Worked examples — Conformations of ethane, butane — Newman projections

This page is the practice ground for the parent topic. We will not re-teach the theory from scratch — instead we hunt down every kind of question this topic can throw at you and solve one of each. If a symbol feels unfamiliar, the parent note built it; here we use it.

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First: what "$>$" means on this page, the range of $\varphi$, and the angle units

Three conventions must be fixed before any example, or later steps become ambiguous.

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The scenario matrix

Every conformation problem falls into one of these cells. The worked examples below are labelled with the cell they hit, so you can see the whole territory is covered.

Cell	Case class	What varies	Example
A	Angle → conformer name	$\varphi =0°,60°,120°,180°$ (all named conformers, minima and maxima)	Ex 1
B	Energy from the ethane formula	plug $\varphi$ into $E(\varphi )=\frac{E_b}{2}(1+\cos 3\varphi )$	Ex 2
C	Degenerate / zero input	ethane where steric strain = 0 (only torsional)	Ex 3
D	Strain bookkeeping (add up contributions)	count eclipsing pairs, sum kJ	Ex 4
E	Boltzmann population	$N_2/N_1=e^{-\Delta E/RT}$, with degeneracy	Ex 5
F	Limiting behaviour	$T\to 0$ and $T\to \infty$ — where do molecules go?	Ex 6
G	Real-world word problem	temperature change shifts the mixture	Ex 7
H	Exam-style twist	staggered ≠ lowest; rank a trick set	Ex 8
I	Draw / read a Newman	build the projection at a given $\varphi$	Ex 9

Prerequisite links you may want open: Boltzmann distribution, Steric strain and van der Waals repulsion, Hyperconjugation, Alkanes — structure and sp3 hybridisation.

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Symbols we will reuse (earned, not assumed)

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Ex 1 — Cell A: angle to conformer name (minima and maxima)

Forecast: guess which of these you'd expect to be highest and lowest in energy before reading on. (Careful: only two of these four are wells.)

1. $\varphi =0°$ → the two CH₃ groups sit directly on top of each other. Name: fully eclipsed (syn). This is an energy maximum (barrier peak, not a well). Strain: torsional (bonds aligned) + steric (methyls clash). Why this step? $\varphi =0$ is the definition of eclipsing; eclipsed conformers are peaks, and two bulky groups aligned is the worst-case cell.
2. $\varphi =60°$ → back bonds peek through gaps → gauche (skew). This is an energy minimum (a well). Strain: mild steric only (methyls just $60°$ apart), torsional strain removed. Why this step? $60°$ is a staggered position → local minimum, so torsional strain vanishes; but bulk still matters.
3. $\varphi =120°$ → aligned again, but now CH₃ sits over an H → eclipsed. Energy maximum (a lower peak than syn). Strain: torsional + smaller CH₃/H steric. Why this step? Every $60°$ step flips between staggered (well) and eclipsed (peak); $120°$ is the second eclipsing peak.
4. $\varphi =180°$ → methyls at opposite ends → anti (trans). This is the global minimum well. Strain: essentially none. Why this step? $180°$ = maximum separation and staggered → deepest well.

Now the symmetry point (why "four wells" is wrong wording): the full turn has four named minima-or-maxima on each side because of $E(\varphi )=E(-\varphi )$. Continuing past $180°$ we hit the mirror copies: $\varphi =240°$ = eclipsed CH₃/H (same peak as $120°$), $\varphi =300°$ = gauche (same well as $60°$), and $\varphi =360°=0°$ = syn again. So over a full turn there are exactly two distinct wells (anti at $180°$, gauche at $60°$ and its twin $300°$) and three distinct peaks (syn at $0°/360°$, eclipsed at $120°$ and its twin $240°$).

