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9476 / 2027
Organic Chemistry

Lesson 1 of 34

A formula should tell you exactly which atoms are connected

Move between molecular, displayed, condensed, skeletal and three-dimensional representations.

A-Level 9476 (2026-2027)

  • I can interpret, name and represent the specified organic families.Syllabus

    Syllabus 9476, 11.1(a). Interpret, name and represent the specified organic families.

A formula should tell you exactly which atoms are connected

Move between molecular, displayed, condensed, skeletal and three-dimensional representations.

A molecular formula tells you how many of each atom there are. It does not tell you which organic compound you have. A structural formula must show clearly which atoms are joined. For example, CH3CH(OH)COOH shows a chain of three carbons, an OH group on the middle carbon and a carboxylic acid group at the end. Its molecular formula is C3H6O3, and its empirical formula is CH2O.

Read each representation without changing the molecule
RepresentationWhat is shown
DisplayedEvery atom and bond, including all C-H bonds. A double bond is two shared pairs, not two different neighbouring atoms.
Condensed structuralGroups such as CH3, CH2 and CH(OH) preserve connectivity; parentheses identify branches.
SkeletalEach unlabelled line end or corner is carbon. Add enough implied H atoms for each carbon to have valency four. Heteroatoms and hydrogens bonded to them are shown.
StereochemicalA solid wedge points towards the viewer; a hashed wedge points away; ordinary lines lie in the page. These encode spatial arrangement.

Read an oxygen-containing skeletal structure

The left line end is CH3, the central junction is CH bearing OH, and the right junction is the carboxyl carbon with a double bond to O and a single bond to OH. This is CH3CH(OH)COOH.

The three carbon positions are implicit; every oxygen is explicit. Count bonds before adding hydrogens. Aromatic rings may be drawn as a hexagon with a circle rather than displaying all six carbon and hydrogen labels.

To name a compound, choose the parent chain that contains the main functional group. Number the chain so that this group gets the lowest number it can. Then give the positions of double bonds and other groups. Examples are butan-2-ol, 2-bromopropane and propan-2-one. Do not pick the longest line on the page if it leaves out the functional group that gives the name.

Worked exampleTurn an ester name into a displayed formula

Draw methyl propanoate, then give its condensed and molecular formulae.

Try it first. Then show the worked stepsHide the worked steps
  1. Propanoate identifies the three-carbon acid-derived part. Include the carbonyl carbon in that count: C-C-C(=O)-O-.
  2. Methyl identifies the one-carbon group attached to the single-bonded oxygen. Join it as C-C-C(=O)-O-C, not as a branch on the carbonyl carbon.
  3. Complete carbon valency four by adding three H to the first carbon, two to the second and three to the last. The carbonyl carbon already has four bonds; neither oxygen has an attached H in this ester.
  4. Show every C-H bond for the displayed formula below. Check that each carbon has four bonds and each oxygen two, counting a double bond twice.
Answer

CH3CH2C(=O)OCH3; molecular formula C4H8O2. A molecular formula alone would not distinguish this ester from its constitutional isomers.

Displayed methyl propanoate: every atom and bond

The chain is H3C-CH2-C(=O)-O-CH3. Each of the eight hydrogen atoms has an explicitly drawn bond to carbon. The carbonyl carbon has no hydrogen, and the bridging oxygen bonds to two different carbons.

The acid-derived three-carbon part is on the left; the alcohol-derived methyl group is on the right. Removing C and attached-H labels while retaining every heteroatom converts this into a skeletal drawing.

Check your understanding

In a skeletal formula, how do you know where the carbon atoms and their hydrogens are?

  1. Each corner is a carbon carrying exactly two H atoms, as in a CH2 group.
  2. Each line end or corner is a carbon; add H atoms until each carbon has four bonds.
  3. Labelled atoms are the carbons; each unlabelled corner is a bond angle with no atom.
Show the answer

Each line end or corner is a carbon; add H atoms until each carbon has four bonds.

Skeletal formulae leave out C and the H atoms on carbon. Count the bonds shown at each corner and fill up to four with implied H. Atoms other than carbon, and H on them, are drawn.