Cambridge International AS & A Level Chemistry 9701 · A Level topic 29

An introduction to A Level organic chemistry

What this chapter covers29.1–29.4

At AS Level, organic chemistry was built on chains: alkanes, alkenes, halogenoalkanes, alcohols, carbonyl compounds and carboxylic acids, each named from its longest carbon chain and each reacting through a handful of mechanisms. The A Level course adds two things. It adds new families — arenes, halogenoarenes, phenols, acyl chlorides, amines, amides and amino acids — and it adds the benzene ring, a flat hexagon of six carbon atoms whose electrons behave quite differently from those in a C=C double bond.

This chapter is the entry point for everything in A Level organic chemistry. It sets out the new functional groups and how they are drawn and named, introduces the two new mechanism names that later chapters rely on, explains the shape and bonding of the benzene ring in terms of σ and π bonds, and extends the AS work on stereoisomerism to the behaviour of optical isomers — including why it matters so much when a drug molecule is chiral.

Functional groups at A Level29.1.1

A functional group is the atom or group of atoms that gives a family of compounds its characteristic reactions. The hydrocarbon skeleton to which it is attached changes physical properties such as boiling point and solubility gradually, but the chemistry — which reagents attack, and where — is set by the functional group. That is why a compound as large as a drug molecule can be understood by picking out its groups one at a time: each group behaves much as it would in a small molecule.

The A Level syllabus adds seven classes of compound to those met at AS Level. They are shown in Figure 29.1 and Table 29.1.

Two distinctions in the table are worth fixing now, because both are examined repeatedly. First, a halogen or an –OH group bonded directly to a benzene ring does not behave like the same group on a chain: chlorobenzene resists the substitution reactions of chloroethane, and phenol is far more acidic than ethanol. In both cases the reason is the same — a lone pair on the atom next to the ring becomes part of the ring's delocalised electron system — and it is developed in the chapters on halogen compounds and hydroxy compounds. A compound such as C₆H₅CH₂OH, where the –OH is on a side-chain carbon, is an alcohol, not a phenol.

Second, amines and amides are easily confused because both contain nitrogen. An amide nitrogen is bonded to a carbonyl carbon: the group is –C(=O)–N. An amine nitrogen is bonded only to carbon atoms of alkyl or aryl groups, or to hydrogen. The distinction matters: amides are hydrolysed by acid or alkali and are almost neutral, whereas amines are bases. An ester, –C(=O)–O–, is not an amide even though it also has a C=O next to a heteroatom.

Formulas for aromatic and nitrogen compounds29.1.2

The four kinds of formula used at AS Level are used unchanged. A general formula describes a whole family (RCOCl, RNH₂); a structural formula shows the arrangement of atoms without drawing every bond (CH₃CH₂COCl); a displayed formula shows every atom and every bond; a skeletal formula shows the carbon skeleton as lines, with carbon atoms at the ends and junctions and hydrogen atoms on carbon left out, but with every other atom — and any hydrogen on O or N — written in.

Three conventions apply to the new compounds.

• The benzene ring is never displayed. Where a benzene ring is part of a molecule, a displayed formula is not expected: the ring is drawn as a hexagon with a circle inside, and only the substituents are displayed. In a structural formula the ring is written C₆H₅– (a phenyl group) or C₆H₄ when two groups are attached.
• Displayed amide and ester links. When a question says a linkage "should be shown displayed", every bond in that group must be drawn, including the C=O, the C–N or C–O bond, and the N–H bond of a secondary amide.
• Condensed groups written in the right order. A carboxylic acid group is –COOH or –CO₂H, never –COHO; an amine on the left of a structure is written H₂N–, and a nitro group on the left O₂N–, so that the bond is drawn to the nitrogen atom.

Naming aliphatic compounds at A Level29.1.3

The rules learned at AS Level still apply: find the longest carbon chain containing the principal functional group, number it to give the principal group the lowest possible locant, name substituents as prefixes in alphabetical order, and use di-, tri- and tetra- for repeated groups. The syllabus limits names to six carbon atoms in a chain, extends the rules to a single ring of up to six carbons, and allows up to six plus six carbons for esters and amides.

