Conformations of ethane, butane — Newman projections: Question bank

Quick vocabulary refresher (every symbol used below is earned here):

$ϕ$ (dihedral / torsion angle) = the angle you see between a front-carbon bond (line from the centre dot) and the nearest back-carbon bond (line from the circle's rim) when you stare down the C–C axis. Picture a clock: $ϕ = 0°$ means the two hands overlap.
Torsional strain = the "bonds too close" penalty that peaks when front and back bonds overlap ( $ϕ = 0°$ ): bonding electron pairs repel and the favourable hyperconjugative overlap is lost.
Steric strain = the "fat groups bumping in space" penalty (van der Waals repulsion), only big for bulky groups like $CH_{3}$ .
Hyperconjugation = a stabilising donation of a filled $σ (C-H)$ bond's electrons into the neighbouring empty $σ^{*} (C-H)$ antibonding orbital; this overlap (and its stabilisation) is best when bonds are staggered and lost when they eclipse. See Hyperconjugation.
$E_{barrier}$ = the height of the energy hill between a low-energy (staggered) valley and the high-energy (eclipsed) peak, in $kJ mol^{- 1}$ ; for ethane $E_{barrier} \approx 12.5 kJ mol^{- 1}$ .
$R$ = the gas constant $= 8.314 J mol^{- 1} K^{- 1}$ ; $T$ = absolute temperature in kelvin. The product $R T$ is the "thermal energy nudge" scale — at room temperature ( $T \approx 298$ K) $R T \approx 2.5 kJ mol^{- 1}$ .

True or false — justify

TF1. "Staggered ethane and staggered butane both have exactly zero total strain."

False. Staggered ethane truly sits at the energy minimum (only tiny H's), but staggered butane comes in a gauche form where the two methyls, only

60°

apart, still clash sterically (

\sim 3.8

kJ/mol above anti).

TF2. "Because conformers look different when drawn, they are structural isomers you could isolate in separate bottles."

False. They interconvert by mere rotation about a σ bond (barrier

<\sim 20

kJ/mol), which thermal energy overcomes billions of times per second — one rapidly-flipping substance, not separable species.

TF3. "Eclipsed always means the single highest-energy point of a rotation profile."

False in general. Butane has two eclipsed types:

CH_{3} / CH_{3}

ϕ = 0°

(fully eclipsed, syn — highest) and

CH_{3} / H

ϕ = 120°

(lower). Both are eclipsed, but only syn is the global maximum.

TF4. "The dihedral angle in a staggered Newman projection is

0°

False. Staggered is

ϕ = 60°

— each back bond sits in the gap between front bonds, forming a clean 6-pointed star.

ϕ = 0°

is eclipsed (bonds overlapping).

TF5. "Ethane's rotation barrier is large enough to trap the molecule in the staggered form at room temperature."

False. The barrier (

E_{barrier} \approx 12.5

kJ/mol) is only a few times the thermal nudge per collision (

R T \approx 2.5

kJ/mol), so ethane spins essentially freely — it just lingers longer in the staggered wells.

TF6. "In butane, anti is favoured only because it has the lowest energy — nothing else matters."

False. Anti does win on energy (both strains minimised), but the actual population mix also depends on the fact that there are two gauche wells (

\pm 60°

) versus one anti well — this degeneracy (entropy) factor, not energy alone, sets the final ratio.

TF7. "Torsional strain and steric strain are two names for the same repulsion."

False. Torsional strain is about bonding electron pairs overlapping (and lost hyperconjugation) as bonds align; steric strain is about whole atom/group electron clouds of bulky substituents bumping. Ethane has torsional but negligible steric.

TF8. "Hyperconjugation makes the eclipsed conformer more stable."

False. The stabilising

σ (C-H) \to σ^{*} (C-H)

overlap is maximal in the staggered geometry; eclipsing spoils that overlap, so hyperconjugation actually explains why staggered is low-energy. See Hyperconjugation.

Spot the error

SE1. Statement: "Butane's gauche conformer has torsional strain from its eclipsed bonds." — find the flaw.

Gauche is staggered, not eclipsed, so it has essentially no torsional strain. Its extra energy over anti is purely steric (

CH_{3}

CH_{3}

60°

SE2. Statement: "The ethane energy formula is

E (ϕ) = \frac{E _{barrier}}{2} (1 + cos ϕ)

." — find the flaw.

The argument must be

cos 3 ϕ

, not

cos ϕ

. Here

E_{barrier}

is the hill height (defined above). The $3 ϕ$ encodes ethane's three-fold symmetry: rotating

120°

gives an identical arrangement, so the energy must repeat three times per turn — plain

cos ϕ

would give only one maximum per

360°

SE3. A student draws a staggered Newman with front and back bonds drawn on top of each other. — what's wrong?

Overlapping bonds is the eclipsed drawing. Staggered must show the six lines $60°$ apart as a 6-pointed star; overlapping them accidentally draws the high-energy conformer.

SE4. Statement: "Since anti butane is most stable, at 298 K essentially 100% of molecules are anti." — find the flaw.

The energy gap is small (

3.8

kJ/mol) and there are two gauche wells, so the Boltzmann mixture is roughly 70% anti / 30% gauche — a substantial gauche population, not ~100% anti. See Boltzmann distribution.

SE5. Statement: "In a Newman projection the front carbon is a circle and the back carbon is a dot." — find the flaw.

It's the reverse (as set out in the notation box above): front = central dot with three lines meeting at the centre; back = circle with three lines from its edge.

