8.2 IUPAC Nomenclature and Isomerism
One molecular formula can stand for many different compounds, so every compound needs a name that pins down its structure. This section teaches the IUPAC rules — parent chain, lowest locants, alphabetical order, functional-group suffixes and priority, substituted benzenes — and then the ways in which compounds with the same formula differ: chain, position and functional isomers, metamers, and the stereoisomers that differ only in space.
Ethanol, the alcohol in hand sanitiser, has the molecular formula $\mathrm{C_{2}H_{6}O}$. So does methoxymethane, a gas used as an aerosol propellant and tested as a clean diesel fuel. One is a liquid that mixes with water and boils at $78\ ^{\circ}\mathrm{C}$; the other boils at $-24\ ^{\circ}\mathrm{C}$. Add two carbons and the formula $\mathrm{C_{4}H_{10}O}$ already describes seven different compounds. A formula is not a name: chemists, pharmacists and customs officers all need a name that tells them exactly which atoms are joined to which.
Older "common" names — acetone, isobutane, wood spirit — were given one by one and say little about structure. In 1892 chemists meeting in Geneva began to build a systematic method, and today the International Union of Pure and Applied Chemistry (IUPAC) maintains the rules. An IUPAC name can be turned back into exactly one structure. This section shows how to build such names, and then how to recognise and count isomers: different compounds with the same molecular formula.
IUPAC Names of Alkanes and Their Derivatives
📘 Definition 8.5 — Parts of an IUPAC name
An IUPAC name is built from:
- a word root giving the number of carbon atoms in the parent chain: meth (1), eth (2), prop (3), but (4), pent (5), hex (6), hept (7), oct (8), non (9), dec (10);
- a primary suffix for the saturation of the chain: -ane (all single bonds), -ene (a $\mathrm{C{=}C}$), -yne (a $\mathrm{C{\equiv}C}$);
- a secondary suffix for the principal functional group: -ol, -al, -one, -oic acid, …;
- prefixes for the substituents (side chains and some groups), each with a locant — the number of the chain carbon that carries it.
A side chain is named as an alkyl group: remove one H from an alkane and change -ane to -yl. Some branched alkyl groups keep common names that are still accepted:
| Group | Structure | Common name | Systematic name |
|---|---|---|---|
| $\mathrm{C_{3}H_{7}{-}}$ | $\mathrm{CH_{3}CH_{2}CH_{2}{-}}$ | propyl ($n$-propyl) | propyl |
| $\mathrm{C_{3}H_{7}{-}}$ | $\mathrm{(CH_{3})_{2}CH{-}}$ | isopropyl | 1-methylethyl |
| $\mathrm{C_{4}H_{9}{-}}$ | $\mathrm{CH_{3}CH_{2}CH(CH_{3}){-}}$ | sec-butyl | 1-methylpropyl |
| $\mathrm{C_{4}H_{9}{-}}$ | $\mathrm{(CH_{3})_{2}CHCH_{2}{-}}$ | isobutyl | 2-methylpropyl |
| $\mathrm{C_{4}H_{9}{-}}$ | $\mathrm{(CH_{3})_{3}C{-}}$ | tert-butyl | 1,1-dimethylethyl |
📐 Rule 8.3 — Naming a saturated chain with substituents
- Parent chain. Choose the longest continuous chain of carbon atoms — it need not be the chain drawn horizontally. If two chains are equally long, choose the one with more substituents.
- Lowest locants. Number the chain from the end that gives the substituents the lowest set of locants. Compare the sets term by term; the first point of difference decides: $\{2,3,5\}$ beats $\{2,4,4\}$ because $3\lt4$.
- Prefixes. Name each substituent with its locant; identical substituents get di-, tri-, tetra- and one locant each ($2,2\text{-dimethyl}$). Halogens are written fluoro, chloro, bromo, iodo; $\mathrm{{-}NO_{2}}$ nitro; $\mathrm{{-}OR}$ alkoxy (methoxy, ethoxy); they are treated exactly like alkyl groups.
- Alphabetical order. Write the prefixes in alphabetical order, ignoring di-, tri-, sec-, tert- but not iso; a complex substituent in brackets is alphabetised under its first letter.
