Drawing and Understanding Enantiomers Revision Notes | A Level AQA Chemistry

1. Why Mirror Images Matter in Chemistry

Your left hand and right hand look identical at a glance, yet you cannot superimpose one on the other — they are mirror images that do not match. Some molecules behave in exactly the same way, and that difference can be the gap between a life-saving drug and a harmful one.

This phenomenon is called optical isomerism, and the two mirror-image molecules are called enantiomers. At A-level, you need to be able to spot when a molecule has this property, draw both forms correctly, and explain what makes them different.

2. The Chiral Centre: Where It All Begins

The origin of optical isomerism is a single carbon atom bonded to four different groups. This carbon is called a chiral centre (sometimes called an asymmetric carbon). Because all four substituents are different, there is no plane of symmetry through that carbon, and two distinct spatial arrangements are possible.

A quick way to check: write out all four groups attached to the carbon in question. If any two are the same, the carbon is not a chiral centre. All four must be distinct. For example, in 2-bromobutane ($CH_3CHBrCH_2CH_3$), the central carbon carries $-H$, $-Br$, $-CH_3$, and $-CH_2CH_3$ — four different groups, so it is a chiral centre.

3. Drawing Enantiomers: Structural and Displayed Formulae

This is the core skill the specification tests. You need to draw both enantiomers of a chiral molecule, showing the three-dimensional arrangement around the chiral centre. The standard convention uses wedge-and-dash notation: a solid wedge ($\wedge$) for a bond coming towards you, and a dashed wedge (- - -) for a bond going away from you, with ordinary lines for bonds in the plane of the paper.

The two enantiomers are drawn as mirror images of each other. Swap the positions of any two groups around the chiral centre and you generate the other enantiomer. The critical check is that no rotation or translation of one structure can make it identical to the other — they are non-superimposable.

Worked example: Drawing the Two Enantiomers of 2-Bromobutane

Draw the structural formulae of both enantiomers of 2-bromobutane, using wedge-and-dash notation to show the arrangement around the chiral centre.

1. $CH_3 - \underset{*}{C}H(Br) - CH_2CH_3$ — Carbon-2 carries four different groups: $-H$, $-Br$, $-CH_3$, and $-CH_2CH_3$. Mark it with an asterisk to keep track.
2. Place $-CH_3$ and $-CH_2CH_3$ as ordinary lines (in the plane of the paper). Draw $-Br$ as a solid wedge (coming towards you) and $-H$ as a dashed wedge (going away from you). — This gives one specific three-dimensional arrangement around the chiral centre. Label it enantiomer 1.
3. Reflect the structure in a vertical mirror plane. $-Br$ is now a dashed wedge (going away) and $-H$ is now a solid wedge (coming towards you). $-CH_3$ and $-CH_2CH_3$ remain as ordinary lines. — Swapping the positions of $-Br$ and $-H$ (or any two groups) generates the second enantiomer. The two structures are mirror images.
4. Attempt mentally to rotate enantiomer 1 to match enantiomer 2. No rotation achieves this — the $-Br$ and $-H$ positions cannot be reconciled without breaking bonds. — This confirms they are genuine enantiomers, not the same compound drawn differently.
5. In a full displayed formula, show every bond explicitly. The wedge-and-dash bonds at the chiral centre must still be present; all C$-$H bonds on other carbons are shown as ordinary lines. — AQA may ask for a displayed formula specifically. The wedge-and-dash notation at the chiral centre is still required to convey the 3D arrangement.

Answer: Two wedge-and-dash structures with $-Br$ and $-H$ swapped between solid and dashed wedges, connected to the same $-CH_3$ and $-CH_2CH_3$ groups in the plane of the paper. The structures are non-superimposable mirror images.

Notice that the only difference between the two drawings is the spatial orientation of the groups around the chiral centre. The connectivity — which atom is bonded to which — is identical in both enantiomers. This is what makes them stereoisomers rather than structural isomers.

4. Properties of Enantiomers: What Is the Same and What Differs

Enantiomers are chemically identical in most respects. They have the same melting point, boiling point, solubility, and density. They react at the same rate with achiral reagents. The one property that distinguishes them is their interaction with plane-polarised light.

