Loav · VCE Chemistry · Unit 4 / Area of Study 1
U4A1-02 · Current 2026 curriculum · Original teaching guide · Study guide
1. Stay inside the course’s naming boundary
VCAA specifies systematic names up to eight carbons, with no more than two functional groups, for non-cyclic hydrocarbons, haloalkanes, primary amines, alcohols, aldehydes, ketones, carboxylic acids and non-branched esters. Recognising a benzene ring, cyclohexane or a primary amide in a structure is also required, but that broader recognition list is not permission to assume every possible cyclic or amide naming problem is prescribed. Use the current VCAA data book’s nomenclature information as a reference, while practising enough that you can translate names confidently.
The carbon-number stems are meth-, eth-, prop-, but-, pent-, hex-, hept- and oct-. An ending identifies the principal family: -ane, -ene, -ol, -amine, -al, -one or -oic acid. Halo substituents use fluoro-, chloro-, bromo- and iodo-. An ester has a two-word name: the alkyl group attached to the single-bonded oxygen, followed by the acid-derived alkanoate name.
2. A reliable naming routine
- Locate all functional groups and choose the principal suffix group. For the combinations here, carboxylic acid takes precedence over aldehyde, ketone, alcohol and amine; an alcohol takes precedence over an amine. A lower-priority OH group becomes hydroxy-, and an amino group becomes amino-.
- Choose an appropriate longest parent chain containing the principal group and required multiple bond. Include the carbon of CHO or COOH in the chain count. An ester is treated as its two component portions.
- Number so the principal suffix group has the lowest locant, then consider multiple bonds and substituents. Do not number merely to make a methyl branch low if this makes the suffix group high.
- Place substituent positions and names before the parent. Use commas between numbers, hyphens between numbers and words, and di- for two identical groups. Alphabetise different substituent names; ignore di- when alphabetising.
- Draw the name back into a structure. Check carbon totals, functional groups and valences.
Examples: CH₃CH₂CH₂OH is propan-1-ol; CH₃CH(OH)CH₃ is propan-2-ol; CH₃COCH₂CH₃ is butan-2-one; CH₃CH₂CHO is propanal; CH₃CH₂COOH is propanoic acid; CH₃CH₂NH₂ is ethanamine. CH₂=CHCH₂CH₃ is but-1-ene. CH₃CHClCH₂CH₃ is 2-chlorobutane. A primary amine has nitrogen bonded to one carbon-containing group; the position of NH₂ does not itself determine whether that nitrogen is primary.
Worked example 1: naming two functional groups
Name CH₃CH(OH)COOH. The carboxyl group has suffix priority. Its carbon is carbon 1, giving a three-carbon parent, propanoic acid. OH is therefore at carbon 2 and appears as hydroxy-. The name is 2-hydroxypropanoic acid. Writing 2-hydroxyethanoic acid would omit the carboxyl carbon. Conversely HOCH₂CH₂OH has two hydroxyl groups and is ethane-1,2-diol.
Worked example 2: split an ester at the right bond
Name CH₃CH₂C(=O)OCH₂CH₃. The oxygen-linked group is CH₂CH₃, so the first word is ethyl. The acid-derived portion includes the carbonyl carbon: CH₃CH₂CO– contains three carbons and gives propanoate. The complete name is ethyl propanoate. Counting all five carbons as a pentanoate parent would erase the ester’s connectivity. Reverse-check: ethanol plus propanoic acid would form this ester and water.
3. Explain boiling using forces between molecules
A liquid boils when its vapour pressure equals external pressure. Molecules must separate substantially from their neighbours. This overcomes intermolecular attractions; it does not normally break the covalent C–C, C–H or O–H bonds inside each molecule. Compare substances at the same external pressure, because boiling temperature depends on pressure.
All molecules experience dispersion forces. In a homologous series, increasing chain length generally increases electron-cloud size and opportunities for intermolecular contact, strengthening dispersion interactions and raising boiling point. Among many alkane structural isomers, branching produces a more compact shape and reduces effective surface contact, often lowering boiling point. This is a structural comparison, not a claim that branched molecules have fewer electrons than their same-formula isomers.
Polar molecules also have permanent dipole interactions. Molecules with O–H or N–H bonds can participate as hydrogen-bond donors, with suitable oxygen or nitrogen lone pairs acting as acceptors. Alcohols and primary amines can hydrogen bond with one another. Aldehydes, ketones and esters have polar bonds but no O–H or N–H donor in their characteristic groups; their molecules cannot form the same self-hydrogen-bond network. They can accept hydrogen bonds from water. Carboxylic acids form strong intermolecular hydrogen-bond associations. Compare similar molecular sizes before attributing a difference to functional group alone: a very large non-polar molecule can have substantial dispersion attractions.
4. Melting point and viscosity need extra care
Melting disrupts the ordered solid arrangement. Intermolecular forces matter, but so does how efficiently molecules pack in a crystal. Symmetry, molecular shape and odd–even chain effects can complicate a simple mass trend. Do not automatically transfer the “more branching lowers boiling point” rule to melting point; a compact, symmetrical structure may pack particularly well. Use supplied data to identify the actual pattern.
