CourseCourseQuestionQuestionDocumentDocument

14. Hydrocarbons

Syllabus
9701–2028–2029
Section
14
Level
AS

14.1 Alkanes

Syllabus
9701–2028–2029
Topic
14.1
Level
AS

Prepare alkanes by hydrogenating alkenes or cracking longer alkanes

Route Reactant and change Reagent / condition Example
hydrogenation H₂ adds across an alkene C=C to give an alkane H₂(g), Pt or Ni catalyst, heat CH₂=CH₂ + H₂ → CH₃CH₃
cracking a longer-chain alkane splits into smaller hydrocarbons, including an alkane product heat with Al₂O₃ C₁₀H₂₂ → C₈H₁₈ + C₂H₄

Hydrogenation preserves the carbon skeleton and removes the C=C by forming two C–H bonds. A proposed cracking equation is acceptable only when its carbon and hydrogen totals balance; cracking can give a mixture rather than one unique product pair.

A catalyst changes the reaction rate but is not consumed in the overall equation. Do not use the hydrogenation conditions for cracking or assume that both cracking products must be alkanes.

Alkanes combust and undergo UV-initiated halogen substitution

Oxygen supply Carbon-containing product(s) Other product
sufficient CO₂ H₂O
limited CO and/or C H₂O

For complete combustion, balance carbon atoms as CO₂, hydrogen atoms as H₂O, then oxygen. For ethane: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O. Incomplete combustion is not one fixed equation because the proportions of CO and carbon depend on the oxygen supply.

In ultraviolet light, Cl₂ or Br₂ reacts with ethane by free-radical substitution: C₂H₆ + Cl₂ → C₂H₅Cl + HCl. A C–H bond is replaced by C–Cl; further substitutions can occur, so excess halogen can produce a mixture.

UV light initiates the radical chain; it is not written as a reactant or catalyst. Halogen substitution replaces H and is not addition across a C=C bond.

Track radicals through initiation, propagation and termination

Initiation: ultraviolet light causes homolytic fission of the halogen bond, so each chlorine atom takes one bonding electron and becomes a radical.

ClX2→UV2 Cl ⋅ \ce{Cl2 ->[UV] 2Cl.}

Propagation: a chlorine radical removes H from ethane, then the ethyl radical removes Cl from another Cl₂ molecule. One radical is consumed and another is produced in each step, so the chain continues.

Cl ⋅ +CX2HX6→HCl+CX2HX5 ⋅ \ce{Cl. + C2H6 -> HCl + C2H5.}

CX2HX5 ⋅ +ClX2→CX2HX5Cl+Cl ⋅ \ce{C2H5. + Cl2 -> C2H5Cl + Cl.}

Termination occurs when two radicals collide and form a molecule: Cl• + Cl• → Cl₂, C₂H₅• + Cl• → C₂H₅Cl, or 2C₂H₅• → C₄H₁₀. No radical is regenerated in a termination step.

If electron movement is drawn, a single-headed (fish-hook) curly arrow represents one electron. A full curly arrow represents an electron pair and must not be substituted into a radical step.

Cracking converts heavy oil fractions into more useful smaller molecules

Cracking breaks C–C bonds in long-chain hydrocarbons to make shorter alkanes and alkenes with lower relative molecular masses.

Demand for petrol and alkene feedstocks can exceed the natural fraction distribution. Choose a cracking equation that balances carbon and hydrogen atoms and includes the required heat/catalyst conditions.

C₁₀H₂₂ can crack to C₈H₁₈ + C₂H₄. The alkane is a fuel-range product and the alkene is a useful petrochemical feedstock.

Cracking is not complete combustion and does not produce one fixed product from every heavy fraction.

Strong, almost non-polar C–H bonds explain alkane unreactivity

Breaking a C–H bond requires substantial energy because it is strong. At ordinary conditions, many possible reactions therefore have a high activation energy and are too slow to observe.

Carbon and hydrogen have similar electronegativities, so a C–H bond has relatively little polarity. An alkane has no strongly δ+ or δ− reaction centre for a polar nucleophile or electrophile to attack.

Alkanes are generally unreactive, not absolutely inert. Combustion supplies heat and oxygen, radical substitution uses UV light to generate radicals, and cracking uses high temperature with Al₂O₃; these routes overcome or bypass the ordinary kinetic barrier.

Do not explain unreactivity merely by saying “alkanes are non-polar”. Both the strength of C–H bonds and their relative lack of polarity are required.

Link engine pollutants to their consequences and catalytic removal

Pollutant from an internal-combustion engine Formation / consequence Catalytic-converter change
CO incomplete combustion; toxic because it reduces the blood's ability to transport O₂ oxidised to CO₂
nitrogen oxides, NOₓ N₂ and O₂ react at high engine temperatures; contribute to respiratory harm, acid rain and photochemical smog reduced to N₂
unburnt hydrocarbons fuel passes through without complete combustion; contributes to photochemical smog oxidised to CO₂ and H₂O

2 CO+2 NO→2 COX2+NX2\ce{2CO + 2NO -> 2CO2 + N2}

A catalytic converter accelerates simultaneous redox: CO and hydrocarbons are oxidised while nitrogen oxides are reduced. The catalyst is not used up and does not make the original fuel use pollution-free.

Catalytic removal changes pollutants after they form; it is different from preventing incomplete combustion or lowering the engine temperature at which NOₓ forms.

