Whiteboard Chemistry with Joe White

Optical Isomerism

Chirality and the chiral centre, how to draw a pair of enantiomers correctly in 3D, why they rotate plane-polarised light in opposite directions, and how a racemic mixture forms.

AQA 7404/7405 Paper 2 A-level only
Hero motif · to come Non-superimposable mirror images
Where this sits

This extends the stereoisomerism you met at AS. There you saw E–Z isomerism, from restricted rotation about a C=C. Optical isomerism is the other kind of stereoisomerism: it comes from a chiral centre, and the two isomers are mirror images. Two things win the marks — drawing the 3D structures correctly, and explaining the effect on plane-polarised light.

The chiral centre & drawing enantiomers

Optical isomerism arises from chirality. A carbon atom is chiral (asymmetric) when it is bonded to four different groups. Such a carbon — the chiral centre — can be arranged in two ways that are non-superimposable mirror images of each other, called enantiomers (optical isomers). At A-level this is limited to molecules with a single chiral centre.

Key definitions

A chiral (asymmetric) carbon is a carbon atom bonded to four different atoms or groups.

Enantiomers (optical isomers) are non-superimposable mirror images of each other.

Drawing enantiomers — get these right
  • Draw the tetrahedral carbon with two normal (in-plane) bonds, then a wedge (coming towards you) and a dash (going back) — and the wedge and dash must be next to each other (adjacent), never on opposite sides. Opposite wedge-and-dash is drawn wrongly and loses the mark.
  • Show the connectivity clearly: the chiral carbon must be joined directly to each group (for example write C–COOH and C–NH2, with the carbon bonded to the carboxyl carbon and to the nitrogen).
  • The second enantiomer is the mirror image — reflect it across a vertical line, keeping every group connected the same way.
Figure F1 · illustration to come Drawing a chiral centre — right vs wrong Wedge & dash adjacent (✓), not opposite (✗); show connectivity.
The wedge and dash sit next to each other — drawing them opposite is a common lost mark.
🧪 Exam-style questions
Q1 [1 mark]

Which compound forms an optically active compound on reduction? Tick (✓) one box.

Q2 [2 marks]

2-Hydroxypropanenitrile, CH3CH(OH)CN, displays optical isomerism. Draw three-dimensional representations of its two enantiomers, showing how they are related.

Show answer

Two tetrahedral carbons drawn as mirror images: the chiral C bonded to OH, CN, CH3 and H, with the wedge and dash adjacent and the two structures reflected across a mirror line. 2 marks

Same connectivity in both; the only difference is the 3D arrangement. Opposite wedge/dash loses a mark.

Source: AQA A-Level Chemistry past papers.

Plane-polarised light & racemates

Enantiomers are almost identical — same bonds, same physical and chemical properties — and differ in only one measurable way: their effect on plane-polarised light. One enantiomer rotates the plane of polarisation clockwise, the other rotates it anticlockwise by the same amount. That is how you distinguish two enantiomers: pass plane-polarised light (in a polarimeter) through each and they rotate it in opposite directions.

Key definition

A racemic mixture (racemate) is a mixture containing equal amounts of the two enantiomers of a compound.

Because a racemate has equal amounts of each enantiomer, their equal and opposite rotations cancel out — a racemate is optically inactive (no net rotation of plane-polarised light).

How a racemate forms

A racemate forms whenever a chiral centre is created by attack on a planar group. The classic case is nucleophilic addition to a C=O: the carbonyl carbon is trigonal planar, so the incoming nucleophile can attack from either side of the plane with equal probability. Equal attack from both sides gives equal amounts of the two enantiomers — a racemic mixture.

Figure F2 · illustration to come Why a racemate forms Equal attack on a planar C=O from both sides.
Attack from both faces of the planar C=O is equally likely, so the two enantiomers form in equal amounts.
🧪 Exam-style questions
Q3 [2 marks]

Describe how separate samples of the two enantiomers of a compound could be distinguished.

Show answer

Pass plane-polarised light (using a polarimeter) through each sample. 1 mark

The two enantiomers rotate the plane of polarised light by equal amounts but in opposite directions (one clockwise, one anticlockwise). 1 mark

Q4 [3 marks]

Ethanal reacts with KCN followed by dilute acid to form a racemic mixture of 2-hydroxypropanenitrile. Explain why a racemic mixture forms.

Show answer

The carbonyl carbon of ethanal is trigonal planar. 1 mark

The cyanide ion (nucleophile) can attack from either side of the plane, with equal probability. 1 mark

So equal amounts of the two enantiomers are formed — a racemic mixture. 1 mark

Q5 [2 marks]

Butanone is reduced by NaBH4 then dilute acid. By considering the mechanism, explain why the product has no effect on plane-polarised light.

Show answer

The hydride ion (from NaBH4) attacks the planar C=O from either side equally, so equal amounts of the two enantiomers of butan-2-ol form — a racemic mixture. 1 mark

The equal and opposite rotations of the two enantiomers cancel, so the mixture is optically inactive. 1 mark

Source: AQA A-Level Chemistry past papers.

Spot it, draw it, explain it

Optical-isomerism questions come in three parts, and each has a reliable method.

The three tasks
  • Spot the chiral centre — find a carbon bonded to four different groups.
  • Draw both enantiomers — tetrahedral carbon, wedge and dash adjacent, explicit connectivity, second one the mirror image.
  • Explain the behaviour — opposite rotation of plane-polarised light; a racemate is optically inactive because the rotations cancel.
Precision points
  • The four groups must be different — a carbon with two identical groups (e.g. two CH3) is not chiral.
  • When you draw the 3D structures, the wedge and dash go next to each other, not opposite — and the connectivity (C–COOH, C–NH2, C–OH…) must be clear.
  • A racemate is optically inactive because the two enantiomers’ rotations cancel — not because the molecules are achiral.
🧪 Capstone question
Q6 [2 marks]

Justify the statement that there are no chiral centres in 3-aminopentane, CH3CH2CH(NH2)CH2CH3.

Show answer

The only carbon that could be a chiral centre (C3) is bonded to two identical ethyl (CH2CH3) groups. 1 mark

A chiral centre needs four different groups; because two are the same, there is no chiral centre. 1 mark

Source: AQA A-Level Chemistry past papers.

3.3.7 Optical isomerism — Quick-reference summary
  • Chiral centre: a carbon bonded to four different groups. It gives two enantiomers — non-superimposable mirror images.
  • Drawing: tetrahedral carbon, two in-plane bonds, then a wedge and dash adjacent to each other (never opposite); keep the connectivity explicit; the second isomer is the mirror image.
  • Optical activity: enantiomers rotate plane-polarised light by equal amounts in opposite directions — this is how you distinguish them.
  • Racemic mixture (racemate): a 50:50 mixture of the two enantiomers. Their rotations cancel, so it is optically inactive.
  • How racemates form: a nucleophile attacks a planar C=O equally from both sides, giving equal amounts of each enantiomer.

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