UNIT 2 CHEMISTRY · PEARSON AAQ · E1.2
Structural Representations of Organic Compounds
FREE CHEMISTRY REVISION · TOPIC 51
Structural Representations of Organic Compounds
Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · E1.2
What you will learn
Read and convert fully displayed, condensed, skeletal and three-dimensional formulae. Each representation selects information about the same molecule. A molecular formula counts atoms, whereas a structural representation also records which atoms are connected. Preserve both atom counts and connectivity when converting.
1. What each representation tells you
| Representation | What is drawn | What must be inferred |
|---|---|---|
| Fully displayed | Every atom symbol and every covalent bond | Three-dimensional shape unless separately indicated |
| Condensed | Groups such as CH₃, CH₂, CH and OH in connectivity order | Individual C–H bonds; brackets mark attached or repeated groups |
| Skeletal | Carbon framework as lines; heteroatoms and their attached H where needed | Unlabelled carbons and hydrogens bonded to carbon |
| Wedge/dashed-line | Bond orientation relative to the page | Perspective from a fixed viewpoint |
A molecular formula such as C₄H₁₀ is not a unique structural formula: different connections can share the same atom counts. A fully displayed formula must not replace a CH₃ group with a single CH₃ label if the question specifically asks to show every bond. Condensed formulae may show C–C single bonds or omit their dashes, but a double bond must remain clear.
2. Valency is your conversion check
For the neutral molecules used here, carbon has total bond order four, hydrogen and each halogen one, and oxygen two. A single bond contributes one, a double bond two. For each carbon, subtract the bond orders to other atoms from four; the remainder is its attached H count. An O in an alcohol has one bond to carbon and one to hydrogen. Lone pairs are not extra bonds and are not usually required in these structural formulae.
Worked conversion 1 — condensed to fully displayed
Start with CH₃CH₂CH₃. Draw a chain of three C atoms linked by single bonds. Add three separate H atoms to each end carbon and two to the middle carbon, each with its own C–H bond, as in the comparison diagram. Check C bond totals: 1+3=4 at either end; 2+2=4 in the middle. Count H: 3+2+3=8. Result: fully displayed propane, C₃H₈.
Worked conversion 2 — fully displayed to condensed
Read the same displayed diagram from the opposite end. Each terminal carbon is CH₃ and the middle one is CH₂, giving CH₃CH₂CH₃ again. Reversing the drawing has not changed connectivity. Do not write CH₃CH₃CH₂: an internal carbon with two C–C bonds cannot also carry three H atoms.
3. Decode skeletal formulae
- Count every unlabelled endpoint and vertex as one carbon. A crossing is a connection only when the drawing actually indicates a bonded junction; draw junctions clearly.
- Each line is a bond; two parallel lines indicate a double bond. An unlabelled line end is a carbon even when no C is printed.
- Supply enough H on each carbon to bring its total bond order to four. Do not automatically add two H to every vertex.
- Write heteroatoms, for example O, Cl and Br, explicitly. A line ending at Cl ends at chlorine; it does not hide a carbon behind the Cl label.
- In an alcohol show the O–H hydrogen explicitly, for example as OH or a drawn O–H bond. Preserve where the group attaches.
Worked conversion 3 — skeletal chain to condensed
The first skeletal example has two ends and two intermediate corners: four carbons. End carbons have one C–C bond each, hence CH₃; internal carbons each have two, hence CH₂. Condensed result: CH₃CH₂CH₂CH₃; molecular formula C₄H₁₀. A zigzag can be an unbranched chain.
Worked conversion 4 — a branch must stay attached
The branched skeleton has a central carbon joined to three terminal carbons. Each terminal carbon is CH₃; the junction has three C–C bonds so only one H. Condensed result: CH₃CH(CH₃)CH₃, also written (CH₃)₃CH. The bracketed CH₃ is attached to the CH immediately before it. Total C=4 and H=3+1+3+3=10. Converting this into an unbranched four-carbon chain would change the compound.