SE6. Statement: "Cyclohexane's chair is just butane's anti conformer, so the same

ϕ = 180°

rule fully describes it." — find the flaw.

A ring closes back on itself, so not every bond can be simultaneously anti; the chair is the best compromise of many dihedrals (all staggered, no eclipsing). It's related in spirit but governed by ring geometry — see Cyclohexane conformations chair and boat.

Why questions

WHY1. Why does butane show four distinct conformers per turn while ethane shows only two extremes?

Ethane's three H's are all identical, so rotation only alternates staggered↔eclipsed. Butane's methyls break the symmetry, so the position of

CH_{3}

CH_{3}

vs H creates four chemically different snapshots (anti, gauche, eclipsed, syn).

WHY2. Why does staggering lower energy even in ethane where the H's barely touch?

Even without steric bumping, aligning bonds (

ϕ = 0°

) forces bonding electron pairs close (repulsion) and destroys the favourable hyperconjugative

σ (C-H) \to σ^{*} (C-H)

overlap; staggering (

ϕ = 60°

) restores both.

WHY3. Why is the fully eclipsed (syn,

ϕ = 0°

) butane higher in energy than the eclipsed at

ϕ = 120°

ϕ = 0°

the two bulky methyls overlap (torsional + big steric), whereas at

ϕ = 120°

a methyl only eclipses a small H (torsional + minor steric), so syn is the more crowded, higher peak.

WHY4. Why must we invoke the Boltzmann factor

e^{- Δ E / R T}

(with

R = 8.314 J mol^{- 1} K^{- 1}

R T \approx 2.5

kJ/mol at 298 K) rather than just declaring "anti is lowest so all molecules are anti"?

Because at finite temperature molecules are constantly kicked by thermal energy of scale

R T

and spread across states; the population ratio depends on the gap

Δ E

relative to

R T

, so a gap comparable to

R T

leaves many molecules in the higher state.

WHY5. Why is steric strain negligible in ethane but decisive in butane?

Steric strain scales with group bulk. Ethane's substituents are tiny H atoms with small electron clouds; butane carries two fat

CH_{3}

groups whose clouds genuinely overlap when brought within

60°

. See Steric strain and van der Waals repulsion.

WHY6. Why do we look down the C–C bond (Newman view) instead of drawing the molecule side-on?

Staring along the bond axis lines the two carbons up so you can directly read the dihedral angle and see exactly which front and back groups are crowding — the side-on picture hides that alignment.

WHY7. Why does the ethane energy curve repeat every

120°

, and why does that force the

cos 3 ϕ

form rather than

cos ϕ

Rotating by

120°

swaps one indistinguishable H for another, giving a geometrically identical arrangement, so energy must repeat with period

120°

(three times per full turn). A cosine repeats every

360°

; to make it repeat every

360°/3 = 120°

you triple the angle inside — hence

cos 3 ϕ

, which delivers exactly three maxima (eclipsed) and three minima (staggered) per turn.

Edge cases

EC1. What happens to the anti-to-gauche population ratio as temperature

T \to

very high?

e^{- Δ E / R T} \to e^{0} = 1

, so gauche and anti wells approach equal occupancy per well; energy differences matter less when thermal energy

R T

swamps them.

EC2. What happens to the same ratio as

T \to 0

e^{- Δ E / R T} \to 0

, so essentially all molecules collapse into the lowest (anti) well — energy fully dominates when there is no thermal kick.

EC3. Is there ever torsional strain in a perfectly staggered conformation (

ϕ = 60°

No — at

ϕ = 60°

every back bond sits in a front-bond gap, giving maximum bond separation and best hyperconjugation, so torsional strain is at its minimum (effectively zero).

EC4. If both carbons of ethane were somehow held rigidly eclipsed, would ethane still be one "compound"?

Yes — it's the same molecule frozen at a high-energy geometry; you'd merely be preventing the natural rotation, not creating a new substance.

EC5. In butane, at exactly

ϕ = 90°

(halfway between the gauche well at

60°

and the eclipse at

120°

), how do the two strains each behave?

Torsional strain rises from its near-zero staggered value toward its eclipsed peak — bonds are now

30°

off perfect alignment, so partial bond-pair repulsion has set in. Steric strain, meanwhile, eases relative to gauche, because the two methyls have swung further apart than the

60°

gauche crowding. So at

90°

you sit on an up-slope dominated by growing torsional strain while methyl–methyl steric strain relaxes — a transient, non-well geometry.

EC6. Could a molecule with a triple bond between the two central carbons show these conformers?

No — a C≡C bond is rigid and cylindrically symmetric; there is no dihedral to vary, so no distinct conformers arise. Conformational analysis needs a rotatable single (σ) bond. See Alkanes — structure and sp3 hybridisation.

EC7. Do gauche(+60°) and gauche(−60°) count as the same conformer for population purposes?

They are mirror-image wells of equal energy but count as two separate states, which is exactly why total gauche population is doubled relative to a single anti well — a degeneracy (entropy) effect. Related idea: Stereochemistry — isomerism overview.

Recall One-line self-test before you leave

Cover everything and answer aloud: "Name the one thing that distinguishes butane's conformational profile from ethane's, and the one experimental quantity that decides how molecules actually distribute." ::: Bulky $CH_{3}$ groups introduce steric strain (giving four conformers), and the Boltzmann factor $e^{- Δ E / R T}$ (with well-degeneracy) decides the actual population mix.