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

The scenario matrix

Before solving anything, let us map the whole territory. Each cell below is a class of situation. If we solve at least one example from every cell, we have covered everything this topic can throw at you.

Cell	Situation class	What makes it tricky
A	Small strained ring ( $n = 3, 4$ )	large positive Baeyer $d$ ; ring cannot pucker away the strain
B	Near-zero strain ( $n = 5$ )	tiny positive $d$ ; the flat formula almost works
C	Baeyer's failure zone ( $n \geq 6$ )	flat formula predicts strain that does not exist
D	Energy → population ( $Δ G$ , $K$ )	turning a strain number into a real ratio via $Δ G = - R T ln K$
E	Sign / direction trap	which way is $Δ G$ ? axial→equatorial vs equatorial→axial
F	Geometry of one chair	counting axial vs equatorial, up vs down at each carbon
G	Ring-flip bookkeeping	what swaps, what stays, after a flip
H	cis / trans stereochemistry	which disubstituted isomer can be $e, e$
I	Degenerate / limiting inputs	$n \to \infty$ , zero-penalty (no substituent / H) groups, symmetric molecules
J	Real-world / exam twist	multi-step reasoning chaining several rules into one answer

We now solve examples pinned to these cells.

Example 1 — Cell A: the most strained ring

Worked example Baeyer strain of cyclopropane and cyclobutane

Compute the per-bond angular deviation $d$ for $n = 3$ and $n = 4$ , and say which is more strained.

Forecast: Guess — will the triangle or the square strain each bond more? By how many degrees?

Step 1. Internal angle of a regular polygon: $θ = \frac{( n - 2 ) \times 180°}{n}$ .

Why this step? A flat $n$ -gon's corners sum to $(n - 2) 180°$ ; sharing equally gives one corner. This is the angle carbon's bonds are forced into if the ring lies flat.
For $n = 3$ : $θ = \frac{( 1 ) 180}{3} = 60°$ . For $n = 4$ : $θ = \frac{( 2 ) 180}{4} = 90°$ .

Step 2. Deviation per bond: $d = \frac{109.5° - θ}{2}$ .

Why this step? Carbon wants $109.5°$ (its tetrahedral angle). The mismatch $109.5° - θ$ is split between the two ring bonds at that carbon, so each bends by half.
$n = 3$ : $d = \frac{109.5 - 60}{2} = + 24.75°$ .
$n = 4$ : $d = \frac{109.5 - 90}{2} = + 9.75°$ .

Step 3. Bigger $d > 0$ = bonds squeezed further inward = more strain.

Cyclopropane ( $+ 24.75°$ ) is far more strained than cyclobutane ( $+ 9.75°$ ). Cyclopropane's bonds are so bent they are called banana bonds and it ring-opens readily.

Verify: $d_{3} - d_{4} = 24.75 - 9.75 = 15°$ extra strain for the triangle. Both positive → both compressed, matching cyclopropane being the more reactive of the two. Units: pure angle (degrees). ✔

Example 2 — Cell B & C: where the flat formula is right vs where it lies

Worked example Cyclopentane (

$n = 5$ ) vs cyclohexane ( $n = 6$ ) by the flat formula Compute $d$ for both. Then state which number to trust and which to throw away.

Forecast: The formula will make one of these look better than the other. Guess which — and guess whether reality agrees.

Step 1. $n = 5$ : $θ = \frac{( 3 ) 180}{5} = 108°$ , so $d = \frac{109.5 - 108}{2} = + 0.75°$ .

Why this step? $108°$ sits almost exactly at $109.5°$ , so the flat pentagon barely strains its bonds — this is Cell B, where the flat model nearly works.

Step 2. $n = 6$ : $θ = \frac{( 4 ) 180}{6} = 120°$ , so $d = \frac{109.5 - 120}{2} = - 5.25°$ .

Why this step? The number is negative — the flat hexagon's angle is "too open." Baeyer read this as strain. This is Cell C.

Step 3. Apply the reality correction.

Why this step? Cyclohexane does not stay flat: it puckers into a chair where every C–C–C angle relaxes back to $\approx 111° \approx 109.5°$ , giving zero angle strain. The $- 5.25°$ is an artefact of the false flat assumption. Cyclopentane cannot fully relax and keeps slight torsional strain (envelope form).