Acyl chlorides and amides

Both are named from the parent carboxylic acid, counting the carbonyl carbon as C1. The ending -oic acid becomes -oyl chloride or -amide: CH₃COCl is ethanoyl chloride, CH₃CH₂COCl is propanoyl chloride, CH₃CONH₂ is ethanamide and CH₃CH₂CONH₂ is propanamide. A substituted amide carries the group on the nitrogen as an N- prefix: CH₃CONHCH₃ is N-methylethanamide.

Esters

An ester RCOOR′ is named with the alkyl group from the alcohol first, as a separate word, and the part from the acid second, ending in -oate. CH₃COOCH₂CH₃ is ethyl ethanoate; CH₃CH₂COOCH₃ is methyl propanoate. The same pattern with a phenyl group gives phenyl ethanoate, CH₃COOC₆H₅, and with the acid part aromatic, methyl benzoate, C₆H₅COOCH₃. Straight chains only are required for esters and nitriles.

Amines

Primary amines are named with the suffix -amine on the alkyl group (CH₃CH₂NH₂, ethylamine) or, when another group takes priority, with the prefix amino- (H₂NCH₂COOH, 2-aminoethanoic acid). The syllabus states that naming secondary and tertiary amines is not required, but you must be able to recognise them: a secondary amine has two carbon groups on the nitrogen, a tertiary amine three.

Cyclic compounds

A single ring of up to six carbon atoms is named with the prefix cyclo-: cyclohexane, cyclohexene, cyclohexanol. Numbering starts at the carbon carrying the principal group, or at one end of the C=C bond in a cycloalkene.

Naming aromatic compounds29.1.4

Aromatic compounds with one benzene ring are named in one of two ways, depending on which groups are attached.

Some groups make the ring part of a special parent name. Three are required: –COOH gives benzoic acid, –OH gives phenol, and –NH₂ gives phenylamine. When one of these groups is present, the carbon carrying it is numbered C1, and the other groups are prefixes. If more than one is present, the priority is COOH, then OH, then NH₂: a ring with –COOH and –OH is a hydroxybenzoic acid, not a carboxyphenol.

Other groups are all prefixes on benzene. Halogens (chloro-, bromo-), nitro (–NO₂) and alkyl groups (methyl-, ethyl-) are named as prefixes: chlorobenzene, nitrobenzene, 1,3-dimethylbenzene. With a single substituent no number is needed. Methylbenzene itself is often treated as a parent: the syllabus refers to the products of chlorinating methylbenzene as 2-chloromethylbenzene and 4-chloromethylbenzene, in which the CH₃ carbon is C1, and names such as 1-chloro-2-methylbenzene describe the same compounds.

The numbering then follows two rules. The substituents take the lowest possible set of locants, going round the ring whichever way gives the lower numbers at the first point of difference. If both directions give the same set, the substituent cited first in alphabetical order takes the lower number. Prefixes are cited alphabetically, ignoring di- and tri-: in 4-chloro-3,5-dimethylphenol, "chloro" comes before "methyl".

Two new mechanism names29.2.1

At AS Level four mechanisms were named: free-radical substitution, electrophilic addition, nucleophilic substitution and nucleophilic addition. Each name has two parts. The first says what attacks — a radical, an electrophile or a nucleophile. The second says what happens overall — something replaces something else (substitution), two molecules become one (addition), or a small molecule is removed (elimination). A Level adds two more names, built in exactly the same way.

Electrophilic substitution

An electrophile is a species that accepts a pair of electrons to form a new covalent bond; it is positively charged or has a positive dipole. In electrophilic substitution an electrophile attacks an electron-rich benzene ring and replaces one of its hydrogen atoms, which leaves as H⁺. The ring survives: the product is still aromatic. Nitration, halogenation and the Friedel–Crafts reactions of arenes all follow this pattern (Figure 29.9a), and the next chapter works through each one.

Addition–elimination

A nucleophile donates a lone pair of electrons to form a new covalent bond. In addition–elimination a nucleophile first adds to the δ+ carbon of a C=O group, breaking the π bond and giving a tetrahedral intermediate; the C=O double bond then re-forms and a small group is eliminated — for an acyl chloride, a chloride ion followed by H⁺, so that HCl is the by-product. The overall result is the substitution of –Cl by the nucleophile (Figure 29.9b). This is the mechanism of all the reactions of acyl chlorides with water, alcohols, phenols, ammonia and amines, and it is described in the chapter on carboxylic acids and their derivatives.

