Intuition What this page is forThe parent note gave you the algorithm . But an algorithm only helps if you have practised it on every kind of molecule it can meet . Here we build a map of every scenario — every tie in numbering, every "which end do I count from", zero-carbon corner cases, ring vs chain, two groups fighting for the suffix, and the stereochemical labels — and then work one example per cell so you never meet a molecule you have not already seen.
Definition What is a "locant"?A locant is simply the position-number we attach to a carbon so we can say where something sits on the chain. If the backbone carbons are counted C 1 , C 2 , C 3 , … exactly one structure. Whenever you see a number in a name (like the 2 in butan-2 -ol), that number is a locant.
Every naming problem in this chapter is really one of these case classes . The columns are the decision you must make ; the cells are the tricky variants of that decision.
#
Case class (the decision)
The variant that trips people
Worked in
C1
Only a chain, one multiple bond — which end?symmetric tie vs asymmetric
Ex 1
C2
Senior group vs unsaturation — who gets low locant?suffix group beats C=C
Ex 2
C3
Two functional groups compete for suffixone becomes a prefix
Ex 3
C4
Longest chain turns a corner most-carbons ≠ straight line
Ex 4
C5
Numbering tie broken by "lowest set of locants" then alphabetequal-sum choices
Ex 5
C6
Degenerate / zero cases — 1-carbon acid, symmetric molecule, no substituentslocant omitted
Ex 6
C7
Ring (aromatic) + substituents — retained name as parentortho/meta/para → locants
Ex 7
C8
Real-world word problem — build structure from a descriptionreverse the algorithm
Ex 8
C9
Exam twist — ester / ether / halide all in one moleculeseniority + prefixes together
Ex 9
C10
Alkene stereochemistry — same connectivity, two shapescis/trans and E/Z
Ex 10
C11
Chiral centre — mirror-image moleculesR/S configuration
Ex 11
We now hit every cell once. Read the Forecast line, cover the steps, and guess before scrolling.
C H 3 − C H 2 − C H = C H − C H 3 Forecast: guess the number that will sit in front of "-ene". Is it 2 or 3?
Step 1 — Find the longest chain.
Trace the carbons: there are 5 in a row. Root = pent .
Why this step? The root word (see parent §1) is always the count of the longest continuous chain — nothing gets named until we know how long the backbone is.
Step 2 — Spot the multiple bond → suffix.
One C=C double bond → suffix -ene (parent §2).
Why this step? The middle suffix reports unsaturation in the chain ; a double bond forces "-ene".
Step 3 — Number to give the double bond the lowest locant.
There is no other functional group , so unsaturation itself decides direction.
Left→right: the double bond starts at C3 .
Right→left: it starts at C2 .
Pick the lower → 2 .
Why this step? With no senior suffix group present, the multiple bond takes the lowest-locant rule (this is the only time C=C wins the low number — contrast Example 2).
Answer: pent-2-ene.
Verify: Rebuild from the name — pent (5 C), double bond starting at C2: C H 3 − C H = C H − C H 2 − C H 3
C H 2 = C H − C H 2 − C H 2 − O H Forecast: does the OH become C1, or does the C=C grab the low number?
Step 1 — List every feature. A C=C and an –OH.
Why this step? You cannot choose a suffix before you know all the candidates.
Step 2 — Pick the senior group. From the seniority table (parent §3), ==alcohol (–OH) outranks a mere C=C==. So –OH is the suffix (-ol ) and the double bond stays inside the chain as -en- .
Why this step? Only the highest-priority group becomes the suffix and earns the lowest locant; everything else is subordinate.
Step 3 — Number so –OH gets the lowest locant.
From the OH end: OH = C1 , double bond spans C3–C4.
From the other end: OH = C4 — worse.
Why this step? The principal characteristic group dictates direction, not the double bond. This is the trap in the parent note.
Answer: but-3-en-1-ol.
Verify: but (4 C), –ol at C1, –en at C3: C H 2 = C H − C H 2 − C H 2 − O H not write "but-1-en-4-ol" — that would give OH the higher locant, violating seniority.
O H C − C H 2 − C H 2 − C O O H Forecast: which of –CHO and –COOH becomes the suffix, and what happens to the loser?
Step 1 — Identify both groups. An aldehyde –CHO and a carboxylic acid –COOH.
Why this step? You cannot choose which group becomes the suffix until you have listed every functional group present; a group you overlook could be the senior one and change the whole name.
