Why is the staggered form of ethane lower in energy than the eclipsed form by approximately 12 kJ mol⁻¹? In ethane, CH₃–CH₃, the two methyl groups rotate relative to each other about the carbon–carbon sigma bond. Look straight down that C–C bond. When the three front C–H bonds lie between the three rear C–H bonds, the arrangement is staggered and relatively stable. When each front bond aligns with a rear bond, the arrangement is eclipsed and higher in energy. The energy difference gives the familiar approximately 12 kJ mol⁻¹ torsional barrier for ethane.
This H3 Chemistry Tutor Hougang guide explains the barrier as a physical feature of electron interactions and orbital overlap, not as a mysterious fact that students must memorise. The common teaching language is that eclipsing bonds produce torsional strain. A fuller orbital account also describes the relative stabilisation of staggered conformations through favourable bonding–antibonding orbital interactions often called hyperconjugation. Both descriptions help explain the trend; neither requires calculating a quantum-mechanical energy surface for the A-Level syllabus.
The 2027 Singapore–Cambridge H3 Chemistry 9813 syllabus explicitly includes interpreting Newman projections, conformational isomerism, barriers to rotation and interconversion. Ethane is a simple foundational example supporting those learning outcomes. A student who reconstructs its threefold rotational energy profile will be better prepared for substituted alkanes and cyclohexane conformations.
- Ethane structure: CH₃–CH₃; formula C₂H₆.
- Lowest-energy arrangement: staggered Newman projection with the nearest front/rear C–H dihedral angles at 60°.
- Highest-energy arrangement: eclipsed Newman projection with front/rear C–H bonds aligned at 0°.
- Barrier: about 12 kJ mol⁻¹, approximately 2.9 kcal mol⁻¹, for ethane rotation.
- One 360° revolution: three equivalent staggered minima and three equivalent eclipsed maxima.
- Key distinction: conformation changes through bond rotation; atoms and covalent connectivity remain unchanged.
- At ordinary temperature: interconversion is rapid and the molecule does not remain locked in one Newman projection.
- Contrast with butane: butane has one anti and two gauche staggered minima of differing energies; ethane’s three staggered minima are equivalent.
- Scientific limit: a 12-kJ barrier is not the same quantity as the equilibrium fraction in an arbitrarily narrow angular interval.
Official SEAB 2027 H3 Chemistry 9813 syllabus · OpenStax: conformations of ethane and the 12-kJ barrier · Organic chemistry teaching resource on torsional strain and orbital interactions · Hougang JC Chemistry hub.
Find: How to draw Newman · Read the dihedral angle · The 360° energy profile · What the 12-kJ barrier means · Why staggered is favoured · Why ethane differs from butane · Rotation at room temperature · Optional energy model · Worked questions · Common errors · Teaching and careers · FAQs.
Step 1: Draw Ethane as a Newman Projection
The condensed formula CH₃CH₃ hides the relative spatial arrangement of the two methyl groups. A Newman projection makes that arrangement visible by looking directly along the C–C bond. The carbon nearer the viewer is called the front carbon, represented by a point or central intersection. The carbon behind it is the rear carbon, represented by a circle around the front point.
Each carbon has three C–H bonds that appear approximately 120° apart in the end-on drawing. The front carbon’s three bonds emerge from the central point; the rear carbon’s three emerge from the circumference. We then rotate one carbon relative to the other to examine how its bonds line up.
The two carbons are still connected by a single sigma bond. Newman projection is a viewing method, not a new chemical reaction. No hydrogen has moved from one carbon to the other, and the molecule remains C₂H₆ throughout.
What does staggered look like?
Rotate the rear methyl so its three bond directions fall exactly between those of the front methyl, producing a 60° dihedral angle between a chosen nearest front/rear C–H pair. In the Newman drawing, rear bonds appear interleaved between front bonds. None of the three front bonds directly eclipses a rear bond.
What does eclipsed look like?
Rotate the rear methyl so each rear C–H bond lies behind one front C–H bond in the chosen projection. The corresponding dihedral angle is 0°. In an honest drawing, the rear bonds can be displaced slightly on the page for visibility, but the intended three-dimensional orientation is aligned.
A common student mistake is to compare two drawing layouts without identifying the viewing axis. If one structure is drawn looking from C1 toward C2 and another from C2 toward C1, differences on the page may be caused by viewpoint reversal. Use a consistent front and rear convention before classifying the conformation.
Step 2: Understand a Dihedral Angle Physically
A dihedral or torsional angle measures the relative orientation of two bonds connected to opposite ends of a bond about which rotation is being considered. In ethane, choose one H–C(front) bond and one H–C(rear) bond. The torsional relationship between them tells us where the front and rear bonds sit when viewed along the central carbon–carbon axis.
