Loav · VCE Chemistry · Unit 4 / Area of Study 1
U4A1-01 · Current 2026 curriculum · Original teaching guide · Study guide
1. Why carbon makes so many compounds
A neutral carbon atom has four valence electrons and usually forms four covalent bonds in the molecules studied here. Count bond order: a single bond contributes one, a double bond two. Thus the carbon in CH₄ has four single bonds, whereas each carbon in CH₂=CH₂ has two C–H single bonds and one C=C double bond. Both satisfy a bond-order total of four.
Carbon forms relatively strong, stable covalent bonds with itself and with elements such as hydrogen, oxygen and nitrogen. Repeated C–C bonding allows long chains, branched chains and rings. Carbon can also form multiple bonds. Combining these possibilities with different attached functional groups produces enormous chemical diversity. Stable does not mean unreactive under every condition: reactions still occur when a suitable pathway and conditions are available. A C=C bond is stronger overall than a C–C bond, but addition can replace its extra bonding component with new bonds; saying “a double bond is simply weaker” is incorrect.
In this course, discussion of degree of unsaturation focuses on carbon–carbon double bonds. Count the relevant C=C bonds in a displayed structure, especially fatty-acid chains. Do not substitute a generic index-of-hydrogen-deficiency formula that also counts rings and C=O bonds. Cyclohexane has a ring but no C=C bonds; a carbonyl group contains C=O, not C=C. Benzene needs special treatment: its ring has delocalised bonding and does not behave like three independent alkene double bonds.
2. Four ways to describe one molecule
A molecular formula gives totals only: C₃H₈O does not locate the oxygen. A structural formula shows which atoms are connected and the bond orders. A condensed or semi-structural formula groups atoms, as CH₃CH₂CH₂OH. Parentheses show branches attached at that point: CH₃CH(OH)CH₃ has OH on the middle carbon. In a skeletal structure, each unlabelled line end and corner represents carbon. Hydrogens attached to carbon are omitted and must be inferred to complete carbon valence. Heteroatoms such as O, N and Cl are written explicitly; show the H in OH or NH₂.
3. Recognise the family from connectivity
R below means a carbon-containing group. It is a placeholder, not the symbol for an element. A functional group is a characteristic arrangement of atoms associated with a family’s reactions. Always inspect the neighbouring atoms: simply spotting O or N is insufficient.
| Family | Diagnostic arrangement and example |
|---|---|
| Alkane / cycloalkane | Only C–C and C–H single bonds; CH₃CH₂CH₃ / cyclohexane. |
| Alkene | C=C; CH₂=CHCH₃. |
| Benzene | Six-carbon aromatic ring, C₆H₆; delocalised bonding. |
| Haloalkane | Halogen attached to saturated carbon; CH₃CH₂Cl. |
| Primary amine | R–NH₂; CH₃CH₂NH₂. Nitrogen has one carbon-containing substituent. |
| Primary amide | R–C(=O)–NH₂; CH₃CONH₂. Nitrogen is directly bonded to a carbonyl carbon. HCONH₂ is the smallest primary amide. |
| Alcohol | C–OH on a saturated carbon; CH₃CH₂OH. |
| Aldehyde | –C(=O)H; CH₃CHO. |
| Ketone | R–C(=O)–R′; CH₃COCH₃. |
| Carboxylic acid | –C(=O)OH; CH₃COOH. |
| Ester | –C(=O)–O–C; CH₃COOCH₂CH₃. The carbonyl carbon can also bond to H, as in methanoate esters. |
An alcohol is primary, secondary or tertiary according to the number of other carbon atoms bonded to the carbon bearing OH: one, two or three. CH₃CH₂OH is primary; CH₃CH(OH)CH₃ is secondary; (CH₃)₃COH is tertiary. Count carbon neighbours of the OH-bearing carbon, not OH groups or total carbons. Methanol is conventionally grouped with primary alcohols despite having no carbon neighbour.
4. Structural isomers: same inventory, different connections
Structural isomers have the same molecular formula but different atom connectivity. Redrawing a bent chain straight, rotating a page or reading the same chain backwards does not make a new isomer. A molecular formula alone therefore cannot determine identity or all physical properties. Chain branching, functional-group position and functional-group identity can each change connectivity.
Worked example 1: count before classifying
For CH₃CH(OH)CH₂CH₃, count four carbons, ten hydrogens including the OH hydrogen, and one oxygen: C₄H₁₀O. The OH-bearing carbon has two carbon neighbours, so this is a secondary alcohol. Its skeletal drawing must contain four carbon vertices/endpoints in a connected chain, with OH attached to the second carbon. The chain contains no C=C bonds.