Past-paper practice · Set 29C · Names, functional groups and mechanism terms

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 29C.1[1]

Answer and marking guidance

4-chloro-3,5-dimethylphenol ✔ (also accepted: 3,5-dimethyl-4-chlorophenol). Phenol is the parent, so the OH carbon is C1; the substituents then take the lowest locants, 3, 4 and 5; chloro is cited before methyl alphabetically. Examiner insight: many answered well; common errors were omitting "di" or numbering incorrectly.

Question 29C.2[1]

Compound Z, referred to in the question, comes from an earlier part; only the structure of W is needed to answer.

Answer and marking guidance

3,4,5-trimethylphenylamine ✔ — the NH₂ carbon is C1 and the three methyl groups take positions 3, 4 and 5. Examiner insight: this discriminated well.

Question 29C.3[3]

Answer and marking guidance

(i) A = ester ✔; B = (secondary) amide ✔. (ii) 2 chiral carbon atoms ✔ — the CH bonded to N (between the ester and the CH₂ of the benzyl group) and the ring CH carrying the CH₃ group next to the ring oxygen. Examiner insight: most gave correct answers; wrong answers included ketone and carboxylic acid for A, and peptide or amine for B. (ii) was answered well.

Question 29C.4[3]

Part (b)(i) is the full mechanism of nitration, developed in the next chapter; part (b)(ii) asks only for its name.

Answer and marking guidance

(i) Curly arrow from inside the hexagon (the delocalised ring) to the N of NO₂⁺ AND curly arrow from the C–H bond back into the ring ✔; intermediate: a horseshoe of delocalisation over five carbons with + inside, H and NO₂ on the sp³ carbon ✔. (ii) electrophilic substitution ✔. Examiner insight: (i) was answered well with many good diagrams of the intermediate, but some did not draw the first curly arrow in the first box. (ii) was answered well.

Benzene: the evidence against three double bonds29.3.1

Benzene, C₆H₆, is a colourless liquid whose six carbon atoms form a ring. The first structure proposed for it, by Kekulé in 1865, was a ring of alternating single and double bonds — cyclohexa-1,3,5-triene. The formula fits, but the chemistry does not. Four independent pieces of evidence show that benzene does not contain three localised C=C bonds.

1 · Benzene does not behave like an alkene

Alkenes decolourise bromine water rapidly at room temperature by electrophilic addition. Benzene does not: it reacts with bromine only when a halogen carrier such as AlBr₃ is present, and even then the product is bromobenzene, formed by substitution. A molecule with three C=C bonds would be expected to add bromine three times over.

2 · The enthalpy change of hydrogenation is too small

Adding hydrogen across one C=C bond in cyclohexene releases 120 kJ mol⁻¹. If benzene had three such bonds, hydrogenating it to cyclohexane should release about three times as much, 360 kJ mol⁻¹. The measured value is only 208 kJ mol⁻¹. Benzene is therefore 152 kJ mol⁻¹ lower in energy — more stable — than the Kekulé structure would be (Figure 29.3). This difference is called the delocalisation energy or stabilisation energy of benzene.

3 · All the carbon–carbon bonds are the same length

Measurements of bond lengths show that every C–C bond in the benzene ring is 140 pm long. A C–C single bond, as in ethane, is 154 pm and a C=C double bond, as in ethene, is 133 pm. Kekulé's structure would have three long and three short bonds, giving an irregular hexagon; the real molecule is a regular hexagon whose bonds are intermediate between single and double (Figure 29.5).

4 · Too few isomers

If the ring had fixed single and double bonds, two substituents on adjacent carbons could be joined either by a C–C bond or by a C=C bond, and these would be different compounds: there would be two 1,2-dibromobenzenes. Only one has ever been isolated, and only three dibromobenzenes exist in total. Kekulé tried to rescue his structure by proposing that the molecule switched rapidly between two arrangements of its double bonds, but the hydrogenation and bond-length evidence rules that out as well. The model below generates every substitution pattern and counts the distinct compounds each structure predicts.

σ and π bonding in benzene29.3.1

The structure that fits all four pieces of evidence treats the six ring electrons that Kekulé placed in three double bonds as delocalised — shared by all six carbon atoms. The orbital model explains how.

