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A-Level Reactions of alkanes, alkenes and arenes
What the A-Level syllabus expects for Reactions of alkanes, alkenes and arenes, and how to practise it.
What the syllabus expects
- Describe alkane chemistry through the following reactions of ethane:
- combustion
- substitution via free radicals when chlorine or bromine is present under ultraviolet light at room temperature
Scope: see also 11.3(c) and 11.3(k) - Describe alkene chemistry through the following reactions of ethene where they apply:
- electrophilic addition of steam (H2O(g) with an H3PO4 catalyst), of hydrogen halides (HX(g)) and of halogens (X2(aq) or X2 in CCl4)
Scope: see also 11.3(d) and 11.3(l) - reduction by catalytic hydrogenation, H2(g) over a Ni catalyst
Scope: see also 8(j) - oxidation by cold alkaline manganate(VII) to give the diol
- oxidation by hot acidified manganate(VII), which severs the carbon-to-carbon double bond and so locates where alkene linkages lie within larger molecules
- Apply Markovnikov's rule when a hydrogen halide adds across an unsymmetrical alkene, and explain the product mix by the stability of the carbocation intermediates.
- Describe benzene-ring chemistry through the following reactions of benzene and methylbenzene:
Scope: also relevant to 11.3(m)(i), to 11.3(e) and to Section 4 - electrophilic substitution with chlorine over an AlCl3 catalyst and with bromine over an AlBr3 catalyst, both being Lewis-acid catalysts
- nitration by a blend of concentrated sulfuric acid and concentrated nitric acid, the mixture kept near 50 C for benzene but near 30 C for methylbenzene, where concentrated sulfuric acid serves as a Bronsted-Lowry acid catalyst
- Friedel-Crafts alkylation with halogenoalkanes over AlCl3 or AlBr3, again Lewis-acid catalysts
- Cover how the alkyl side-chain on a benzene ring reacts, illustrated by these reactions of methylbenzene:
- free-radical substitution driven by chlorine or by bromine while ultraviolet light shines at room temperature
- full oxidation to benzoic acid, either with hot alkaline KMnO4 then dilute acid, or with hot acidified KMnO4
- Predict, according to the conditions, whether an arene is halogenated on its side-chain or on the ring itself.
- Use knowledge of directing effects to place substituents in the electrophilic substitution of mono-substituted arenes.
- Recognise the environmental fallout from:
- carbon monoxide, nitrogen oxides and unburnt hydrocarbons put out by the internal combustion engine, and how they are stripped out catalytically
- gases that intensify the enhanced greenhouse effect
How it's examined
About 5% of the past-paper style questions in Rae's bank for this subject sit in this topic.
Worked examples
Example 1 (3 marks)
The preparation of G begins by reacting benzene with propene over an acid catalyst; this yields D as the main product along with the minor products E and F. Work out the structures of D, E and F using the information below: D: composition by mass C 89.94 %, H 10.06 %; its ¹³C NMR spectrum shows 6 signals. E: composition by mass C 88.82 %, H 11.18 %; its ¹³C NMR spectrum shows 4 signals. F: composition by mass C 88.82 %, H 11.18 %; its ¹³C NMR spectrum shows 5 signals. The underlying ideas of ¹³C NMR resemble those of ¹H NMR spectroscopy.
Show the worked answer
Benzene + propene over an acid catalyst is Friedel-Crafts alkylation, adding isopropyl group(s). D: C 89.94%, H 10.06% matches C9H12 (cumene: C 90.0%, H 10.0%). Isopropylbenzene has 6 distinct carbons (4 ring signals: ipso, ortho, meta, para; plus CH and equivalent CH3), i.e. 6 signals. So D = isopropylbenzene (cumene). E and F: C 88.82%, H 11.18% matches C12H18 (diisopropylbenzene: C 88.9%, H 11.1%), i.e. a second isopropyl added. Signal count fixes the substitution pattern: - 1,4- (para): highly symmetric, 2 aromatic + CH + CH3 = 4 signals -> E. - 1,2- (ortho): 3 aromatic + CH + CH3 = 5 signals -> F. (1,3-meta would give 6, and is not one of these.) So E = 1,4-diisopropylbenzene, F = 1,2-diisopropylbenzene.
Example 2 (2 marks)
(e) The ethyl radical produced in (d) reacts easily with chlorine gas to yield chloroethane, and 1,2-dichloroethane can also be formed. Starting from chlorine gas and the ethyl radical, set out the mechanistic steps that lead to 1,2-dichloroethane. [2] [Total: 20]
Show the worked answer
Propagation continues by radical chain steps. First the ethyl radical reacts with chlorine to give chloroethane and a chlorine radical: CH3CH2* + Cl2 -> CH3CH2Cl + Cl*. A chlorine radical then abstracts a hydrogen from chloroethane to give the 2-chloroethyl radical: Cl* + CH3CH2Cl -> *CH2CH2Cl + HCl. This radical reacts with chlorine to give 1,2-dichloroethane and regenerate a chlorine radical: *CH2CH2Cl + Cl2 -> ClCH2CH2Cl + Cl*.
Example 3 (2 marks)
Both acetophenone and benzene are nitrated by nitric acid, yet the two conversions demand unlike conditions to give S and T (acetophenone with conc HNO3/conc H2SO4 above 55 °C yields 3-nitroacetophenone, S; benzene with the same acids at 55 °C yields nitrobenzene, T). Account for why the conditions must differ.
Show the worked answer
Both undergo electrophilic substitution (nitration) by the NO2+ electrophile. In acetophenone the -COCH3 (acetyl) group is electron-withdrawing: it draws electron density out of the ring, deactivating it and lowering the rate of attack by the electrophile. The ring is therefore less reactive than benzene, so a higher temperature (above 55 C) is needed to make the reaction proceed. Benzene has no such withdrawing group, so its ring is more electron-rich and more reactive, allowing nitration at the milder 55 C.
More worked questions on this topic
- In Reaction II, benzaldehyde (C6H5CHO) is transformed into nitrobenzaldehyde. Give the reagents (2 marks)
More A-Level H2 Chemistry topics
The make-up of the atom and how its electrons are arranged · How atoms bond and how that governs a substance's behaviour · Ideal gases and working with gas mixtures · Competing definitions of acids and bases · Trends in the elements across a period and down a group · The mole and reacting-quantity calculations · all of A-Level H2 Chemistry