OxfordAQA A-Level Chemistry: Optical isomerism (9620)

This study guide teaches section 3.3.7, Optical isomerism, from OxfordAQA’s International AS and A-level Chemistry specification (9620), Version 5.3, for International AS and A-level exams from May/June 2020. Its only subsection is 3.3.7.1. The specification places it in the International A2 organic content, so everything here is International A-level only. Unit 4 (Organic 2 and Physical 2) assesses sections 3.3.7 to 3.3.16, and the practical and synoptic Unit 5 can draw on any content.

Pair it with the revision notes and the practice questions. Background you need: the introduction to organic chemistry (stereoisomers and E–Z isomerism) and shapes, polarity and intermolecular forces (the tetrahedral shape). For the wider course, use the hub page, the printable checklist and the free diagnostics.

What this section covers

Section	What you must know or do	Stage
3.3.7.1	Optical isomerism is a form of stereoisomerism caused by chirality, limited to molecules with a single chiral centre	International A-level only
3.3.7.1	An asymmetric carbon atom is chiral; it gives two enantiomers, mirror-image forms that cannot be superimposed and that affect plane-polarised light differently	International A-level only
3.3.7.1	A racemic mixture (racemate) is a mixture of equal amounts of enantiomers	International A-level only
3.3.7.1	Draw the structural formulas and displayed formulas of enantiomers	International A-level only
3.3.7.1	Understand how racemates are formed and why they are optically inactive	International A-level only

Section 3.3.7.1 lists no practical activity. Two other sections use the same ideas. Section 3.3.8.1 returns to these ideas when you explain the enantiomer mixture formed from KCN and then dilute acid, and section 3.2.5.3 covers optical isomerism in octahedral complexes with bidentate ligands (see the transition metals guide).

Where optical isomerism fits

Section 3.3.1.3 asks you to define a stereoisomer: a compound with the same structural formula as another but a different arrangement of its atoms in space. E–Z isomerism, from restricted rotation about a planar C=C bond, is the first kind you meet.

Optical isomerism is the second kind. Its cause is different. A carbon atom with four single bonds is tetrahedral, with bond angles of about 109.5°. When the four atoms or groups on that carbon are all different, there are two ways to arrange them in space. The two arrangements are mirror images, and no amount of turning makes one fit exactly on top of the other.

Asymmetric carbon atoms

A carbon atom is asymmetric when its four bonds go to four different atoms or groups. Chemists also call it a chiral centre, and you mark it with an asterisk (C*). In this course the specification limits optical isomerism to molecules with one chiral centre, so every molecule you are asked to draw will have exactly one.

How to find it

1. Rule out any carbon in a double or triple bond. It has only three or two groups around it.
2. Rule out CH₃ and CH₂ carbons. Each has at least two identical hydrogen atoms.
3. For every carbon left, write out its four groups in full and compare them group by group to the end of each chain.
4. A carbon with no hydrogen at all can still be chiral, as long as its four groups differ.

Worked example 1: which molecules are chiral?

Decide whether each compound has an asymmetric carbon atom, and give the four groups on it.

(a) 3-methylpentan-1-ol, HOCH₂CH₂CH(CH₃)CH₂CH₃ (b) 2-methylbutane, (CH₃)₂CHCH₂CH₃ (c) 1-chloro-1-fluoroethane, CH₃CHClF (d) 2-bromo-2-chlorobutane, CH₃CBrClCH₂CH₃

(a) Only C3 has a single hydrogen. Its groups are H, CH₃, CH₂CH₃ and CH₂CH₂OH. The last two both start with CH₂, but one ends in OH and the other in H, so all four differ. C3 is chiral.

(b) C2 carries H, CH₃, CH₃ and CH₂CH₃. Two methyl groups are identical, so there is no chiral centre.

(c) C1 carries H, Cl, F and CH₃. C1 is chiral. A carbon with three different atoms and one group counts in exactly the same way as one with four groups.

(d) C2 carries Br, Cl, CH₃ and CH₂CH₃. There is no hydrogen on this carbon, but the four groups are different, so C2 is chiral.

Enantiomers and plane-polarised light

Any molecule with a single asymmetric carbon has two forms, the optical isomers or enantiomers. Each is the mirror image of the other, and no rotation lets one sit exactly on the other: they are non-superimposable mirror images. According to the specification, the pair also behave differently towards plane-polarised light.

Ordinary light has waves vibrating in every plane at right angles to its direction of travel. A polarising filter removes all but one of these planes, giving plane-polarised light. When this light passes through a solution of a single enantiomer, the plane of vibration is turned through an angle. Its partner, tested at the same concentration and under the same conditions, turns the plane through an equal angle the other way. We call a substance optically active if it turns the plane of polarisation.

Apart from this, the two enantiomers of a compound have the same melting point, boiling point and solubility, and the same chemical reactions with most reagents. That is why you cannot tell them apart with a melting point apparatus but can with polarised light.

Drawing enantiomers

A flat formula such as CH₃CHClCOOH is the same for both enantiomers, so a drawing has to show how the four groups are arranged in space around the chiral carbon. The usual conventions are:

• plain line: the bond lies flat on the page
• wedge (▼ below): the bond points out of the page, towards the viewer
• hashed bond (⋮ below): the bond points into the page, away from the viewer

Give the chiral carbon two flat bonds, a wedge and a hashed bond. Then draw a mirror line and reflect the molecule: anything on the left moves to the right, and wedges stay wedges.

Worked example 2: the two enantiomers of 2-chloropropanoic acid

Step 1: find the chiral centre. C2 in CH₃CHClCOOH carries H, Cl, CH₃ and COOH. All four differ, so C2 is the asymmetric carbon.

