Related topics: Amino Acids, Proteins & DNA
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). You draw the 3D structures for molecules with a single chiral centre, but you can be asked to count the optical isomers of a molecule with several chiral centres: the 2n rule below. It matters biologically because an enzyme’s active site is stereospecific: it binds one enantiomer and not its mirror image.
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.
Precision points: drawing enantiomers
- 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.
- A quicker way to draw the mirror image: keep the same tetrahedron and swap any two groups over. Swapping two groups on a chiral centre inverts it: you get the other enantiomer. The one thing never to do is swap two groups and then mirror it: that is two inversions, which cancel, and you are back to the isomer you started with.
Interactive: spot the chiral centre
Interactive: try to superimpose the enantiomers
Molecule A is fixed. Turn molecule B any way you like. Can you ever make all four groups line up at once?
A: one enantiomer
B: its mirror image
Worked example: is this carbon chiral?
Butanone, CH3CH2COCH3, is reduced to butan-2-ol, CH3CH(OH)CH2CH3. Does butan-2-ol show optical isomerism?
Step 1. List the four groups on the candidate carbon: the carbon that gained the OH is bonded to OH, H, CH3 and CH2CH3.
Step 2. Compare them: all four are different, so the carbon is a chiral centre and butan-2-ol shows optical isomerism. Its separate enantiomers are optically active, but ordinary reduction with achiral reagents forms equal amounts of them: the racemic mixture is optically inactive.
Step 3. Check the near-miss: reduce propanone instead and you get propan-2-ol, CH3CH(OH)CH3: the central carbon carries two identical CH3 groups, so it is not chiral and the product is not optically active. Writing out all four groups is what earns the mark either way.
Counting optical isomers: the 2n rule
A molecule can have more than one chiral centre. Each chiral centre can be arranged two ways, and the arrangements are independent, so the number of optical isomers doubles for every chiral centre:
number of optical isomers = 2n, where n = number of chiral centres
So one chiral centre gives 2 isomers, two give 4, three give 8, four give 16. The whole skill is finding every chiral centre (each carbon with four different groups) and not missing one. The trap is a group like threonine’s side chain, where the carbon bearing the –OH is also chiral.
Exam questions
Q1a
[2 marks]
2-Hydroxypropanenitrile displays optical isomerism. Draw three-dimensional representations of the two enantiomers of 2-hydroxypropanenitrile, showing how the two structures are related to each other.
Show answer
One 3D enantiomer. 1 mark
Second enantiomer correctly drawn as 3D mirror image of first. 1 mark
Each tetrahedral carbon needs two in-plane bonds plus a wedge and a dash drawn adjacent. AQA’s schemes for this drawing require at least one wedge and one dash in each structure, with no two in-plane bonds at 180° to each other.
Q1b
[2 marks]
Describe how separate samples of each of these enantiomers could be distinguished.
Show answer
Plane-polarised light. 1 mark
Rotated in opposite directions. 1 mark
Q2a
[1 mark]
This question is about isomers with the molecular formula C5H10O. Draw the skeletal formula of a branched chain aldehyde with molecular formula C5H10O that is optically active.
Show answer
2-Methylbutanal, CH3CH2CH(CH3)CHO: the carbon bonded to the CHO, CH3, C2H5 and H is the chiral centre. 1 mark
Q2b
[2 marks]
Describe how you distinguish between separate samples of the two enantiomers of the branched chain aldehyde C5H10O.
Show answer
Use plane polarised light. 1 mark
Rotates (the plane of) it in opposite directions. 1 mark
Q3
[1 mark]
Compounds V, W, X and Y are isomers with the molecular formula C5H10O2. Isomers V and W are carboxylic acids with formulas that can be written as C4H9COOH. Isomer V has an asymmetric carbon atom. Deduce the structure of V.
Show answer
CH3CH2CH(CH3)COOH, 2-methylbutanoic acid. 1 mark
Work backwards from the definition: the asymmetric carbon needs four different groups, so put the COOH, a CH3, an ethyl group and an H on the same carbon.
Q4
[1 mark]
Justify the statement that there are no chiral centres in 3-aminopentane.
