H3 Molecular Stereochemistry (9813)

Learn H3 stereochemical projections, conformations, configurations, optical activity and transition-metal stereochemistry through seven ordered lessons.

GCE A-Level H3 Chemistry 9813-2027

Move between two-dimensional drawings and three-dimensional molecules, then use symmetry, priority rules and energy to explain conformations, configurations and complex-ion stereochemistry. The lessons run from projections to conformations, then configurations and optical activity, and finish with the stereochemistry of complex ions.

Prerequisites

Tetrahedral, square-planar and octahedral geometry.
Structural, displayed and wedge–dash formulae, including chiral centres from H2 organic chemistry.
Bond rotation, steric repulsion and transition-metal ligand notation.

Lessons (Recommended Order)

Follow the seven lessons in order on a first pass: reading projections accurately comes before every later judgement. For revision, go straight to the lesson for the statement you need.

2.1 Molecular Stereochemistry

Common Exam Traps

Changing connectivity while redrawing a projection or reversing the viewing direction.
Using visual size instead of CIP priority, or assuming R means positive optical rotation.
Calling a coordination complex chiral without testing its complete mirror image for superimposability.

Learning goals

Use Newman and other stereochemical projections
Interpret stereochemical projections
Explain conformational isomerism and rotational barriers
Assign cis/trans and E/Z configurations
Apply R/S, optical activity and optical-purity reasoning
Identify cis/trans isomerism in square-planar and octahedral complexes
Identify enantiomerism in octahedral complexes

Syllabus statements covered

use stereochemical projections, including Newman projections, to represent molecules
interpret stereochemical projections of molecules [knowledge of Fischer projections is not required]
apply their understanding of the following types of isomerism to explain the stereochemistry of molecules, including saturated ring systems: — conformational isomerism, including energy barriers to rotation and interconversion
apply their understanding of the following types of isomerism to explain the stereochemistry of molecules, including saturated ring systems: — cis-trans isomerism, including E, Z nomenclature
apply their understanding of the following types of isomerism to explain the stereochemistry of molecules, including saturated ring systems: — enantiomerism and diastereomerism - R, S configuration - optical activity optical purity as the excess of one enantiomer, including calculation of optical purity by the equation: [α ]obs optical purity = × 100% [α ]pure material
recognise that transition element complexes can also exhibit stereoisomerism: — cis-trans isomerism, e.g. square planar complexes such as [Pt(NH3)2Cl2] and octahedral complexes such as [Co(NH3)4(H2O)2]2+
recognise that transition element complexes can also exhibit stereoisomerism: — enantiomerism, e.g. [Ni(H2NCH2CH2NH2)3]2+ and [Ni(H2NCH2CH2NH2)2(H2O)2]2+ [identification of fac-mer isomerism is not required]

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What does a Newman projection of ethane represent?

The molecule viewed straight along one C–C bond: the near carbon atom is the point where three lines meet, the far carbon atom is the circle behind it, and the angle between a front bond and a back bond is the dihedral angle.
The molecule viewed perpendicular to the C–C bond, with the circle representing the electron cloud of the bond.
A flat map of the molecule in which every bond angle is drawn as 90°.
A representation in which bold wedges point towards the viewer and dashes away from it.

Why is the staggered conformation of ethane lower in energy than the eclipsed conformation?

In the staggered conformation the C–H bonding pairs on the two carbon atoms are as far apart as possible, so repulsion between them is least; eclipsing brings each front C–H bond in line with a back C–H bond and raises the energy.
The staggered conformation has stronger C–C bonding, because the p orbitals overlap better at a dihedral angle of 60°.
The eclipsed conformation is a different compound with weaker bonds.
The staggered conformation is lower in energy because the hydrogen atoms attract each other at 60°.

Why is the E/Z system used for alkenes rather than the older cis/trans labels?

cis and trans need an identical group on each doubly bonded carbon atom to refer to; E/Z instead ranks the two groups on each carbon atom by Cahn-Ingold-Prelog priority, so it applies even when all four groups differ.
cis and trans apply only to rings, while E/Z applies only to double bonds.
Z always means the same as trans and E the same as cis, so the new labels are simply clearer names.
E/Z describes the conformation about the double bond, while cis/trans describes the configuration.

What makes a carbon atom a stereocentre, and what does it mean to say a substance is optically active?

The carbon atom carries four different groups, so its mirror image cannot be superimposed on it; a single pure enantiomer of such a substance rotates the plane of plane-polarised light passed through it.
The carbon atom carries four different groups, and optical activity means the substance is coloured when viewed through a polarising filter.
The carbon atom is part of a double bond, and optical activity means the substance exists as E and Z forms.
The carbon atom carries at least two different groups, and optical activity means the molecule absorbs polarised light.

The square-planar complex [Pt(NH₃)₂Cl₂] exists as two distinct compounds with different properties. Why?

The two chloride ligands can occupy adjacent positions, 90° apart, giving the cis isomer, or opposite positions, 180° apart, giving the trans isomer; the ligands cannot move between these arrangements without breaking bonds.
One compound is square planar and the other tetrahedral.
The two forms are non-superimposable mirror images of each other.
One form has platinum in oxidation state +2 and the other in +4.

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A sawhorse drawing shows the C1–C2 bond of butane drawn diagonally across the page, with the substituents on each carbon atom splayed out. What happens when the same conformation is redrawn as a Newman projection?