The σ framework

Each carbon atom in benzene is bonded to three other atoms: two carbons and one hydrogen. It uses three sp² hybrid orbitals, formed by mixing its 2s orbital with two of its 2p orbitals. The three sp² orbitals lie in one plane at 120° to each other. Each C–C σ bond is formed by end-on (head-on) overlap of an sp² orbital from each carbon; each C–H σ bond by end-on overlap of an sp² orbital of carbon with the 1s orbital of hydrogen. Because every carbon is trigonal planar and every bond angle is 120°, the six carbons and six hydrogens all lie in one plane: benzene is a flat, regular hexagon (Figure 29.4a).

The delocalised π system

Each carbon has one electron left in its third 2p orbital, which was not used in hybridisation. This unhybridised p orbital is perpendicular to the plane of the ring (Figure 29.4b). Adjacent p orbitals are close enough to overlap sideways, and they overlap equally with the neighbours on both sides — so the overlap continues all the way round the ring. The result is a π system of six electrons in two ring-shaped regions of electron density, one above and one below the plane (Figure 29.4c). No electron pair belongs to one particular C–C bond: the six π electrons are delocalised over all six carbons.

Delocalisation explains every observation in the previous section. The bonds are identical because each has the same share of the π electrons — in effect one and a half bonds. The molecule is more stable than the Kekulé structure because spreading electrons over a larger region lowers their energy, and that extra stability is the 152 kJ mol⁻¹ measured by hydrogenation. The π electron density is spread over the whole ring rather than concentrated between two carbons, so benzene is a weaker attractor of electrophiles than an alkene — which is why bromine water is not decolourised. The circle drawn inside the hexagon represents the delocalised π system.

Other aromatic molecules

The same model applies whenever a benzene ring is present: the ring carbons are sp², the ring is planar and the six π electrons are delocalised. What happens at the substituent depends on its own bonding.

• Methylbenzene. The ring is planar; the CH₃ carbon has four single bonds, so it is sp³ and tetrahedral. The C–H bonds of the methyl group are not in the plane of the ring.
• Phenol and phenylamine. The oxygen or nitrogen atom next to the ring has a lone pair in a p-type orbital that can overlap with the ring's π system. This partial delocalisation of the lone pair into the ring is the origin of phenol's acidity, phenylamine's low basicity, and the high reactivity of both rings towards electrophiles — ideas developed in later chapters.
• Pyridine, C₅H₅N. One CH of benzene is replaced by N. The nitrogen is also sp² hybridised: two sp² orbitals form σ bonds to carbon and the third holds its lone pair, in the plane of the ring. Its remaining p orbital, with one electron, joins the five carbon p orbitals in a six-electron delocalised π system exactly like benzene's.
• Nitrobenzene and benzoic acid. The N of –NO₂ and the C of –COOH are also sp², with trigonal planar geometry around them.

Why benzene is substituted rather than added to29.3.1, 30.1.2

An alkene reacts with an electrophile by addition because the product is more stable than the starting alkene: a weaker π bond is replaced by two stronger σ bonds. For benzene the balance is different. If an electrophile added across the ring and the intermediate then picked up a nucleophile, the product would have lost its delocalised π system — and with it the 152 kJ mol⁻¹ of delocalisation energy. The intermediate avoids this by losing a proton from the carbon that was attacked; the two electrons from that C–H bond return to the ring and the full delocalised system is restored. Substitution keeps the aromatic stabilisation; addition would throw it away.

The same stability explains the conditions needed. Because the π electrons are spread over six atoms, the ring is a less concentrated source of electrons than a C=C bond. Most electrophiles that attack alkenes — bromine molecules, for example — are not strong enough to attack benzene. A catalyst is used to generate a much stronger, positively charged electrophile: NO₂⁺ from concentrated nitric and sulfuric acids, Br⁺ or Cl⁺ from the halogen and its aluminium halide, and carbocations or acylium ions from halogenoalkanes and acyl chlorides. Hydrogenation, which does add across the ring, needs hot hydrogen and a Pt or Ni catalyst, and cannot be stopped part-way because the partly hydrogenated rings are no longer aromatic.