At 0°, the selected C–H bonds are eclipsed. At 60°, they are staggered. Continue rotating through 120°, 180°, 240°, 300° and 360° and the pattern repeats every 60° between an alternating maximum and minimum.
This is fundamentally different from measuring the C–C–H bond angle inside one carbon’s tetrahedral bond arrangement. The tetrahedral bond angles remain near their ordinary values while the dihedral angle changes with rotation about the sigma bond.
Step 3: Build the Threefold Energy Profile
In one full 360° rotation, the rear methyl alternates between eclipsed and staggered orientations. Because all hydrogens are identical and the two methyl groups are equivalent, there are three energetically equivalent eclipsed maxima and three equivalent staggered minima.
- 0°: eclipsed, energy maximum.
- 60°: staggered, energy minimum.
- 120°: eclipsed, another equivalent maximum.
- 180°: staggered, another equivalent minimum.
- 240°: eclipsed, third equivalent maximum.
- 300°: staggered, third equivalent minimum.
- 360°: same geometry as 0° after one full revolution.
A plotted potential-energy curve therefore has a threefold repeating pattern with alternating high and low points. The total number of conformations is not just six: there is a continuum of intermediate torsional angles between the idealised maxima and minima. The six positions are convenient landmarks.
Students sometimes confuse the three staggered minima with three constitutional isomers. They are not different compounds. They are rotationally related conformations of the same ethane molecule, and ethane’s high symmetry makes their energies equivalent.
Step 4: Interpret the 12 kJ mol⁻¹ Energy Barrier
The approximately 12-kJ-mol⁻¹ torsional barrier refers to the potential-energy difference between an eclipsed maximum and a staggered minimum in the standard ethane model. It does not mean that one C–C sigma bond requires 12 kJ mol⁻¹ to break; bond dissociation is a fundamentally different process with a much larger energy scale.
If a question draws the staggered minimum at 0 relative energy and the eclipsed maximum at +12 kJ mol⁻¹, that is a useful teaching reference. Rotation from staggered toward eclipsed climbs the barrier, and continuing rotation reaches the next equivalent staggered minimum.
The barrier is small enough for ethane to interconvert rapidly between conformations under ordinary room-temperature conditions. At the molecular level, thermal energy is distributed across many modes and collisions; molecules continually explore torsional orientations, rather than remaining permanently locked at a minimum.
What does approximately 2.9 kcal mol⁻¹ mean?
The same rotational barrier may be reported in kcal mol⁻¹. Using 1 kcal ≈ 4.184 kJ, 12 kJ mol⁻¹ corresponds to about 2.87 kcal mol⁻¹, usually rounded to 2.9. Both values describe an energy difference per mole of molecules in the chosen model.
Step 5: Explain the Physical Origin of Torsional Strain
In an eclipsed ethane Newman projection, three front C–H bonds align with three rear C–H bonds. The electron distributions associated with the bonds are in an unfavourable relative arrangement, commonly described as torsional or eclipsing strain. Each eclipsing H/H relationship contributes to the higher potential energy.
A simplistic account may say the hydrogens bump into one another. That is an intuitive picture of crowding, but modern orbital interpretations emphasise differences in electron delocalisation and overlap. The staggered form benefits from favourable interactions between occupied C–H sigma-bonding orbitals and appropriate antibonding orbitals on the adjacent carbon, often described using hyperconjugation.
The exact partition of an energy difference into steric repulsion, hyperconjugation and other terms depends on the theoretical analysis used. H3 candidates do not need a quantitative orbital decomposition to answer a Newman projection question, but they should understand that the observed barrier is an electronic energy effect rather than an arbitrary geometrical penalty.
There are no large methyl–methyl groups competing across the central ethane bond, unlike butane. In ethane, the comparison isolates fundamental interactions between C–H bond orientations. That is why it is such a good starting example.
Step 6: Why This Is Different from Anti and Gauche Butane
In n-butane viewed along C2–C3, each central carbon bears one methyl group and two hydrogens. Staggered orientations now differ in how far apart the two methyl groups are. The anti conformer with methyl groups 180° apart is normally lower in energy than either gauche minimum with methyl groups 60° apart.
Ethane has no such substituent difference: every substituent on each end is H. The three staggered minima are therefore equal in energy. In butane, one anti and two gauche minima are not all equal in energy. A student who reads ‘all staggered forms have equal energy’ from ethane and applies it to butane has missed the structural reason for the difference.
The complementary anti versus gauche butane Newman projection and population guide explains those unequal minima and how degeneracy influences their thermal populations. The present article focuses on the more fundamental threefold ethane torsional barrier.