Verify: the named angles alternate staggered/well ($60,180$) and eclipsed/peak ($0,120$), and the whole turn closes with mirror partners at $240°$ and $300°$ — so wells:peaks over $360°$ = "anti + two gauche" vs "syn + two eclipsed." Ranking by strain magnitude, in words (largest strain first): syn has the most strain, then eclipsed at $120°$, then gauche, and anti has the least — note we deliberately do not write this with "$>$", since on this page "$>$" is reserved for stability. ✓

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Ex 2 — Cell B: energy from the ethane formula

Forecast: more or less than half the barrier, i.e. above or below $6.25$?

1. Fix the symbol: $E_b$ and $E_{\text{barrier}}$ are the same number, $12.5{\text{kJ mol}}^{-1}$. Substitute it into the formula. Why this step? The formula was written with $E_b$; the data was given as $E_{\text{barrier}}$ — we must state they are one quantity before plugging in.
2. Compute the inside angle in degrees: $3\varphi =3\times 30°=90°$. Why this step? The $3\varphi$ encodes the three-fold symmetry — we must feed the tripled angle to $\cos$, keeping it in degrees as declared.
3. $\cos 90°=0$ (degrees). Why this step? $90°$ is the exact crossover where cosine passes through zero — a clean midpoint value.
4. $E=\frac{12.5}{2}(1+0)=\frac{12.5}{2}=6.25{\text{kJ mol}}^{-1}$. Why this step? Substitute the cosine value; the $(1+0)$ leaves half the barrier.

Verify: at $\varphi =30°$ we are exactly halfway in angle between a minimum ($60°$) and a maximum ($0°$), and we got exactly half the barrier — the cosine curve is symmetric, so this is the expected midpoint. Units: kJ/mol ✓.

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Ex 3 — Cell C: degenerate case (steric strain = 0)

Forecast: if 3 pairs eclipse at once, how big is one pair's share?

1. Ethane's substituents are all H atoms — small, no bulky groups to clash → steric (van der Waals) term $\approx 0$. Why this step? This is the degenerate input: strip away one of the two competing effects, so only torsion remains. See Steric strain and van der Waals repulsion.
2. At $\varphi =0°$ (eclipsed), three H–H pairs align simultaneously and share the full barrier. Why this step? Three front H's each face a back H — the strain is a sum over pairs.
3. Per pair: $\frac{12.5}{3}=4.17{\text{kJ mol}}^{-1}\approx 4.2{\text{kJ mol}}^{-1}$. Why this step? Divide the total by the number of equal contributors.

Verify: the parent note quotes "≈ 4.2 kJ per eclipsing H–H pair" — matches. Because steric is zero, $E(60°)=\frac{12.5}{2}(1+\cos 180°)=0$: the staggered well truly sits at zero, unlike butane's gauche. ✓

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Ex 4 — Cell D: strain bookkeeping (add up contributions)

Forecast: will simple addition land inside the accepted $\sim 19\text{–}20{\text{kJ mol}}^{-1}$?

1. Get the H–H eclipsing energy from Ex 3: one eclipsing H–H pair $=4.17{\text{kJ mol}}^{-1}$ (derived, not assumed). Why this step? We only allow numbers this page has already justified — the H–H value came out of ethane in Ex 3.
2. Name the one literature anchor we lean on: the experimentally measured syn-butane barrier is $\approx 19{\text{kJ mol}}^{-1}$ (this is the target we are reproducing, quoted from measurement — the parent note lists $\sim 19\text{–}20$). Using it plus the gauche well ($3.8$, a given datum defined earlier), we derive the CH₃/CH₃ eclipsing term. The two H/H eclipses in syn contribute $2\times 4.17=8.34$, so the CH₃/CH₃ eclipsing term $\approx 19-8.34=10.7\approx 11{\text{kJ mol}}^{-1}$. Why this step? This earns the $\approx 11$ figure by back-solving it from quantities already established or explicitly named as given, instead of dropping it in unexplained.
3. Reassemble the total as a forward check: $\underset{\text{CH₃/CH₃}}{\underbrace{11}}+\underset{\text{H/H}}{\underbrace{4.17}}+\underset{\text{H/H}}{\underbrace{4.17}}=19.3{\text{kJ mol}}^{-1}$. Why this step? Total strain is the sum of independent eclipsing interactions; rebuilding it confirms internal consistency.
4. Compare: $19.3$ sits right inside the accepted $19\text{–}20$ range. Why this step? Sanity-check the additive model against the measured value.