Verify: Confirmed by heat of combustion per CH₂: cyclohexane burns at $658.7 kJ/mol$ per CH₂ — the lowest (most stable) of the common rings, beating cyclopentane. So the negative $d$ must be ignored for $n \geq 6$ . ✔

Example 3 — Cell I: the limiting case $n \to \infty$

Worked example What does the flat formula predict for a giant ring, and why is it nonsense?

Evaluate $d$ as $n$ grows without bound.

Forecast: Does $d$ blow up, settle to a number, or shrink? Guess the limiting value.

Step 1. Simplify the internal angle: $θ = \frac{( n - 2 ) 180}{n} = 180° - \frac{360°}{n}$ .

Why this step? Rewriting exposes the $n$ -dependence cleanly: the $\frac{360}{n}$ term is what shrinks as the ring grows.

Step 2. As $n \to \infty$ , $\frac{360}{n} \to 0$ , so $θ \to 180°$ and $d \to \frac{109.5 - 180}{2} = - 35.25°.$

Why this step? A huge flat polygon looks locally like a straight line ( $180°$ ), so the flat model claims each bond is bent a fixed $- 35.25°$ forever.

Step 3. Reality: large rings pucker freely into zig-zag 3D shapes, keeping $\approx 109.5°$ everywhere. Strain does not grow with $n$ ; it stays near zero.

Why this step? This is the sharpest demonstration of Baeyer's flaw — his model diverges to a fixed wrong answer while nature stays comfortable.

Verify: $θ (n = 6) = 180 - 60 = 120°$ ✔ matches Example 2, so the rewritten form is correct; the limit $- 35.25°$ is a mathematical fact about the flat model and a physical falsehood about real molecules. ✔

Example 4 — Cell D & E: energy to population, and the sign trap

Worked example Methylcyclohexane: turn strain into a percentage

Axial-CH₃ suffers two $1, 3$ -diaxial clashes of $\approx 3.8 kJ/mol$ each. Find $K = [eq] / [ax]$ at $298 K$ and the equatorial percentage.

Forecast: Guess the equatorial:axial ratio — closer to $2 : 1$ , $20 : 1$ , or $200 : 1$ ?

Step 1. Total axial penalty $= 2 \times 3.8 = 7.6 kJ/mol$ , so equatorial is lower by that much.

Why this step? Each $1, 3$ -diaxial interaction is an independent steric clash; they add.

Step 2. Fix the sign (this is Cell E). For the process axial → equatorial (uphill in stability backwards, so downhill in energy): $Δ G_{ax \to eq} = - 7.6 kJ/mol = - 7600 J/mol .$

Why this step? Going to the more stable state releases energy, so $Δ G$ is negative. Get this sign wrong and $K$ inverts.

Step 3. Use $\Delta G=-RT\ln K$ rearranged to $K = e^{- Δ G / R T}$ , with $R = 8.314 J mol^{- 1} K^{- 1}$ , $T = 298 K$ : $K = e^{- (- 7600) / (8.314 \times 298)} = e^{+ 7600/2477.6} = e^{3.067} \approx 21.5.$

Why this step? $Δ G < 0 \Rightarrow$ exponent positive $\Rightarrow K > 1 \Rightarrow$ equilibrium favours equatorial.

Step 4. Percentage equatorial $= \frac{K}{K + 1} = \frac{21.5}{22.5} = 0.955 = 95.5%$ .

Why this step? $K$ is a ratio eq:ax $= 21.5 : 1$ ; the fraction that is equatorial is $K / (K + 1)$ .

Verify: $R T = 8.314 \times 298 = 2477.6 J/mol$ ; $7600/2477.6 = 3.067$ ; $e^{3.067} \approx 21.5$ ; $21.5/22.5 \approx 0.955$ . Ratio $\approx 21 : 1$ , i.e. $\sim 95%$ equatorial — matches the parent's answer. Units cancel: $J/mol \div J/mol$ = dimensionless exponent. ✔

Example 5 — Cell E again: the deliberate sign flip

Worked example Same molecule, wrong direction — catch the error

A student writes $Δ G_{eq \to ax} = + 7.6 kJ/mol$ and computes $K = [ax] / [eq]$ . What do they get, and is it consistent?

Forecast: Should this $K$ be $> 1$ or $< 1$ ? Guess before computing.

Step 1. Now the process is equatorial → axial, which is uphill, so $Δ G = + 7600 J/mol$ .

Why this step? Reversing the reaction flips the sign of $Δ G$ .