Step 7: Why Conformations Are Not the Same as Stable Configurational Isomers
Ethane’s covalent connectivity is unchanged throughout its rotation. All carbon–hydrogen bonds remain attached to the same atoms. No stereogenic centre is inverted and no E/Z double bond is transformed. Only the dihedral angle about the C–C single bond changes.
Consequently, staggered and eclipsed ethane are conformations, not separable constitutional or configurational stereoisomers under ordinary conditions. Their interconversion is rapid relative to typical laboratory handling. A Newman projection is an instantaneous structural model that helps describe energy, not a guarantee of an isolable bottle of eclipsed ethane.
A learner should distinguish a rotational barrier from an absolute prohibition on rotation. Ethane has an energy barrier, but a small one. Restricted rotations in other molecules can be much slower or even give isolable atropisomers under special circumstances. This broader chemistry does not change the simple ethane result.
Optional Enrichment: A Simple Threefold Torsional Energy Formula
For a purely illustrative energy profile with the eclipsed orientation at 0° and a barrier height V≈12 kJ mol⁻¹, one can write a simple periodic model E(φ)=V/2 [1+cos(3φ)], with φ measured as the dihedral angle and the cosine evaluated consistently. This is a compact mathematical representation of the threefold repeat, not a quantitative H3 9813 requirement.
At φ=0°, cos 0=1, so E=V. At φ=60°, cos 180°=−1, so E=0. At φ=120°, cos 360°=1, so E=V again. The repeating maxima and minima match the qualitative Newman projection energy diagram.
One must distinguish potential-energy profiles from conformer populations. A simple Boltzmann factor e^(−ΔE/RT) compares idealised state weights, but an actual torsional angle is continuous, and the fraction in a region depends on its angular width, molecular entropy and the full potential. Dividing 12 kJ by RT and calling the result an exact observed fraction of eclipsed ethane would be unjustified.
The value of this mathematical example is interpretive: it helps a student see how threefold molecular symmetry appears in an energy graph. They do not need to memorise the formula to understand or answer the syllabus’s qualitative conformational questions.
Worked H3-Style Questions
Case A: Identify an energy minimum
An original question presents an ethane Newman projection with front/rear H–C bond directions 60° apart. Which type of conformation is shown, and is it an energy maximum or minimum?
Answer: Staggered and normally a minimum. Its bonds are offset rather than aligned.
Case B: Interpret 0°
A second drawing has corresponding front and rear C–H bond directions eclipsing at 0°. What energy difference is commonly quoted relative to staggered?
Answer: Approximately +12 kJ mol⁻¹ for the standard ethane rotational barrier.
Case C: Three cycles
A graph of potential energy is shown from 0° through 360°. The learner counts three high points and three low points. Explain the pattern.
Answer: Ethane has threefold symmetry around the C–C rotation because both methyl groups bear three equivalent H atoms. Eclipsed orientations repeat every 120°, with staggered minima halfway between.
Case D: Compare to butane
An unknown alkane has one lowest staggered conformation and two higher staggered conformations. Could it be a standard Newman projection of ethane?
Answer: Not in the ordinary ethane model, where all three staggered minima have equal energy. A substituted chain such as butane is a better comparison because unequal substituent orientations create anti/gauche energy differences.
Nine Errors That a Tutor Should Correct
- Assuming C–C sigma rotation is perfectly free. Ethane has a small but real torsional barrier.
- Calling eclipsed the minimum. The eclipsed arrangement is usually the high-energy orientation.
- Confusing dihedral and bond angle. Rotation changes torsion without changing the tetrahedral bond angles in the same way.
- Counting three staggered conformations as three chemical formulas. They are conformers of one ethane molecule.
- Claiming the 12-kJ barrier breaks C–C. It concerns relative rotational potential energy.
- Calling all bond eclipsing merely large methyl steric clashes. Ethane has only H substituents and the barrier has an orbital/electronic basis.
- Assuming any staggered orientation of a substituted alkane has identical energy. Butane anti and gauche are different.
- Interpreting an energy profile as a measured exact room-temperature percentage without statistical assumptions. The torsion is continuous.
- Believing the molecule freezes in a Newman drawing. Ordinary thermal motion gives rapid interconversion.
Sixteen Practice Questions With Answers
1. What is the formula of ethane?
Answer: C₂H₆.
2. What bond is viewed in the Newman projection?
Answer: The central C–C sigma bond.
3. Which carbon is represented by a circle?
Answer: The rear carbon in the conventional end-on Newman drawing.
4. What is the staggered dihedral angle?
Answer: 60° between the nearest front and rear C–H bond directions in the reference drawing.