Step 2 — Rank them. From the table: carboxylic acid (rank 1) > aldehyde (rank 5) . So –COOH is the suffix (-oic acid ); the aldehyde is demoted to the prefix oxo- .
Why this step? Only one suffix is allowed. The senior group wins; every other group is cited as a prefix, and the aldehyde's prefix form is "oxo".
Step 3 — Number. The acid carbon of –oic acid is by rule C1 (it is the end carbon). Then the chain is 4 carbons → root but , and the –CHO sits on C4 .
Why this step? The acid carbon is always the terminal carbon and always the lowest locant — this is automatic, not a choice.
Answer: 4-oxobutanoic acid.
Verify: butanoic acid = C H 3 C H 2 C H 2 C O O H O H C − C H 2 − C H 2 − C O O H
Definition Figure 1 — what you are looking atA blueprint-style skeleton : white dots are carbon atoms, thick cyan lines are the obvious horizontal 3-carbon row, and the semi-transparent amber line traces a different path that turns the corner and is actually 4 carbons long. The amber arrow points to where the "corner" happens. The whole point of the picture is to make you see that the longest chain is not the flat row you first notice.
Worked example Name the branched molecule in the figure: a 3-carbon horizontal chainC H 3 − C H ( − ) − C H 3 − C H 2 − C H 3 up .
Forecast: guess the longest chain length — is it 3 (the drawn row) or something bigger?
Step 1 — Trace EVERY path, not just the straight row. Look at the figure: the obvious horizontal path is 3 carbons (cyan). But a path that turns the corner — starting at a bottom methyl, through the centre, up the branch — is 4 carbons (amber).
Why this step? "Longest chain" means longest continuous path through bonds; the page is 2-D but the molecule is not a straight line. Always follow the turns.
Step 2 — Root from the true longest chain. 4 carbons → root but . The remaining carbon (the other bottom methyl) is a 1-carbon branch → methyl .
Why this step? Choosing the genuinely longest chain minimises the number of substituents and gives the unique correct name.
Step 3 — Number for lowest substituent locant. The methyl branch sits on the second carbon of the 4-chain from either end → 2 .
Why this step? No functional group, so we minimise substituent locants; 2 < 3.
Answer: 2-methylbutane (this molecule is the branched pentane also called isopentane ).
Verify: 2-methylbutane = C H 3 − C H ( C H 3 ) − C H 2 − C H 3 C 5 H 12 3 + 1 + 1 = 5 not confuse it with isobutane (C 4 H 10 wrong — you must pick the 4-chain. See Isomerism for why C 5 H 12
Definition What "lowest set of locants" meansWhen two numbering directions give different lists of locants, write each list in increasing order , then compare them term by term from the left . The set that has the smaller number at the first point where they differ wins. Example: { 2 , 3 , 6 } { 2 , 4 , 5 } 2 = 2 3 < 4
C H 3 − C H ( C H 3 ) − C H 2 − C H ( C H 3 ) − C H 3 Forecast: the two methyls sit symmetrically. Will the locant list be { 2 , 4 }
Step 1 — Longest chain. 5 carbons in the main row → pent .
Step 2 — Substituents. Two methyl branches → dimethyl .
Step 3 — Number both directions and compare the lists term-by-term.
Left→right: methyls on C2 and C4 → list { 2 , 4 }
Right→left: methyls on C2 and C4 → list { 2 , 4 } 2 = 2 4 = 4 Why this step? This is the exact rule defined in the callout above; because the molecule is symmetric, both directions collapse to the same list.
Step 4 — Alphabetise & assemble. Only one kind of substituent (methyl); the multiplying prefix "di" is not counted for alphabetical order (parent §6). So we alphabetise under m for "methyl", not under d for "di". With just one substituent-type there is nothing to reorder, so we simply prepend the count.
Why this step? Alphabetical order fixes the sequence in which substituents are written, and IUPAC deliberately ignores multiplying prefixes (di, tri, tetra) so that "dimethyl" still files under m — otherwise the same substituent would sort differently depending on how many copies there are, breaking uniqueness.
Answer: 2,4-dimethylpentane.
Verify: locant list { 2 , 4 } C 7 H 16
Worked example Name three "corner" molecules: (a)H C O O H C H 3 C O C H 3 C H 3 C H 3 Forecast: for each, guess whether a locant number even appears.
(a) H C O O H
Step 1: senior group –COOH → -oic acid. Step 2: the whole molecule is a single carbon (the acid carbon itself). Root = meth .
