IUPAC nomenclature for organic compounds explained: HSC Chemistry Module 7

Syllabus dot point

“Apply IUPAC rules to name and represent the structural formula of organic compounds including alkanes, alkenes, alkynes, alcohols, aldehydes, ketones, carboxylic acids, esters, and amines”

11 min readUpdated 22 Jun 2026

Quick answer

Identify the principal functional group to set the suffix, find the longest chain that contains it, number that chain so the principal group gets the lowest locant, then list substituents alphabetically with their locants in front of the parent name.

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In plain English

Think of an IUPAC name as a very literal, no-nonsense street address for a molecule. First you find the longest street (the carbon chain) that passes the house with the most important feature on it (the principal functional group), and that feature decides the "suffix" at the end of the name, like -ol for an alcohol or -oic acid for an acid. Then you number the houses along that street starting from whichever end gets the important feature the lowest number, the same way street numbering usually starts from the end nearest the main road. Any other features (branches, other groups) get listed as prefixes in front, in alphabetical order, each with its own house number. Once you know the algorithm, you can turn ANY structure into a name, or any name back into a structure, the same way you could give a stranger walking directions to any house on a numbered street as long as you know the numbering rule.

What this dot point is asking

NESA wants you to take any organic structural formula in the HSC scope and produce the IUPAC name, and to take any IUPAC name and draw the structural formula. The compounds in scope are alkanes, alkenes, alkynes, alcohols, aldehydes, ketones, carboxylic acids, esters and amines, with branches up to about C6. This is the foundation for every other Module 7 dot point.

The answer

The five-step IUPAC algorithm

Identify the principal functional group to set the suffix. Priority (high to low): carboxylic acid > ester > amide > nitrile > aldehyde > ketone > alcohol > amine > alkene/alkyne > alkane. Only the highest-priority group becomes the suffix; others become prefixes.
Find the longest continuous carbon chain that contains the principal functional group. The carbon count sets the parent name root (meth, eth, prop, but, pent, hex, hept, oct, non, dec).
Number the chain so that the principal functional group gets the lowest possible locant. If the principal group is at a chain end (acid, ester, aldehyde), it is automatically C1. For unsaturation, give the first $C=C$ or $C \equiv C$ the lowest locant. For substituents only, use the lowest set of locants.
Identify substituents (branches and lower-priority groups), assign their locants, and list them alphabetically.
Assemble the name: locants of substituents, substituent names (alphabetised), parent root, locant of principal group, suffix.

Suffix and general formula for each series

Series	General formula	Suffix	Example
Alkane	$C_nH_{2n+2}$	-ane	propane $CH_3CH_2CH_3$
Alkene	$C_nH_{2n}$	-ene	propene $CH_3CH=CH_2$
Alkyne	$C_nH_{2n-2}$	-yne	propyne $CH_3C \equiv CH$
Alcohol	$C_nH_{2n+1}OH$	-ol	propan-2-ol $CH_3CH(OH)CH_3$
Aldehyde	$C_nH_{2n}O$	-al	propanal $CH_3CH_2CHO$
Ketone	$C_nH_{2n}O$	-one	propan-2-one $CH_3COCH_3$
Carboxylic acid	$C_nH_{2n}O_2$	-oic acid	propanoic acid $CH_3CH_2COOH$
Ester	$RCOOR'$	-oate	methyl propanoate $CH_3CH_2COOCH_3$
Amine	$C_nH_{2n+1}NH_2$	-amine	propan-1-amine $CH_3CH_2CH_2NH_2$
Amide	$RCONH_2$	-amide	propanamide $CH_3CH_2CONH_2$

Priority order and locant checkpoints

Principal group priority (high to low): carboxylic acid > ester > amide > nitrile > aldehyde > ketone > alcohol > amine > alkene/alkyne > alkane. Only the top-ranked group present gets the suffix; every lower-ranked group becomes a prefix.
Always C1 by definition: the carbon of $-COOH$ (acid), $-COOR$ (ester's acyl carbon) and $-CHO$ (aldehyde) never needs a locant number in front of the suffix.
Lowest locant rule: number the chain so the principal group gets the lowest possible number first; only if that is tied does the lowest SET of substituent locants (compared term by term) break the tie.
Carbon-count roots: meth(1), eth(2), prop(3), but(4), pent(5), hex(6), hept(7), oct(8), non(9), dec(10).