Viscosity is resistance to flow. Stronger intermolecular attractions and longer chains that interfere with one another’s movement generally increase viscosity at the same temperature. Several OH groups can create extensive hydrogen bonding, increasing resistance to flow. Raising temperature usually reduces a liquid’s viscosity because molecules can move past neighbours more readily. Never compare viscosity measurements at different temperatures without acknowledging this variable.
Practice
- Name CH₃CH₂CH₂CHO and CH₃COOCH₂CH₂CH₃.
- Write a condensed formula for 3-methylbutan-1-ol and classify the alcohol.
- Explain why ethanol generally boils above a non-polar molecule of comparable size.
- A compact isomer has a lower boiling point but a higher melting point than its straight-chain isomer. Is that impossible?
- Why must a comparison of two liquids’ viscosities control temperature?
Answers with reasoning
- Butanal: include the aldehyde carbon in the four-carbon chain. Propyl ethanoate: propyl is attached through oxygen; the acid-derived portion has two carbons.
- HOCH₂CH₂CH(CH₃)CH₃. It is primary because the OH-bearing CH₂ has one carbon neighbour. There are five carbons overall but only four in the parent chain.
- Both have dispersion forces, but ethanol also forms hydrogen bonds between its molecules. More energy is generally needed to separate them. Do not say that its covalent O–H bond must break during boiling.
- No. Compactness may reduce liquid intermolecular contact and lower boiling point, while favourable symmetry and solid packing raise melting point. The two changes measure different processes.
- Temperature itself changes molecular motion and resistance to flow. Without controlling it, a measured difference cannot be attributed confidently to molecular structure.
Go deeper: build a defensible comparison
A good property explanation has a chain of reasoning: structural feature → intermolecular consequence → energy or motion consequence → measured trend. For example, adding CH₂ units to members of the same alcohol series increases the size of the electron cloud and opportunities for contact. The resulting stronger dispersion interactions act in addition to hydrogen bonding, so more energy is generally needed to separate the molecules. “It is bigger, therefore higher boiling point” omits the physical mechanism.
A homologous series shares a characteristic functional group and general structural pattern, with successive members differing by CH₂. Do not confuse this with isomerism: successive members have different molecular formulas, whereas structural isomers have the same formula. Comparing two members of a series helps control functional-group type. Comparing same-formula isomers helps control composition but changes shape and sometimes functional-group location.
Do not use an unconditional ladder claiming any hydrogen-bonding substance must boil above every substance without hydrogen bonding. There are many dispersion contacts in a large non-polar molecule. Likewise, the phrase “has hydrogen atoms” is not enough to establish hydrogen bonding: the donor H must be bonded to a sufficiently electronegative atom, such as O or N in these organic families. In an ester, the oxygen atoms can accept hydrogen bonds from water, but the ester alone does not supply the required O–H or N–H donor.
Worked example 3 · Apply: diagnose two independent errors
A student names HOCH₂CH(NH₂)CH₂CH₃ as “3-hydroxybutan-2-amine” and says it boils by breaking its O–H bonds. First trace the longest four-carbon chain. The OH group has priority over the amino group, so use the alcohol suffix and number from its end. OH is on carbon 1 and NH₂ on carbon 2: 2-aminobutan-1-ol. Its formula is C₄H₁₁NO, from 2 + 1 + 2 + 3 carbon-bound H plus one OH H and two NH₂ H.
Second, both OH and NH₂ offer opportunities for intermolecular hydrogen bonding. Boiling separates molecules while preserving their covalent structure, so the O–H bond is not the bond being broken. The correct explanation refers to overcoming attractions between molecules. This conclusion does not by itself establish an exact boiling point; numerical data would still be needed.
Application practice: justify the rule you use
- Name CH₃CH₂CH(Br)CH₂OH and identify the principal group that determines numbering.
- Two pure liquids of comparable molecular size are an ester and an alcohol. Explain a plausible reason for the alcohol’s higher boiling point, and one limit on this prediction.
- Two measurements show liquid A is more viscous at 15°C than liquid B at 35°C. A student concludes A has stronger intermolecular forces. Evaluate that conclusion and improve the comparison.
Application answers and checks
- 2-bromobutan-1-ol. The alcohol OH receives suffix priority and the lowest locant. Reading from the right puts Br on carbon 2. The formula is C₄H₉BrO; replacing one H in a saturated C₄ alcohol by Br reduces the hydrogen count by one.
- The alcohol molecules can form hydrogen bonds with one another; the ester has permanent dipole and dispersion attractions but no intrinsic O–H or N–H donor. For similar sizes, this can raise the alcohol’s boiling point. Differences in size, branching or additional groups can change the comparison, so a universal ordering or exact numerical gap is unsupported.
- The conclusion is not established because temperature differs. Raising temperature commonly lowers viscosity. Repeat at the same temperature, with comparable measurement technique and pure samples; then relate the controlled result to structure. Equal temperature removes one confounding factor but does not make every possible structural cause identical.