Why this step? The acid carbon is C1 and there is nothing else; a 1-carbon chain still uses "meth".
Answer: methanoic acid (trivial name: formic acid). No locant — there is only one carbon.
(b) C H 3 C O C H 3
Step 1: senior group is the ketone >C=O → -one. Step 2: 3-carbon chain → prop . Step 3: the C=O is on C2 from either end (symmetric).
Why this step? By symmetry both numberings give C2, so the position is fixed. Under the 2013 IUPAC recommendations the locant is mandatory whenever it is needed to specify the group's position — so the preferred IUPAC name explicitly cites it.
Answer: propan-2-one (acetone). The "2" is required, not optional, in current IUPAC practice.
(c) C H 3 C H 3
Step 1: no suffix group; all single bonds → -ane. Step 2: 2 carbons → eth . No substituents, no locant.
Answer: ethane.
Verify: (a) H C O O H C 3 H 6 O C 2 H 6
Definition Figure 2 — what you are looking atA blueprint hexagon = the benzene ring; the inner white circle is the shorthand for its shared (aromatic) electrons. Each corner is a numbered carbon (1–6). The amber labels mark the two substituents — an –OH pinned to C1 and a –CH₃ at C3. The side legend shows the three neighbour-relationships: ortho (1,2), meta (1,3), para (1,4). Reading the figure tells you the methyl is meta to the OH.
Worked example Name the benzene ring in the figure carrying an –OH on C1 and a –CH₃ on the carbon two positions away.Forecast: what is the parent name — benzene, or a retained name?
Step 1 — Pick the parent. –OH on benzene has the retained IUPAC name phenol (parent §5, Ex D). So the ring is named as a phenol, and –OH defines C1 .
Why this step? IUPAC keeps a short list of retained aromatic names (Aromatic Compounds — Benzene Derivatives ); phenol is one, and its –OH is automatically C1.
Step 2 — Locate the methyl. Look at the figure: OH is at position 1; the methyl is at position 3 (the "meta" position — one carbon skipped). The three classic relationships are labelled: ortho = 1,2 (adjacent), meta = 1,3 (one apart), para = 1,4 (opposite).
Why this step? On a ring you number around it, giving the substituent the lowest locant consistent with C1 = the ring's defining group.
Step 3 — Assemble. Methyl at C3, parent phenol → 3-methylphenol (older name: m -cresol).
Why this step? Substituent prefix + locant + parent; "meta" is a colloquial synonym for the 1,3 relationship.
Answer: 3-methylphenol.
Verify: phenol = C 6 H 5 O H C 7 H 8 O
Worked example A flavour chemist writes in her notebook:"a straight 4-carbon acid — the sour note in butter." Build the structure and give the IUPAC name.
Forecast: guess the name before reading — 4 carbons, an acid.
Step 1 — "Acid" → –COOH, and it defines C1.
Why this step? Carboxylic acid is the senior suffix and its carbon is always C1.
Step 2 — "Straight 4-carbon" → root but, all single bonds → -an-. Chain: C H 3 − C H 2 − C H 2 − C O O H Why this step? "Straight" means no branches; 4 carbons → butanoic backbone.
Step 3 — Assemble. No substituents, no unsaturation. Glue the pieces in slot order: root but + saturation -an- + suffix -oic acid → butanoic acid.
Why this step? IUPAC fixes the order of the name slots (prefix + root + unsaturation + suffix, parent §4); assembling in that order is what guarantees a reader rebuilds exactly this molecule and no other.
Answer: butanoic acid (trivial name: butyric acid — the smell of rancid butter).
Verify: butanoic acid = C 4 H 8 O 2 C H 3 C H 2 C H 2 C O O H
Definition Punctuation & parentheses in complex namesWhen several locants and prefixes pile up, IUPAC punctuation is strict: commas separate numbers from numbers (2 , 4 hyphens separate numbers from letters (2 -chloro alphabetical order with no spaces (chloromethoxy…), and the whole is one word up to the suffix. Parentheses are only needed when a substituent itself carries locants (e.g. a "(1-methylpropyl)" group) so the inner numbers can't be confused with the main-chain numbers. In this example each prefix is simple, so no parentheses are required — but the comma/hyphen rule still governs every join.
C H 3 − O − C H 2 − C H C l − C O O − C H 3 Forecast: which of the three groups (ester, ether, halide) becomes the suffix?