An illustrative skeletal diagram shows how the locant numbering runs along the parent chain once the principal group and longest chain are fixed:

Locant rules in detail

Alcohols, aldehydes, ketones, amines, alkenes, alkynes all take a locant before the suffix: butan-2-ol, pent-2-ene, hex-3-yne, butan-2-one, butan-2-amine. Aldehydes and carboxylic acids do not need a locant because they are always C1.

Lowest locant rule. When two numbering directions are possible, choose the one that gives the lowest locant to the principal group. If the principal group has the same locant in both directions, choose the direction that gives the lowest locants to the substituents, taken as a set (compare term by term).

Substituent prefixes. Halogens are $fluoro$ , $chloro$ , $bromo$ , $iodo$ . Alkyl groups are $methyl$ , $ethyl$ , $propyl$ , etc. An $-OH$ becomes a $hydroxy-$ prefix only when not the principal group (e.g. in 2-hydroxypropanoic acid, where the acid outranks the alcohol). An $-NH_2$ becomes $amino-$ when downgraded.

Worked example: name a complex structure

Structure: $CH_3CH_2CH(CH_3)CH_2CH_2OH$ .

Principal group: $-OH$ (alcohol), suffix $-ol$ .
Longest chain containing $-OH$ : 5 carbons (pent).
Number to give $-OH$ lowest locant: from the OH end, OH at C1.
Substituent: methyl at C3.
Assemble: 3-methylpentan-1-ol.

Worked example: draw from name

Name: 4-amino-2-methylpentanoic acid.

Parent: pentanoic acid means a 5 carbon chain with $-COOH$ at C1.
Methyl at C2, amino at C4.

HOOC - CH(CH_3) - CH_2 - CH(NH_2) - CH_3

The acid outranks the amine, so $-NH_2$ is a prefix.

Naming esters specifically

Esters are named alkyl alkanoate in two words.

Alkyl part: comes from the alcohol, named as a substituent off the ester oxygen. Count the carbons attached to the single-bonded $O$ .
Alkanoate part: comes from the carboxylic acid, includes the $C=O$ . Count those carbons.

For $CH_3COOCH_2CH_3$ : $CH_3CO-$ has 2 carbons (ethanoate), $-OCH_2CH_3$ has 2 carbons (ethyl). Name: ethyl ethanoate.

For $CH_3CH_2CH_2COOCH_3$ : 4 carbons on the acid side (butanoate), 1 carbon on the alcohol side (methyl). Name: methyl butanoate.

Naming amines

A primary amine $R-NH_2$ takes the suffix $-amine$ with a locant for the nitrogen-bearing carbon. Secondary and tertiary amines are named by the largest parent amine, with the other substituents named with the $N-$ locant prefix. Example: $CH_3 - N(CH_3) - CH_2CH_3$ is $N,N$ -dimethylethanamine (parent is ethanamine, two methyls on nitrogen).

Full five-step walkthroughs

Worked example 1: a branched alcohol with a competing locant choice.

Name $CH_3CH(CH_3)CH_2CH(OH)CH_3$ .

Principal group: $-OH$ (alcohol), suffix $-ol$ . No higher-priority group present.
Longest chain containing $-OH$ : 5 carbons (pent).
Number both ways: from the OH end, OH is at C2, methyl at C4 (locants 2, 4). From the other end, OH is at C4, methyl at C2 (locants 4, 2). The principal group must get the lower locant, so numbering from the OH end wins: OH at C2.
Substituent: methyl at C4.
Assemble: 4-methylpentan-2-ol.

Marker's note: this is the exact trap NESA tests most in nomenclature questions, the "same pair of locants either way" trick where students think it does not matter which end they start from. It always matters: the PRINCIPAL group's locant is compared first, before the substituent's.

Worked example 2: a carboxylic acid with a downgraded amine.

Name $CH_3CH(NH_2)CH_2COOH$ .

Principal group: $-COOH$ (carboxylic acid, outranks amine), suffix $-oic$ $acid$ . The carboxyl carbon is automatically C1.
Longest chain containing $-COOH$ : 4 carbons (but).
Numbering is fixed: $-COOH$ carbon is C1, so counting along the chain the $-NH_2$ falls on C3.
Substituent: amino at C3 (the $-NH_2$ becomes the prefix $amino-$ because the acid outranks it).
Assemble: 3-aminobutanoic acid.

Marker's note: full marks require recognising that $-NH_2$ is DOWNGRADED to a prefix once a higher-priority group is present, and that the prefix used is $amino-$ , not $-amine$ .

