Alkenes

1. Overview

Alkenes are hydrocarbons with a C=C double bond, and that bond makes them far more reactive than alkanes. This section covers how alkenes are made, the addition reactions that turn them into alkanes, alcohols, halogenoalkanes and polymers, and their oxidation by acidified potassium manganate(VII), which can show where a double bond sits in a molecule. It also explains the electrophilic addition mechanism and why an unsymmetrical alkene gives one main product. Alkenes link alkanes, alcohols and halogenoalkanes, so they turn up in many organic synthesis questions.

Key Definitions

• Alkene: a hydrocarbon whose molecules contain a carbon-carbon double bond, C=C; an alkene with one C=C and no ring has the general formula CₙH₂ₙ.
• Electrophile: a species that accepts a pair of electrons to form a new covalent bond; it is attracted to regions of high electron density, such as a C=C bond.
• Electrophilic addition: a reaction in which an electrophile attacks the electron-rich C=C bond, so that two atoms or groups add across the double bond, which becomes a single bond, and one product forms.
• Carbocation: an ion in which a carbon atom carries a positive charge and forms only three covalent bonds; it is primary, secondary or tertiary when that carbon is bonded to one, two or three alkyl groups.
• Positive inductive effect: the pushing of electron density along sigma bonds by alkyl groups towards a neighbouring atom; on a carbocation it spreads out the positive charge and makes the ion more stable.
• Markovnikov addition: when HX or H₂O adds to an unsymmetrical alkene, the major product is the one formed through the more stable carbocation, so the H atom usually joins the C=C carbon that already has more hydrogen atoms.
• Elimination reaction: a reaction in which a small molecule such as HX or H₂O is removed from a larger molecule, leaving a double bond (in this section, C=C); removing water from an alcohol is called dehydration.
• Cracking: breaking longer-chain alkanes into shorter molecules, including alkenes and a shorter alkane or hydrogen, by heating them, usually with a catalyst such as Al₂O₃.
• Addition polymerisation: the joining together of many alkene monomer molecules, each C=C opening into a single bond, to form one long saturated chain with no other product.

Content

The C=C bond: why alkenes react

A C=C double bond is one σ bond plus one π bond (topic 13.3). The π electrons lie above and below the plane of the molecule, further from the carbon nuclei than the σ electrons, so they are held less tightly. The C=C bond is therefore a region of high electron density that attracts electrophiles (topic 13.2).

The π bond is also weaker than the σ bond (the data booklet gives C=C as 610 kJ mol⁻¹, less than twice C-C, 350 kJ mol⁻¹). So alkenes take part in addition reactions: the π bond breaks, an atom or group joins each of the two carbons, and a single product forms.

How alkenes are made

Method	Start from	Reagents and conditions
Elimination of HX	halogenoalkane	NaOH dissolved in ethanol, heat
Dehydration	alcohol	heated Al₂O₃ catalyst, or heat with concentrated H₂SO₄
Cracking	longer-chain alkane	heat, with an Al₂O₃ catalyst

Elimination. Overall, HX is removed: the halogen atom and an H atom from the next carbon. The NaOH turns it into NaX and water:

CH₃CH₂CH₂CH₂Br + NaOH → CH₃CH₂CH=CH₂ + NaBr + H₂O (NaOH in ethanol, heat)

The solvent matters. With aqueous NaOH, the halogenoalkane undergoes substitution and gives an alcohol instead (topic 15.1).

Dehydration. The OH group and an H atom from the next carbon leave as water. Concentrated H₃PO₄ can replace the H₂SO₄:

CH₃CH₂OH(g) → CH₂=CH₂(g) + H₂O(g) (heated Al₂O₃)

Cracking. Heating a long alkane with a catalyst breaks C-C bonds in many different places, so a mixture forms. For example:

C₉H₂₀(g) → C₆H₁₄(g) + C₃H₆(g) (nonane → hexane + propene)

Alkenes are CₙH₂ₙ, so they cannot carry all the hydrogen of a CₙH₂ₙ₊₂ alkane. At least one product must be an alkane or hydrogen. Topic 14.1 covers why cracking is useful.

More than one alkene. In elimination and dehydration, the H atom can come from any carbon next to the C-X or C-OH carbon, provided that carbon has an H atom. For heptan-3-ol, CH₃CH₂CH(OH)CH₂CH₂CH₂CH₃:

• losing H from carbon 2 gives hept-2-ene, CH₃CH=CHCH₂CH₂CH₂CH₃
• losing H from carbon 4 gives hept-3-ene, CH₃CH₂CH=CHCH₂CH₂CH₃

Both have cis and trans isomers (topic 13.4), so four alkenes can form.