Show answer
No carbon atom is attached to 4 different groups. 1 mark
Also accepted: the central carbon has two (identical) alkyl groups, or the molecule is symmetrical. In CH3CH2CH(NH2)CH2CH3 the carbon carrying the NH2 is bonded to two identical ethyl groups.
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.
Precision points: saying what is planar
Mark schemes credit “planar carbonyl group” or “trigonal planar around the carbonyl carbon”, and have refused “the molecule is planar” (the rest of the molecule isn’t) and “a planar bond” (a bond can’t be planar). Pin the word planar to the carbonyl group.
Worked example: the racemate explanation, stage by stage
Propanal, CH3CH2CHO, reacts with KCN then dilute acid. Explain why the product mixture has no effect on plane-polarised light. Extended answers like this are built from three stages:
Stage 1. Formation of the product: the carbonyl group of propanal is planar, so the CN− nucleophile attacks the carbonyl carbon from either side of the plane, with equal probability.
Stage 2. Nature of the product: the product, 2-hydroxybutanenitrile CH3CH2CH(OH)CN, has a carbon bonded to four different groups (a chiral centre), so it exists as two enantiomers, formed in equal amounts: a racemic mixture.
Stage 3. Optical activity: the enantiomers rotate plane-polarised light equally but in opposite directions, so in the 50:50 mixture the rotations cancel: no effect on plane-polarised light.
These products are named as hydroxynitriles, and the nitrile carbon counts as carbon 1: CH3CH(OH)CN has a three-carbon chain, so it is 2-hydroxypropanenitrile: the OH sits on C-2. The same counting names any relative made from a longer aldehyde.
This mechanism is nucleophilic addition: it sits on the organic synthesis map (aldehyde/ketone → hydroxynitrile, KCN then dilute acid), and you meet it drawn out step by step in Aldehydes & Ketones (3.3.8).
Exam questions
Q5a
[1 mark]
The aldehyde CH3CH2CH2CH2CHO reacts with KCN followed by dilute acid to form a racemic mixture of the two stereoisomers of CH3CH2CH2CH2CH(OH)CN. Give the IUPAC name of CH3CH2CH2CH2CH(OH)CN.
Show answer
2-Hydroxyhexanenitrile. 1 mark
Count the nitrile carbon as C-1: six carbons in the chain, OH on C-2.
Q5b
[2 marks]
Describe how you would distinguish between separate samples of the two stereoisomers of CH3CH2CH2CH2CH(OH)CN.
Show answer
(Plane) polarised light. 1 mark
Enantiomers would rotate light in opposite directions. 1 mark
“Different” alone is not enough: you must say opposite directions.
Q5c
[3 marks]
Explain why the reaction produces a racemic mixture.
Show answer
Planar carbonyl group. 1 mark
Attack from either side. 1 mark
With equal probability, OR produces equal amounts (of the two isomers / enantiomers). 1 mark
Not accepted for the first mark: “planar molecule”, “planar bond”, “planar C=O”.
Q5d
[2 marks]
An isomer of CH3CH2CH2CH2CHO reacts with KCN followed by dilute acid to form a compound that does not show stereoisomerism. Draw the structure of the compound formed and justify why it does not show stereoisomerism.
Show answer
Structure: CH3CH2C(OH)(CN)CH2CH3, from pentan-3-one, the ketone isomer of pentanal. 1 mark
Does not contain a chiral centre, OR does not contain a C attached to 4 different groups, OR contains two identical (ethyl) groups, OR the product is symmetrical. 1 mark
The justification mark depends on a correct structure: no structure scores zero. Ethyl groups may be written as C2H5, and skeletal formulae are accepted.
Q6a
[1 mark]
Ethanal reacts with potassium cyanide, followed by dilute acid, to form 2-hydroxypropanenitrile. Name the mechanism for the reaction between potassium cyanide and ethanal.
Show answer
Nucleophilic addition. 1 mark
Both words needed, and no additional names.
Q6b
[5 marks]
The 2-hydroxypropanenitrile formed by the reaction in part (a) is a mixture of equal amounts of two isomers. State the name of this type of mixture. Explain how the structure of ethanal leads to the formation of two isomers. Draw 3D representations of the two isomers to show the relationship between them.