Exam technique for naming questions

Work the five steps in order every time, out loud in your head: principal group, longest chain containing it, number for lowest locant on the principal group first, list substituents alphabetically, assemble. When two numbering directions give the SAME locant to the principal group, only then compare the substituent locants as a set. For "draw the structure" questions, draw the parent chain first with its correct length, place the principal group using its locant, then add substituents at their stated locants, and always add the correct number of H atoms to fill each carbon's four bonds.

Common traps

Forgetting the lowest locant rule: Always renumber from both ends and choose the direction that minimises the principal group locant first.
Naming the longest chain when it does not contain the functional group: The chain must contain the principal group. A longer chain that bypasses the $-OH$ is not the parent.
Ester order confusion: Alkyl (from alcohol) is named first, alkanoate (from acid) second. The alkanoate carbons include the carbonyl.
Including the carbonyl carbon twice: When counting carbons in an aldehyde or carboxylic acid, the $C$ in $-CHO$ or $-COOH$ is C1, not an additional carbon.
Alphabetising with prefixes: Use only the letter of the substituent name itself, ignoring multipliers (di, tri, tetra). So $ethyl$ comes before $dimethyl$ alphabetically (e by e).

Examples in context

Example 1. Labelling industrial chemicals at Orica Botany. Orica's Botany facility ships hundreds of products labelled with IUPAC names rather than common names because the Safety Data Sheet schema demands an unambiguous identifier. A tank truck of 2-chloropropan-1-ol is labelled with the IUPAC name plus the structural diagram, and the locant 2 makes clear that the chlorine sits on the central carbon while the OH sits on carbon 1. A misread label that swapped the locants would produce 1-chloropropan-2-ol, a different compound with different toxicology. The HSC five-step naming algorithm matches the SDS regulator's expectations one-to-one and is used by Orica safety officers when reviewing labels.

Example 2. NSW HSC depth study naming task. A common Stage 6 written task gives students 10 structural diagrams and asks for IUPAC names. A diagram showing $CH_3CH_2CH(OH)CH(CH_3)CH_2CH_3$ has a 6-carbon parent chain numbered from the right (to give OH the lowest locant); the principal group is the OH (suffix -ol) at position 3, with a methyl substituent at position 4, giving 4-methylhexan-3-ol. A student who counts from the left mislocates both groups and loses 2 marks. NESA marking schemes specifically reward "lowest set of locants" reasoning. Industrial scientists at the Australian Synchrotron also use these rules to log NMR samples.

Try this

Q1. Apply IUPAC rules to name each structure: (a) $CH_3CH_2CH=CH_2$ , (b) $CH_3CH(OH)CH_2CH_3$ , (c) $HCOOCH_2CH_3$ . [3 marks]

Cue. (a) But-1-ene. (b) Butan-2-ol. (c) Ethyl methanoate.

Q2. Draw the structural formula and write the molecular formula for 3-methylpentan-2-ol. [3 marks]

Cue. 5-carbon parent chain with OH at C2 and methyl at C3; $CH_3CH(OH)CH(CH_3)CH_2CH_3$ ; $C_6H_{14}O$ .

Q3. Provide IUPAC names for: (a) the addition product of HBr with prop-1-ene (Markovnikov); (b) the ester from propanoic acid and butan-1-ol; (c) the primary amine derived from propan-1-ol by direct ammonia substitution. [2+2+2 marks]

Cue. (a) 2-bromopropane. (b) Butyl propanoate. (c) Propan-1-amine.

Exam-style questions

Questions in the style of NESA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2022 HSC-style3 marks

Give the IUPAC name of the compound with the structural formula CH₃CH(OH)CH₂CH(CH₃)CH₃, and draw the structural formula of 3-methylbutan-2-one.

Show worked answer →

A 3 mark answer needs the correct IUPAC name with locants, plus the drawn structure.

Part 1: Name of $CH_3CH(OH)CH_2CH(CH_3)CH_3$ .

Longest chain containing the OH group is 5 carbons (pentane). Numbering from one end gives OH at C2 and the methyl branch at C4; numbering from the other end gives OH at C4 and the methyl branch at C2. Since the principal group (OH) must receive the lower locant, numbering starts from the end nearer the OH, placing it at C2 and the methyl branch at C4.

Name: 4-methylpentan-2-ol.

Part 2: Structure of 3-methylbutan-2-one.