Addition reactions

Reagent and conditions	Product type
H₂(g), Ni or Pt catalyst, heat	alkane
H₂O(g), H₃PO₄ catalyst	alcohol
HX(g), room temperature	halogenoalkane
X₂, room temperature	1,2-dihalogenoalkane

With ethene:

CH₂=CH₂(g) + H₂(g) → CH₃CH₃(g)

CH₂=CH₂(g) + H₂O(g) → CH₃CH₂OH(g)

CH₂=CH₂(g) + HBr(g) → CH₃CH₂Br(l)

CH₂=CH₂(g) + Br₂(l) → CH₂BrCH₂Br(l)

• Hydrogenation turns the C=C into C-C. It is how alkanes are made from alkenes (topic 14.1).
• Steam must be written as H₂O(g), not just "water". Industrially, high temperature and pressure are used.
• Halogens (Cl₂, Br₂) add at room temperature with no UV light and no catalyst, unlike their reactions with alkanes, which need UV light. One halogen atom ends up on each of the two former C=C carbons.

All four are addition reactions. The mechanism you must describe (below) uses Br₂ with ethene and HBr with propene; steam adds in a similar way, with H⁺ from the acid catalyst as the electrophile. For hydrogenation, learn the reagents, conditions and product only. An unsymmetrical alkene such as propene can add HX or H₂O two ways round: see Markovnikov addition below.

Oxidation by acidified potassium manganate(VII)

In both reactions the purple manganate(VII) is reduced to Mn²⁺, so the solution turns from purple to colourless. The conditions decide how far the oxidation goes. In equations, [O] stands for one oxygen atom supplied by the oxidising agent.

Cold, dilute, acidified KMnO₄ adds an OH group to each carbon of the C=C, giving a diol. The carbon chain stays whole:

CH₂=CH₂ + [O] + H₂O → HOCH₂CH₂OH (ethane-1,2-diol)

Hot, concentrated, acidified KMnO₄ breaks the C=C completely. Each former C=C carbon becomes a C=O carbon, and what you get depends on what was attached to it:

End of the C=C	Product from that end
=CH₂	CO₂ and H₂O
=CHR (one H)	carboxylic acid, RCOOH
=CRR′ (no H)	ketone, RCOR′

An end with one H first forms an aldehyde, which the hot concentrated oxidant oxidises on to a carboxylic acid. A =CH₂ end goes on through methanoic acid to CO₂, so you may see effervescence. A ketone has no H on its C=O carbon, so it is not oxidised further. For example:

CH₃CH₂CH₂CH=CH₂ + 5[O] → CH₃CH₂CH₂COOH + CO₂ + H₂O

Locating a double bond. Work backwards. For a hydrocarbon, every C=O carbon in the products was one end of a C=C, so join those carbons in pairs with a double bond. Three points help:

• An open-chain molecule with two C=C bonds splits into three pieces; the middle piece has a C=O group at each end.
• A ring with one C=C gives one molecule, not two: 1-methylcyclopentene gives CH₃CO(CH₂)₃COOH.
• Cis and trans isomers give the same products, so this test cannot tell them apart.

Worked example 1: finding the position of a double bond

Question: Alkene Q, C₇H₁₄, is heated with hot, concentrated, acidified KMnO₄. The only organic products are butanone, CH₃COCH₂CH₃, and propanoic acid, CH₃CH₂COOH. Deduce the structure and name of Q, and give its product with cold, dilute, acidified KMnO₄.

Step 1: read each product back to an end of the C=C. Butanone is a ketone, so that carbon had two alkyl groups and no H: =C(CH₃)CH₂CH₃. Propanoic acid is a carboxylic acid, so that carbon had one H: =CHCH₂CH₃.

Step 2: join the two ends. CH₃CH₂C(CH₃)=CHCH₂CH₃

Step 3: check. Carbons: 4 + 3 = 7, and the formula is C₇H₁₄. The equation balances:

CH₃CH₂C(CH₃)=CHCH₂CH₃ + 3[O] → CH₃COCH₂CH₃ + CH₃CH₂COOH

Step 4: name it. The longest chain containing the C=C has six carbons. The double bond starts at carbon 3 from either end, so number from the end that gives the methyl group the lower number, 3: 3-methylhex-3-ene.

