow[\text{2. Zn}/H_{2}O]{\text{1. }O_{3}}(CH_{3})_{2}C{=}O+HCHO}$.
(ii) 2-Bromo-2-methylpropane, $\mathrm{(CH_{3})_{3}CBr}$ (Markovnikov, via the 3° carbocation).
(iii) 1-Bromo-2-methylpropane, $\mathrm{(CH_{3})_{2}CHCH_{2}Br}$ (anti-Markovnikov, via the 3° radical).
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Food regulators and nutritionists describe cooking oils by their "iodine value" — the mass of iodine, in grams, taken up by $100\ \mathrm{g}$ of oil. What does this number measure, and how can you predict it from the structure of the fat molecules?

Why it matters. Oils rich in C=C bonds (sunflower, soyabean, linseed) are liquid and considered healthier, but they go rancid faster when oxygen attacks the double bonds; hydrogenation (Section 9.1) hardens them into fats such as vanaspati and can create harmful trans fats. The iodine value is a quick laboratory measure of unsaturation used by food-safety agencies from India's FSSAI to the EU.

What to do.

  1. Each C=C bond adds one molecule of halogen, as in the bromine test. Model three fats as triglycerides: tristearin $\mathrm{C_{57}H_{110}O_{6}}$ (no C=C), triolein $\mathrm{C_{57}H_{104}O_{6}}$ (3 C=C) and trilinolein $\mathrm{C_{57}H_{98}O_{6}}$ (6 C=C). Check the number of C=C from the hydrogen deficit.
  2. Calculate the molar mass of each and its iodine value: $\text{IV}=\dfrac{(\text{number of C=C})\times253.8}{M}\times100$.
  3. Look up typical iodine values of coconut, groundnut, olive, sunflower and linseed oil. Estimate the average number of C=C per triglyceride molecule in each, taking $M\approx880\ \mathrm{g\,mol^{-1}}$.
  4. Which oils would you expect to be most stable on storage, and which best for a paint that dries by reacting with air?

Tools: Molar masses (Section 1.2), addition of halogens (this section), a food-composition database or regulator's standard for edible oils.

Hand in: A table of IV for the three model fats and five real oils, with your estimates of C=C per molecule, and a 150-word explanation for a consumer magazine.

Going further. How much hydrogen (in kg) is needed to fully harden one tonne of an oil with an iodine value of $130$?

Instructor note: compare $\mathrm{C_{57}H_{110}O_{6}}$ (saturated) — each 2 H fewer is one C=C. $M$: tristearin $891.4$, triolein $885.4$, trilinolein $879.4\ \mathrm{g\,mol^{-1}}$; IV $0$, $86.0$, $173.2$. Typical IVs: coconut ≈ 7–11, olive ≈ 75–94, groundnut ≈ 85–100, sunflower ≈ 120–140, linseed ≈ 170–200 (≈ 0.3, 2.9, 3.2, 4.5, 6.4 C=C per molecule for mid-range values). Linseed (most unsaturated) dries fastest in paint; coconut is most stable. Going further: $130\ \mathrm{g}$ $\mathrm{I_{2}}$ per $100\ \mathrm{g}$ = $0.512\ \mathrm{mol}$ C=C per $100\ \mathrm{g}$ → $5.12\times10^{3}\ \mathrm{mol}$ per tonne → $\mathrm{H_{2}}$ $5.12\times10^{3}\times2.016\ \mathrm{g}\approx10.3\ \mathrm{kg}$.

🧪 Self-check quiz

0/5 correct

Q1. Which alkene exists as cis and trans isomers?

Q2. Propene reacts with HBr in the presence of benzoyl peroxide to give mainly

Q3. Baeyer's reagent is

Q4. Ozonolysis of 2-methylpropene gives

Q5. 2-Bromopropane is converted into propene by heating it with

← 9.1 Alkanes9.3 Alkynes →