Butan-2-one is $CH_3COCH_2CH_3$ . Add a methyl branch at C3:

CH_3 - CO - CH(CH_3) - CH_3

Markers reward (1) correct longest chain identification, (2) lowest locant rule for the principal group, (3) correct branch placement.

2019 HSC-style2 marks

State the IUPAC name of CH₃CH₂COOCH₂CH₃ and identify the homologous series to which it belongs.

Show worked answer →

The compound has a $-COO-$ linkage, so it is an ester.

An ester is named alkyl alkanoate. The alkyl portion (right of $O$ ) comes from the alcohol that formed the ester, and the alkanoate portion (left of $O$ , including $C=O$ ) comes from the carboxylic acid.

Alcohol-derived group: $-OCH_2CH_3$ , so ethyl.
Acid-derived group: $CH_3CH_2COO-$ (3 carbons including the $C=O$ ), so propanoate.

Name: ethyl propanoate. Homologous series: esters (general formula $RCOOR'$ ).

Markers reward (1) correct identification of the ester linkage and direction, (2) the full name with the alkyl group named first.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marks

Give the IUPAC name of each: (a)

CH_3CH_2CH_2CH_3

; (b)

CH_3CH_2CH=CH_2

; (c)

CH_3CH_2CH_2COOH

Show worked solution →

Apply the five-step algorithm to each structure.

(a) $CH_3CH_2CH_2CH_3$: Four carbons, no functional group other than the alkane chain itself. Suffix -ane. Name: butane.
(b) $CH_3CH_2CH=CH_2$: Four carbons with a $C=C$ at one end. Suffix -ene. Numbering from the double-bond end gives the double bond the lowest locant, C1. Name: but-1-ene.
(c) $CH_3CH_2CH_2COOH$: Four carbons including the carboxyl carbon, which is always C1. Suffix -oic acid. Name: butanoic acid.

Marking criteria: 1 mark per correct name; a name missing a required locant (e.g. "butene" instead of "but-1-ene") loses that mark.

foundation4 marks

Draw the structural formula for each named compound: (a) 2-methylpropan-1-ol; (b) 3-methylbutanal; (c) ethyl methanoate.

Show worked solution →

(a) 2-methylpropan-1-ol. Parent chain propan-1-ol is $CH_3CH_2CH_2OH$ with OH at C1. A methyl branch at C2 gives:

HOCH_2 - CH(CH_3) - CH_3

(b) 3-methylbutanal. Parent chain butanal is $CH_3CH_2CH_2CHO$ with the aldehyde carbon as C1. A methyl branch at C3 gives:

OHC - CH_2 - CH(CH_3) - CH_3

(c) ethyl methanoate. Methanoate (acid side) has 1 carbon including the carbonyl; ethyl (alcohol side) has 2 carbons:

HCOOCH_2CH_3

Marking criteria: 1 to 2 marks per structure for correct chain length, correct branch/functional-group placement, and correct bonding (e.g. the carbonyl oxygen double-bonded, not single-bonded, in (b) and (c)).

core5 marks

A 4.60 g sample of a straight-chain primary alcohol,

C_nH_{2n+1}OH

, is fully combusted. Given the sample is entirely propan-1-ol (

M = 60.10\ \text{g mol}^{-1}

), calculate the number of moles present, to 3 significant figures, and hence state the IUPAC name and molecular formula of the alcohol used.

Show worked solution →

Step 1: identify the given data.

Mass $m = 4.60\ \text{g}$ , molar mass $M = 60.10\ \text{g mol}^{-1}$ for propan-1-ol, $CH_3CH_2CH_2OH$ .

Step 2: apply $n = m / M$ .

n = \frac{m}{M} = \frac{4.60\ \text{g}}{60.10\ \text{g mol}^{-1}} = 0.076539\ \text{mol}

Step 3: round to 3 significant figures (matching 4.60 g, which has 3 s.f.).

n(\text{propan-1-ol}) = 0.0765\ \text{mol}

Step 4: name and formula.

IUPAC name: propan-1-ol (the OH is on C1 of a 3-carbon chain, so no ambiguity in locant, but the locant 1 is still conventionally written). Molecular formula: $C_3H_8O$ (equivalently $C_3H_7OH$ ).