Worked example 2: test an isomer claim
CH₃CH₂CH₂OH and CH₃CH(OH)CH₃ both total C₃H₈O. In the first, O is bonded to an end carbon; in the second it is bonded to the middle carbon. They are structural isomers. CH₃CH₂CHO instead totals C₃H₆O and cannot be a structural isomer of either, because the hydrogen totals differ.
Check your understanding
- Find the molecular formula and family of CH₃CH₂CONH₂.
- Explain why cyclohexane and hex-1-ene can share C₆H₁₂ yet have different C=C counts.
- Classify the alcohol (CH₃)₂CHCH₂OH.
- Are CH₃CH₂CH₂CH₃ and CH₃CH(CH₃)CH₃ structural isomers? Justify.
Answers with reasoning
- C₃H₇NO: the hydrogen count is 3 + 2 + 2. It is a primary amide because NH₂ is bonded directly to the carbonyl carbon, not a primary amine.
- Cyclohexane closes a chain into a ring with six single C–C bonds and has zero C=C bonds. Hex-1-ene has one C=C bond in an open chain. Their different connectivities give the same atom totals.
- Primary. The CH₂ bearing OH has only one carbon neighbour. Branching elsewhere does not change the classification.
- Yes. Both are C₄H₁₀, but one has an unbranched chain and the other a carbon bonded to three other carbons. A rotation cannot change those connections.
Go deeper: translate, then test the structure
A drawing is useful only if you can recover its atom inventory and connectivity. Start at a labelled heteroatom or a functional group, trace each carbon–carbon connection once, and count branch endpoints separately. For each carbon, subtract its existing bond-order total from four to find the attached hydrogens. A carbonyl carbon with C=O and two single bonds has no spare valence for H. By contrast, an aldehyde carbonyl carbon has one single bond to H and one to its remaining substituent.
The following original representation panel uses butan-2-one. Its molecular formula is C₄H₈O. The expanded connectivity formula CH₃–C(=O)–CH₂–CH₃ tells you the oxygen is double-bonded to the second carbon. In its skeletal form the same carbon is an unlabelled corner with a double line to O. Neither the bend in the drawing nor the direction in which it faces changes the compound. The bond lengths in a teaching diagram are not measurements.
Use a similar translation for every family. CH₃C(=O)OH and CH₃C(=O)NH₂ share a carbonyl but differ in the atom attached beside it: O in the acid, N in the amide. H–C(=O)–O–CH₂CH₃ is an ester even though the carbonyl carbon is bonded to H. The entire O–C(=O) arrangement decides the family; the presence of a carbonyl carbon bonded to H does not make every such molecule an aldehyde.
Worked example 3 · Apply: isomer search without duplicates
Find all unbranched saturated alcohol structures with four carbons and one OH group, then explain why moving OH to carbon 3 does not add a new one. Begin with CH₃CH₂CH₂CH₃. Replacing an end H with OH gives CH₃CH₂CH₂CH₂OH. Replacing an inner H gives CH₃CH(OH)CH₂CH₃. Both have C₄H₁₀O. Labelling OH at carbon 3 gives CH₃CH₂CH(OH)CH₃, which becomes the second structure when read from the opposite end. Similarly carbon 4 duplicates carbon 1. There are therefore two structures under the stated unbranched restriction. Allowing branching would add other alcohol structures, so you must state the search boundary.
The first alcohol is primary and the second secondary. Their different connectivity can change reactions as well as physical properties: ordinary oxidation can take the primary alcohol to a carboxylic acid but takes the secondary alcohol to a ketone. This is why a molecular formula alone is too weak to specify a synthesis target.
Application practice: read before you name
- For CH₃CH₂C(=O)OCH₃, give the molecular formula and explain why it is neither a ketone nor a carboxylic acid.
- A skeletal drawing has a four-carbon chain and an OH label attached to the second corner from an end. A student counts that O as a fifth carbon. Explain the error and give the correct molecular formula.
- Compare CH₂=CHCH₂COOH and CH₃CH₂CH₂COOH: count C=C bonds and oxygen atoms, then explain whether they are structural isomers.
Application answers and checks
- C₄H₈O₂: 3 + 2 + 0 + 3 = 8 H. The carbonyl carbon is directly bonded to an oxygen that is bonded onward to a carbon group, establishing an ester. A ketone requires carbon groups on both sides of C=O; an acid requires the carboxyl OH.
- The explicitly labelled O is an oxygen atom attached to a chain carbon. Only unlabelled line ends and corners count as carbon atoms; the attached OH does not add another carbon. Four chain carbon atoms remain and the formula is C₄H₁₀O. Clarify which carbon bears OH by numbering the full chain, not by counting only the corners and forgetting endpoints.
- The first has one C=C and two O atoms; the second has zero C=C and two O atoms. Their formulas are C₄H₆O₂ and C₄H₈O₂. They are not structural isomers because their hydrogen totals differ. The C=O in each acid is not counted as a C=C.