When plane-polarised light passes through a solution of one pure enantiomer, the plane of polarisation rotates by a fixed angle. One enantiomer rotates it clockwise — described as dextrorotatory and given the symbol (+). The other rotates it anticlockwise — described as laevorotatory and given the symbol (−). The magnitude of rotation is the same for both; only the direction differs.

Many laboratory synthesis reactions produce a racemic mixture because the reagent attacks the planar intermediate (such as a carbonyl carbon) from either face with equal probability. Biological systems, by contrast, typically produce only one enantiomer because enzymes are themselves chiral and only accept one spatial arrangement.

5. Biological and Pharmaceutical Significance

The distinction between enantiomers matters enormously in medicine. Biological molecules such as enzymes and receptors are chiral, so they interact differently with each enantiomer of a drug. One enantiomer may be therapeutically active while the other is inactive, or — in serious cases — harmful.

The most cited historical example is thalidomide. One enantiomer relieved morning sickness as intended, but the other enantiomer caused severe birth defects. Crucially, even administering the pure therapeutic enantiomer would not have prevented harm, because the compound undergoes racemisation in the body — converting between enantiomers in vivo. This case drove the modern pharmaceutical requirement to test and often isolate individual enantiomers.

Amino acids provide another key example: naturally occurring amino acids are almost exclusively the L-form (L-configuration). Enzymes that synthesise proteins are chiral and will only incorporate the L-enantiomer. The D-form of an amino acid is not incorporated into human proteins, illustrating how chirality governs biological function at the molecular level.

6. Exam Technique: Drawing and Explaining Enantiomers

AQA questions on optical isomerism tend to fall into three categories: identifying a chiral centre, drawing both enantiomers, and explaining properties. Each requires a slightly different approach, and the marks are lost in predictable places.

When drawing enantiomers, the most common error is drawing two structures that are actually the same compound rotated — not genuine mirror images. Always draw a mirror line between your two structures and check that each group on the left has its counterpart reflected on the right. If you swap only one pair of groups (e.g. $-Br$ and $-H$) while keeping the other two in the plane of the paper, you will always generate the correct mirror image.

Worked example: Answering an AQA 'Draw the Enantiomers' Question

Draw the two enantiomers of lactic acid ($CH_3CH(OH)COOH$), using wedge-and-dash notation. [2 marks]

1. Carbon-2 in lactic acid carries: $-H$, $-OH$, $-CH_3$, and $-COOH$. All four groups are different. — Confirming the chiral centre before drawing prevents errors. Mark it with an asterisk in your working.
2. Draw $-CH_3$ and $-COOH$ as ordinary lines in the plane of the paper. Draw $-OH$ as a solid wedge (towards you) and $-H$ as a dashed wedge (away from you). — This is one valid spatial arrangement. [1 mark for a correct 3D representation with wedge-and-dash notation]
3. Reflect: $-OH$ becomes a dashed wedge (away from you) and $-H$ becomes a solid wedge (towards you). $-CH_3$ and $-COOH$ remain as ordinary lines. — [1 mark for the second structure being a genuine non-superimposable mirror image of the first]
4. Place a vertical dashed line between the two structures to indicate the mirror relationship. — AQA mark schemes reward clarity. The mirror line signals to the examiner that you understand the relationship between the two structures.

Answer: Two wedge-and-dash structures of lactic acid with $-OH$ and $-H$ swapped between solid and dashed wedges, separated by a mirror line. Both structures show $-CH_3$ and $-COOH$ in the plane of the paper.

When an exam question asks you to explain why a molecule shows optical isomerism, the mark scheme expects two elements: identification of the chiral centre (a carbon with four different groups) and the statement that the two arrangements are non-superimposable mirror images. Stating only one of these typically earns one mark out of two.

For 2-bromobutane, the structural formula CH3CH(Br)CH2CH3 tells you the connectivity. A displayed or wedge-and-dash formula is needed to show the three-dimensional arrangement at the chiral carbon.