Marking criteria: 1 mark for correctly substituting into $n = m/M$ , 1 mark for the correct unrounded value, 1 mark for rounding to 3 significant figures with correct units, 1 mark for the correct IUPAC name, 1 mark for the correct molecular formula.

core4 marks

The IR spectrum below is an illustrative spectrum of an organic unknown, Y, with molecular formula

C_4H_8O

. Using the two labelled peaks, identify whether Y is an aldehyde or a ketone, and suggest one structure and IUPAC name consistent with the data.

Show worked solution →

Reading the spectrum: One dominant feature is labelled: a strong, sharp peak near 1715 $cm^{-1}$ . No broad O-H stretch (2500 to 3300 or 3200 to 3550 $cm^{-1}$ ) is present, and no distinct aldehydic C-H peak near 2720 to 2850 $cm^{-1}$ is marked as present in this illustrative trace.
Interpretation: The single strong sharp peak near 1715 $cm^{-1}$ confirms a carbonyl ( $C=O$ ) group. The absence of any O-H stretch rules out a carboxylic acid or alcohol. With formula $C_4H_8O$ (one degree of unsaturation, consistent with one $C=O$ and no ring or extra double bond), and no O-H present, the compound is either an aldehyde or a ketone; IR alone cannot fully distinguish these two without the weak aldehydic C-H doublet, so a wet chemical test (Tollens' or Fehling's) would be needed to confirm which.
Structure: A ketone fit for $C_4H_8O$ is $CH_3COCH_2CH_3$ , butan-2-one. (An aldehyde fit would be $CH_3CH_2CH_2CHO$ , butanal; either is a valid structural answer if justified.)

Marking criteria: 1 mark for identifying the carbonyl peak and its region, 1 mark for correctly noting the absence of an O-H stretch (ruling out acid/alcohol), 1 mark for a structure consistent with $C_4H_8O$ , 1 mark for the correct IUPAC name of that structure.

core4 marks

Name

CH_2=CHCH_2COOH

using the five-step IUPAC algorithm, showing your reasoning for the principal group, chain length and numbering at each step.

Show worked solution →

Step 1: principal group: Two functional features are present, a $C=C$ double bond and a $-COOH$ group. Carboxylic acid outranks alkene in the priority order, so $-COOH$ is the principal group and sets the suffix $-oic$ $acid$ .
Step 2: longest chain containing $-COOH$: Counting the carbons $HOOC-CH_2-CH=CH_2$ gives 4 carbons, so the parent root is but.
Step 3: numbering: The carboxyl carbon is automatically C1 (it is a chain-terminal principal group), so numbering runs $COOH$ = C1, then C2, C3, with the double bond between C3 and C4.
Step 4: locant for the double bond: The double bond sits between C3 and C4, so it is cited as "3-ene" ("but-3-enoic acid"), since by convention the lower of the two double-bond carbons is used as the locant.
Step 5: assemble: No substituents to alphabetise. Name: but-3-enoic acid.

Marking criteria: 1 mark for correctly identifying the acid as the principal group over the alkene, 1 mark for the correct 4-carbon parent chain, 1 mark for correctly fixing the numbering from the $-COOH$ end, 1 mark for the fully assembled name including the double-bond locant.

exam6 marks

A structural isomer of

C_5H_{10}O_2

contains an ester linkage. (a) Draw and name TWO different possible esters with this molecular formula, showing full working for how you assigned the alkyl and alkanoate portions. (b) Explain why IUPAC nomenclature, rather than a common name, is essential when communicating such isomers in an industrial safety data sheet.

Show worked solution →

(a) Two isomeric esters of $C_5H_{10}O_2$ .

An ester $RCOOR'$ with 5 carbons total (one degree of unsaturation from the $C=O$ ) can split the 5 carbons across the acid and alcohol portions in more than one way.

Split 3 (acid) : 2 (alcohol). Acid side $CH_3CH_2COO-$ (propanoate, 3 carbons including carbonyl), alcohol side $-OCH_2CH_3$ (ethyl, 2 carbons): $CH_3CH_2COOCH_2CH_3$ , ethyl propanoate.
Split 4 (acid) : 1 (alcohol). Acid side $CH_3CH_2CH_2COO-$ (butanoate, 4 carbons including carbonyl), alcohol side $-OCH_3$ (methyl, 1 carbon): $CH_3CH_2CH_2COOCH_3$ , methyl butanoate.

Both fit $C_5H_{10}O_2$ : each has 5 carbons, 10 hydrogens and 2 oxygens once the full structure is drawn out and the hydrogens counted.

