Exercises — Cycloalkanes — Baeyer's strain theory; cyclohexane chair - boat, axial vs equatorial

A points straight up, parallel to the vertical axis of the ring → this is the axial bond. B points outward, roughly along the "equator" of the ring → this is the equatorial bond. How to never mix them up: axial = along the axis (up/down). The word literally contains "axis". Everything else is equatorial.

The chair. Reason in one line: every neighbouring C–H pair is staggered (no torsional strain) and every carbon angle is $\approx 111^{\circ }$ (essentially no angle strain), and there is no flagpole crowding.

He assumed every ring is a flat regular polygon. Reality: rings larger than 4 carbons pucker (fold into 3D) to escape strain, which his flat model ignored.

Step 1 (WHAT): internal angle of a square. ${\theta }_{\text{internal}}=\frac{(4-2)\times 180^{\circ }}{4}=\frac{360^{\circ }}{4}=90^{\circ }.$ Step 2 (WHAT): strain per bond. $d=\frac{109.5^{\circ }-90^{\circ }}{2}=\frac{19.5^{\circ }}{2}=+9.75^{\circ }.$ WHY it matters: $d>0$ means the bonds are squeezed inward by nearly $10^{\circ }$ each — cyclobutane really is strained (though it also puckers a little to trade some angle strain for less torsional strain).

${\theta }_{\text{internal}}=\frac{(5-2)\times 180^{\circ }}{5}=\frac{540^{\circ }}{5}=108^{\circ }.$ $d=\frac{109.5^{\circ }-108^{\circ }}{2}=\frac{1.5^{\circ }}{2}=+0.75^{\circ }.$ Comment: $108^{\circ }$ is a whisker from $109.5^{\circ }$, so angle strain is almost nothing. The flat model is nearly fine here — but cyclopentane still puckers into an "envelope" to kill the leftover torsional strain (bonds eclipsing), which the angle formula cannot see.

${\theta }_{\text{internal}}=\frac{(6-2)\times 180^{\circ }}{6}=\frac{720^{\circ }}{6}=120^{\circ }.$ $d=\frac{109.5^{\circ }-120^{\circ }}{2}=\frac{-10.5^{\circ }}{2}=-5.25^{\circ }.$ Why misleading: the negative $d$ makes the flat model predict strain. But cyclohexane is not flat — the chair fold restores each angle to $\approx 111^{\circ }$, giving zero angle strain. The flat formula fails for $n\ge 6$ exactly because those rings pucker.

Chair (left figure): sighting down a ring bond, the front and back C–H bonds are staggered (offset by $60^{\circ }$, like staggered ethane). Staggered = electron clouds far apart = low torsional strain. Boat (right figure): along the two "side" bonds the C–H's are eclipsed (lined up) → torsional strain. Worse, the two hydrogens at the prow and stern — the flagpole hydrogens — point directly at each other and crowd (steric strain). Conclusion: boat carries both eclipsing and flagpole penalties; chair carries neither. Chair is lower in energy by $\approx 29\text{kJ/mol}$.

Ring flip: the chair inverts through a twist-boat into the other chair. Every axial bond becomes equatorial and every equatorial becomes axial. Fractions: for plain cyclohexane the two chairs are identical (all H's), so $\Delta G=0$. Then $K=e^{-0/RT}=e^0=1$ → a perfect 50 : 50 mix. No preference, because there is nothing to prefer.

Step 1 (WHAT — count the penalty): an axial methyl points up into the space of the two other axial groups on the same face → two 1,3-diaxial clashes. $E_{\text{ax penalty}}=2\times 3.8=7.6\text{kJ/mol}.$ Step 2 (WHY the sign): going axial→equatorial removes that penalty, so the system drops in energy: $\Delta G=-7.6\text{kJ/mol}=-7600\text{J/mol}.$ Negative $\Delta G$ = spontaneous, equatorial is favoured. Step 3 (WHAT — plug into $K=e^{-\Delta G/RT}$): $K=e^{-(-7600)/(8.314\times 298)}=e^{+7600/2477.6}=e^{+3.068}\approx \mathrm{21.5.}$ Interpret: $K\approx 21.5$ means equatorial:axial $\approx 21.5:1$, i.e. $\frac{21.5}{22.5}\approx 95.6\%$ of molecules sit in the equatorial-methyl chair. Bulky group → outward → happy.

Setup: on a chair, carbons 1 and 4 sit across the ring. Because of the alternating up/down pattern, two substituents that are trans (opposite faces) can occupy the pair of positions that are both equatorial, whereas cis (same face) forces one axial, one equatorial.

• trans-1,4: both methyls equatorial (e,e) → no 1,3-diaxial strain → more stable.
• cis-1,4: one methyl is forced axial (a,e) → suffers 1,3-diaxial clashes → less stable. Winner: trans-1,4-dimethylcyclohexane. (Note: which isomer gets e,e flips as you go 1,2 → 1,3 → 1,4 — memorise the reasoning, not a rule.)

(a) $K=e^{-(-22800)/(8.314\times 298)}=e^{+22800/2477.6}=e^{+9.203}\approx 9.94\times 10^3.$ (b) Axial fraction $=\frac{1}{K+1}=\frac{1}{9941}\approx 1.006\times 10^{-4}=0.010\%$. Practically locked equatorial. (c) tert-Butyl is a bulky, three-methyl ball; axial it slams into both 1,3-diaxial partners far harder than a lone methyl does, so its energy penalty (and hence $\ln K$) is roughly three times larger — turning a $\sim 95\%$ preference into a $\sim 99.99\%$ one.

(a) Set $d=0$: $109.5=\frac{(n-2)180}{n}$. Multiply out: $109.5n=180n-360\Rightarrow 360=70.5n\Rightarrow n=\frac{360}{70.5}\approx \mathrm{5.106.}$ Since $n$ must be a whole number, no ring is exactly strain-free in the flat model; $n=5$ ($d=+0.75^{\circ }$) is the closest. (b) For $n=6$: $\theta =120^{\circ }$, $d=\frac{109.5-120}{2}=-5.25^{\circ }$. Reconciliation: the flat model assumes a $120^{\circ }$ corner, but real cyclohexane refuses to stay flat — it puckers into the chair, restoring each angle to $\approx 111^{\circ }$ and dissolving the predicted strain entirely.

Internal angle formula for a flat $n$-gon? ::: $\theta =\frac{(n-2)180^{\circ }}{n}$ Baeyer strain per bond formula? ::: $d=\frac{109.5^{\circ }-\theta }{2}$ $d$ for cyclopropane ($n=3$)? ::: $+24.75^{\circ }$ $d$ for cyclobutane ($n=4$)? ::: $+9.75^{\circ }$ On a ring flip, axial becomes...? ::: equatorial (and equatorial becomes axial) $K$ formula linking energy gap to the equilibrium ratio? ::: $K=e^{-\Delta G/RT}$ Methylcyclohexane equatorial:axial ratio at 298 K? ::: about $21.5:1$ ($\approx 95\%$ eq) Why is Baeyer wrong for $n\ge 6$? ::: real rings pucker into 3D, escaping angle strain