Alkanes: Conformations

Why rotation is (almost) free. A $\sigma$ bond is formed by head-on overlap of orbitals along the bond axis, so it is cylindrically symmetric about that axis. Spinning one $\text{CH}_3$ group relative to the other does not reduce this overlap at all — the orbital picture along the axis looks identical at every angle of rotation. That is the structural reason rotation around a C–C single bond costs (almost) no energy, unlike rotation around a C=C double bond, where the sideways $\pi$ overlap would be destroyed by twisting.

What a conformation is — and is not. As the far $\text{CH}_3$ group turns, ethane sweeps through infinitely many transient shapes called conformations (or conformers). They are not isomers: interconversion needs no bond to break or re-form, only rotation about the existing C–C $\sigma$ bond, and it happens millions of times a second at room temperature. You can never isolate "staggered ethane" in a bottle separate from "eclipsed ethane" — they are the same compound caught at different rotational instants.

Reading the dihedral angle. The angle between a front C–H bond and the nearest back C–H bond, measured looking straight down the C–C axis, is the dihedral (or torsional) angle $\theta$. Because each carbon carries three C–H bonds spaced $120^\circ$ apart, the two special families of conformation recur every $60^\circ$:

where: $\theta$ is the dihedral angle between a chosen front C–H bond and the nearest back C–H bond, read directly off a Newman projection. Every eclipsed geometry sits exactly $60^\circ$ from its neighbouring staggered geometry, so rotating ethane by $60^\circ$ at a time alternates between the two families forever.

The two projections you must be able to draw.

• Sawhorse projection: a diagonal, 3-D-style view of the C–C bond showing both carbons and all six hydrogens explicitly, front carbon at lower left, back carbon at upper right.
• Newman projection: a view straight down the C–C bond axis. The front carbon is a dot with three bonds radiating out at $120^\circ$; the back carbon is a circle with three bonds emerging just inside its rim. This is the projection examiners use to test whether you can identify staggered versus eclipsed at a glance.

The staggered form is the more stable arrangement of ethane, and the reasoning stands on a single idea: electron-pair repulsion.

Step 1. Every C–H bond is a cloud of bonding electron density. Two such clouds on adjacent carbons repel each other, and that repulsion is smallest when the clouds are as far apart in angle as they can be, and largest when they point in almost the same direction.

Step 2. In the staggered conformation, each back C–H bond sits exactly midway between two front C–H bonds — the maximum possible angular separation ($60^\circ$) available on a three-fold-symmetric carbon. Electron-pair repulsion is minimised, so potential energy is lowest here.

Step 3. In the eclipsed conformation, each back C–H bond lines up directly behind a front C–H bond ($0^\circ$ apart). The electron clouds crowd close together, repulsion rises, and this extra energy is called torsional strain. Eclipsed ethane is therefore a real, existing conformation — just a higher-energy, less populated one, not a forbidden shape.

Step 4. Because the energy difference between staggered and eclipsed is small, thermal motion at room temperature easily supplies enough energy to rotate through every intermediate angle continuously. Ethane is never "stuck" in one conformation; it spends more time nearer the staggered geometries on average, which is exactly what "staggered is more stable" means for a freely rotating molecule — a population preference, not a static structure.

Q. Draw and label the Newman projection of ethane in its staggered conformation, viewed along the C1–C2 axis, and explain why this arrangement is energetically preferred over the eclipsed form.

Step 1. Set up the front atom: draw a dot for C1 with three lines radiating out at $120^\circ$ to each other, representing its three C–H bonds pointing toward the viewer.

Step 2. Set up the back atom: draw a circle around the dot for C2, with three more lines emerging from just inside the circle's rim, representing C2's three C–H bonds pointing away from the viewer.

Step 3. Fix the staggered geometry: rotate the back set of bonds so each one bisects the angle between two front bonds. A front bond and its nearest back bond are then $60^\circ$ apart, never $0^\circ$.

Step 4. Justify the stability: in the staggered form every C–H bonding pair on C2 sits as far as geometrically possible from every C–H pair on C1, minimising electron-pair repulsion and lowering potential energy. In the eclipsed form, front and back bonds line up directly, raising repulsion (torsional strain) and energy.

Answer: the staggered Newman projection has back bonds at $60^\circ$, $180^\circ$, and $300^\circ$ relative to the front bonds, and it is the more stable conformation because it minimises repulsion between adjacent C–H bonding electron pairs — though ethane keeps rotating freely through all intermediate angles at room temperature, without breaking any bond.

Competitive stretch

Q. A student claims staggered and eclipsed ethane are structural isomers. Is this correct?

Step 1. Isomers are different compounds with the same molecular formula that require breaking and re-forming bonds to interconvert.

Step 2. Only the dihedral angle around the C–C $\sigma$ bond changes between staggered and eclipsed ethane; no bond breaks and connectivity is unchanged.

Step 3. The claim is incorrect — staggered and eclipsed ethane are conformations of one molecule, interconverting by free rotation, not distinct isomeric compounds.

• Calling staggered and eclipsed ethane "isomers" — they are conformations of one compound, interconvertible by free rotation, not distinct compounds with different molecular connectivity.
• Assuming eclipsed ethane cannot exist or is "forbidden" — it is a real, higher-energy, less populated conformation, not an impossible geometry.
• Trying to quantify the rotational energy barrier in kJ/mol — this syllabus tests only the qualitative comparison (staggered more stable than eclipsed), never a numeric barrier height.
• Mixing up the Newman convention or the dihedral angles — the front carbon is always the dot, the back carbon the circle, eclipsed bonds overlap at $0^\circ$, and staggered bonds sit $60^\circ$ apart, not the other way round.

Ask the tutor: Not sure whether a given drawing shows a staggered or eclipsed Newman projection, or want more practice sketching conformations — ask the AI tutor to walk through it interactively.