An introduction to AS Level organic chemistry

• The functional group decides the chemistry; the name tells you the structure; every curly arrow starts at an electron pair; cis/trans needs two different groups on each C=C carbon; a chiral centre needs four different groups.

• 13.1.1 I can define — define the term hydrocarbon as a compound made up of C and H atoms only
• 13.1.2 I can understand — understand that alkanes are simple hydrocarbons with no functional group
• 13.1.3 I can understand — understand that the compounds in the table on pages 29 and 30 contain a functional group which dictates their physical and chemical properties
• 13.1.4 I can interpret — interpret and use the general, structural, displayed and skeletal formulas of the classes of compound stated in the table on pages 29 and 30
• 13.1.5 I can understand — understand and use systematic nomenclature of simple aliphatic organic molecules with functional groups detailed in the table on pages 29 and 30, up to six carbon atoms (six plus six for esters, straight chains only for esters and nitriles)
• 13.1.6 I can deduce — deduce the molecular and/or empirical formula of a compound, given its structural, displayed or skeletal formula
• 13.2.1 I can interpret — interpret and use the following terminology associated with types of organic compounds and reactions: (a) homologous series (b) saturated and unsaturated (c) homolytic and heterolytic fission (d) free radical, initiation, propagation, termination (e) nucleophile, electrophile, nucleophilic, electrophilic (f) addition, substitution, elimination, hydrolysis, condensation (g) oxidation and reduction (in equations for organic redox reactions, the symbol [O] can be used to represent one atom of oxygen from an oxidising agent and the symbol [H] to represent one atom of hydrogen from a reducing agent)
• 13.2.2 I can understand — understand and use the following terminology associated with types of organic mechanisms: (a) free-radical substitution (b) electrophilic addition (c) nucleophilic substitution (d) nucleophilic addition (in organic reaction mechanisms, the use of curly arrows to represent movement of electron pairs is expected; the arrow should begin at a bond or a lone pair of electrons)
• 13.3.1 I can describe — describe organic molecules as either straight-chained, branched or cyclic
• 13.3.2 I can describe — describe and explain the shape of, and bond angles in, molecules containing sp, sp² and sp³ hybridised atoms
• 13.3.3 I can describe — describe the arrangement of σ and π bonds in molecules containing sp, sp² and sp³ hybridised atoms
• 13.3.4 I can understand — understand and use the term planar when describing the arrangement of atoms in organic molecules, for example ethene
• 13.4.1 I can describe — describe structural isomerism and its division into chain, positional and functional group isomerism
• 13.4.2 I can describe — describe stereoisomerism and its division into geometrical (cis/trans) and optical isomerism (use of E/Z nomenclature is acceptable but is not required)
• 13.4.3 I can describe — describe geometrical (cis/trans) isomerism in alkenes, and explain its origin in terms of restricted rotation due to the presence of π bonds
• 13.4.4 I can explain — explain what is meant by a chiral centre and that such a centre gives rise to two optical isomers (enantiomers) (Candidates should appreciate that compounds can contain more than one chiral centre, but knowledge of meso compounds, or nomenclature such as diastereoisomers is not required.)
• 13.4.5 I can identify — identify chiral centres and geometrical (cis/trans) isomerism in a molecule of given structural formula, including cyclic compounds
• 13.4.6 I can deduce — deduce the possible isomers for an organic molecule of known molecular formula

Units and significant figures are marked. The syllabus states that failure to quote units, the inclusion of units in quantities defined as ratios, and answers given to an inappropriate number of significant figures are all liable to be penalised. Give a calculated answer to the same number of significant figures as the least precise data, or one more; keep full precision in the working and round only at the end. A fifth of the qualification is experimental: Papers 3 and 5 test AO3 only, and their questions may be set in contexts outside the syllabus content, so the practical work in this chapter is set out as procedure, recording and evaluation rather than as theory.