Past-paper practice · Set 29A · The shape of benzene: σ and π bonds

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 29A.1[4]

Answer and marking guidance

Any four of: the molecule is a planar (regular) hexagon ✔; all C–C bonds are the same length (between single and double); all bond angles are 120°; every carbon atom is sp² hybridised ✔; C–H bonds are formed by overlap of an sp² orbital of carbon with the 1s orbital of hydrogen ✔; C–C σ bonds come from sp²–sp² overlap ✔; the π bonding comes from sideways p–p overlap; σ and π each used correctly at least once. Examiner insight: found very difficult. Many gave the hexagonal shape, 120°, equal C–C lengths and sp² for partial credit, but good descriptions of orbital overlap — s with sp² for C–H — were very rarely seen. Some knew the bonds are equal in length but did not say so clearly, and some confused p (an orbital) with π (a bond made by p-orbital overlap).

Question 29A.2[3]

Answer and marking guidance

M1 bond angle 120° AND every carbon is sp² ✔. M2 σ bonds are formed by end-on (head-on, linear) overlap of orbitals ✔. M3 π bonds are formed by sideways overlap of p orbitals ✔. Examiner insight: found difficult. The sentence "explain how orbital overlap leads to the formation of σ and π bonds" was the signal: an answer had to say which orbitals overlap and how, not just name the bonds.

Question 29A.3[3]

Answer and marking guidance

M1 the nitrogen AND the carbon atoms are sp² hybridised ✔. M2 σ bonds (C–H, C–C, C–N) are formed by end-on overlap of orbitals ✔. M3 π bonds are formed by sideways overlap of p orbitals (on C and N) ✔ — the same delocalised six-electron system as benzene. Examiner insight: performance was poor. Good answers made two points: hybridisation describes atoms (here C and N are sp²), and bonds come from overlap — σ bonds from head-on overlap of sp² orbitals, π bonds from sideways overlap of unhybridised p orbitals.

Question 29A.4[2]

Parts (b) onward of this question test NMR spectroscopy, basicity and synthesis; only part (a) is reproduced here.

Answer and marking guidance

(i) phenylamine (aryl amine), (tertiary) amine AND ester ✔ — note there is no amide: the C=O is bonded to O, not N. (ii) sp = 0, sp² = 7, sp³ = 6 ✔ (six ring carbons and the ester C=O carbon are sp²; the six carbons of the CH₂CH₂N(CH₂CH₃)₂ chain are sp³). Examiner insight: identifying all the functional groups was difficult and "amide" was a common wrong answer; in (ii) many answers did not add up to 13.

Chirality and enantiomers29.4 (from 13.4)

At AS Level you met stereoisomers — molecules with the same structural formula but a different arrangement of their atoms in space — and two kinds of them. Cis–trans isomers arise from restricted rotation about a C=C bond. Optical isomers arise when a molecule is chiral: not superimposable on its own mirror image, in the way a left hand is not superimposable on a right hand.

The commonest cause of chirality in organic molecules is a chiral centre: a carbon atom bonded to four different atoms or groups. The four groups are arranged tetrahedrally, and there are exactly two ways of arranging four different groups round a tetrahedron. The two arrangements are mirror images of each other, and no rotation turns one into the other (Figure 29.6). The two optical isomers of a chiral molecule are called enantiomers.

Finding chiral centres in larger molecules

Drug molecules and natural products often contain several chiral centres, and questions regularly ask how many a molecule contains. Work carbon by carbon and eliminate the ones that cannot be chiral:

• any carbon in a C=C or C=O double bond, or in a benzene ring (it has only three groups attached);
• any CH₂ or CH₃ carbon (two or three identical hydrogen atoms);
• any carbon carrying two identical groups, such as the central carbon of propan-2-ol, (CH₃)₂CHOH.

For each carbon that remains, compare the four groups as whole groups, not just the first atom. In butan-2-ol, CH₃CH(OH)CH₂CH₃, the central carbon carries H, OH, CH₃ and CH₂CH₃: two groups begin with carbon, but they are different groups, so the carbon is chiral. Ring carbons count too: a CH in a ring is chiral if the two paths round the ring from it are different.

A molecule with n chiral centres can have up to 2ⁿ stereoisomers. The syllabus expects you to appreciate that a molecule can have more than one chiral centre; the special cases in which internal symmetry reduces the number (meso compounds) and the term diastereoisomer are not required.

How enantiomers differ: plane-polarised light29.4.1, 29.4.2, 29.4.3

Two enantiomers have the same structural formula and the same bonds. Their intermolecular forces with any non-chiral molecule are identical, so they have the same melting point, boiling point, density, solubility in water and ethanol, and spectra. They react at the same rate with non-chiral reagents to give the same products. For almost every purpose in the laboratory, the two enantiomers of a substance are indistinguishable.