Step 5: cold, dilute KMnO₄. An OH group adds to each C=C carbon, and the chain stays whole: CH₃CH₂C(OH)(CH₃)CH(OH)CH₂CH₃.

Answer: Q is 3-methylhex-3-ene, CH₃CH₂C(CH₃)=CHCH₂CH₃ (cis or trans; the products cannot tell which). With cold, dilute, acidified KMnO₄ it gives the diol 3-methylhexane-3,4-diol.

Addition polymerisation

Under suitable conditions, many alkene molecules (monomers) join into one long chain (a polymer). Each C=C opens into a C-C single bond, and the monomers link end to end:

n CH₂=CH₂ → -[CH₂-CH₂]ₙ- (poly(ethene))

n CH₂=CHCH₃ → -[CH₂-CH(CH₃)]ₙ- (poly(propene))

• The repeat unit is the part in brackets: the two carbons that formed the C=C, with their groups.
• In poly(propene), CH₃ is a side group on every second carbon. The main chain contains only the former C=C carbons.
• The polymer is the only product, and it is saturated.

Topic 20.1 covers repeat units from other monomers, finding the monomer from a polymer section, and disposal problems.

Testing for a C=C bond with aqueous bromine

Add a few drops of aqueous bromine (yellow to orange) to the sample and shake at room temperature.

Sample	Observation
contains C=C (alkene)	orange colour disappears quickly: the mixture turns colourless
no C=C (alkane)	no change: stays orange (without UV light)

The bromine adds across the C=C, so it is used up and its colour goes. An alkane reacts with bromine only by free-radical substitution, which needs UV light (topic 14.1).

For the product of an alkene with bromine, give the 1,2-dibromo compound, as in the addition equations above (CH₂BrCH₂Br from ethene). In aqueous bromine, water molecules are in large excess and can also bond to the carbocation in step 2 of the mechanism (below), so a bromo-alcohol such as CH₂BrCH₂OH forms as well. The colour change is the same.

Some other functional groups met later, such as phenols at A Level, also decolourise aqueous bromine, so a positive result is strong evidence for C=C rather than proof.

One Br₂ molecule adds to each C=C, so the amount of bromine used can count the C=C bonds in a molecule.

Worked example 2: counting and locating C=C bonds

Question: A sample of hydrocarbon Z, C₁₀H₁₈ ($M_{\text{r}}$ = 138.0), of mass 0.414 g, reacts with exactly 40.0 cm³ of a 0.150 mol dm⁻³ solution of bromine in an organic solvent: any extra bromine leaves the mixture coloured. With hot, concentrated, acidified KMnO₄, Z gives only propanoic acid and butanedioic acid, HOOCCH₂CH₂COOH, in a 2 : 1 mole ratio. Deduce the structure of Z.

Step 1: moles of Z. $$n(\text{Z}) = \frac{0.414}{138.0} = 0.00300\ \text{mol}$$

Step 2: moles of bromine. $$n(\text{Br}_2) = 0.150 \times \frac{40.0}{1000} = 0.00600\ \text{mol}$$

Step 3: C=C bonds per molecule. One Br₂ adds to each C=C: $$\frac{0.00600}{0.00300} = 2.00$$

Step 4: rings? An open-chain alkane with 10 carbons is C₁₀H₂₂. Each C=C (or ring) means 2 fewer H atoms. Z has 4 fewer, which the two C=C bonds account for, so Z has no ring.

Step 5: the pieces. Each propanoic acid comes from a CH₃CH₂CH= end. Butanedioic acid has a COOH group at each end, so it is the middle piece, =CHCH₂CH₂CH=. Join them:

CH₃CH₂CH=CHCH₂CH₂CH=CHCH₂CH₃

Check: 3 + 4 + 3 = 10 carbons, formula C₁₀H₁₈.

Answer: Z has two C=C bonds per molecule and no ring. It is deca-3,7-diene, CH₃CH₂CH=CHCH₂CH₂CH=CHCH₂CH₃. One molecule of Z gives two molecules of propanoic acid and one of butanedioic acid.

The mechanism of electrophilic addition

A curly arrow shows a pair of electrons moving. It starts at a bond or a lone pair (topic 13.2) and ends at the atom that gains a bond or a lone pair.

Bromine and ethene. Br₂ is non-polar, but as it approaches the C=C, the π electrons repel the electrons in the Br-Br bond. This induces a dipole: the nearer Br atom becomes δ+ and the further one δ−.

