Structural Representations

1. Complete (Lewis) structural formula. Every atom is shown, every bond — including every C–H bond — is drawn explicitly as a line, and lone pairs on heteroatoms (O, N, halogens) are shown as dots. This is the most information-dense form and the slowest to draw, but it is the only form that leaves nothing to infer: useful for small molecules or when a mechanism needs every electron pair visible.

2. Condensed structural formula. Bonds to hydrogen are omitted; atoms are written next to whichever atom they are attached to, and branches go in parentheses. Butane condenses to $\text{CH}_3{-}\text{CH}_2{-}\text{CH}_2{-}\text{CH}_3$, or even $\text{CH}_3(\text{CH}_2)_2\text{CH}_3$; isobutane is $\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_3$. It keeps the chain readable in one line without drawing every bond.

3. Bond-line (skeletal) formula. The fastest notation, and the one you will use most once molecules get larger. Carbon atoms are never written — every line-end and every vertex where lines meet is a carbon — and hydrogens on carbon are never written either. Heteroatoms and functional groups (–OH, –Cl, –NH₂, C=O, and so on) are always written explicitly, since they cannot be inferred. The number of implicit hydrogens at any carbon vertex follows directly from tetravalence:

where: $H_{\text{implicit}}$ is the number of hydrogen atoms silently attached at that vertex, and "bonds already drawn" counts every line touching the vertex, with a double or triple bond line counted according to how many $\sigma$-connections it represents at that carbon (a C=C carbon has three $\sigma$-connections to draw around, so it takes only one more line — to H or to another atom — to reach $4$ total bonding positions). A chain-end vertex with a single line touching it therefore carries $4 - 1 = 3$ hydrogens; a mid-chain vertex with two lines carries $4 - 2 = 2$; a branch point with three lines carries $4 - 3 = 1$. A related, separate idea is the three-dimensional (wedge-and-dash) representation, used once the actual spatial arrangement — not just connectivity — matters. A bold wedge shows a bond coming out of the plane of the paper toward the viewer; a dashed or hashed wedge shows a bond going behind the plane; plain lines stay in the plane. At Class 11 level this is used only to depict the tetrahedral ($sp^3$), trigonal planar ($sp^2$), or linear ($sp$) shapes already fixed by hybridisation — full stereoisomer analysis (R/S configuration, chirality) is a later topic.

The bond-line rule is not arbitrary shorthand — it is a direct consequence of carbon's fixed valency of four. Once you commit to always drawing carbon with four bonding connections, an unlabelled vertex carries no ambiguity: whatever bonds are visibly drawn to it, the rest must be C–H bonds, because there is no other atom carbon could be silently bonded to. This is why heteroatoms must always be written explicitly — a silent O or N would break the one rule (carbon's tetravalence) that makes the whole shorthand decodable. The same logic extends the other way: at a double bond, a carbon already has three $\sigma$-connections (two single bonds plus the one to its double-bond partner), so only one more bonding position remains, and hydrogen count drops accordingly. Condensed formulas work by a looser version of the same idea — you trust the reader to know that "CH₃" means a carbon already saturated with three hydrogens — while the complete Lewis formula makes no such demand, at the cost of taking far longer to draw. Choosing a representation is therefore a trade-off between speed and self-evidence: complete formulas for teaching or mechanism-level detail, condensed formulas for compact chain notation, bond-line formulas for everyday use once a student is fluent in tetravalence. Wedge-and-dash notation adds a further layer on top of any of these: it does not change which atoms are bonded to which (connectivity), only where in space those bonds point, which is why it is drawn only when the actual 3-D shape — tetrahedral, trigonal planar, or linear — needs to be shown rather than assumed.

Convert the condensed formula $\text{CH}_3{-}\text{CH}(\text{OH}){-}\text{CH}_3$ (propan-2-ol) into (a) a complete Lewis formula and (b) a bond-line formula, then verify the bond-line formula by counting hydrogens with the implicit-H rule.

Step 1. Identify the skeleton. Three carbons in a row, C1–C2–C3, with an –OH group on the middle carbon, C2.

Step 2. Build the complete Lewis formula. C1 bonds to three H atoms and to C2. C2 bonds to C1, C3, one H, and the O of –OH. C3 bonds to three H atoms and to C2. The oxygen bonds to C2, one H, and carries two lone pairs. Every one of these bonds is drawn as an explicit line in the complete formula, giving $\text{H}_3\text{C}{-}\text{CH}(\text{OH}){-}\text{CH}_3$ with all fourteen atoms shown.

Step 3. Draw the bond-line formula. A two-segment zig-zag represents the three-carbon chain (C1–C2–C3), with "OH" written off the middle vertex — no carbon or hydrogen symbols appear anywhere.

Step 4. Verify with the implicit-H rule. C1 is a chain end with one line drawn to it, so $H = 4 - 1 = 3$ hydrogens. C2 has two chain bonds plus the line to OH, three connections drawn, so $H = 4 - 3 = 1$ hydrogen. C3, a chain end, again has $H = 4 - 1 = 3$ hydrogens. Total hydrogens on carbon: $3 + 1 + 3 = 7$, plus the one H on oxygen gives $8$ hydrogens overall.

Step 5. Cross-check against the molecular formula. Propan-2-ol is $\text{C}_3\text{H}_8\text{O}$ — three carbons, eight hydrogens, one oxygen. The bond-line count of $8$ hydrogens matches exactly, confirming all three representations describe the same molecule.

Competitive stretch

For a six-membered ring drawn in bond-line notation with one vertex labelled N and all ring bonds single, apply the same rule vertex by vertex: each of the five carbon vertices has two ring bonds drawn, so $H = 4 - 2 = 2$ hydrogens each, giving $10$ H from carbon; the nitrogen vertex has two ring bonds and nitrogen's valency is $3$, so it carries $1$ H. Totalling gives $\text{C}_5\text{H}_{11}\text{N}$ — piperidine, a saturated nitrogen-containing ring you will classify as a heterocycle in the next topic.

• Forgetting that in a bond-line formula, every unlabelled vertex or line-end is a carbon, and its hydrogen count must be computed from $H = 4 - (\text{bonds drawn})$, never assumed to be zero.
• Miscounting hydrogens at a double or triple bond — a carbon in a C=C already has three $\sigma$-connections, so it takes only one more line to reach four, not two.
• Writing a condensed formula without showing bond order, so that $\text{CH}_2\text{CH}_2$ is ambiguous; if the bond is a double bond it must be written $\text{CH}_2{=}\text{CH}_2$.
• Treating wedge-and-dash as advanced or optional at this level — here it is simply a tool to depict the already-known tetrahedral, trigonal, or linear shape of a carbon centre, not a stereochemistry problem to solve.

Ask the tutor: unsure how many hydrogens to add at a branch point or ring vertex in a bond-line structure — ask your AI tutor to walk through the implicit-H count for any molecule you're stuck on.