Visual walkthrough — Conformations of ethane, butane — Newman projections

Everything below rests on one idea from Alkanes — structure and sp3 hybridisation: a carbon in an alkane points its four bonds toward the corners of a tetrahedron, so the three bonds we can see in a Newman projection sit $120°$ apart. Keep that "three spokes, $120°$ apart" picture ready.

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Step 1 — Look straight down the bond

WHAT. We take the C2–C3 bond of butane and put our eye on the axis of that bond, looking down it like looking down a drinking straw. The near carbon hides the far carbon.

WHY. Rotation happens around this bond axis. If we look at any other angle, the twisting is hard to measure. Looking straight down turns a 3-D twist into a flat 2-D angle we can actually read off — this is the whole reason Newman projections exist.

PICTURE. Below, the near carbon becomes a dot with three spokes; the far carbon becomes a circle with three spokes coming off its rim. On C2 (front) one spoke is a ${\text{CH}}_3$ (burnt orange), two are H. Same on C3 (back).

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Step 2 — The two "moods" of a bond: eclipsed vs staggered

WHAT. Spin the back circle. At some angles a back spoke hides directly behind a front spoke (eclipsed). At other angles each back spoke peeks through the gap between two front spokes (staggered).

WHY. These are the two situations where electron clouds are either forced together or held apart. Because the front carbon has three spokes $120°$ apart, eclipsing repeats every $120°$, and staggering sits exactly halfway between, at $60°$ offsets. This $120°$ repeat is the seed of the whole curve.

PICTURE. Left = eclipsed (spokes overlap, drawn slightly fanned so you can see both). Right = staggered (clean six-pointed star, spokes $60°$ apart).

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Step 3 — Cost #1: torsional strain (present even in ethane)

WHAT. Every time two spokes eclipse, the bonding electron pairs are squeezed close and we lose the favourable overlap of a filled bond with an empty one. This costs energy. We call it torsional strain.

WHY this tool — a cosine. We need a function of $\varphi$ that is high when eclipsed and low when staggered, and that automatically repeats every $120°$. The cheapest honest choice is a cosine of $3\varphi$: cosine already oscillates smoothly between $+1$ and $-1$, and the "$3$" packs three full waves into one $360°$ turn — one bump per eclipse. That is exactly the ethane law from the parent note:

$$E_{\text{tors}}(\varphi )=\underset{\text{half the peak height}}{\underbrace{\frac{E_{\text{b}}}{2}}}(1+\underset{+1\text{ eclipsed},-1\text{ staggered}}{\underbrace{\cos 3\varphi }})$$

• $E_{\text{b}}$ — the barrier height (peak minus valley). For an all-H bond it is small.
• $\frac{E_{\text{b}}}{2}$ — scales the wave so its peak is $E_{\text{b}}$ and its valley is $0$.
• $\cos 3\varphi$ — the shape: three humps per turn.
• $1+\cos 3\varphi$ — lifts the valleys down to exactly $0$ (no negative energy).

PICTURE. The teal wave: peaks at $\varphi =0°,120°,240°$ (eclipsed), valleys at $60°,180°,300°$ (staggered). This is all of ethane's story — three equal humps.

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Step 4 — Cost #2: steric strain (the new thing in butane)

WHAT. Butane's front and back carbons each carry a fat ${\text{CH}}_3$. When those two methyls swing close in space, their electron clouds bump — this is van der Waals repulsion. It depends on how close the two methyls are, i.e. on $\varphi$ directly, not on eclipsing.

WHY a second, separate term. Torsional strain (Step 3) only cares whether spokes line up. Steric strain cares which bulky groups are near, regardless of lining up. These are two independent physical effects, so we model them as two separate curves and add them. Steric cost is largest at $\varphi =0°$ (methyls on top of each other) and smallest at $\varphi =180°$ (methyls on opposite sides).

PICTURE. The plum curve: one big hump centred at $\varphi =0°$ (methyls closest), a small residual bump near $\varphi =\pm 60°$ (methyls only $60°$ apart — the famous gauche squeeze), and a deep trough at $\varphi =180°$ where the methyls are farthest apart.

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Step 5 — Add the two costs: the full butane curve

WHAT. Total strain at any angle is the sum of the two independent costs:

$$E_{\text{total}}(\varphi )=\underset{\text{teal, Step 3}}{\underbrace{E_{\text{tors}}(\varphi )}}+\underset{\text{plum, Step 4}}{\underbrace{E_{\text{steric}}(\varphi )}}$$

WHY just add them. The two effects come from different electrons feeling different things; to first approximation they do not interfere, so their energies stack. Adding a curve with three equal humps to a curve with one giant hump at $0°$ breaks the symmetry — the hump at $\varphi =0°$ becomes the tallest, and the valleys stop being equal.