Step 1. The π electrons attack the δ+ Br atom. The Br-Br bond breaks heterolytically: both of its electrons go to the δ− Br, which leaves as Br⁻. This gives a carbocation, CH₂BrCH₂⁺. The positive charge is on the carbon that did not gain the Br atom.

Step 2. A lone pair on Br⁻ forms a bond to the positive carbon, giving 1,2-dibromoethane, CH₂BrCH₂Br.

Arrow	Starts at	Ends at
1 (step 1)	middle of the C=C bond	the δ+ Br atom
2 (step 1)	middle of the Br-Br bond	the δ− Br atom
3 (step 2)	a lone pair on Br⁻	the C⁺ atom

Hydrogen bromide and propene. H-Br already has a permanent dipole, Hδ+ Brδ−, because bromine (electronegativity 2.6) is more electronegative than hydrogen (2.1). The three arrows follow the table above, with the δ+ H atom in place of the δ+ Br atom: arrow 1 goes from the C=C bond to the H.

If the H atom bonds to the end CH₂ carbon, the intermediate is CH₃C⁺HCH₃ and the product is 2-bromopropane, CH₃CHBrCH₃. If it bonds to the middle carbon, the intermediate is CH₃CH₂CH₂⁺ and the product is 1-bromopropane. The next subsection explains why the first route wins.

Carbocation stability and Markovnikov addition

Carbocations are classified by how many alkyl groups are bonded to the positive carbon:

Type	Alkyl groups on C⁺	Example
primary	1	CH₃CH₂CH₂⁺
secondary	2	CH₃C⁺HCH₃
tertiary	3	(CH₃)₃C⁺

Alkyl groups have a positive inductive effect: they push electron density along the σ bonds towards the C⁺. This reduces the positive charge on the C⁺ and spreads it over more atoms, which makes the ion more stable. The more alkyl groups on C⁺, the larger the effect, so the order of stability is:

tertiary > secondary > primary

A more stable carbocation forms more easily: the step that makes it has a lower activation energy, so it is faster. Most molecules react by that route, and its product is the major product. The other product still forms, in smaller amounts.

For propene and HBr, the secondary carbocation CH₃C⁺HCH₃ (two alkyl groups) is more stable than the primary CH₃CH₂CH₂⁺ (one alkyl group). So 2-bromopropane is the major product and 1-bromopropane the minor one. Steam behaves the same way: propene gives mainly propan-2-ol, CH₃CH(OH)CH₃.

This pattern is Markovnikov addition: the H atom usually joins the C=C carbon that already has more H atoms, and the Br or OH goes to the carbon with more alkyl groups. The rule describes the result; the carbocation stability explains it.

Two limits:

• A symmetrical alkene, such as ethene or but-2-ene, gives only one structural product with HX or steam.
• If both possible carbocations are of the same type, as with pent-2-ene (both secondary), expect similar amounts of the two products.

Worked example 3: predicting and explaining the major product

Question: 2-Methylpent-1-ene, CH₂=C(CH₃)CH₂CH₂CH₃, reacts with HBr(g) at room temperature. Two structurally different products form. Name the major product and explain why it is the major one.

Step 1: draw both carbocations. The H atom can join either carbon of the C=C.

• H to carbon 1 (the CH₂ end): the positive charge is on carbon 2, giving (CH₃)₂C⁺CH₂CH₂CH₃. Carbon 2 is bonded to three alkyl groups (two CH₃ and one CH₂CH₂CH₃), so this is a tertiary carbocation.
• H to carbon 2: the positive charge is on carbon 1, giving ⁺CH₂CH(CH₃)CH₂CH₂CH₃. Carbon 1 is bonded to one alkyl group, so this is a primary carbocation.

Step 2: compare their stability. Three alkyl groups push electron density towards the tertiary C⁺ (positive inductive effect), against one alkyl group on the primary C⁺. The positive charge is spread more in the tertiary ion, so it is more stable.

Step 3: link stability to the product. The more stable tertiary carbocation forms more easily (lower activation energy), so most molecules react this way. Br⁻ then bonds to that C⁺.

Answer: the major product is 2-bromo-2-methylpentane, (CH₃)₂CBrCH₂CH₂CH₃, formed through the more stable tertiary carbocation. The minor product is 1-bromo-2-methylpentane, BrCH₂CH(CH₃)CH₂CH₂CH₃, formed through the primary carbocation. The explanation rests on the stability of the intermediates, not of the products.