Worked conversion 5 — a written oxygen is not a carbon
The alcohol skeleton has three carbon positions followed by labelled O–H. Along the carbon chain the H counts are 3, 2 and 2. The O–H supplies one more H. Condensed result: CH₃CH₂CH₂OH; molecular formula C₃H₈O. The last carbon has one C–C and one C–O bond, leaving two C–H bonds.
4. Preserve multiple bonds and substituents
Worked conversion 6 — condensed alkene to skeletal
For CH₂=CHCH₃, draw a three-carbon open chain with two parallel bond lines between carbon positions 1 and 2. The double-bond end carbon has bond order two to C and therefore two H. The middle carbon has bond orders 2+1 to C and therefore one H. The last carbon is CH₃. Atom check: C₃H₆. A second line along a bond changes bond order; it is not a wedge.
Worked conversion 7 — halogen position survives conversion
For CH₃CH(Cl)CH₃, draw a three-carbon skeleton and attach a clearly labelled Cl to the middle vertex with a single bond. That carbon has two C–C bonds plus one C–Cl bond, leaving one H. The end carbons are CH₃. Formula: C₃H₇Cl. Attaching Cl to an end carbon instead would give different connectivity. The Cl is a single chlorine atom, not C followed by a separate symbol l.
When checking two representations, follow connections atom by atom. Compare carbon-chain length, branch attachment, multiple-bond positions and heteroatom positions before checking the molecular formula. Matching formulae alone do not prove they represent the same compound.
5. Wedges and dashed bonds show direction
A plain line lies approximately in the plane of the page. A solid wedge projects towards the observer, with the narrow end at the atom in the plane. A hashed wedge (often described as a dashed bond) projects away behind the page. Use a clear convention consistently; a dashed perspective bond is not a second bond line.
Worked conversion 8 — wedge diagram to atom connectivity
In the methane diagram, count four separate C–H single bonds: two plain, one solid wedge and one hashed wedge. Carbon is bonded to four H atoms, so the condensed/molecular formula is CH₄. A fully displayed connectivity diagram can show the same four single C–H bonds in a cross, but does not then communicate the tetrahedral directions. It does not mean methane is square planar.
Conversely, converting an ordinary displayed methane diagram to a wedge diagram adds an explicit tetrahedral perspective. A two-dimensional connectivity formula does not uniquely prescribe a viewing angle. In molecules with a specified spatial arrangement, preserve that arrangement rather than inventing it.
Practice with answers
| Task | Answer and check |
|---|---|
| Convert CH₃CH₂OH to a displayed formula | Two C joined once; first C has three separate H, second C two; C–O–H explicit. C₂H₆O. |
| Read a four-sided skeletal ring with only single bonds | Four carbons, each joined to two C and bearing two H: C₄H₈, not C₄H₁₀. |
| Convert CH₃CH=CHCH₃ to skeletal | Four C positions; double bond between the two middle positions. H counts 3,1,1,3 → C₄H₈. |
| Does a solid wedge mean a double bond? | No. It gives the orientation of one bond; two parallel lines indicate a double bond. |
Common misconceptions
- A corner is hydrogen: unlabelled corners and ends represent carbon.
- An OH endpoint also hides carbon: the written O is the atom at that position.
- All vertices are CH₂: bond orders, branches and heteroatoms determine H count.
- A bracketed group can move anywhere: parentheses preserve an attachment position.
- A wedge and a double bond are interchangeable: direction and bond order are different information.
- A flat displayed diagram proves a flat molecule: connectivity drawings do not necessarily show geometry.
Examination tips
- Label every H and draw its bond when asked for a fully displayed formula.
- Count carbon positions before adding implicit H to a skeletal diagram.
- Check each atom’s bond total, then check the complete molecular formula.
- Retain the positions of double bonds and substituents during every conversion.
- Explain what is lost when a 3D drawing is reduced to a connectivity-only formula.
Check your understanding
Choose one answer, select Check Answer, then read the explanation. Try Again resets the question. Use Tab between controls, arrow keys within a radio group, and Enter or Space on buttons. Answers are not saved.