Step 2. $K^{'} = e^{- (+ 7600) / (8.314 \times 298)} = e^{- 3.067} \approx 0.0465.$

Why this step? Positive $Δ G \Rightarrow$ exponent negative $\Rightarrow K^{'} < 1 \Rightarrow$ the axial-rich side is disfavoured — correct!

Step 3. Consistency check: $K^{'}$ (ax:eq) should be the reciprocal of $K$ (eq:ax) from Example 4.

Why this step? Reversing a reaction inverts its equilibrium constant.

Verify: $1/21.5 = 0.0465$ ✔. Both routes agree; the only thing that changes is which species sits on top of the ratio. This is why fixing the direction of $Δ G$ before plugging in is non-negotiable. ✔

Example 6 — Cell F: geometry of a single chair

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

Worked example Count axial and equatorial bonds; identify up vs down

In one chair of cyclohexane, how many axial bonds point up, how many point down, and where do equatorial bonds go?

Forecast: Six carbons, each with two ring-H bonds. Guess the up/down split of the axial set.

Step 1. Each carbon has exactly one axial and one equatorial C–H bond → $6$ axial + $6$ equatorial total.

Why this step? Every ring carbon is equivalent by symmetry; it must carry one of each type.

Step 2. Axial bonds alternate around the ring: up, down, up, down, up, down.

Why this step? Alternate carbons are puckered up vs down (that is what "chair" means), so their vertical axial bonds alternate too. Look at the figure: the yellow axial arrows point straight up on carbons 1,3,5 and straight down on 2,4,6.

Step 3. Each equatorial bond points outward, tilted opposite to its own carbon's axial bond (a carbon with an "up" axial has a slightly-down equatorial). See the green equatorial arrows fanning outward in the figure.

Why this step? The two bonds on one carbon share the tetrahedral budget; one goes vertical (axial), the other splays outward and slightly the other way (equatorial).

Verify: $3$ axial-up $+ 3$ axial-down $= 6$ axial ✔; $6$ equatorial ✔; total ring-H $= 12$ , matching $C_{6} H_{12}$ 's twelve hydrogens. ✔

Example 7 — Cell G: ring-flip bookkeeping

Worked example Track a substituent through a ring flip

A methyl group sits axial-up on carbon 1. After a chair→chair ring flip, where is it?

Forecast: Will the methyl become equatorial-up, equatorial-down, axial-down, or stay put?

Step 1. Rule: a ring flip converts every axial bond to equatorial and vice versa — but keeps each substituent on the same face (up stays up-ish / a group above the mean plane stays above).

Why this step? Flipping pushes up-carbons down and down-carbons up; the vertical axial direction and the outward equatorial direction trade roles on each carbon.

Step 2. So axial-up methyl becomes equatorial after the flip. Its "up" character is preserved as an equatorial-pointing-slightly-up bond.

Why this step? Only axial↔equatorial swap; the face label is conserved because bonds don't detach.

Step 3. Because equatorial is lower energy (Example 4), the flipped chair is the favoured one — the molecule spends $\sim 95%$ of its time here.

Why this step? The system relaxes toward the low-strain chair, consistent with the $K \approx 21$ we computed.

Verify: In the figure, before the flip the red methyl is axial (vertical); after, it is equatorial (outward). Everything axial↔equatorial swapped, the face stayed the same. Consistent with Example 4's $\sim 95%$ equatorial. ✔

Example 8 — Cell H & I: cis/trans and a symmetric degenerate case

Worked example cis- vs trans-1,4-dimethylcyclohexane, plus the symmetric limit

Which of cis- and trans-1,4-dimethylcyclohexane can place both methyls equatorial? What about trans-1,4 after a ring flip (the degenerate/symmetric check)?

Forecast: Guess which isomer is more stable and whether flipping trans-1,4 changes anything measurable.

Step 1. cis vs trans fixes the up/down faces of the two methyls: trans = opposite faces, cis = same face.

Why this step? The face relationship is a fixed molecular property; it does not change on ring flip.

Step 2. On a chair, the $1, 4$ carbons have opposite axial directions (one axial-up, one axial-down). For both methyls to be equatorial they need the face pattern that lines up with two equatorial slots.

trans-1,4 → can be equatorial,equatorial ( $e, e$ ) → most stable. ✔
cis-1,4 → forced into axial,equatorial ( $a, e$ ) → carries $1, 3$ -diaxial strain → less stable.
Why this step? You cannot rearrange faces without breaking bonds; only trans-1,4 geometry matches the double-equatorial pattern.

