Alkenes and Electrophilic Addition

AS · 15 min

Alkenes, $\ce{C_{n}H_{2n}}$ , contain a carbon–carbon double bond, and that one feature makes them far more reactive than alkanes. The C=C bond's $\pi$ electrons attract electrophiles, so alkenes take part in addition reactions that turn them into alkanes, alcohols, halogenoalkanes and diols, and into polymers. This note covers how alkenes are made, every addition reaction on the syllabus with its conditions, the oxidation of alkenes by potassium manganate(VII) (used to locate the double bond), the bromine test, and the electrophilic addition mechanism with Markovnikov's rule. Expect at least one alkene mechanism or product-prediction question on every Paper 2.

Structure and reactivity

In an alkene each carbon of the C=C is sp² hybridised: it forms three $\sigma$ bonds in a plane at $120^\circ$ , and its remaining p orbital overlaps sideways with the p orbital on the other carbon to form a $\pi$ bond. The $\pi$ electrons sit above and below the plane of the molecule, away from the nuclei.

Key result

Alkenes are much more reactive than alkanes because:

The C=C bond is a region of high electron density, which attracts electrophiles (electron-pair acceptors).
The $\pi$ bond is weaker than a $\sigma$ bond (the sideways overlap is less effective) and its electrons are more exposed, so it breaks relatively easily, leaving the $\sigma$ bond intact.

The typical reaction of an alkene is electrophilic addition: the $\pi$ bond breaks and two new $\sigma$ bonds form, one to each carbon.

Because the $\pi$ bond prevents rotation, alkenes with two different groups on each carbon of the C=C show cis/trans isomerism (see Isomerism).

Making alkenes

Key result

method	reagents and conditions	example
elimination of HX from a halogenoalkane	NaOH in ethanol, heat (reflux)	$\ce{CH3CH2Br + NaOH -> CH2=CH2 + NaBr + H2O}$
dehydration of an alcohol	heated $\ce{Al2O3}$ (pass alcohol vapour over it), or concentrated $\ce{H2SO4}$ (or $\ce{H3PO4}$ ), heat	$\ce{CH3CH2OH -> CH2=CH2 + H2O}$
cracking of long-chain alkanes	heat with $\ce{Al2O3}$ or zeolite catalyst	$\ce{C10H22 -> C8H18 + C2H4}$

Elimination and dehydration remove an H from the carbon next to the carbon carrying the halogen or OH. If the functional group is in the middle of an unsymmetrical chain, the H can come from either side, so more than one alkene forms. Butan-2-ol gives but-1-ene and but-2-ene (as both cis and trans isomers): three alkenes in all.

Addition reactions

Key result

reagent	conditions	product (from ethene)	equation
hydrogen, $\ce{H2}$	Ni catalyst and heat (about $150\ ^\circ\text{C}$ ), or Pt catalyst at room temperature	alkane: ethane	$\ce{CH2=CH2 + H2 -> CH3CH3}$
steam, $\ce{H2O(g)}$	$\ce{H3PO4}$ catalyst, about $300\ ^\circ\text{C}$ , $6\ \text{MPa}$	alcohol: ethanol	$\ce{CH2=CH2 + H2O -> CH3CH2OH}$
hydrogen halide, HX (HCl, HBr, HI)	gas (or concentrated solution), room temperature	halogenoalkane: bromoethane	$\ce{CH2=CH2 + HBr -> CH3CH2Br}$
halogen, $\ce{X2}$ ( $\ce{Cl2}$ , $\ce{Br2}$ )	room temperature, in the dark (often dissolved in an organic solvent)	dihalogenoalkane: 1,2-dibromoethane	$\ce{CH2=CH2 + Br2 -> CH2BrCH2Br}$
cold dilute acidified $\ce{KMnO4}$	room temperature	diol: ethane-1,2-diol	$\ce{CH2=CH2 + [O] + H2O -> HOCH2CH2OH}$
hot concentrated acidified $\ce{KMnO4}$	heat	C=C split (see below)
itself (many molecules)	high pressure and heat, or a catalyst	addition polymer: poly(ethene)	$\ce{nCH2=CH2 -> [-CH2-CH2-]_{n}}$

Addition of steam is the industrial route to ethanol from ethene (from cracking). The reaction is reversible and only about $5\%$ of ethene is converted per pass, so unreacted ethene is recycled. Polymerisation is covered in Addition polymers.

