Exercises — Conformations of ethane, butane — Newman projections

Staggered. The clue is the dihedral angle $\varphi$ — the angle between a front bond (line from the centre dot) and the nearest back bond (line from the circle's edge).

• If $\varphi =0°$ the back bonds would hide directly behind the front bonds (three lines only, they overlap) → eclipsed.
• Here the back bonds sit in the gaps at $\varphi =60°$, giving a clean six-pointed star → staggered.

What we did: measured the front-to-back bond gap. Why: the gap is the single number that names the conformer.

$\varphi =180°$ means the two methyls point in exactly opposite directions — the maximum possible separation. This is the anti (trans) conformer, the global energy minimum. All bonds are staggered and the bulky groups are as far apart as geometry allows, so neither torsional nor steric strain bites.

What the formula says: $\cos 3\varphi$ tracks the three-fold symmetry — three H's mean the pattern repeats every $120°$. Plug in $\varphi =30°$, so $3\varphi =90°$: $\cos 90°=0\Rightarrow E=\frac{12.5}{2}(1+0)=6.25{\text{kJ mol}}^{-1}$ Interpretation: $\varphi =30°$ is exactly halfway between staggered ($60°$, $E=0$) and eclipsed ($0°$, $E=12.5$). The molecule is climbing the barrier, sitting at half-height. Makes physical sense ✓.

The three pairs are identical by symmetry, so they share the barrier equally: $\frac{12.5}{3}\approx 4.17{\text{kJ mol}}^{-1}\text{per H–H pair}$ Why divide: symmetry guarantees each pair costs the same, so total $=$ 3 × (one pair). This ~4.2 kJ/mol number is the building block chemists quote per eclipsing interaction.

Which tool and why: the Boltzmann distribution answers "given an energy gap, what fraction of molecules climb into the higher state?" That is exactly our question. $\frac{N_{\text{gauche}}}{N_{\text{anti}}}=e^{-\Delta E/RT}$ Keep units consistent — convert kJ to J: $\Delta E=3800{\text{J mol}}^{-1}$. $\frac{\Delta E}{RT}=\frac{3800}{8.314\times 298}=\frac{3800}{2477.6}=1.533$ $\frac{N_{\text{gauche}}}{N_{\text{anti}}}=e^{-1.533}\approx 0.216$ Meaning: for every 100 anti molecules, about 22 sit in each gauche well.

Why count two wells: $+60°$ and $-60°$ are distinct arrangements of equal energy — a degeneracy that doubles gauche's share (an entropy effect). Total gauche relative population $=2\times 0.216=0.432$. Treat anti's population as $1$. Then the fractions are: $\text{anti}=\frac{1}{1+0.432}=\frac{1}{1.432}=0.698\approx 70\%$ $\text{gauche (both)}=\frac{0.432}{1.432}=0.302\approx 30\%$ So room-temperature butane is roughly 70% anti, 30% gauche ✓ — the classic textbook mixture.

Read the curve left to right. Each feature comes from adding two kinds of strain — torsional (from eclipsing) + steric (from bulky ${\text{CH}}_3$ clashes). See Steric strain and van der Waals repulsion.

Stability order: $\text{Anti}>\text{Gauche}>\text{Eclipsed}>\text{Syn}$. Key insight: the two eclipsed peaks are unequal. At $0°$ methyl sits over methyl (fat over fat) → worst. At $120°$ methyl sits over H (fat over thin) → milder. Ethane would show three equal peaks; butane's methyl breaks that symmetry.

Compute how many "thermal units" the barrier is worth: $\frac{E_{\text{barrier}}}{RT}=\frac{19}{2.48}\approx 7.7$ The barrier is under 8 $RT$. Barriers below roughly $80-100{\text{kJ mol}}^{-1}$ are crossed billions of times per second at room temperature (the Boltzmann tail always supplies enough molecules with $E>E_{\text{barrier}}$). Conclusion: anti and gauche interconvert far too fast to isolate — they are conformers, not isomers you can put in separate flasks. This is the boundary drawn in Stereochemistry — isomerism overview.

Set up the equation. With two gauche wells of ratio $r=e^{-\Delta E/RT}$ each: $\text{fraction anti}=\frac{1}{1+2r}=0.65$ Solve for $r$: $1+2r=\frac{1}{0.65}=1.538\Rightarrow 2r=0.538\Rightarrow r=0.269$ Invert the Boltzmann relation. Take $\ln$: $\ln r=-\frac{\Delta E}{RT}\Rightarrow T=-\frac{\Delta E}{R\ln r}=-\frac{3800}{8.314\times \ln (0.269)}$ $\ln (0.269)=-1.313\Rightarrow T=-\frac{3800}{8.314\times (-1.313)}=\frac{3800}{10.92}\approx 348\text{K}$ Physical direction: $348\text{K}>298\text{K}$ — it is hotter. Heating supplies more thermal energy, so more molecules climb into the higher-energy gauche wells; the anti fraction falls from 70% toward 65%. Raising $T$ always pushes populations toward the higher states (the exponential $e^{-\Delta E/RT}$ rises toward 1 as $T\to \infty$).

Build the formula from symmetry. The exponent inside cosine counts maxima per $360°$: three-fold → $\cos 3\varphi$ (3 peaks). Two-fold → replace 3 with 2: $E(\varphi )=\frac{E_b}{2}(1+\cos 2\varphi )$ Check: maxima where $\cos 2\varphi =+1$, i.e. $2\varphi =0°,360°$ → $\varphi =0°,180°$. That is 2 maxima per full turn. Real-molecule link. Butane's dominant clash is between the two methyl groups (one on each carbon). As you rotate, the big methyl–methyl eclipse recurs strongly with a period tied to those two groups meeting — a roughly two-fold pattern in the steric term, superimposed on the three-fold torsional background. That is precisely why butane's curve has two unequal big features (syn at $0°$, plus the milder eclipse at $120°$) rather than three identical ones like ethane. Why this matters: the cosine's integer coefficient is literally a symmetry counter — read the molecule's rotational symmetry and you can predict the number of energy peaks before doing any chemistry.

Why is butane's $\varphi =0°$ peak higher than its $\varphi =120°$ peak? ::: At $0°$ methyl eclipses methyl (both bulky → torsional + large steric); at $120°$ methyl eclipses only H (torsional + small steric). What single number in $\cos n\varphi$ tells you the peak count? ::: $n$ — the rotational symmetry order equals the number of maxima per full turn. Heating butane shifts population which way? ::: Toward the higher-energy gauche conformers, so anti fraction drops.