Show answer
M1: racemic (mixture) / racemate. 1 mark
M2: planar (around) carbonyl / C=O. 1 mark
M3: (equal chance of) attack from each side (by CN−). 1 mark
M4: a correct structure of 2-hydroxypropanenitrile (any correct 2D or 3D structure). 1 mark
M5: correct 3D representations of both isomers: 1 mark
M5 must show at least one wedge bond and one dash bond in each structure, and any bonds in the plane cannot be at 180° to each other. The second structure can be the mirror image of the first, or the same orientation with two groups swapped round.
Q7a
[1 mark]
Butanone is reduced in a two-step reaction using NaBH4 followed by dilute hydrochloric acid. Write an overall equation for the reduction of butanone using [H] to represent the reductant.
Show answer
CH3CH2COCH3 + 2[H] → CH3CH2CH(OH)CH3
1 mark
Q7b
[6 marks]
By considering the mechanism of the reaction, explain why the product has no effect on plane polarised light. This question is marked using levels of response.
Show answer
Stage 1. Formation of product: nucleophilic attack; planar carbonyl group; H− attacks from either side (stated or drawn). stage 1
Stage 2. Nature of product: product of step 1 shown (butan-2-ol); this exists in two chiral forms (stated or drawn); equal amounts of each enantiomer / racemic mixture formed. stage 2
Stage 3. Optical activity: optical isomers / enantiomers rotate the plane of polarised light equally, in opposite directions; with a racemic / equal mixture the effects cancel. stage 3
How it is marked (levels of response):
- Level 3 (5–6): all stages covered, each generally correct and virtually complete; coherent, logical progression from stage 1 to 3.
- Level 2 (3–4): all stages covered with gaps or inaccuracies, OR two stages covered and virtually complete; mainly coherent.
- Level 1 (1–2): two stages incomplete, OR one stage virtually complete; isolated statements.
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. Mark schemes demand at least one wedge and one dash in each structure, with no two in-plane bonds at 180° to each other.
- A racemate is optically inactive because the two enantiomers’ rotations cancel, not because the molecules are achiral.
Capstone quiz: four past-paper questions
Four real AQA multiple-choice questions on this topic, in the style that opens Paper 3. Pick one answer each. The reasoning appears once you commit.
Q1
[1 mark]
Which compound forms optically active compounds on reduction?
Butanone is reduced to butan-2-ol, CH3CH(OH)CH2CH3. The carbon bonded to OH carries four different groups (OH, H, CH3, CH2CH3), so butan-2-ol is chiral and its separate enantiomers are optically active. Ordinary reduction with achiral reagents produces an optically inactive racemic mixture. Propanone gives propan-2-ol (two identical CH3 groups); the alkenes are reduced to 3-methylpentane and 2-methylbutane, where every candidate carbon carries two identical alkyl groups.
Q2
[1 mark]
Which compound does not show stereoisomerism?
Both dichloropropenes have two different groups on each carbon of the C=C, so they show E–Z isomerism. In 1,2-dichloropropane, CH2ClCHClCH3, the middle carbon carries Cl, H, CH3 and CH2Cl: four different groups, a chiral centre. 1,3-Dichloropropane, ClCH2CH2CH2Cl, has no C=C and no carbon with four different groups: no stereoisomerism of either kind.
Q3
[1 mark]
Which pair of compounds does not form a racemic mixture when the compounds react?
Propanone + HCN gives (CH3)2C(OH)CN. The new carbon carries two identical CH3 groups, so no chiral centre forms and there are no enantiomers to make a racemate. The other three all create a chiral centre on a planar carbon attacked equally from both sides: but-2-ene and but-1-ene both give 2-chlorobutane with HCl, and propanal + HCN gives 2-hydroxybutanenitrile.
Q4
[1 mark]
Which does not contain an asymmetric carbon atom?
In 2-methylbutane the branch carbon carries two identical CH3 groups (plus H and CH2CH3), so it is not asymmetric, and no other carbon comes close. The other three each have a carbon with four different groups: CH3/H/C2H5/C3H7 in 3-methylhexane, OH/H/CH3/CH2OH in propane-1,2-diol, and Cl/H/CH3/C2H5 in 2-chlorobutane.
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.