Verify: $19.3\in [19,20]$ ✓, and the derived CH₃/CH₃ term ($\approx 11$) is by construction consistent with syn = 19. The dominant term is the CH₃/CH₃ overlap ($11$ of $19$) — which is exactly why syn is less stable than the $\varphi =120°$ eclipsed (where CH₃ only meets H). ✓

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Ex 5 — Cell E: Boltzmann population with degeneracy

Forecast: with $\Delta E\approx 1.5RT$, roughly what fraction of one gauche well vs anti — a fifth? a half?

1. Convert energy to joules: $\Delta E=3800{\text{J mol}}^{-1}$; compute $\frac{\Delta E}{RT}=\frac{3800}{8.314\times 298}=1.534$. Why this step? The exponent in $e^{-\Delta E/RT}$ must be dimensionless — J on top, J on bottom (see Boltzmann distribution).
2. One well: $\frac{N_{\text{gauche}}}{N_{\text{anti}}}=e^{-1.534}=0.216$. Why this step? The Boltzmann factor gives the per-state population ratio.
3. Two gauche wells ($+60°$ and $-60°$, equal by $E(\varphi )=E(-\varphi )$): total gauche $=2\times 0.216=0.431$. Why this step? Anti is one state; gauche is two mirror-equivalent states — the $\pm \varphi$ symmetry from the range definition is why the degeneracy is exactly 2.
4. Fractions: anti $=\frac{1}{1+0.431}=0.699$, gauche $=\frac{0.431}{1.431}=0.301$. Why this step? Normalise so all fractions sum to 1.

Verify: $\approx 70\%$ anti, $30\%$ gauche — matches the parent note. Energy alone (one well) would say $18\%$ gauche; the ×2 degeneracy pushes it to $30\%$, showing entropy matters. ✓

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Ex 6 — Cell F: limiting behaviour ($T\to 0$ and $T\to \infty$)

Forecast: where do all the molecules pile up when frozen? When scorching?

1. $T\to 0$: exponent $\frac{\Delta E}{RT}\to +\infty$, so $e^{-\infty }\to 0$ → no gauche. Why this step? Cold means no thermal energy to climb even a $3.8$ kJ hill; everything falls into the lowest well (anti).
2. Numeric at $T=1\text{K}$: exponent $=3800/8.314=457$, $e^{-457}\approx 0$, anti fraction $\approx 1$. Why this step? Confirms the limit with a concrete tiny temperature.
3. $T\to \infty$: exponent $\to 0$, $e^0=1$ → every state equally likely. Why this step? Infinite heat erases the energy gap; only counting matters.
4. With 1 anti + 2 gauche equally likely: anti fraction $=\frac{1}{3}=0.333$. Why this step? At the hot limit populations reflect pure degeneracy (3 states total).

Verify: monotonic story — anti fraction runs from $1$ (cold) down toward $1/3$ (infinitely hot), passing $0.699$ at $298$ K, which lies between $1/3$ and $1$. ✓

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Ex 7 — Cell G: real-world word problem

Forecast: heat generally lets molecules explore higher-energy states — up or down for gauche?

1. At $400\text{K}$: exponent $=\frac{3800}{8.314\times 400}=1.143$; per well $e^{-1.143}=0.319$. Why this step? Recompute the Boltzmann factor at the new temperature — only $T$ changed.
2. Total gauche count $=2\times 0.319=0.638$. Why this step? Same degeneracy factor as before; the physics of two wells is unchanged.
3. Gauche fraction $=\frac{0.638}{1+0.638}=0.389$, i.e. $\approx 39\%$. Why this step? Normalise against the single anti state.
4. Compare to $30\%$ at $298\text{K}$: it rose by about $9$ percentage points. Why this step? Answer the "which way / how much" question directly.