Testing for a C=C bond with bromine water

Method

Test for unsaturation

Add a few drops of bromine water (orange-brown) to the sample and shake.
If a C=C bond is present, the bromine adds across it and the solution is decolourised (orange-brown to colourless) at room temperature, quickly.
An alkane gives no change in the dark: the bromine water stays orange.

Tip

In bromine water, the carbocation intermediate is more likely to meet a water molecule than a bromide ion, so the main product is actually a bromo-alcohol (for ethene, 2-bromoethanol, $\ce{CH2BrCH2OH}$ ) together with some 1,2-dibromoethane. For the test you only need the colour change, and examiners accept $\ce{CH2BrCH2Br}$ as the product in the equation.

Oxidation by potassium manganate(VII)

Acidified potassium manganate(VII) is a strong oxidising agent. What it does to an alkene depends on the conditions. In both cases the purple solution is decolourised, as purple $\ce{MnO4-}$ is reduced to almost colourless $\ce{Mn^2+}$ .

Cold, dilute, acidified: making a diol

At room temperature, dilute $\ce{KMnO4}$ adds an OH group to each carbon of the double bond:

\ce{CH3CH=CH2 + [O] + H2O -> CH3CH(OH)CH2OH}

The product from propene is propane-1,2-diol. The $\pi$ bond breaks but the $\sigma$ bond stays: the carbon chain is unchanged.

Hot, concentrated, acidified: splitting the C=C bond

With hot concentrated $\ce{KMnO4}$ , the whole double bond (both $\sigma$ and $\pi$ ) is broken, and each carbon of the old C=C becomes a C=O. What happens next depends on how many hydrogens that carbon carried:

Key result

group at one end of the C=C	first product	final product with hot conc. $\ce{KMnO4}$
$\ce{=CH2}$ (two H)	methanal, then methanoic acid	carbon dioxide (and water)
$\ce{=CHR}$ (one H)	aldehyde RCHO	carboxylic acid RCOOH
$\ce{=CRR'}$ (no H)	ketone RCOR′	ketone (not oxidised further)

\ce{CH3CH=CHCH3 + 4[O] -> 2CH3COOH}

\ce{CH2=CHCH2CH3 + 5[O] -> CH3CH2COOH + CO2 + H2O}

\ce{(CH3)2C=CHCH3 + 3[O] -> (CH3)2CO + CH3COOH}

This reaction was historically used to find where the double bond is in an unknown alkene: identify the products, then "join them back together" at their C=O carbons.

Method

Deducing an alkene from its oxidation products

Write each organic product with its C=O carbon marked. $\ce{CO2}$ came from a $\ce{=CH2}$ end; a carboxylic acid from a $\ce{=CHR}$ end; a ketone from a $\ce{=CRR'}$ end.
Remove the =O (and the OH of an acid) from each, and join the two marked carbons with a double bond.
Check that the carbon count of the alkene equals the total carbon count of the products.

The electrophilic addition mechanism

Ethene and bromine

Bromine is a non-polar molecule, yet it acts as an electrophile. This is how:

As a $\ce{Br2}$ molecule approaches the C=C, the high electron density of the $\pi$ bond repels the electrons in the Br–Br bond, inducing a dipole: the nearer bromine atom becomes $\delta+$ and the further one $\delta-$ .
A curly arrow from the C=C double bond to the $\delta+$ bromine atom: the $\pi$ electrons form a new C–Br bond.
At the same time, a curly arrow from the Br–Br bond to the $\delta-$ bromine atom: the Br–Br bond breaks heterolytically and a bromide ion, $\ce{Br-}$ , is released.
This leaves a carbocation intermediate, $\ce{CH2BrCH2+}$ : the carbon that did not gain the bromine has lost its share of the $\pi$ electrons and carries a positive charge.
A curly arrow from a lone pair on the bromide ion to the positive carbon forms the second C–Br bond. The product is 1,2-dibromoethane, $\ce{CH2BrCH2Br}$ .

Propene and hydrogen bromide

Hydrogen bromide is permanently polar, $\ce{H^{\delta+}-Br^{\delta-}}$ , so it needs no induced dipole.

Electrophilic addition of hydrogen bromide to propene. Step 1: a curly arrow from the C=C bond to the δ+ hydrogen, and a curly arrow from the H–Br bond to bromine, give a secondary carbocation and a bromide ion. Step 2: a curly arrow from a lone pair on Br⁻ to the positive carbon forms the C–Br bond.