Step 2: draw one enantiomer. Put CH₃ on the left and COOH on the right, both in the plane. Put Cl on a wedge and H on a hashed bond.

Step 3: reflect it. Swap left and right. Cl stays on the wedge and H stays on the hashed bond.

These are structural formulas of the two enantiomers:

             Cl          :          Cl
             ▼           :          ▼
    H3C  --  C  --  COOH : HOOC --  C  --  CH3
             ⋮           :          ⋮
             H           :          H

       enantiomer A    mirror   enantiomer B

Step 4: the displayed formulas. A displayed formula shows every atom and every bond. Keep exactly the same arrangement around C2, and draw out the bonds inside CH₃ and COOH.

Enantiomer A:

         H     Cl    O
         |     ▼    //
    H -- C --- C -- C
         |     ⋮     \
         H     H      O -- H

Enantiomer B (A reflected left to right):

      O      Cl     H
      \\      ▼     |
        C --- C --- C -- H
      /       ⋮     |
H -- O        H     H

If you swapped Cl and H as well as reflecting, you would have drawn enantiomer A again, turned round. Only reflect.

Racemic mixtures

When the two enantiomers are present in equal amounts, the sample is a racemic mixture, also called a racemate.

Why a racemate is optically inactive

The two enantiomers turn the plane of polarisation through equal angles but in opposite senses. A racemate holds as many molecules of one as of the other, so their effects cancel and the mixture produces no overall rotation. Every molecule in the mixture is still chiral. It is the mixture, not the molecules, that is optically inactive.

How racemates form

A racemate forms when a reaction creates a new chiral centre at a carbon that was planar in the reactant or in an intermediate. The attacking species can approach either face of the planar carbon, and each face is equally likely to be attacked. Attack on one face gives one enantiomer and attack on the other face gives the other, so they form in equal amounts.

You meet two such reactions in this course.

Nucleophilic addition to C=O (section 3.3.8.1). The carbonyl carbon is trigonal planar. With KCN followed by dilute acid, the cyanide ion adds to that carbon from above or below the plane. If the carbon ends up with four different groups, the hydroxynitrile is formed as a racemate. That is the reason section 3.3.8.1 links aldehydes and unsymmetrical ketones with mixtures of enantiomers.

Electrophilic addition to an unsymmetrical alkene (section 3.3.4.2). Hex-1-ene reacts with HBr through a secondary carbocation, CH₃CH₂CH₂CH₂C⁺HCH₃. The positive carbon is planar, so Br⁻ bonds to either face equally. The major product, 2-bromohexane, has a chiral C2 (H, Br, CH₃ and C₄H₉), and it forms as a racemate. The minor product, 1-bromohexane, has no chiral centre.

Worked example 3: racemate or not?

Each compound below reacts with KCN followed by dilute acid. Decide whether the product is a racemate.

Carbonyl compound	Hydroxynitrile formed	Groups on the C–OH carbon	Racemate?
Butanal, CH₃CH₂CH₂CHO	2-hydroxypentanenitrile, CH₃CH₂CH₂CH(OH)CN	H, OH, CN, CH₂CH₂CH₃	Yes
Heptan-4-one, (CH₃CH₂CH₂)₂CO	2-hydroxy-2-propylpentanenitrile, (CH₃CH₂CH₂)₂C(OH)CN	OH, CN and two identical propyl groups	No chiral centre, so no
4-methylpentan-2-one, CH₃COCH₂CH(CH₃)₂	2-hydroxy-2,4-dimethylpentanenitrile, (CH₃)₂CHCH₂C(OH)(CH₃)CN	OH, CN, CH₃, CH₂CH(CH₃)₂	Yes

Heptan-4-one is a symmetrical ketone, so the product has two identical groups on the new C–OH carbon. In the last product, C4 carries two CH₃ groups, so C2 is the only chiral centre.

Worked example 4: counting isomers including enantiomers

How many isomers have the formula C₃H₆BrCl, counting enantiomers separately?

Step 1: list the structural isomers. With a three-carbon chain there are five:

Isomer	Structure	Chiral carbon?
1-bromo-1-chloropropane	CHBrClCH₂CH₃	C1: H, Br, Cl, C₂H₅
1-bromo-2-chloropropane	CH₂BrCHClCH₃	C2: H, Cl, CH₃, CH₂Br
2-bromo-1-chloropropane	CH₂ClCHBrCH₃	C2: H, Br, CH₃, CH₂Cl
1-bromo-3-chloropropane	CH₂BrCH₂CH₂Cl	none
2-bromo-2-chloropropane	CH₃CBrClCH₃	none (two CH₃ groups)

Step 2: add the enantiomers. Three structural isomers each exist as a pair of enantiomers, so each adds one extra isomer.

Total: 5 + 3 = 8 isomers.

Common errors

• Choosing a carbon whose groups match only at first glance. In 3-methylpentan-1-ol, CH₂CH₃ and CH₂CH₂OH both begin with CH₂ but are different groups.
• Assuming a chiral centre must carry a hydrogen. 2-bromo-2-chlorobutane has none on C2.
• Leaving out “non-superimposable”: any molecule, chiral or not, has a mirror image.
• Drawing both enantiomers with every bond in the plane. Without a wedge and a hashed bond, the two drawings show nothing different.
• Reflecting the molecule and also swapping two groups, which gives the original enantiomer again.
• Saying a racemate is inactive because its molecules “are not chiral”. They are chiral; their rotations cancel.
• Explaining racemate formation without the three key points: a planar carbon, attack from either side, and equal probability of each.

Where to go next

Use the revision notes for a condensed version of this topic, then work through the practice questions. The free diagnostics help you choose what to revise next, and the course hub links every OxfordAQA Chemistry page.

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.3.7 Optical isomerism (3.3.7.1), International A2 content.