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How do alkanes, alkenes and haloalkanes react, and by what mechanisms?

Alkanes as fuels and free-radical substitution, the reactions of alkenes including electrophilic addition and Markownikoff's rule, addition polymerisation, and the nucleophilic substitution and elimination reactions of haloalkanes.

A CCEA A-Level Chemistry answer on alkanes, alkenes and haloalkanes, covering alkanes as fuels and free-radical substitution, electrophilic addition to alkenes and Markownikoff's rule, addition polymerisation, and the nucleophilic substitution and elimination reactions of haloalkanes.

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  1. What this dot point is asking
  2. Alkanes
  3. Alkenes
  4. Haloalkanes
  5. Examples in context
  6. Try this

What this dot point is asking

CCEA wants you to describe alkanes as fuels and their free-radical substitution, the electrophilic addition reactions of alkenes with Markownikoff's rule, addition polymerisation, and the nucleophilic substitution and elimination reactions of haloalkanes, including the mechanisms.

Alkanes

Alkanes are saturated hydrocarbons with the general formula CnH2n+2\text{C}_n\text{H}_{2n+2}. They are used as fuels, burning in oxygen to release energy. With halogens in ultraviolet light they undergo free-radical substitution:

Taking the chlorination of methane as the model, the three stages are written as: initiation Cl2β†’2Clβ‹…\text{Cl}_2 \rightarrow 2\text{Cl}\cdot (homolytic fission, each atom keeps one electron); propagation Clβ‹…+CH4β†’HCl+CH3β‹…\text{Cl}\cdot + \text{CH}_4 \rightarrow \text{HCl} + \text{CH}_3\cdot then CH3β‹…+Cl2β†’CH3Cl+Clβ‹…\text{CH}_3\cdot + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}\cdot; and termination such as CH3β‹…+Clβ‹…β†’CH3Cl\text{CH}_3\cdot + \text{Cl}\cdot \rightarrow \text{CH}_3\text{Cl} or 2CH3β‹…β†’C2H62\text{CH}_3\cdot \rightarrow \text{C}_2\text{H}_6. Because a chlorine radical is regenerated in propagation, a single initiation event sets off a long chain. Further substitution can replace more than one hydrogen, giving CH2Cl2\text{CH}_2\text{Cl}_2, CHCl3\text{CHCl}_3 and CCl4\text{CCl}_4 as well, which is why a mixture of products always forms. Burning alkanes completely gives carbon dioxide and water, for example CH4+2O2β†’CO2+2H2O\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}, while incomplete combustion gives toxic carbon monoxide or soot.

Alkenes

Alkenes are unsaturated, with a reactive C=C\text{C}=\text{C} double bond and general formula CnH2n\text{C}_n\text{H}_{2n}. The high electron density of the double bond attracts electrophiles, so alkenes undergo electrophilic addition.

Alkenes also undergo addition polymerisation, where many monomers join through their double bonds to form a long-chain polymer such as poly(ethene): n CH2=CH2β†’βˆ’(CH2βˆ’CH2)βˆ’nn\,\text{CH}_2{=}\text{CH}_2 \rightarrow {-}(\text{CH}_2{-}\text{CH}_2){-}_n. The double bond opens up and the monomers link end to end, with no other product. The reaction of an alkene with bromine water is also the standard test for unsaturation: the orange bromine is decolourised.

Haloalkanes

Reactivity increases as the carbon-halogen bond gets weaker down the group, so iodoalkanes react fastest and fluoroalkanes slowest. It is the bond strength that controls the rate, not the bond polarity: although the Cβˆ’F\text{C} - \text{F} bond is the most polar, it is also the strongest and hardest to break, so fluoroalkanes are the least reactive. The choice of nucleophile decides the product: OHβˆ’\text{OH}^- in warm aqueous solution gives an alcohol, CNβˆ’\text{CN}^- in ethanol gives a nitrile (adding a carbon to the chain), and excess ammonia in a sealed tube gives an amine. Switching to hot ethanolic hydroxide and removing the water favours elimination instead, removing HX\text{HX} to form an alkene.

Examples in context

Example 1. CFCs and the ozone layer. Chlorofluorocarbons were once used as refrigerants because they are unreactive and non-toxic. High in the stratosphere, ultraviolet light breaks the weak Cβˆ’Cl\text{C} - \text{Cl} bond homolytically (an initiation step exactly like free-radical substitution), releasing chlorine radicals that catalyse the breakdown of ozone in a chain reaction. CCEA expects candidates to connect the radical chemistry of alkanes to this real environmental problem and to explain why the more inert Cβˆ’F\text{C} - \text{F} bonds are not the culprits.

Example 2. Margarine from vegetable oils. Unsaturated vegetable oils are hardened into margarine by addition of hydrogen across their C=C\text{C}{=}\text{C} double bonds using a nickel catalyst, the same electrophilic-style addition reaction (here catalytic hydrogenation) that adds bromine or water to a simple alkene. Controlling how many double bonds are hydrogenated sets the melting point and spreadability of the product, a direct industrial use of alkene addition chemistry.

Try this

Q1. Name the three stages of free-radical substitution. [3 marks]

  • Cue. Initiation, propagation and termination.

Q2. State the type of reaction when an alkene reacts with bromine. [1 mark]

  • Cue. Electrophilic addition.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA 20216 marksDescribe the mechanism for the reaction of bromine with ethene, naming the type of mechanism and showing the movement of electron pairs with curly arrows. State what you would observe.
Show worked answer β†’

Markers want the mechanism named, the curly arrows described correctly, the intermediate, and the observation.

The mechanism is electrophilic addition. As a bromine molecule approaches the electron-rich C=C\text{C}{=}\text{C} double bond, the double bond induces a dipole in the bromine, making the near bromine slightly positive.

A curly arrow goes from the Ο€\pi bond of the double bond to the slightly positive bromine atom, forming a new Cβˆ’Br\text{C} - \text{Br} bond. At the same time a curly arrow goes from the Brβˆ’Br\text{Br} - \text{Br} bond to the far bromine, which leaves as a bromide ion Brβˆ’\text{Br}^-. This produces a carbocation intermediate.

A second curly arrow then goes from a lone pair on the bromide ion to the positive carbon, forming the second Cβˆ’Br\text{C} - \text{Br} bond and giving 1,2-dibromoethane.

Observation: the orange or yellow-brown bromine is decolourised, which is the standard test for unsaturation.

Markers reward naming electrophilic addition, the induced dipole, the three curly arrows (or the two stages), the carbocation intermediate, and the decolourising of bromine.

CCEA 20194 marks1-bromobutane reacts with warm aqueous sodium hydroxide. Name the mechanism, name the organic product, and explain why 1-iodobutane reacts faster than 1-bromobutane under the same conditions.
Show worked answer β†’

A nucleophilic substitution question linking rate to bond strength.

The mechanism is nucleophilic substitution. The hydroxide ion is the nucleophile; it attacks the slightly positive carbon bonded to the halogen and displaces the halide ion. The organic product is butan-1-ol:

CH3CH2CH2CH2Br+OHβˆ’β†’CH3CH2CH2CH2OH+Brβˆ’\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} + \text{Br}^-.

1-iodobutane reacts faster because the Cβˆ’I\text{C} - \text{I} bond is weaker than the Cβˆ’Br\text{C} - \text{Br} bond (the iodine atom is larger, so the shared electrons are further from the nuclei). The weaker bond breaks more easily, so the substitution is faster. Bond strength, not bond polarity, controls the rate.

Markers reward naming nucleophilic substitution, the product butan-1-ol, and the explanation in terms of the weaker carbon-iodine bond.

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