Step 3. Degenerate check (Cell I): flip the $e, e$ trans chair. Both methyls become $a, a$ .

Why this step? Ring flip swaps axial↔equatorial for both. So trans-1,4 has two chairs: $e, e$ and $a, a$ . The $e, e$ one dominates hugely because $a, a$ has two axial methyls (heavy $1, 3$ -diaxial strain).

Verify: $e, e$ has $0$ axial-methyl clashes; $a, a$ has clashes from both methyls; cis $a, e$ has exactly one axial methyl ( $\approx 7.6 kJ/mol$ ). Stability order: trans- $ee$ > cis- $a e$ > trans- $aa$ . The lowest-energy accessible form is trans- $ee$ , so trans-1,4 is the more stable isomer ✔ — matching the parent note.

Example 9 — Cell J: the multi-step exam twist

Worked example Chain four rules into one answer

tert-Butylcyclohexane has one axial tert-butyl $1, 3$ -diaxial penalty totalling $\approx 22.8 kJ/mol$ (much bulkier than methyl). (a) Which chair is favoured? (b) What is $K = [eq] / [ax]$ at $298 K$ ? (c) What percentage sits axial? (d) Why is tert-butylcyclohexane called "conformationally locked"?

Forecast: Guess the axial percentage — will it be a few percent, or vanishingly small?

Step 1. Identify the favoured chair (uses the $1, 3$ -diaxial rule). A big group strongly prefers equatorial, so the equatorial-tert-butyl chair is favoured.

Why this step? Larger groups suffer far worse axial clashes, so the energy gap is much bigger than for methyl.

Step 2. Set the sign (uses the direction rule). For axial → equatorial, $Δ G = - 22800 J/mol$ (negative = favourable).

Why this step? Same logic as Example 4/5: going to the stable side releases energy.

Step 3. Convert to $K$ (uses $\Delta G=-RT\ln K$): $K = e^{- (- 22800) / (8.314 \times 298)} = e^{9.203} \approx 9930.$

Why this step? A large negative $Δ G$ makes a huge exponent → an enormous $K$ .

Step 4. Percentage axial (uses the population rule): fraction axial $= \frac{1}{K + 1} = \frac{1}{9931} \approx 0.0001 = 0.01%$ .

Why this step? $[ax] / ([ax] + [eq]) = 1/ (K + 1)$ .

Step 5. (d) Because $< 0.02%$ ever sits axial, the molecule is effectively frozen in the equatorial chair — "conformationally locked." That is why tert-butyl is used as a holding group to fix ring geometry in exam problems.

Why this step? It ties the number back to a qualitative, exam-friendly conclusion.

Verify: $R T = 2477.6$ ; $22800/2477.6 = 9.203$ ; $e^{9.203} \approx 9930$ ; $1/ (9930 + 1) \approx 1.007 \times 1 0^{- 4} \approx 0.01%$ . Compare methyl's $\sim 5%$ axial (Example 4): the bulkier group is $\sim 500 \times$ more locked. ✔

Coverage check

Recall Did we hit every cell? (reveal)

A (small strained ring) → Ex 1
B (near-zero strain, $n = 5$ ) → Ex 2
C (Baeyer failure zone) → Ex 2
D (energy → population) → Ex 4
E (sign / direction trap) → Ex 4, Ex 5
F (geometry of a chair) → Ex 6
G (ring-flip bookkeeping) → Ex 7
H (cis/trans) → Ex 8
I (degenerate / limiting) → Ex 3, Ex 8
J (real-world / exam twist) → Ex 9

Recall Rapid self-test

Flat-ring $d$ formula? ::: $d = \frac{109.5 - ( n - 2 ) 180/ n}{2}$ $K$ for methylcyclohexane axial⇌equatorial at 298 K? ::: $\approx 21$ , so $\sim 95%$ equatorial Limit of flat $d$ as $n \to \infty$ ? ::: $- 35.25°$ (a false prediction — rings pucker) After a ring flip, axial-up methyl becomes? ::: equatorial (same face) Which 1,4-dimethyl isomer is $e, e$ ? ::: trans-1,4 Why is tert-butylcyclohexane "locked"? ::: axial fraction $\sim 0.01%$ , so it stays equatorial