A curly arrow from the C=C double bond to the $\delta+$ hydrogen atom of H–Br: the $\pi$ electrons form a new C–H bond.
A curly arrow from the H–Br bond to the $\delta-$ bromine atom: heterolytic fission releases $\ce{Br-}$ .
A carbocation forms. Because the hydrogen bonded to the end carbon ( $\ce{CH2}$ ), the positive charge is on the middle carbon: $\ce{CH3CH+CH3}$ , a secondary carbocation.
A curly arrow from a lone pair on $\ce{Br-}$ to the positive carbon forms the C–Br bond. The major product is 2-bromopropane.

Markovnikov's rule: which product is major?

Propene is unsymmetrical. In step 1 the hydrogen could instead bond to the middle carbon, putting the positive charge on the end carbon to give $\ce{CH3CH2CH2+}$ , a primary carbocation. That route leads to 1-bromopropane. In practice 2-bromopropane is the major product and 1-bromopropane is minor.

Key result

Markovnikov's rule: when HX adds to an unsymmetrical alkene, the hydrogen atom attaches to the carbon of the C=C that already has more hydrogen atoms, and X attaches to the carbon with fewer hydrogen atoms.

Why: this route goes through the more stable carbocation. Alkyl groups are electron-donating (positive inductive effect) and stabilise a positive charge, so stability is tertiary > secondary > primary. The more stable carbocation forms faster (lower activation energy), so its product is the major one.

The same reasoning applies to steam: propene and steam give mainly propan-2-ol, with some propan-1-ol.

Worked examples

Routine: reactions of propene

Give the structural formula and name of the organic product when propene reacts with (a) hydrogen and nickel, heated; (b) bromine at room temperature; (c) cold dilute acidified potassium manganate(VII); (d) hydrogen chloride (major product).

Solution

(a) $\ce{CH3CH2CH3}$ , propane.

(b) $\ce{CH3CHBrCH2Br}$ , 1,2-dibromopropane.

(d) $\ce{CH3CHClCH3}$ , 2-chloropropane (H adds to the $\ce{CH2}$ carbon, which has more hydrogens; the secondary carbocation is more stable).

Routine: distinguishing cyclohexane and cyclohexene

Describe a simple chemical test to distinguish cyclohexene from cyclohexane, giving the observations and an equation.

Solution

Add bromine water to each and shake at room temperature.

Cyclohexene: the orange-brown colour disappears quickly (decolourised). Cyclohexane: remains orange-brown.

\ce{C6H10 + Br2 -> C6H10Br2}

(The product is 1,2-dibromocyclohexane.)

Alternatively, acidified potassium manganate(VII) is decolourised (purple to colourless) by cyclohexene but not by cyclohexane.

Standard: describing the mechanism with bromine

Describe, with numbered steps, the mechanism of the reaction between but-2-ene and bromine. Name the mechanism and the product.

Solution

Electrophilic addition.

The $\pi$ electrons of the C=C repel the electrons in Br–Br, inducing a dipole: $\ce{Br^{\delta+}-Br^{\delta-}}$ , with the $\delta+$ end nearer the alkene.
Curly arrow from the C=C bond to the $\delta+$ Br: a new C–Br bond forms on C2.
Curly arrow from the Br–Br bond to the $\delta-$ Br: heterolytic fission gives $\ce{Br-}$ .
Intermediate: the carbocation $\ce{CH3CHBrCH+CH3}$ , with the positive charge on C3 (a secondary carbocation).
Curly arrow from a lone pair on $\ce{Br-}$ to the positive C3 forms the second C–Br bond.

Product: $\ce{CH3CHBrCHBrCH3}$ , 2,3-dibromobutane.

Standard: predicting and explaining the major product

2-methylpropene, $\ce{(CH3)2C=CH2}$ , reacts with hydrogen bromide. (a) Draw the two possible carbocations and name the major product. (b) Explain why it is the major product.

Solution

(a) H adds to the $\ce{CH2}$ carbon: $\ce{(CH3)3C+}$ , tertiary. H adds to the central carbon: $\ce{(CH3)2CHCH2+}$ , primary. The major product is from the tertiary carbocation: $\ce{(CH3)3CBr}$ , 2-bromo-2-methylpropane.

(b) The tertiary carbocation has three electron-donating methyl groups bonded to the positive carbon; their positive inductive effect reduces the charge density on the carbon and stabilises the ion. The primary carbocation has only one alkyl group. The more stable tertiary carbocation forms faster, so its product predominates.