14.2 Alkenes

Syllabus
9701–2028–2029
Topic
14.2
Level
AS

Prepare alkenes by elimination, dehydration or cracking

Starting material Change Reagent and conditions Example
halogenoalkane eliminate HX NaOH in ethanol, heat CH₃CH₂Br → CH₂=CH₂
alcohol eliminate H₂O (dehydration) heated Al₂O₃ catalyst, or concentrated H₂SO₄ and heat CH₃CH₂OH → CH₂=CH₂ + H₂O
longer-chain alkane crack C–C bonds heat, with Al₂O₃ C₁₀H₂₂ → C₈H₁₈ + C₂H₄

The first two routes form C=C by removing atoms from neighbouring carbon atoms; cracking redistributes the atoms of a longer alkane into smaller hydrocarbons. In every route, check that the proposed equation conserves C, H and any halogen or oxygen atoms.

Solvent is part of the reagent choice: hot ethanolic NaOH favours elimination from a halogenoalkane, whereas aqueous NaOH is associated with nucleophilic substitution. Do not interchange these conditions.

The C=C bond supports addition, oxidation and polymerisation

Alkene reaction Reagent and conditions Product change
hydrogenation H₂(g), Pt or Ni catalyst, heat alkane
hydration H₂O(g), H₃PO₄ catalyst alcohol
hydrogen-halide addition HX(g), room temperature halogenoalkane
halogen addition X₂ 1,2-dihalogenoalkane
mild oxidation cold, dilute, acidified KMnO₄ two –OH groups replace C=C, forming a diol
oxidative cleavage hot, concentrated, acidified KMnO₄ C=C breaks and each alkene carbon becomes an oxidised product
Groups originally attached to one C of C=C Final fragment from that carbon
two carbon groups, no H ketone
one carbon group and one H carboxylic acid
two H (terminal =CH₂) CO₂ and H₂O

Apply the cleavage rule separately to the two C=C carbon atoms. The carbon skeletons retained in the products reveal which two atoms were joined by the original double bond; conversely, joining the carbonyl or carboxyl carbon atoms reconstructs the alkene position.

nCHX2=CHX2→[−CHX2−CHX2X−]Xnn\ce{CH2=CH2 -> [-CH2-CH2-]_{n}}

nCHX2=CHCHX3→[−CHX2−CH(CHX3)X−]Xnn\ce{CH2=CHCH3 -> [-CH2-CH(CH3)-]_{n}}

In addition polymerisation, many alkene monomers open their π bonds and form new C–C σ bonds. No small molecule is eliminated, and the propene CH₃ side group remains attached to every other backbone carbon in the repeat unit.

Aqueous bromine tests for a reactive C=C bond

Shake the sample with aqueous bromine. An alkene changes the reagent from orange to colourless at room temperature because Br₂ adds across C=C; this observable decolourisation is the positive result.

CHX2=CHX2+BrX2→CHX2BrCHX2Br\ce{CH2=CH2 + Br2 -> CH2BrCH2Br}

An alkane does not rapidly decolourise aqueous bromine under these conditions: alkane bromination requires ultraviolet light for free-radical substitution. Keep the reagent, conditions and observation together when interpreting the test.

A positive result shows that bromine has reacted under the test conditions; it does not count how many C=C bonds are present. Aqueous bromine is not a colourless bromide solution.

Electrophilic addition follows electron-pair movement from the π bond

The alkene π bond is an electron-pair donor. A full curly arrow therefore starts at the π bond and points to the electrophilic atom; a second arrow moves the reagent's bonding pair to the atom that leaves as an anion.

With ethene and Br₂, the nearby π electrons induce Brδ+–Brδ−. Draw π → Brδ+ and Br–Br → Brδ−, giving a positively charged carbon intermediate and Br⁻; then draw the Br⁻ lone pair to the positive carbon to form 1,2-dibromoethane.

With propene and HBr, Hδ+ is the electrophile. Draw π → H and H–Br → Br⁻. Proton attachment can place the positive charge on either alkene carbon; the favoured route forms the more stable secondary carbocation, which Br⁻ attacks to give mainly 2-bromopropane.

Every curly-arrow tail must begin at a bond or lone pair, and every head must show where that electron pair goes. Do not start an arrow at H⁺ or a positive carbon, and do not use radical fish-hook arrows for these mechanisms.

Alkyl-group induction favours the more stable carbocation pathway

Carbocation carbon bonded to Class Relative stability
three alkyl groups tertiary, 3° highest
two alkyl groups secondary, 2° intermediate
one alkyl group primary, 1° lowest

An alkyl group has a positive inductive effect: electron density in its σ bonds is displaced towards the electron-deficient, positively charged carbon. More attached alkyl groups spread and reduce the intensity of the positive charge, lowering the carbocation's energy.

When HX adds to an unsymmetrical alkene, compare the carbocations that could form after protonation. The lower-energy pathway is favoured, so H bonds in the direction that leaves the more substituted carbocation; X then bonds to that carbon. This gives the Markovnikov major product.

CHX3CH=CHX2+HBr→CHX3CHBrCHX3(major)\ce{CH3CH=CH2 + HBr -> CH3CHBrCH3}\quad\text{(major)}

The stability order predicts a favoured pathway and major product, not an absolutely exclusive product. The positive charge belongs to a reaction intermediate and is absent from the final halogenoalkane.