The viewpoint moves to look straight down the C1–C2 bond, so C1 becomes the central point, C2 becomes the circle behind it, and the dihedral angles between the substituents are preserved and shown directly.
The dihedral angles change, because a Newman projection always shows the staggered arrangement.
The molecule is redrawn flat, so all the substituents end up in the plane of the paper.
C2 becomes the central point and C1 the circle, because the Newman projection is always drawn from the far end.

The energy barrier to rotation about the C–C bond of ethane is about 12 kJ mol⁻¹. What follows at room temperature?

Molecules have more than enough thermal energy to cross the barrier many times a second, so the conformations interconvert constantly and no individual conformer can be isolated.
The staggered and eclipsed forms can be separated by fractional distillation, because they differ in energy.
Rotation stops completely, because 12 kJ mol⁻¹ is a large amount of energy on the molecular scale.
Every molecule sits permanently in the eclipsed conformation, because that is where rotation is blocked.

In one stereoisomer of 2-bromobut-2-ene, CH₃–CBr=CH–CH₃, the bromine atom and the methyl group attached to C3 lie on the same side of the double bond. What is its label?

Z, because bromine outranks the methyl group on C2 and the methyl group outranks hydrogen on C3, and the two higher-priority groups are on the same side.
E, because the two methyl groups lie on opposite sides of the double bond.
Z, because the two methyl groups lie on the same side.
Neither, because the four groups on the double bond are all different so no label applies.

In a molecule of bromochlorofluoromethane, CHFClBr, the hydrogen atom points directly away from the viewer, and the sequence Br → Cl → F appears clockwise. What is the configuration?

R, because the priority order by atomic number is Br > Cl > F > H, the lowest-priority atom points away, and the remaining three run clockwise.
S, because clockwise corresponds to S and anticlockwise to R.
R, but only because hydrogen happens to be the lightest atom; the direction of view does not matter.
Neither, because a carbon atom bonded to three halogens cannot be a stereocentre.

The octahedral complex ion [Co(NH₃)₄Cl₂]⁺ exists as two geometrical isomers. How do they differ?

In the cis isomer the two chloride ligands occupy adjacent vertices, with a Cl–Co–Cl angle of 90°; in the trans isomer they occupy opposite vertices, with an angle of 180°.
In one isomer the cobalt ion is in oxidation state +3 and in the other +2.
One isomer has four ammonia ligands and the other has two ammonia and two water ligands.
They are non-superimposable mirror images of each other.

Practise after feedback

After a check, practise the skills it showed you need to work on.

About 10 minutes

Questions are picked at random each time you start. You'll see the answer after each question. It's for practice only and doesn't count towards mastery.

Recent attempts

History is stored only in this browser.

No completed attempts are saved yet.

This check needs JavaScript to record answers. The reviewed questions are listed below for study.

The viewpoint moves to look straight down the C1–C2 bond, so C1 becomes the central point, C2 becomes the circle behind it, and the dihedral angles between the substituents are preserved and shown directly.
The dihedral angles change, because a Newman projection always shows the staggered arrangement.
The molecule is redrawn flat, so all the substituents end up in the plane of the paper.
C2 becomes the central point and C1 the circle, because the Newman projection is always drawn from the far end.

The energy barrier to rotation about the C–C bond of ethane is about 12 kJ mol⁻¹. What follows at room temperature?

Molecules have more than enough thermal energy to cross the barrier many times a second, so the conformations interconvert constantly and no individual conformer can be isolated.
The staggered and eclipsed forms can be separated by fractional distillation, because they differ in energy.
Rotation stops completely, because 12 kJ mol⁻¹ is a large amount of energy on the molecular scale.
Every molecule sits permanently in the eclipsed conformation, because that is where rotation is blocked.

In one stereoisomer of 2-bromobut-2-ene, CH₃–CBr=CH–CH₃, the bromine atom and the methyl group attached to C3 lie on the same side of the double bond. What is its label?

Z, because bromine outranks the methyl group on C2 and the methyl group outranks hydrogen on C3, and the two higher-priority groups are on the same side.
E, because the two methyl groups lie on opposite sides of the double bond.
Z, because the two methyl groups lie on the same side.
Neither, because the four groups on the double bond are all different so no label applies.

In a molecule of bromochlorofluoromethane, CHFClBr, the hydrogen atom points directly away from the viewer, and the sequence Br → Cl → F appears clockwise. What is the configuration?

R, because the priority order by atomic number is Br > Cl > F > H, the lowest-priority atom points away, and the remaining three run clockwise.
S, because clockwise corresponds to S and anticlockwise to R.
R, but only because hydrogen happens to be the lightest atom; the direction of view does not matter.
Neither, because a carbon atom bonded to three halogens cannot be a stereocentre.

The octahedral complex ion [Co(NH₃)₄Cl₂]⁺ exists as two geometrical isomers. How do they differ?

In the cis isomer the two chloride ligands occupy adjacent vertices, with a Cl–Co–Cl angle of 90°; in the trans isomer they occupy opposite vertices, with an angle of 180°.
In one isomer the cobalt ion is in oxidation state +3 and in the other +2.
One isomer has four ammonia ligands and the other has two ammonia and two water ligands.
They are non-superimposable mirror images of each other.

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Answer 5 questions. You'll see your score, the answers and explanations at the end. Your result can count towards your course progress.

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Check my progress

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H3 Molecular Stereochemistry (9813)

Prerequisites

Lessons (Recommended Order)

2.1 Molecular Stereochemistry

Common Exam Traps

Syllabus statements covered

Check what I know

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Practise

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Practise after feedback

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Check my progress

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