Exam-hard: locating a double bond

An alkene A, $\ce{C5H10}$ , reacts with HBr to give mainly 2-bromo-2-methylbutane. When A is heated with concentrated acidified potassium manganate(VII), it gives butanone and carbon dioxide. Identify A, and explain why its isomer B, which also gives 2-bromo-2-methylbutane with HBr, gives different products with hot $\ce{KMnO4}$ .

Solution

$\ce{CO2}$ comes from a $\ce{=CH2}$ end. Butanone, $\ce{CH3COCH2CH3}$ , comes from a $\ce{=C(CH3)(CH2CH3)}$ end (a carbon with no H). Joining the two carbons: $\ce{CH2=C(CH3)CH2CH3}$ , 2-methylbut-1-ene. Carbon check: $1 + 4 = 5$ .

With HBr, H adds to the $\ce{CH2}$ end, giving the tertiary carbocation $\ce{CH3C+(CH3)CH2CH3}$ , then 2-bromo-2-methylbutane. Consistent.

B is 2-methylbut-2-ene, $\ce{(CH3)2C=CHCH3}$ : HBr adds H to the CH carbon, again via a tertiary carbocation, giving the same product. But hot $\ce{KMnO4}$ splits it into a $\ce{=C(CH3)2}$ end (giving propanone) and a $\ce{=CHCH3}$ end (giving ethanoic acid):

\ce{(CH3)2C=CHCH3 + 3[O] -> (CH3)2CO + CH3COOH}

Exam-hard: all the products of dehydration

Butan-2-ol is heated with concentrated sulfuric acid. (a) Give the structures and names of all the alkenes formed. (b) Explain why there is more than one product. (c) Which product would give only ethanoic acid when heated with acidified potassium manganate(VII)?

Solution

(a) Removing OH from C2 and H from C1 gives but-1-ene, $\ce{CH2=CHCH2CH3}$ . Removing H from C3 gives but-2-ene, $\ce{CH3CH=CHCH3}$ , which exists as cis- and trans-but-2-ene. Three alkenes.

(b) The OH is on C2, which has two neighbouring carbons (C1 and C3) each carrying hydrogen atoms, so the H can be eliminated from either side. But-2-ene has two different groups on each carbon of the C=C, so it also forms two geometrical isomers.

(c) But-2-ene (either isomer): $\ce{CH3CH=CHCH3 + 4[O] -> 2CH3COOH}$ . But-1-ene would give propanoic acid and $\ce{CO2}$ .

Watch out

Hydrogen in the wrong place. The H of HX adds to the carbon with more hydrogens. Saying "the Br goes to the carbon with more H" reverses Markovnikov's rule.
Curly arrow from the wrong place in step 1. The first arrow starts from the C=C bond (the $\pi$ electrons), not from a carbon atom, and ends on the $\delta+$ atom of the electrophile.
Forgetting the dipole on $\ce{Br2}$ . Show $\delta+$ and $\delta-$ on the Br–Br bond, and explain it as induced by the C=C.
Lone pair on $\ce{Br-}$ . The final arrow must start at a lone pair on the bromide ion, and the ion must carry a negative charge.
Cold vs hot $\ce{KMnO4}$ . Cold dilute gives a diol (no C–C bond broken). Hot concentrated splits the molecule. Do not mix them up.
Bromine water and alkanes. Alkanes only decolourise bromine in UV light, slowly, by substitution. The bromine water test for C=C is done at room temperature without UV.

Exam tip

The full electrophilic addition mechanism usually earns 4 to 5 marks: dipole on the electrophile, arrow from C=C to $\delta+$ atom, arrow from the bond to the $\delta-$ atom, correct carbocation with + charge, arrow from lone pair on $\ce{X-}$ to $\ce{C+}$ (and the product).
"Explain why 2-bromopropane is the major product" needs: secondary carbocation is more stable than primary, because of the electron-donating (positive inductive) effect of the alkyl groups.
Reagents and conditions are often a separate mark: steam needs $\ce{H3PO4}$ catalyst; hydrogenation needs Ni (or Pt); cold dilute vs hot concentrated acidified $\ce{KMnO4}$ .
Observation marks: bromine water "orange (or yellow/brown) to colourless"; $\ce{KMnO4}$ "purple to colourless". "Clear" is not a colour.