They differ in two ways only. The first is their effect on plane-polarised light. Ordinary light consists of waves vibrating in every plane at right angles to the direction of travel. A polarising filter lets through only the waves vibrating in one plane: the light that emerges is plane-polarised. When plane-polarised light passes through a solution of a single enantiomer, the plane of polarisation is rotated. The instrument that measures this is a polarimeter (Figure 29.7): light passes through a polariser, through the sample tube, and then through a second filter, the analyser, which is turned until the light is extinguished again. The angle through which it is turned is the angle of rotation.

The two enantiomers of a compound always rotate the plane by the same angle but in opposite directions, under the same conditions of concentration, path length and temperature. The isomer that rotates it clockwise, as seen by the observer, is labelled (+); the other is (−). Both are optically active. A 50:50 mixture — a racemic mixture — is not optically active: every molecule that rotates the plane one way is matched by one that rotates it back, so the net rotation is zero.

Chirality and drugs29.4.1, 29.4.4

The second difference between enantiomers is in their biological activity. Most molecules in living things are chiral: amino acids, sugars, and therefore proteins, enzymes and the receptor sites on cell surfaces. A receptor or enzyme active site is a three-dimensional pocket, and a drug molecule acts by fitting into it, with several of its groups interacting with complementary groups in the pocket. A chiral site can distinguish between two enantiomers just as a right-handed glove distinguishes between hands: one enantiomer fits, and its mirror image, with two groups in swapped positions, cannot make the same contacts. Frequently only one of a pair of enantiomers is biologically effective.

The consequences for a drug are that the other enantiomer may be:

• inactive — it adds nothing, so a racemic drug must be given in twice the dose to deliver the same amount of the active isomer;
• active in a different way — it binds to a different receptor and causes side effects;
• harmful — in the worst cases, toxic.

The problem with ordinary synthesis

Most laboratory and industrial reactions use achiral reagents and achiral conditions. When such a reaction creates a new chiral centre, it has no reason to favour one arrangement over the other, and the product is a racemic mixture. For example, when a nucleophile adds to a planar C=O group, it can attack from either face with equal probability, giving equal amounts of the two enantiomers. If only one enantiomer is wanted, a racemic product has two drawbacks: at most half of it is useful, so the yield of the active isomer is lower; and the enantiomers must be separated. Separation is difficult and expensive precisely because enantiomers have identical physical properties — they cannot be separated by distillation or ordinary crystallisation.

Making a single enantiomer

The alternative is to make only the wanted isomer in the first place. A chiral catalyst — a catalyst that is itself a single enantiomer — provides a chiral environment in which the two possible products are formed at different rates, so that one enantiomer predominates. Enzymes, which are naturally chiral protein catalysts, do the same. Both approaches are summarised in Figure 29.8.

Past-paper practice · Set 29B · Optical isomerism and chiral drugs

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 29B.1[1]

Answer and marking guidance

A mixture containing equal amounts (equimolar) of the two optical isomers (enantiomers) ✔. Examiner insight: generally well answered; some did not state that the two enantiomers are present in equal concentrations.

Question 29B.2[3]

Answer and marking guidance

(i) one chiral carbon: the CH carrying the NH₂ group ✔ (the CH₂ and the ester carbon each have two identical atoms or a double bond). (ii) Both enantiomers rotate the plane of plane-polarised light ✔; by the same angle but in opposite directions ✔. Examiner insight: (i) was usually correct. In (ii) many knew the plane would be rotated by both isomers, but that the rotation is by equal amounts in opposite directions was less well known.

Question 29B.3[3]

The structure of tulobuterol is given at the start of the question; parts (a)–(d) are used in the chapters on arenes and synthesis.

Answer and marking guidance

(i) Molecules that are non-superimposable mirror images of each other (or: they rotate plane-polarised light in opposite directions) ✔. (ii) The optical isomers must be separated to obtain the pure active isomer; or the other enantiomer may have reduced or different biological activity; or a lower yield of the active molecule ✔. (iii) Use a chiral catalyst, or an enzyme ✔. Examiner insight: the definition in (i) was not well known — incomplete answers such as "they have non-superimposable molecules" were common. (ii) was usually correct; a way of adapting the synthesis to give one enantiomer, (iii), was not well known.