Step 1 — Rank the groups. Ester (rank 2) > ether (never a suffix) and halide (never a suffix). So the ester is the suffix (-oate ); the ether becomes methoxy- and the chlorine becomes chloro- .
Why this step? Only the senior group is a suffix; ethers and halides are always prefixes (parent §3).
Step 2 — Name the ester's two parts. An ester R − C O O − R ′ "(R' group) (acid-chain)oate" . Here R ′ = C H 3 methyl ; the acid chain is the carbons ending in C O O Why this step? An ester is formally the acid's –OH replaced by an –OR′ group, so its name is built as two words: the alcohol-derived group R ′ both halves of the ester unambiguously.
Step 3 — Number the acid chain. The carbonyl carbon of the ester is C1 . Then C2 carries –Cl, and C3 carries the –OCH₃ (methoxy). Chain length = 3 → propanoate .
Why this step? The senior group (ester carbonyl) is C1; number outward from it so the senior group gets the lowest locant.
Step 4 — Alphabetise prefixes & punctuate. "chloro" (c) before "methoxy" (m); each prefix gets its locant joined by a hyphen, and the alcohol group "methyl" is written as a separate leading word.
Why this step? Prefixes are cited in alphabetical order (parent §4); the punctuation rules in the callout above then fix every comma and hyphen so the name is machine-readable.
Answer: methyl 2-chloro-3-methoxypropanoate.
Verify: acid chain C 3 C 1 C 2 C 3 why the ester outranks the others mechanistically, see Priority and Seniority Rules (CIP vs IUPAC) .
Definition Figure 3 — what you are looking atTwo blueprint sketches of the same molecule, but-2-ene. On the left (amber) both C H 3 same side of the locked double bond; on the right (cyan) they point to opposite sides. The H's fill the remaining corners. Because the C=C bar cannot rotate, these are two genuinely different, non-interconvertible molecules — the picture is showing you why a shape-label is needed .
Intuition WHY a stereo-label is needed at allA C=C double bond cannot rotate (unlike a single bond). So the two groups on each end are locked on one side or the other. Two molecules with the same connectivity — same name so far — can be different shapes . The name must therefore carry an extra label saying which shape . This is Isomerism of the "geometric" kind.
Definition cis/trans vs E/Z
cis = the two like (or reference) groups are on the same side ; trans = on opposite sides . This simple label only works cleanly when each carbon has one H.E/Z is the general rule using CIP priorities : on each carbon rank the two attached groups by atomic number. If the two higher-priority groups are on the same side → Z (German zusammen = together); on opposite sides → E (entgegen = opposite).
Worked example Name the two shapes ofC H 3 − C H = C H − C H 3 Forecast: which shape is "Z" — the one with both methyls up together, or the one with them across?
Step 1 — Base name first. 4-carbon chain, C=C at C2 → but-2-ene (from Example 1's rules).
Why this step? Stereo-labels are added to an otherwise-complete name; you never skip the connectivity step.
Step 2 — Rank the groups on each double-bond carbon (CIP). On C2 the two competitors are C H 3 H C H 3 Why this step? E/Z is defined by where the higher-priority groups sit, so we must rank first.
Step 3 — Read the figure.
Left molecule (amber): both C H 3 same side → Z → this is (Z)-but-2-ene = cis -but-2-ene.Right molecule (cyan): the C H 3 E → (E)-but-2-ene = trans -but-2-ene.
Why this step? The locked double bond makes these two genuinely different, non-interconvertible molecules.
Answers: (Z)-but-2-ene (cis) and (E)-but-2-ene (trans).
Verify: for but-2-ene the higher-priority group on each carbon is C H 3 C 4 H 8 Degree of Unsaturation ).
Definition Figure 4 — what you are looking atA blueprint sketch of butan-2-ol 's central carbon. Three bonds lie in the plane of the page (drawn as solid lines) and one — the lowest-priority H behind the page (a dashed wedge). The four groups are labelled with their CIP priority numbers 1–4. The amber curved arrow traces the path 1 → 2 → 3 anticlockwise , which is exactly what the R/S rule reads to give the answer.
A carbon bonded to four different groups is like a left vs right hand: its mirror image cannot be superimposed on the original. Same connectivity, same name so far — yet two distinct molecules (Isomerism , "optical"). The label R or S tells you which hand .
Definition The R/S procedure (CIP)
Rank the four groups by CIP priority (highest atomic number = highest), call them 1 > 2 > 3 > 4 .