(b) Why IUPAC naming matters for safety data sheets.

An SDS must uniquely and unambiguously identify a chemical so any reader worldwide, regardless of trade name or local convention, retrieves the correct hazard and handling information. "Ethyl propanoate" and "methyl butanoate" are both plausible esters of the same molecular formula $C_5H_{10}O_2$ but are chemically distinct compounds with different boiling points, flash points and toxicological profiles; a common or trade name could be applied inconsistently or omitted, whereas the systematic IUPAC name is derived algorithmically from the structure itself and cannot be confused between the two isomers.

Marker's note: full marks in (a) require BOTH the drawn/written structure and the reasoning for how carbons were split between the alkyl and alkanoate portions, not just the final name. In (b), top answers explicitly link the STRUCTURAL distinction between the two named isomers back to the SAFETY consequence (different hazard properties), not just a general statement that "names should be unambiguous".

exam7 marks

Justify why systematic IUPAC nomenclature, rather than common or trade names, is essential for both academic communication and industrial safety, using the pair of isomers 1-chloropropan-2-ol and 2-chloropropan-1-ol as your worked example.

Show worked solution →

This is a 7-mark JUSTIFY: markers reward a reasoned argument built on a correctly worked structural comparison, not just a list of facts.

Band 6 PLAN.

Thesis: systematic IUPAC names are essential (not merely convenient) because they encode structure unambiguously and algorithmically, whereas common names are arbitrary labels that cannot distinguish close structural isomers and therefore cannot be trusted for either chemical communication or hazard identification.
Worked structural comparison: draw and name both isomers of $C_3H_7ClO$ that carry a chlorine and a hydroxyl on adjacent carbons of a 3-carbon chain. Apply the five-step algorithm to each: principal group is $-OH$ (alcohol outranks the halogen, which is always a prefix); chain is 3 carbons (propan); numbering gives $-OH$ the lowest locant.
- Isomer A: $ClCH_2CH(OH)CH_3$ . Numbering from the OH end gives OH at C2, Cl at C1: 1-chloropropan-2-ol.
- Isomer B: $CH_3CH(OH)CH_2Cl$ read the other way, i.e. $HOCH_2CH(Cl)CH_3$ . Numbering from the OH end gives OH at C1, Cl at C2: 2-chloropropan-1-ol.
Link to academic communication: the two isomers have the same molecular formula ( $C_3H_7ClO$ ) and even very similar common-sense descriptions ("chloropropanol"), but the locants alone distinguish which carbon carries which group; a chemist reading only "chloropropanol" cannot draw a unique structure, whereas the full IUPAC name can only be interpreted one way.
Link to industrial safety: the two isomers have different physical and toxicological properties because the electronegative chlorine and hydroxyl sit on different carbons, changing the local electron distribution and reactivity; a Safety Data Sheet, shipping label or emergency-response guide that used only a common name or an incomplete/incorrect locant could direct a responder to treat the wrong substance, with potentially serious consequences.
Judgement: because IUPAC names are generated by a fixed, universal algorithm (the same five steps every time) rather than by historical convention, they remove the ambiguity that trade and common names inevitably carry, making them essential rather than optional for both scientific literature and regulatory/safety documentation.

Model paragraph (excerpt). The isomers 1-chloropropan-2-ol and 2-chloropropan-1-ol share the same molecular formula, $C_3H_7ClO$ , and would likely attract the same casual description of "chloropropanol", yet the IUPAC locants reveal they are structurally distinct: in the first, the hydroxyl sits on the central carbon and chlorine on a terminal carbon, while in the second, the hydroxyl is terminal and the chlorine is central. Because the position of a halogen relative to a hydroxyl group changes the compound's polarity, boiling point and reactivity, a safety data sheet that relied on an ambiguous common name could not reliably communicate which isomer's hazard data applied, whereas the fully locant-specified IUPAC name resolves the ambiguity completely and identically for any reader, in any country, applying the same nomenclature rules.

Marker's note: top-band answers (1) correctly derive BOTH isomer names using the five-step algorithm shown as working, not asserted, (2) explicitly explain why the locant difference corresponds to a genuine structural and property difference, (3) connect this to a concrete safety or communication consequence rather than a vague assertion that "names should be clear", and (4) close with an explicit judgement rather than a neutral summary.

Practise this

Sources & how we know this

Chemistry Stage 6 Syllabus (2017) — NESA
Chemistry HSC exam papers — NESA