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

Step 1 — What "comfortable" means for a carbon atom

WHAT. A carbon atom in an alkane forms four bonds. Those four bonds don't point randomly — they point to the four corners of a tetrahedron (a triangular pyramid). The angle between any two of them is $109.5°$ .

WHY this angle. Carbon uses == $s p^{3}$ hybridization== (see sp3 Hybridization and Tetrahedral Geometry): its four bonding "arms" are identical and push away from each other as far as possible. Four things spreading out equally in 3D land at exactly $109.5°$ apart. Any bending away from $109.5°$ costs energy — the arms don't want to be squeezed.

PICTURE. Look at the red central atom: its four black arms splay out symmetrically, and the marked angle is $109.5°$ . Remember this number — it is the "happy" angle every carbon wants.

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

Step 2 — Baeyer's assumption: rings are flat polygons

WHAT. Take $n$ carbons and join them in a ring. Baeyer (1885) assumed the ring lies flat, as a regular polygon — a shape with $n$ equal sides and $n$ equal corners.

WHY he assumed it. In 1885 no one could see molecules in 3D. A flat, symmetric shape was the simplest guess. If the ring is flat, the angle at each corner is fixed by geometry alone — the carbon has no choice, it must bend its bonds to match the corner.

PICTURE. Three flat polygons: a triangle ( $n = 3$ ), a square ( $n = 4$ ), and a hexagon ( $n = 6$ ). At each corner sits a carbon (red dot). The interior corner angle is marked. Notice the triangle's corner is a tight $60°$ — nowhere near $109.5°$ .

Polygon corner angles

60°

for a triangle,

90°

for a square,

108°

for a pentagon,

120°

for a hexagon.

Step 3 — Turning the mismatch into a strain number

WHAT. Compare the corner angle the ring forces ( $θ_{internal}$ ) with the angle carbon wants ( $109.5°$ ). The gap between them is the strain. Baeyer wrote it as a strain per bond, $d$ .

WHY divide the gap in half. At each corner, two bonds meet. When the corner is squeezed, the total squeeze is shared: each of the two bonds bends by half the mismatch. So the deviation felt by one single bond is half the gap.

PICTURE. A close-up of one corner. The dashed black lines show where the bonds would point at the happy $109.5°$ ; the solid red lines show where the flat ring forces them. The two red arrows are the bend of each bond — equal, each equal to $d$ .

Step 4 — Plot it and watch Baeyer's prediction go wrong

WHAT. Compute $d$ for $n = 3, 4, 5, 6, 7, \dots$ and plot it.

WHY plot. A picture of $d$ versus ring size shows the pattern Baeyer trusted — and exactly where reality diverges from it.

PICTURE. The black curve is $d$ from the flat-ring formula. It starts high at $n = 3$ (very strained), crosses zero near $n = 5$ , and then keeps sliding below zero for $n \geq 6$ — Baeyer's claim that big rings get more and more strained. The red dots are the real measured strain: they stay near zero for $n \geq 6$ . The black curve and the red dots agree for small rings and split apart for large rings.

Ring $n$	forced angle	$d$ (flat)	reality
3	$60°$	$+ 24.75°$	very strained ✔
4	$90°$	$+ 9.75°$	strained ✔
5	$108°$	$+ 0.75°$	nearly strain-free ✔
6	$120°$	$- 5.25°$	strain-FREE ✗ Baeyer wrong

Step 5 — Escape route: the ring puckers into a chair

WHAT. A six-carbon ring does not stay flat. It folds: alternate carbons lift up and drop down, so three point up and three point down. This folded shape is the chair.

WHY folding helps. When flat, each carbon was forced to $120°$ . By letting carbons rise and fall out of the plane, the ring can adjust every C–C–C angle back to about $111°$ — essentially the happy $109.5°$ . Angle strain vanishes. The puckering "spends" the third dimension to buy back the ideal angle.

PICTURE. On the left, the flat hexagon with its forced $120°$ corner. On the right, the same six carbons folded into a chair (red), with the restored $\approx 111°$ angle marked. The alternating up/down carbons are labelled U and D.

Step 6 — Look down a bond: staggered means no torsional strain

WHAT. Torsional strain appears when the C–H bonds on two neighbouring carbons line up (eclipsed). When they sit in the gaps between each other (staggered), there is no such strain. To see which one we have, we look straight down a C–C bond — this is a Newman projection.