Point the lowest group (4 ) away from you.
Trace 1 → 2 → 3 R (Latin rectus , right); anticlockwise → S (sinister , left).
Worked example Name the chiral molecule in the figure: a central carbon bonded to− O H − C H 3 − C H 2 C H 3 − H Forecast: with H pointing away and the arrow going anticlockwise, will it be R or S?
Step 1 — Base name. 4-carbon chain, –OH senior at C2 → butan-2-ol (exactly Example A of the parent note).
Why this step? R/S is appended to a complete connectivity name; C2 is the chiral centre.
Step 2 — Assign CIP priorities to the four groups on C2.
− O H priority 1 (highest).− C H 2 C H 3 priority 2 .− C H 3 priority 3 (loses to ethyl at the second atom: C beats H).− H priority 4 (lowest).
Why this step? When two first atoms tie (both C), you compare what they are bonded to — this is the CIP tie-break.
Step 3 — Orient and trace. In the figure the lowest group (H back (dashed wedge). Trace O H ( 1 ) → ethyl ( 2 ) → methyl ( 3 ) anticlockwise → S .
Why this step? Direction of 1 → 2 → 3 is the definition of R vs S.
Answer: (S)-butan-2-ol (its mirror image, with the arrow reversed, is (R)-butan-2-ol).
Verify: priority order OH(O=8) > ethyl(C→C,H,H) > methyl(C→H,H,H) > H(1); ethyl beats methyl at the second-atom comparison (C > H). Anticlockwise with #4 behind = S ✅.
Recall Did we hit every cell of the matrix?C1 → Ex 1 · C2 → Ex 2 · C3 → Ex 3 · C4 → Ex 4 · C5 → Ex 5 · C6 → Ex 6 (three sub-cases) · C7 → Ex 7 · C8 → Ex 8 · C9 → Ex 9 · C10 → Ex 10 (E and Z) · C11 → Ex 11 (R and S). Every decision-class, every degenerate corner, and both flavours of stereochemistry are now solved examples. ✅
Recall Q: When does the double bond get the low locant, and when does it lose?It wins only when there is no senior suffix group (Ex 1). The moment an –OH, =O, –COOH, ester… appears, that senior group takes the low locant and the C=C is secondary (Ex 2).
Recall Q: In an ester like Ex 9, which carbon is C1?The carbonyl (C=O) carbon of the acid part is C1; number outward from it.
Recall Q: In (Z)-but-2-ene, what does "Z" tell you?The two higher-CIP-priority groups (here both methyls) are on the same side of the locked double bond (Z = zusammen = together = cis here).
Recall Q: To assign R or S, which group must point away from you?The lowest-priority group (#4); then 1 → 2 → 3
Recall Q: What is a "locant"?The position-number attached to a carbon (C1, C2, …) that tells you where a group, bond, or substituent sits on the chain.
IUPAC name of CH3CH2CH=CHCH3 pent-2-ene
IUPAC name of CH2=CHCH2CH2OH but-3-en-1-ol
IUPAC name of OHC-CH2-CH2-COOH 4-oxobutanoic acid
IUPAC name of the isopentane skeleton (C5H12 branched) 2-methylbutane
IUPAC name of the symmetric 2,4-dimethyl pentane 2,4-dimethylpentane
IUPAC name of HCOOH methanoic acid
IUPAC name of CH3COCH3 propan-2-one
IUPAC name of CH3CH3 ethane
Retained parent name for benzene + OH phenol
IUPAC name of phenol with a methyl at position 3 3-methylphenol
1,3-disubstituted benzene relationship is called meta
IUPAC name of CH3CH2CH2COOH butanoic acid
IUPAC name of CH3-O-CH2-CHCl-COOCH3 methyl 2-chloro-3-methoxypropanoate
When does the C=C get the lowest locant only when no senior suffix group is present
Which carbon is C1 in a carboxylic acid or ester the carboxyl/carbonyl carbon
Z means the higher-priority groups are on the same side of the double bond (cis-like)
E means the higher-priority groups are on the opposite sides of the double bond (trans-like)
In R/S, which group points away from you the lowest-priority group (#4)
1 to 2 to 3 clockwise (lowest behind) gives R configuration
Is the locant in propan-2-one optional under 2013 IUPAC rules no, it is mandatory
2-methylbutane is C5H12 (isopentane); isobutane is C4H10 (2-methylpropane), one carbon fewer
What is a locant the position-number telling where a group sits on the chain
