8.1 · Conjugation & colour
Line-bond structures
The chemist's shorthand for organic molecules.
By the end you should be able to:
- Read and interpret line-bond (skeletal) structures
Key idea
Reading a skeletal structure
- Each line is a bond; a double line is a double bond and a triple line a triple bond.
- Every vertex (corner) and every line end without a label is a carbon atom.
- Hydrogens on carbon are not drawn: each carbon carries enough H atoms to make four bonds in total.
- Heteroatoms (O, N, S, halogens) are always written, together with any H attached to them (OH, NH₂).
- Chains are drawn as zig-zags to reflect the real bond angles (about 109.5° or 120°); lone pairs are usually left out.
So a zig-zag of three lines is butane, : two line ends and two vertices make 4 carbons.
Formula
Counting hydrogens on each carbon
A double bond counts as 2 and a triple bond as 3.
| Carbon | Bonds drawn | Hydrogens |
|---|---|---|
| line end, single bond | 1 | 3 (a CH₃ group) |
| chain vertex, two single bonds | 2 | 2 (CH₂) |
| vertex with one double and one single bond | 3 | 1 (CH) |
| branch point with three single bonds | 3 | 1 |
| four bonds drawn | 4 | 0 |
Check with the unsaturation count: rings + π bonds , where counts halogens. Each ring or π bond removes 2 H from the saturated formula .
Method
Writing the molecular formula
- Count the carbons: every vertex and every unlabelled line end.
- For each carbon, H = 4 − (bonds drawn), counting double bonds as 2; add these up.
- Add the H atoms drawn explicitly on heteroatoms (OH, NH₂ …).
- Count each heteroatom.
- Write C first, then H, then the other elements alphabetically, e.g. .
- Check with the unsaturation count (rings + π bonds).
Common mistake
Skeletal-structure traps
- Wrong: adding implicit H to O or N. Right: only carbon gets implicit hydrogens. H on heteroatoms is always drawn, so an unlabelled –O– in a chain (an ether) has no H.
- Wrong: forgetting that line ends are carbons. Right: a line that ends with no label is a CH₃ group.
- Wrong: treating a C=C as one bond when counting H. Right: a double bond uses two of carbon's four bonds.
- Wrong: counting a labelled end as an extra carbon. Right: where a line ends at "OH", that end is the oxygen, not a carbon.
Worked example
Worked example: aspirin
Aspirin is drawn as a benzene ring (a hexagon with three alternating double bonds) carrying two neighbouring groups: –COOH and –O–C(=O)–CH₃.
- Ring: 6 C. The 4 unsubstituted ring carbons each have 3 bonds drawn (one double, one single), so 1 H each: 4 H. The 2 substituted ring carbons have 4 bonds: 0 H.
- –COOH: 1 C with 4 bonds drawn (C=O, C–OH, C–ring), so 0 H on carbon; 2 O; 1 H drawn on the O.
- –O–C(=O)–CH₃: 2 C (the carbonyl C has 0 H; the line-end CH₃ has 3 H); 2 O.
Totals: C = 6 + 1 + 2 = 9; H = 4 + 1 + 3 = 8; O = 2 + 2 = 4. The formula is .
Check: rings + π bonds = (18 + 2 − 8)/2 = 6 = 1 ring + 3 ring C=C + 2 C=O.
Worked example
Worked example: 1,3-butadiene and cyclohexanol
1,3-Butadiene (a zig-zag of three lines; the first and last are double): 4 C. Each end carbon has a double bond (2 bonds drawn), so 2 H; each inner carbon has a double and a single bond (3 drawn), so 1 H. H = 2 + 1 + 1 + 2 = 6, giving .
Cyclohexanol (a hexagon with OH on one vertex): 6 C. The carbon carrying OH has 3 bonds drawn, so 1 H; the other five ring carbons are CH₂, giving 10 H; plus the H drawn on O. H = 1 + 10 + 1 = 12, giving .
Check yourself
Fresh questions every time you visit. Answers count toward your progress.