WHY a Newman projection. It is the only view that reveals front-vs-back bond alignment. The front carbon's three bonds and the back carbon's three bonds are drawn on one circle; if they overlap → eclipsed → strain; if they alternate → staggered → happy. This is the exact same eclipsed/staggered story you met in Conformations of Ethane and Butane.

PICTURE. Two Newman circles. Left = the chair: front bonds (black) sit neatly in the gaps between back bonds (red) → staggered, marked $60°$ offset. Right = the boat's side: front and back bonds overlap → eclipsed, marked $0°$ offset, with a red "clash" flag.

Step 7 — The runner-up: the boat and its flagpole clash

WHAT. If instead of alternating up/down you fold two carbons up on the same side, you get the boat. It looks symmetric but hides two problems.

WHY it's worse. (1) Along its two sides the bonds are eclipsed (Step 6, right circle) → torsional strain. (2) The two carbons that point up ("prow" and "stern") carry hydrogens that aim straight at each other — the flagpole hydrogens — and crowd into the same space → steric strain (see sp3 Hybridization and Tetrahedral Geometry for why atoms resist overlap). The boat sits about $29 kJ/mol$ above the chair.

PICTURE. The boat (red frame) with the two upward carbons labelled, and the two flagpole H's drawn pointing at one another with a red repulsion arrow between them.

Step 8 — Axial vs equatorial, and why bulky groups pick equatorial

WHAT. Back in the winning chair, each carbon has two outward C–H bonds of different types:

Axial: points straight up or down, parallel to the ring's vertical axis (they alternate up–down around the ring).
Equatorial: points outward around the ring's "equator".

WHY it matters for big groups. An axial group on one carbon points into the same narrow column as the axial groups two carbons away on the same face — the 1,3-diaxial positions. A bulky group jammed there clashes with them (steric strain). Swing it to equatorial and it points safely outward into open space. Hence bulky substituents prefer equatorial.

PICTURE. A chair with all axial bonds drawn vertical (red) and equatorial bonds drawn outward (black). Three axial positions on the top face are highlighted, and dashed red arcs show the 1,3-diaxial crowding a big axial group would suffer.

The one-picture summary

This single figure carries the whole walkthrough: the happy angle (Step 1) → flat rings force a bad angle (Steps 2–4) → puckering restores it (Step 5) → the chair beats the boat on both strain types (Steps 6–7) → axial vs equatorial (Step 8).

Recall Feynman retelling — the whole story in plain words

Every carbon wants its four arms spread at a comfy $109.5°$ . Baeyer said "let's lay rings out flat like coins" — and for a flat triangle the arms get crushed to $60°$ , which really does hurt, so tiny rings are strained. But he then claimed big rings hurt too, because a flat hexagon forces $120°$ . That's where he tripped: big rings don't stay flat. A six-carbon ring folds like a lawn chair — three carbons up, three down — and by folding it quietly steers every angle back to $\approx 111°$ , almost perfect. Even better, when you sight down any bond in that chair, the neighbouring arms sit in each other's gaps (staggered), so nothing rubs. The alternative fold, the boat, lets two hydrogens on top point right at each other and lets the sides line up badly — so the boat loses. In the winning chair, some arms stick straight up or down (axial) and some poke outward (equatorial). Wearing a big backpack? Point it outward (equatorial) so it doesn't bonk the arms above and below you — that's why bulky groups go equatorial, and the numbers say about 21-to-1 in favour.

Recall Quick self-test

Formula for flat-ring strain per bond ::: $d = \frac{109.5° - ( n - 2 ) 180°/ n}{2}$ Why was Baeyer wrong for $n \geq 6$ ::: rings pucker into 3D, restoring $\approx 109.5°$ and escaping strain. Two reasons the chair beats the boat ::: staggered bonds (no torsional strain) and no flagpole clash (no steric strain). What a ring flip does to axial/equatorial labels ::: swaps them completely. $K$ for equatorial:axial methylcyclohexane ::: about $21 : 1$ (~95% equatorial).

Related: Heat of Combustion as a Stability Measure, Geometrical Isomerism cis-trans.