(a) Introduction
- Syllabus
- 2024
- Topic
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- Level
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A hydrocarbon is a compound that contains hydrogen and carbon only.
| Substance | Hydrocarbon? | Reason |
|---|---|---|
| methane, CHX4 | yes | it is a compound containing only carbon and hydrogen |
| ethene, CX2HX4 | yes | it is a compound containing only carbon and hydrogen |
| ethanol, CX2HX5OH | no | it also contains oxygen |
| carbon dioxide, COX2 | no | it contains no hydrogen and also contains oxygen |
Check every element symbol in the formula. Both C and H must be present, and any additional element means the substance is not a hydrocarbon.
A compound is not a hydrocarbon merely because it contains carbon and hydrogen. The word ‘only’ excludes alcohols, carboxylic acids and every other compound containing an additional element.
Different formula types answer different questions about an organic substance: atom ratio, actual atom count, homologous-series pattern, connectivity, or every bond.
| Formula type | What it shows | Example for ethene |
|---|---|---|
| empirical | simplest whole-number ratio of atoms | CHX2 |
| molecular | actual number of each atom in one molecule | CX2HX4 |
| general | algebraic pattern shared by a homologous series | CXnHX2n for alkenes |
| structural | how atoms or groups are connected, with bonds condensed | CHX2=CHX2 |
| displayed | every atom and every covalent bond | draw both C atoms, four H atoms and all six bonds |
To obtain an empirical formula, divide every molecular subscript by their highest common factor. To obtain a molecular formula from a displayed or structural formula, count each atom; a general formula is not the formula of one particular molecule until a value of n is chosen.
A molecular formula does not show connectivity, so different structures can share it. A structural formula may group bonds such as CHX3CHX2OH; a displayed formula must show the O–H bond and every other bond explicitly.
Homologous series, functional group and isomerism describe three different relationships: membership of a family, the reactive part of a molecule, and alternative structures for one molecular formula.
| Term | Meaning | Diagnostic question |
|---|---|---|
| homologous series | a family with the same functional group and general formula, similar chemical reactions, a trend in physical properties, and successive members differing by CHX2 | do the compounds follow one family pattern? |
| functional group | the atom or group of atoms responsible for a compound’s characteristic chemical reactions | which part controls the typical reactions? |
| isomerism | the existence of compounds with the same molecular formula but different structural formulae | is the atom count unchanged but the connectivity different? |
\ce{CH3CH2CH2CH3} \qquad \ce{CH3CH(CH3)CH3}
The two formulae above both have molecular formula CX4HX10 but different carbon connectivity, so they are structural isomers.
Isomers must have the same molecular formula, not merely the same empirical or general formula. Members of a homologous series have similar chemical properties and a trend in physical properties; they do not all have identical physical properties.
An IUPAC name combines a carbon-chain stem, any substituent prefixes and a suffix that identifies the principal bond or functional group. Number the chain so the important feature receives the lowest possible locant.
| Carbon atoms in the longest relevant chain | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| stem | meth- | eth- | prop- | but- | pent- | hex- |
| Feature | Naming pattern | Example |
|---|---|---|
| alkane | stem + -ane | CHX3CHX2CHX3: propane |
| alkene | lowest C=C locant + -ene | CHX2=CHCHX2CHX3: but-1-ene |
| alcohol | lowest –OH locant + -ol | CHX3CHX2CHX2OH: propan-1-ol |
| carboxylic acid | stem + -anoic acid; acid carbon is C1 | CHX3CHX2COOH: propanoic acid |
| halogen or branch | numbered prefix before the parent name | CHX3CHClCHX3: 2-chloropropane |
Choose the longest chain containing the principal functional group or C=C bond, number it from the nearer important feature, identify substituents, then assemble locants, prefixes, stem and suffix. For an ester, name the alkyl group attached to oxygen first and the acid-derived alkanoate second, for example methyl ethanoate.
Count the parent chain, not every carbon in a branch. Do not omit a locant when more than one position is possible, and keep this specification’s naming work to compounds containing no more than six carbon atoms.
A molecular formula fixes the number of each atom but not their connectivity. Generate possible structures systematically while satisfying normal valencies: carbon forms four bonds, oxygen two, hydrogen and halogens one.
| Step | Check |
|---|---|
| 1 | write the exact inventory of C, H and any other atoms |
| 2 | draw distinct carbon skeletons, beginning with the longest chain and then branching |
| 3 | place any multiple bond or functional group in each genuinely different position |
| 4 | add H atoms so every atom has its normal valency |
| 5 | recount atoms and remove drawings that differ only because they are rotated or reversed |
| 6 | convert each valid structural formula to a displayed formula by showing every atom and bond |
\ce{CH3CH2CH2CH3} \qquad \ce{CH3CH(CH3)CH3}
For CX4HX10, the only carbon skeletons are a four-carbon chain and a three-carbon chain with one methyl branch. These give butane and methylpropane; redrawing either from the opposite end does not create another isomer.
A valid drawing must match the molecular formula exactly and obey every valency. Different orientations of the same connectivity are not different structures; different atom connectivity is required.
Classify an organic reaction from its overall change: replacement is substitution, joining across an unsaturated bond is addition, and reaction with oxygen that releases energy is combustion.
| Type | Overall pattern | Example clue |
|---|---|---|
| substitution | one atom or group in an organic molecule is replaced by another | alkane + halogen under ultraviolet light gives a haloalkane and hydrogen halide |
| addition | atoms add across a C=C bond and two reactants form one organic product | alkene + bromine gives a dibromoalkane |
| combustion | an organic compound burns in oxygen | oxygen is a reactant and energy is released |
\ce{CH4 + Cl2 ->[UV] CH3Cl + HCl}
\ce{CH2=CH2 + Br2 -> CH2BrCH2Br}
\ce{CH4 + 2O2 -> CO2 + 2H2O}
Use reactants and products to identify the overall pattern. Addition does not replace an atom and normally gives one organic product; substitution retains the carbon framework while replacing a group. Reaction mechanisms are not required.