• “The curly arrow goes from the hydroxide ion to the carbon.” Correct Every curly arrow starts at a bond or a lone pair, because it shows a pair of electrons moving. It starts at the lone pair on the O of :OH⁻, never at the minus sign, never at the letter O and never in empty space.
• “But-1-ene has a double bond, so it has cis and trans isomers.” Correct Each carbon of the C=C must carry two different groups. C1 of but-1-ene carries two hydrogens, so swapping them changes nothing. But-2-ene has cis and trans isomers; but-1-ene and 2-methylbut-2-ene do not.
• “A hydrocarbon is an organic compound containing carbon and hydrogen.” Correct A hydrocarbon is made up of C and H atoms only. Ethanol and chloromethane contain C and H but are not hydrocarbons.
• “C₃H₇OH is the structural formula of propan-1-ol.” Correct A structural formula must be unambiguous. C₃H₇OH could be propan-1-ol or propan-2-ol; write CH₃CH₂CH₂OH or CH₃CH(OH)CH₃.
• “CH₃CH₂COOCH₃ is ethyl propanoate.” Correct The alkyl group on the single-bonded O comes from the alcohol and is named first. Here it is methyl, and the acid part has three carbons: methyl propanoate.
• “2-methylbutan-2-ol has four groups on C2, so C2 is a chiral centre.” Correct The four groups must be different. C2 carries OH, C₂H₅ and two CH₃ groups, so it is not a chiral centre.
• “Ethene is planar because it has a double bond.” Correct Ethene is planar because each carbon is sp² hybridised, with three σ bonds at 120° in one plane. The π bond lies above and below that plane.
• “A hydrocarbon is any compound containing carbon.” Repair A hydrocarbon contains C and H only. Ethanol and chloromethane are not hydrocarbons.
• “The structural formula of propan-1-ol is C₃H₇OH.” Repair A structural formula must be unambiguous: CH₃CH₂CH₂OH. C₃H₇OH could equally be propan-2-ol.
• “Displayed formula of ethanol: H₃C–CH₂–OH, with –OH drawn as one unit.” Repair A displayed formula shows every bond, including every C–H and the O–H.
• “Skeletal formula of propan-1-ol: a zig-zag ending in O, with CH₂ written at the vertices.” Repair No C or H is written on the carbon skeleton, and every H on a heteroatom is shown: the end is OH, not O.
• “CH₃CH(C₂H₅)CH₂CH₂CH₃ is 2-ethylpentane.” Repair The longest chain runs through the ethyl group and has six carbons: 3-methylhexane. Find the longest chain before naming any branch.
• “CH₃CH₂CH(OH)CH₃ is butan-3-ol; CH₃CH₂CH=CHCH₃ is pent-3-ene.” Repair Number from the end that gives the functional group the lowest locant: butan-2-ol and pent-2-ene.
• “methyl-2-butane; 2,methylbutane.” Repair Locant, hyphen, prefix, stem: 2-methylbutane. Commas go only between numbers (2,2-dimethylpropane).
• “CH₃CH₂COOCH₃ is ethyl propanoate.” Repair The alkyl group on the single-bonded O comes from the alcohol and is named first: methyl propanoate. The acid part has three carbons, counting the C=O carbon.
• “Ethanenitrile has two carbons, so CH₃CH₂CN is ethanenitrile.” Repair The C of C≡N counts in the chain: CH₃CH₂CN has three carbons and is propanenitrile.
• “A nucleophile is a negative ion.” Repair A nucleophile is an electron-pair donor. NH₃ and H₂O are neutral nucleophiles; what they need is a lone pair.
• “CH₃CH₂OH + [O] → CH₃CHO.” Repair Balance every element: the two H removed leave with the [O] as water. CH₃CH₂OH + [O] → CH₃CHO + H₂O.
• “A curly arrow drawn from the negative sign of OH⁻, or from the letter C, to the atom being attacked.” Repair Every curly arrow begins at a bond or a lone pair, because it shows a pair of electrons moving.
• “Ethene is planar because it has a double bond.” Repair Ethene is planar because each carbon is sp² hybridised, with three σ bonds at 120° in one plane; the π bond lies above and below that plane.
• “But-1-ene has cis and trans isomers.” Repair C1 carries two hydrogens. Each carbon of the C=C must carry two different groups; but-2-ene does, but-1-ene does not.
• “2-methylbutan-2-ol has a chiral centre because C2 has four groups.” Repair The four groups must be different. C2 carries two methyl groups, so it is not a chiral centre.
• “An enantiomer drawn by turning the first structure round on the page.” Repair The second structure must be the mirror image across a drawn mirror line: left and right swap, wedge stays wedge, hash stays hash. A rotation gives the same molecule.
• “C₄H₈ has five isomers: cis- and trans-but-2-ene are the same compound.” Repair They are different compounds that cannot interconvert without breaking the π bond. C₄H₈ has six isomers.