Summary

Alkenes $\ce{C_{n}H_{2n}}$ : sp² carbons, planar around the C=C, a $\sigma$ and a $\pi$ bond. The $\pi$ electrons make the C=C electron-rich and reactive towards electrophiles.
Made by elimination of HX (NaOH in ethanol, heat), dehydration of alcohols (heated $\ce{Al2O3}$ or conc. $\ce{H2SO4}$ ), and cracking.
Additions: $\ce{H2}$ (Ni, heat) → alkane; steam ( $\ce{H3PO4}$ ) → alcohol; HX (room temp.) → halogenoalkane; $\ce{X2}$ (room temp.) → dihalogenoalkane; cold dilute acidified $\ce{KMnO4}$ → diol.
Hot concentrated acidified $\ce{KMnO4}$ splits the C=C: $\ce{=CH2}$ → $\ce{CO2}$ ; $\ce{=CHR}$ → RCOOH; $\ce{=CRR'}$ → ketone.
Bromine water: orange-brown to colourless with C=C.
Electrophilic addition: C=C attacks the $\delta+$ end of the electrophile, carbocation intermediate, then attack by $\ce{X-}$ .
Markovnikov: H adds to the carbon with more H, via the more stable carbocation (tertiary > secondary > primary, due to the inductive effect of alkyl groups).

Practice

Question

Explain why alkenes undergo addition reactions with electrophiles but alkanes do not.
Give the reagents and conditions for converting propene into (a) propane, (b) propan-2-ol, (c) propane-1,2-diol.
Write an equation for the reaction of but-1-ene with bromine, and name the product.
Describe, with numbered steps naming each curly arrow, the mechanism for the reaction of ethene with hydrogen bromide.
Explain why but-1-ene reacts with hydrogen chloride to give mainly 2-chlorobutane rather than 1-chlorobutane.
Give the organic products when each alkene is heated with concentrated acidified potassium manganate(VII): (a) pent-2-ene, (b) 2-methylbut-1-ene, (c) cyclohexene.
An alkene $\ce{C6H12}$ gives propanone and propanoic acid with hot concentrated acidified $\ce{KMnO4}$ . Identify the alkene and name it.
2-bromobutane is heated with sodium hydroxide in ethanol. Name all the organic products, and explain why there is more than one.
$0.140\ \text{g}$ of an alkene reacts exactly with $25.0\ \text{cm}^3$ of a $0.100\ \text{mol dm}^{-3}$ solution of bromine. (a) Calculate the $M_r$ of the alkene and its molecular formula. (b) The alkene shows cis/trans isomerism. Identify it and give the product of its reaction with HBr. (c) Explain why this alkene gives only one structural isomer with HBr, whereas its isomer 2-methylpropene could in principle give two.
Limonene is a natural product with the molecular formula $\ce{C10H16}$ and two C=C bonds in a molecule that also contains one ring. (a) Calculate the volume of hydrogen at room temperature and pressure that reacts with $6.80\ \text{g}$ of limonene in the presence of a nickel catalyst. (b) What mass of bromine would react with the same mass of limonene? (c) Explain why the electrophilic addition mechanism goes through a carbocation, and why the carbocation formed at a C=C carbon carrying two alkyl groups is preferred. ( $A_r$ : H 1.0, C 12.0, Br 79.9; molar gas volume $24.0\ \text{dm}^3\ \text{mol}^{-1}$ )