PICTURE. Burnt-orange = the sum. Read it left to right and name every feature:

• $\varphi =0°$ — fully eclipsed (syn): torsional peak plus the huge steric peak → the global maximum (~19–20 kJ mol⁻¹).
• $\varphi =60°$ — gauche: torsional valley, but a small steric bump remains → a shallow local minimum (~3.8 kJ mol⁻¹).
• $\varphi =120°$ — eclipsed (CH₃ over H): torsional peak, moderate steric → an intermediate maximum (~16 kJ mol⁻¹).
• $\varphi =180°$ — anti: torsional valley and methyls farthest apart → the global minimum, set to 0.

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Step 6 — Edge case: turn the methyls off and you recover ethane

WHAT. Set the steric term to zero (imagine shrinking both methyls back to H). The burnt-orange curve collapses back onto the teal one — three equal humps, all valleys at zero.

WHY show this. A good model must reproduce the simpler case it grew from. Ethane is butane with the fat groups deleted; if our sum did not reduce to the three-equal-hump ethane curve when steric $\to 0$, the model would be wrong. It does, so we trust it.

PICTURE. The plum curve fades to a flat line at zero; the sum snaps onto the symmetric teal ethane curve. Notice the $0°$ peak drops from ~20 down to ethane's ~12.5 kJ mol⁻¹ — exactly the steric contribution we removed.

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Step 7 — From curve to crowd: who sits in each valley?

WHAT. The curve tells us the energy of each valley; the Boltzmann distribution tells us the fraction of molecules that actually sit there. Deeper valley → more crowded with molecules.

WHY this tool. Energy alone does not give populations — temperature spreads molecules over the wells. The Boltzmann ratio answers "for two wells differing by $\Delta E$, how many times more molecules sit in the lower one?"

$$\frac{N_{\text{gauche}}}{N_{\text{anti}}}=\underset{-\stackrel{3.8\text{ kJ}}{\overbrace{\Delta E}}/\stackrel{8.314\times 298}{\overbrace{RT}}}{\underbrace{e}}=e^{-1.534}\approx 0.216$$

• $\Delta E$ — how much higher the gauche well sits above anti.
• $RT$ — the thermal "budget" ($\approx 2.48{\text{kJ mol}}^{-1}$ at $298$ K); it sets how easily molecules climb.
• the minus sign — higher well ⇒ fewer molecules ⇒ ratio below 1.

PICTURE. Two anti wells' worth of depth vs the shallower gauche wells, with molecule-dots piled in proportion. Remember there are two gauche wells ($\pm 60°$) but only one anti well, so multiply the gauche share by 2.

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The one-picture summary

Everything on this page in a single frame: the teal torsional wave plus the plum steric hump equals the burnt-orange butane curve, with the four Newman snapshots pinned under their angles and the anti/gauche populations shown as molecule-dots.

Recall Feynman retelling — the whole walkthrough in plain words

Point your eye down the middle bar joining the two carbons. Each carbon shows three arms; one arm on each is a chubby methyl ball, the others are little hydrogen dots. Now slowly spin the back set.

Two things make the molecule uncomfortable. First, whenever a back arm hides exactly behind a front arm, their electrons crowd — that's the springy "torsional" discomfort, and because there are three arms it flares up three times per full spin, evenly (that's the teal wave — and that's the whole story for ethane, which has no chubby balls).

Second, the two chubby methyl balls hate being near each other in space — pure elbow-room complaint. That's biggest when they sit right on top of each other ($0°$) and smallest when they're on opposite sides ($180°$) — that's the plum hump.

Stack the two complaints and you get the real butane curve (burnt orange): a giant spike at $0°$ (both complaints screaming — syn), a tall-but-smaller spike at $120°$ (methyl over hydrogen), a comfy deep valley at $180°$ (methyls far apart, arms staggered — anti, the winner), and a shallow dip at $60°$ that should be perfect (arms staggered) but isn't, because the balls are still a bit too close — gauche.

Finally, drop a crowd of molecules onto this landscape. They pool in the valleys, more of them in the deeper one. Because anti is deepest they mostly sit there — about 70% anti, 30% gauche at room temperature. Delete the chubby balls and the whole thing relaxes back into ethane's three identical gentle humps. That's conformation, start to finish.

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Connections

• Steric strain and van der Waals repulsion — the physics behind the plum curve (Step 4).
• Hyperconjugation — the electronic reason staggered lowers the teal curve (Step 3).
• Boltzmann distribution — turns valley depths into populations (Step 7).
• Cyclohexane conformations chair and boat — the same energy-vs-twist reasoning applied to a ring.
• Alkanes — structure and sp3 hybridisation — why the three spokes sit $120°$ apart.
• Stereochemistry — isomerism overview — where conformers sit among the other "same-formula, different-shape" relationships.