• Read the command word before you decide how much to write. This syllabus has twenty-two of them: analyse, calculate, compare, consider, contrast, deduce, define, demonstrate, describe, determine, discuss, evaluate, examine, explain, give, identify, justify, predict, show (that), sketch, state and suggest. Comment, estimate, name and outline are not among them: where a question wants something named it says identify, which the syllabus glosses as “name/select/recognise”. State and give want a fact and nothing more. Describe wants the points or the features. Explain wants the reasons and the relationships — a describe-level answer to an explain question is incomplete however well written it is. Deduce and determine want a conclusion reached from the information given, with the reasoning visible.
• Identify the group, then predict. Faced with an unfamiliar molecule, first find every functional group in it. Its reactions will be the reactions of those groups, and its boiling point will be set mainly by whether it can hydrogen-bond.
• Interleave with the chapters that use this one. Topic 14 (alkanes and alkenes): re-answer “why does an alkene attract electrophiles?”. Topic 15 (halogenoalkanes): redraw the two curly arrows of hydroxide attacking a halogenoalkane. Topics 16 and 17 (alcohols, carbonyls): write every [O] equation with its water. Topic 22 (analysis): find a molecular formula from a mass spectrum, then list its isomers. Topic 29 at A Level: revisit the chiral centre when optical activity is taught. Recalling a topic inside a new context is worth more than another pass over this chapter on its own; at A Level, Paper 4 assumes the whole of the AS content, so nothing here is ever finished with.

• Cambridge International AS & A Level Chemistry 9701 has five components. Topic 13 is AS Level content, so it is examined in Papers 1, 2 and 3. AS Level content: examined in Paper 1 (multiple choice), Paper 2 (AS structured) and, as practical context, Paper 3. Assumed knowledge for Papers 4 and 5. AS Level candidates take Papers 1, 2 and 3; A Level candidates take all five, either staged over two years (Papers 1–3 in year one, Papers 4 and 5 in year two) or together in one series. Examinations are available in the June and November series, and in March in India.
• Across both the AS Level and the A Level the assessment objectives are weighted AO1 40% (knowledge and understanding), AO2 40% (handling, applying and evaluating information) and AO3 20% (experimental skills and investigations). AS candidates are graded a–e; A Level candidates A*–E.
• A Paper 1 item on this topic can turn on one exact rule: the name of a given structure, the number of isomers of a formula, the number of σ and π bonds, whether a molecule shows cis/trans or has a chiral centre, or which term fits a reaction. In Paper 2 the same ideas become structured parts: define a term, identify a functional group, draw a displayed or skeletal formula, deduce a structure from information, and explain a shape or restricted rotation in terms of σ and π bonds.
• The only calculations are formula work: counting atoms from a structure for the molecular formula, dividing by the highest common factor for the empirical formula, and turning percentage composition by mass into an empirical and then a molecular formula (the method of topic 2). The relative atomic masses come from the Periodic Table in the Data section; no other constant is used in this chapter.
• Topic 13 has no experiment of its own. Its practical value is the reasoning that the Paper 3 observational problem and a Paper 5 planning question use: list the possible isomers of a formula first, then choose the functional-group tests (from topics 14, 16 and 17) that tell them apart, in an order that isolates one compound at each step, and say plainly when no test can separate two of them.
• Read the command word before you decide how much to write. This syllabus has twenty-two of them: analyse, calculate, compare, consider, contrast, deduce, define, demonstrate, describe, determine, discuss, evaluate, examine, explain, give, identify, justify, predict, show (that), sketch, state and suggest. Comment, estimate, name and outline are not among them: where a question wants something named it says identify, which the syllabus glosses as “name/select/recognise”. State and give want a fact and nothing more. Describe wants the points or the features. Explain wants the reasons and the relationships — a describe-level answer to an explain question is incomplete however well written it is. Deduce and determine want a conclusion reached from the information given, with the reasoning visible.