Answers

The C=C bond contains a $\pi$ bond, a region of high electron density above and below the plane of the molecule, which attracts electrophiles; the $\pi$ bond is also relatively weak and easily broken. Alkanes have only strong, almost non-polar $\sigma$ bonds and no electron-rich region, so electrophiles are not attracted.
(a) $\ce{H2}$ , nickel catalyst, heat (about $150\ ^\circ\text{C}$ ). (b) Steam, phosphoric acid catalyst, about $300\ ^\circ\text{C}$ and $6\ \text{MPa}$ (gives mainly propan-2-ol). (c) Cold, dilute, acidified potassium manganate(VII).
$\ce{CH2=CHCH2CH3 + Br2 -> CH2BrCHBrCH2CH3}$ : 1,2-dibromobutane.
(1) H–Br is polar, $\ce{H^{\delta+}-Br^{\delta-}}$ . (2) Curly arrow from the C=C bond to the $\delta+$ H; a C–H bond forms. (3) Curly arrow from the H–Br bond to the Br; heterolytic fission forms $\ce{Br-}$ . (4) Carbocation intermediate $\ce{CH3CH2+}$ . (5) Curly arrow from a lone pair on $\ce{Br-}$ to the positive carbon; product bromoethane, $\ce{CH3CH2Br}$ .
If H adds to C1 (the $\ce{CH2}$ end, which has more H), the carbocation is $\ce{CH3CH2CH+CH3}$ , which is secondary; if H adds to C2, the carbocation is $\ce{CH3CH2CH2CH2+}$ , primary. The secondary carbocation is more stable, because two alkyl groups donate electron density to the positive carbon (positive inductive effect) rather than one. It forms faster, so $\ce{Cl-}$ attacks C2 and 2-chlorobutane is the major product.
(a) $\ce{CH3CH=CHCH2CH3}$ : both ends are $\ce{=CHR}$ , so ethanoic acid, $\ce{CH3COOH}$ , and propanoic acid, $\ce{CH3CH2COOH}$ . (b) $\ce{CH2=C(CH3)CH2CH3}$ : $\ce{CO2}$ (and water) and butanone, $\ce{CH3COCH2CH3}$ . (c) Both ends of the C=C are $\ce{=CHR}$ and are joined by the ring, so the ring opens to give one molecule with a COOH at each end: hexanedioic acid, $\ce{HOOC(CH2)4COOH}$ .
Propanone, $\ce{(CH3)2C=O}$ , came from a $\ce{=C(CH3)2}$ end; propanoic acid, $\ce{CH3CH2COOH}$ , from a $\ce{=CHCH2CH3}$ end. Joining: $\ce{(CH3)2C=CHCH2CH3}$ , 2-methylpent-2-ene. Carbon check: $3 + 3 = 6$ .
But-1-ene, cis-but-2-ene and trans-but-2-ene. The Br is on C2; the H that is eliminated with it can come from C1 (giving but-1-ene) or from C3 (giving but-2-ene), and but-2-ene has two different groups on each C=C carbon, so it forms cis and trans isomers.
(a) $n(\ce{Br2}) = 0.100 \times 25.0 / 1000 = 2.50 \times 10^{-3}\ \text{mol}$ . One C=C reacts with one $\ce{Br2}$ , so $n(\text{alkene}) = 2.50 \times 10^{-3}\ \text{mol}$ and $M_r = 0.140 / 2.50 \times 10^{-3} = 56.0$ . $\ce{C_{n}H_{2n}}$ : $14n = 56$ , $n = 4$ : $\ce{C4H8}$ . (b) But-2-ene, $\ce{CH3CH=CHCH3}$ ; with HBr it gives 2-bromobutane, $\ce{CH3CHBrCH2CH3}$ . (c) But-2-ene is symmetrical: whichever carbon the H adds to, the carbocation is the same (secondary, on the other carbon) and the product is the same. 2-methylpropene is unsymmetrical, so H could add to either carbon, giving a tertiary or a primary carbocation and two different products (2-bromo-2-methylpropane major, 1-bromo-2-methylpropane minor).
(a) $M_r(\ce{C10H16}) = 136.0$ ; $n = 6.80 / 136.0 = 0.0500\ \text{mol}$ . Two C=C per molecule, so $n(\ce{H2}) = 0.100\ \text{mol}$ , volume $= 0.100 \times 24.0 = 2.40\ \text{dm}^3$ . (The ring is not affected.) (b) $n(\ce{Br2}) = 0.100\ \text{mol}$ ; $M_r(\ce{Br2}) = 159.8$ ; mass $= 16.0\ \text{g}$ . (c) In the first step, the $\pi$ electrons form a bond to the electrophile, so the other carbon of the former double bond loses its share of the $\pi$ electrons and is left with only three bonds and a positive charge: a carbocation. A carbocation in which the positive carbon carries more alkyl groups is more stable, because alkyl groups are electron-donating and reduce the positive charge density; so the electrophile bonds to the carbon that leaves the positive charge on the more substituted carbon.

Structure and reactivity

Making alkenes

Addition reactions

Testing for a C=C bond with bromine water

Oxidation by potassium manganate(VII)

Cold, dilute, acidified: making a diol

Hot, concentrated, acidified: splitting the C=C bond

The electrophilic addition mechanism

Ethene and bromine

Propene and hydrogen bromide

Markovnikov's rule: which product is major?

Worked examples

Practice

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