Conformational analysis of ethane

Why rotation is even possible. The C–C bond in ethane is a single $\sigma$ bond formed by head-on overlap of two $sp^3$ hybrid orbitals, one from each carbon. A $\sigma$ bond is cylindrically symmetric about the internuclear axis, so rotating one $\text{CH}_3$ group relative to the other does not disturb this overlap at all — the bond stays intact through the entire $360^\circ$ turn. This is precisely why conformations are not isomers: isomers differ in bonding and must be separated by breaking bonds, while conformations differ only in the dihedral (torsional) angle and interconvert freely.

Drawing what you see: Sawhorse and Newman. A Sawhorse projection draws the C–C bond as a diagonal viewed slightly from the side, with three substituent bonds fanning out from each carbon — intuitive to sketch but clumsy for reading off angles. The Newman projection solves that by looking straight down the C–C axis: the front carbon becomes a point with three C–H bonds meeting at $120^\circ$, and the back carbon becomes a circle with its three C–H bonds emerging from just behind the circle's rim. The angle between a front bond and its nearest back bond, measured in this view, is the dihedral angle $\theta$.

The two named conformations. In the staggered conformation, every back C–H bond sits exactly midway between two front C–H bonds, so $\theta = 60^\circ$ for each adjacent pair. In the eclipsed conformation, every back C–H bond sits directly behind a front C–H bond, so $\theta = 0^\circ$ for each pair. Because the molecule has three-fold symmetry, both patterns repeat every $120^\circ$ of rotation — turning the back carbon by $120^\circ$ returns the same conformation.

Why the angle costs energy. As $\theta$ shrinks from $60^\circ$ toward $0^\circ$, the bonding electron pairs of a front C–H bond and the nearest back C–H bond move closer together. Being like-charged electron clouds, they repel, and this repulsion — called torsional strain — raises the molecule's potential energy. Chemists summarise the resulting energy profile with a simple periodic model:

where: $V(\theta)$ is the torsional (potential) energy of the molecule at dihedral angle $\theta$, $V_0$ is the height of the energy barrier between a staggered minimum and the adjacent eclipsed maximum, and the factor $3$ inside the cosine encodes the three-fold repeat of the C–H bonds on each carbon. You are never asked to compute $V_0$ in kJ at this level — the equation is here only to show why the energy pattern repeats every $120^\circ$ with minima at the staggered angles and maxima at the eclipsed angles, matching the qualitative picture above exactly.

The stability difference between staggered and eclipsed ethane is a direct consequence of a single physical fact: electron pairs in bonds repel other nearby electron pairs, and this repulsion grows sharply as the pairs get closer, exactly as VSEPR reasoning already predicts for electron pairs around one atom, only here the two sets of C–H bonding pairs belong to different carbons and are being pushed together purely by rotation rather than by geometry around a single centre.

Step 1. At $\theta = 60^\circ$ (staggered), each front C–H bond is angularly as far as it can get from every back C–H bond while the molecule keeps its $sp^3$ tetrahedral geometry at both carbons, so the bonding electron clouds are maximally spread out and repulsion between them is at its smallest possible value for this molecule.

Step 2. As rotation carries $\theta$ toward $0^\circ$, the nearest front and back C–H bonds swing toward direct alignment; the distance between their electron clouds shrinks continuously, so the repulsive (torsional strain) contribution to the total energy rises continuously along with it.

Step 3. At $\theta = 0^\circ$ (eclipsed) the alignment is exact for all three pairs simultaneously, giving the maximum possible repulsion and therefore the highest point on the energy profile — a shape captured qualitatively by the $\cos 3\theta$ pattern above, whose minima sit at the staggered angles and whose maxima sit at the eclipsed angles.

Step 4. Because this barrier is small compared with the thermal energy available at room temperature, molecules do not get stuck at any one angle; they rotate rapidly and continuously, spending more time near the energy minima (staggered) than near the maxima (eclipsed) but never settling permanently — which is exactly why staggered and eclipsed ethane cannot be bottled separately as two different substances.

Problem: A Newman projection of ethane is drawn looking down the C1–C2 axis. Starting from the staggered conformation, the back carbon (C2) is rotated clockwise through $180^\circ$ in one continuous motion. List, in order, every conformation (staggered or eclipsed) the molecule passes through, state the dihedral angle at each, and identify which of these is the highest-energy point reached.

Step 1. Fix the starting point: staggered, so $\theta = 60^\circ$, and this is a torsional-energy minimum (least repulsion, most stable).

Step 2. Rotate through the first $60^\circ$ of the turn: $\theta$ decreases from $60^\circ$ to $0^\circ$, and at $\theta = 0^\circ$ the molecule is eclipsed — the front and back C–H bonds are directly aligned, giving a torsional-energy maximum.

Step 3. Continue rotating; because the pattern repeats every $60^\circ$ due to the three-fold symmetry of each $\text{CH}_3$ group, the next $60^\circ$ of rotation carries $\theta$ back up through the sequence and the molecule returns to a staggered arrangement (its mirror image relative to the first), i.e. staggered again at the point corresponding to a total rotation of $120^\circ$.

Step 4. The third $60^\circ$ segment (from $120^\circ$ to $180^\circ$ of total rotation) again passes through an eclipsed arrangement at the $180^\circ$ mark, since $180^\circ$ is an odd multiple of $60^\circ$ just like $0^\circ$ and $60^\circ$ were within the first cycle.

Step 5. Collect the sequence: staggered ($0^\circ$ rotated) to eclipsed ($60^\circ$ rotated) to staggered ($120^\circ$ rotated) to eclipsed ($180^\circ$ rotated). All three eclipsed points reached along a full $360^\circ$ turn are equally high in energy by the three-fold symmetry, so the molecule passes through two eclipsed (highest-energy) conformations during this $180^\circ$ rotation, at $60^\circ$ and $180^\circ$ of total rotation, alternating with staggered (lowest-energy) conformations at $0^\circ$ and $120^\circ$.

Competitive stretch

If asked how many staggered and how many eclipsed conformations exist in one complete $360^\circ$ rotation, use the $120^\circ$ repeat directly: a full turn contains exactly three staggered conformations (at $60^\circ$, $180^\circ$, $300^\circ$ of total rotation, all equivalent by symmetry) and exactly three eclipsed conformations (at $0^\circ$, $120^\circ$, $240^\circ$), giving six alternating energy extrema per revolution — three minima and three maxima.

• Calling staggered and eclipsed ethane "isomers" — they are conformations, related only by bond rotation, not by breaking and re-forming bonds, so they are never isomers.
• Forgetting the syllabus wants a qualitative stability argument (torsional strain from electron-pair repulsion); do not try to quote or derive an actual barrier height in kJ mol⁻¹.
• Drawing the Newman back-carbon bonds as emerging from the centre of the circle instead of from behind its edge, which erases the front/back distinction the projection exists to show.
• Assuming ethane is "frozen" in one conformation at room temperature — it is a rapidly interconverting mixture dominated by, but not restricted to, the staggered form.

Ask the tutor: if the Newman-projection geometry or the idea of torsional strain still feels fuzzy, ask the AI tutor to sketch the staggered-to-eclipsed rotation step by step with you.