Question 29B.4[5]

Answer and marking guidance

(a)(i) 5 ✔. (ii) Benefit: higher biological activity, fewer side effects, or a smaller dose needed ✔. Disadvantage: the racemic mixture must be separated, the yield of the active isomer is lower, or an (expensive) chiral catalyst or enzyme is needed ✔. (b)(i) carboxylic acid, ester, amide, (secondary) amine — any two ✔, all four ✔✔. Examiner insight: the number of chiral centres was often correct. (a)(ii) proved difficult: vague answers such as "higher yield" or "expensive" gained no credit — the benefit must be about activity or side effects, and the cost must be tied to separation or a chiral catalyst. In (b)(i), carbonyl and arene were common errors.

Question 29B.5[1]

Answer and marking guidance

The mirror image of isomer 1: the same four groups on the central carbon, with two of them exchanged in position (for example H and CH₃ swapped, or the drawing reflected in a vertical mirror), keeping one bond in the plane, one wedge and one hashed bond ✔. Examiner insight: this was usually correct.

Misconceptions and how the topic is assessed29.1–29.4

Topic 29 is rarely examined as a question of its own. Its ideas are embedded in longer organic questions built round a drug or natural product: name the functional groups, count the chiral centres or the sp² carbons, describe the bonding in the ring, name the mechanism. The table below collects the misconceptions behind the commonest lost marks.

Self-test29.1–29.4

Twelve questions on the whole chapter. Each gives its reason once you answer.

Definitions to learn29.1–29.4

Summary

Essential knowledge

• New A Level classes: arenes, halogenoarenes, phenols, acyl chlorides, amines (1°, 2°, 3°), amides (1°, 2°, 3°) and amino acids. Each functional group sets the chemistry; the skeleton changes physical properties gradually.
• A benzene ring is drawn as a hexagon with a circle and is never displayed. Esters are named alkyl alkanoate; acyl chlorides -oyl chloride; amides -amide; primary amines -amine or amino-.
• Aromatic names: –COOH, –OH, –NH₂ give benzoic acid, phenol, phenylamine with that carbon as C1; halogen, nitro and alkyl groups are prefixes; lowest locants; alphabetical order; di-/tri-.
• Electrophilic substitution: electrophile replaces H on an arene, ring restored. Addition–elimination: nucleophile adds to C=O of an acyl chloride, C=O re-forms, HCl eliminated.
• Benzene is planar with 120° angles; every carbon sp²; σ bonds by end-on overlap (sp²–sp², sp²–s); six p electrons overlap sideways into a delocalised π system above and below the ring.
• Evidence against the Kekulé structure: no addition with bromine water; ΔH_{hydrogenation} −208 not −360 kJ mol⁻¹ (152 kJ mol⁻¹ delocalisation energy); all C–C bonds 140 pm; only three dibromobenzenes.
• Substitution, not addition, keeps the delocalisation energy; strong electrophiles generated by catalysts are needed.
• Enantiomers: identical except for the direction in which they rotate plane-polarised light (same angle, opposite directions) and their biological activity. A racemic mixture (equal amounts) is not optically active.
• Chiral drugs: the other enantiomer may be inactive or harmful; racemic products must be separated (lower yield, costly); chiral catalysts or enzymes make a single enantiomer directly.

Examination checklist

• Can I name every functional group in an unfamiliar drug molecule, distinguishing ester from amide and amine from amide?
• Can I name a substituted benzene with the right parent, numbering and alphabetical order?
• Can I name esters, acyl chlorides and amides with up to six carbons in each part?
• Can I state which mechanism is electrophilic substitution and which is addition–elimination, and say what attacks what?
• Can I describe the shape of benzene, including sp², 120°, end-on σ overlap, sideways p overlap and delocalisation?
• Can I give three pieces of evidence against the Kekulé structure, including the calculation from hydrogenation data?
• Can I count sp, sp² and sp³ carbons in a molecule and check that the total is right?
• Can I find every chiral centre in a large molecule, including ring carbons?
• Can I draw two enantiomers in three dimensions?
• Can I describe the effect of two enantiomers and a racemic mixture on plane-polarised light?
• Can I give a precise benefit and disadvantage of single-enantiomer drugs and name a way to make one?

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