5. What is the eclipsed dihedral angle?
Answer: 0° for aligned corresponding front/rear bonds.
6. Which conformation is more stable?
Answer: Staggered.
7. What is the usual barrier?
Answer: Approximately 12 kJ mol⁻¹, or about 2.9 kcal mol⁻¹.
8. How many minima in 360°?
Answer: Three equivalent staggered minima.
9. How many maxima in 360°?
Answer: Three equivalent eclipsed maxima.
10. Does the C–C bond break during rotation?
Answer: No. Covalent connectivity remains unchanged.
11. Are the three minima different constitutional isomers?
Answer: No. They are related by rotation and symmetry.
12. What is torsional strain?
Answer: The excess energy associated with unfavourable eclipsing bond orientations relative to staggered arrangements.
13. Why is ethane simpler than butane?
Answer: All substituents around the rotating bond are H, so its three staggered minima are energetically equivalent.
14. Does the 12-kJ barrier prevent room-temperature rotation?
Answer: No. Ethane interconverts rapidly under ordinary conditions.
15. Is φ=180° eclipsed or staggered if φ=0° is eclipsed?
Answer: Staggered in the conventional alternating 60° sequence.
16. Is the cosine model a compulsory 9813 formula?
Answer: No. It is optional quantitative enrichment; the prescribed goal is conformational reasoning and barriers.
17. What physical insight does hyperconjugation add?
Answer: It describes favourable bonding-to-antibonding orbital interactions that help stabilise the staggered form in an electronic analysis.
How I Would Teach the Energy Profile in a Small Group
First, each learner draws an ethane Newman projection and rotates the rear group in 60° increments. We record the torsional angle and label whether the bonds are staggered or eclipsed. The student should be able to explain why a 360° rotation gives three maxima and three minima without seeing a finished energy graph.
Next we plot relative energy on the vertical axis and dihedral angle on the horizontal axis, marking +12 kJ mol⁻¹ at the eclipsed positions and zero at staggered minima. The learner must describe the meaning of a barrier and distinguish it from a bond dissociation energy. An incorrect graph is usually evidence of a misidentified starting angle or a confused Newman drawing.
In a final round, we introduce butane to reveal why identical H/H eclipsing interactions are not the complete story for substituted alkanes. A student who can explain why butane has an anti preference whereas ethane’s staggered minima are equivalent has transferred the model to new chemistry.
Families enquiring from Hougang can consult the eduKate JC2 H3 Chemistry small-group tutorial guide. A useful specialist tutor should correct students’ angle, projection and energy reasoning rather than merely hand over a familiar graph. Verify actual class availability, timetable and teaching venue near Sixth Avenue MRT; this Hougang title is not a claim of an in-neighbourhood classroom.
Why Rotational Energy Barriers Matter Beyond JC
Internal rotation affects molecular shape, conformational populations, and the orientations that can undergo specific chemical reactions. Some organic reaction mechanisms require particular relative positions of bonds, such as anti-periplanar alignment in E2 elimination. In larger molecules, conformational changes can affect molecular recognition and reaction selectivity.
Ethane is the simplest useful case because the geometry is symmetric and its torsional barrier is small. The physical habit it teaches—relate geometry to potential energy, consider a model’s assumptions and then predict possible observations—continues into organic synthesis, computational chemistry and materials science.
Ethane Conformation FAQs
Is staggered or eclipsed ethane more stable?
Staggered ethane is the lower-energy form in the conventional model.
What is the barrier to rotation?
About 12 kJ mol⁻¹, or 2.9 kcal mol⁻¹.
How many staggered forms occur in one revolution?
Three equivalent minima over 360°.
Does ethane rotate freely?
It rotates rapidly at ordinary temperatures but encounters a finite torsional energy barrier.
Is staggered ethane a different stereoisomer from eclipsed ethane?
They are interconverting conformations, not different constitutional or configurational compounds.
Why are butane’s staggered minima different?
Butane’s methyl substituents introduce anti and gauche geometries with different nonbonded interactions.
Is this relevant to the official H3 syllabus?
Yes. The 2027 9813 syllabus specifies Newman projections, conformational isomerism, rotational energy barriers and interconversion.
A useful Chemistry lesson is one that allows students to construct the result from the molecule. With ethane, three equivalent hydrogen substituents produce a threefold torsional profile; staggered arrangements are more stable; eclipsed arrangements produce an energy barrier. Once that chain of reasoning is secure, the apparently complicated Newman projections of other molecules become manageable.
Hougang JC Chemistry hub · H3 Chemistry 9813 study guide · Anti and gauche butane conformers · Contact eduKate Singapore