Key Equations

General formula of an open-chain alkene with one C=C: CₙH₂ₙ. An open-chain alkane is CₙH₂ₙ₊₂; each C=C or ring means 2 fewer H atoms.

Addition across C=C (A-B = H-H, H-OH, H-X or X-X, where X is a halogen; conditions in the addition table):

R₂C=CR₂ + A-B → R₂CA-CR₂B

Cold, dilute, acidified KMnO₄ ([O] = one oxygen atom from the oxidising agent; R = an alkyl group or H):

R₂C=CR₂ + [O] + H₂O → R₂C(OH)C(OH)R₂

Hot, concentrated, acidified KMnO₄: =CH₂ → CO₂ + H₂O; =CHR → RCOOH; =CRR′ → RCOR′.

Counting C=C bonds with bromine ($m$ = mass in g; $M$ = molar mass in g mol⁻¹; $c$ = concentration in mol dm⁻³; $V$ = volume in cm³): $$n = \frac{m}{M} \qquad n(\text{Br}_2) = c \times \frac{V}{1000}$$ Number of C=C bonds per molecule: $$\frac{n(\text{Br}_2)}{n(\text{compound})}$$

Cracking: CₙH₂ₙ₊₂ → shorter alkane (or H₂) + one or more alkenes.

Common Mistakes to Avoid

• Mistake: writing aqueous NaOH, or "ethanoic", for the elimination. Fix: NaOH dissolved in ethanol, with heat. Heated with aqueous NaOH, 1-bromobutane gives butan-1-ol, not but-1-ene.
• Mistake: dilute acid for dehydration, or no conditions. Fix: concentrated H₂SO₄ and heat, or alcohol vapour over heated Al₂O₃. If asked for the role of the concentrated acid, call it a dehydrating agent: it removes H₂O from the alcohol.
• Mistake: mixing up the two KMnO₄ reactions. Fix: cold and dilute keeps the chain and adds two OH groups; hot and concentrated breaks it. Pent-1-ene gives pentane-1,2-diol with cold, dilute KMnO₄, but butanoic acid and CO₂ with hot, concentrated KMnO₄.
• Mistake: counting one of each product when a molecule has more than one C=C. Fix: split the molecule at every C=C. In Worked example 2, one molecule of Z gives two molecules of propanoic acid, not one.
• Mistake: full charges on H-Br, a first arrow drawn from the HBr towards the alkene, or curly arrows starting from an atom or a charge sign. Fix: write Hδ+ Brδ−. Arrow 1 goes from the middle of the C=C bond to the H, arrow 2 from the middle of the H-Br bond to the Br, and arrow 3 from a lone pair on Br⁻ to the C⁺.
• Mistake: explaining the major product by "Markovnikov's rule" alone, or by saying the product is more stable. Fix: compare the two carbocations. For 2-methylpent-1-ene, the tertiary carbocation (three alkyl groups, larger positive inductive effect) is more stable than the primary one, so 2-bromo-2-methylpentane is the major product.
• Mistake: putting the CH₃ group of propene into the polymer's main chain, as -CH₂-CH₂-CH₂-. Fix: only the two C=C carbons join the chain. The repeat unit of poly(propene) is -CH₂-CH(CH₃)-.
• Mistake: writing "turns clear" for the bromine test. Fix: clear is not a colour, and decolourise means "becomes colourless". Write "orange to colourless" for an alkene and "stays orange" for an alkane.

Exam Tips

• State the reagents and conditions: give both, with states and concentrations where they matter, for example "H₂O(g) with an H₃PO₄ catalyst" or "cold, dilute, acidified KMnO₄".
• Describe the mechanism: name it (electrophilic addition). Show the C=C with all four groups, the δ+ and δ− dipole, three curly arrows from the correct starting points, a full + charge on the right carbon of the carbocation, and Br⁻ with its charge and a lone pair.
• Explain a major product in terms of the actual molecules in the question: draw both carbocations, classify them as primary, secondary or tertiary, and link the number of alkyl groups to the inductive effect, stability and rate of formation.
• Describe observations as a colour change from one colour to another, for example "purple to colourless" for KMnO₄, and add "effervescence" when CO₂ forms.
• Suggest all the alkenes an elimination or dehydration can give: check every neighbouring carbon that has an H atom, then look for cis/trans isomers.