Verify: heating moves population toward the higher-energy gauche state ($30\%\to 39\%$), heading toward the hot limit of $2/3\approx 67\%$. Direction and magnitude both sensible. ✓

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Ex 8 — Cell H: exam-style twist (staggered ≠ lowest)

Forecast: spot the hidden assumption before reading the fix.

1. Identify the trap: the claim assumes "staggered ⇒ lowest" is a universal law. Why this step? Exam twists usually over-generalise a true statement (see Steric strain and van der Waals repulsion).
2. Counter: staggering only removes torsional strain. Gauche ($\varphi =60°$) is staggered but its two methyls sit only $60°$ apart → residual steric strain $\approx 3.8{\text{kJ mol}}^{-1}$. Why this step? Butane has bulky groups; ethane does not — the two cases differ precisely here.
3. Anti ($\varphi =180°$) is staggered and methyls maximally separated → true minimum, $\Delta E=0$. Why this step? Only anti minimises both strains simultaneously.
4. Correct stability ranking (recall "$>$" = more stable, defined at the top of this page): Anti $>$ Gauche $>$ Eclipsed($120°$) $>$ Fully eclipsed/syn($0°$). Why this step? Rank by total strain = torsional + steric; most stable (least strain) is written first, matching our "$>$" convention. Note that reading this list right-to-left gives the strain-magnitude order (syn most strained) — but we express that reversal in words, never by flipping the "$>$".

Verify: gauche sits $3.8$ kJ/mol above anti (Ex 5), so anti ≠ gauche — the claim's equality is false. Ranking matches the parent's AGES mnemonic. ✓

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Ex 9 — Cell I: draw / read a Newman

Forecast: if anti is $\varphi =180°$ and we rotate one methyl by $120°$, where does it land?

1. Anti: front carbon (dot) carries CH₃ pointing up, two H's lower-left/lower-right. Back carbon (circle) carries CH₃ pointing down, staggered in the gaps → ${\varphi }_{\text{CH₃–CH₃}}=180°$. Why this step? Anchor the reference geometry: front star, back star offset by $60°$, methyls opposite.
2. Rotate back carbon by $+120°$: the back CH₃ that pointed down swings up by $120°$. Why this step? Rotating in $120°$ steps keeps everything staggered (each back bond lands in a new gap).
3. New dihedral: $180°-120°=60°$ → this is a gauche projection. Why this step? Subtract the rotation from the starting dihedral; $60°$ is the gauche signature.
4. Confirm it is a well: $60°$ is a staggered angle (bonds in gaps), so torsional strain is zero → a local energy minimum, matching the gauche well of Ex 1. Why this step? Reading a Newman is only useful if we can also say whether that geometry is stable — closing the loop with the energy picture.

Verify: $180°\to 60°$ is a $120°$ change; still staggered (bonds in gaps, 6-pointed star), but the methyls are now only $60°$ apart → gauche, consistent with Ex 1 and Ex 5. If instead we had rotated only $+60°$ we would have hit $\varphi =120°$, an eclipsed peak — so the $120°$ step size is what keeps us landing on wells. ✓

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Recall Which cell does each example cover?

Ex 1 → A (angle→name) ::: Ex 2 → B (ethane formula) Ex 3 → C (degenerate, steric=0) ::: Ex 4 → D (strain bookkeeping) Ex 5 → E (Boltzmann) ::: Ex 6 → F (limits T→0, T→∞) Ex 7 → G (word problem) ::: Ex 8 → H (exam twist) Ex 9 → I (draw Newman) ::: — every matrix cell hit ✓

Connections

• Boltzmann distribution — the engine behind Ex 5–7.
• Steric strain and van der Waals repulsion — why gauche isn't zero (Ex 3, 8).
• Hyperconjugation — the electronic reason staggered wins.
• Cyclohexane conformations chair and boat — the same strain logic on a ring.