- Tie-break. If both directions give the same set of locants, the lower number goes to the substituent cited first in alphabetical order.
- Punctuation. Commas separate numbers, hyphens separate numbers from words, and the name is one word: $\mathrm{3\text{-}ethyl\text{-}2,2\text{-}dimethylpentane}$.
✏️ Example 1 — Naming branched alkanes and haloalkanes
Give the IUPAC name of (a) $\mathrm{CH_{3}CH(CH_{3})CH_{2}CH(CH_{3})_{2}}$; (b) $\mathrm{CH_{3}CH_{2}CH(C_{2}H_{5})CH_{2}CH_{2}CH_{3}}$; (c) $\mathrm{(CH_{3})_{3}CCH_{2}CH(CH_{3})_{2}}$ ("iso-octane", the reference fuel for octane ratings); (d) $\mathrm{CH_{3}CH_{2}CH(CH_{3})CH(CH_{2}CH_{3})CH_{2}CH_{2}CH_{3}}$; (e) $\mathrm{CH_{3}CHClCH_{2}CH(CH_{3})CH_{2}CH_{3}}$; (f) $\mathrm{CH_{3}CHBrCH_{2}CHClCH_{3}}$.
💡 Show solution▶
(a) Longest chain: 5 carbons; methyls on C2 and C4 from either end. 2,4-dimethylpentane.
(b) Write the formula out: the longest chain has 6 carbons. (A path through the ethyl group, from its end via C3 to C6, also has 6 carbons and again carries an ethyl group on C3 — the same name results.) Hexane with ethyl on C3 from either end: 3-ethylhexane.
(c) Chain of 5; from the left the methyls are at $\{2,2,4\}$, from the right at $\{2,4,4\}$. First point of difference $2\lt4$: 2,2,4-trimethylpentane.
(d) Longest chain: the 7-carbon chain from the first $\mathrm{CH_{3}}$ to the last (the path through the ethyl branch has only 6). From the left: methyl 3, ethyl 4 → $\{3,4\}$; from the right $\{4,5\}$. Alphabetical order puts ethyl first: 4-ethyl-3-methylheptane.
(e) Chain of 6; from the left $\{2\ (\mathrm{Cl}),4\ (\mathrm{CH_{3}})\}$, from the right $\{3,5\}$. 2-chloro-4-methylhexane.
(f) Both directions give $\{2,4\}$: the tie-break gives the lower locant to bromo, first in alphabetical order. 2-bromo-4-chloropentane, not 4-bromo-2-chloropentane.
Check Count carbons in each name: (c) pent (5) + 3 methyl = 8 = $\mathrm{C_{8}H_{18}}$ ✓; (d) hept (7) + ethyl (2) + methyl (1) = 10 ✓.
Functional Groups, Multiple Bonds and Benzene Rings
When a compound has a functional group, the group is named as a suffix on the parent chain if it is the principal group, or as a prefix if a more senior group is present. Halogens, nitro and alkoxy groups are only ever prefixes.
| Class | Group | Suffix | Prefix | Example |
|---|---|---|---|---|
| carboxylic acid | $\mathrm{{-}COOH}$ | -oic acid | carboxy | $\mathrm{CH_{3}CH_{2}COOH}$ propanoic acid |
| sulphonic acid | $\mathrm{{-}SO_{3}H}$ | -sulphonic acid | sulpho | $\mathrm{C_{6}H_{5}SO_{3}H}$ benzenesulphonic acid |
| ester | $\mathrm{{-}COOR}$ | alkyl …-oate | alkoxycarbonyl | $\mathrm{CH_{3}COOC_{2}H_{5}}$ ethyl ethanoate |
| acid chloride | $\mathrm{{-}COCl}$ | -oyl chloride | chlorocarbonyl | $\mathrm{CH_{3}COCl}$ ethanoyl chloride |
| amide | $\mathrm{{-}CONH_{2}}$ | -amide | carbamoyl | $\mathrm{CH_{3}CH_{2}CONH_{2}}$ propanamide |
| nitrile | $\mathrm{{-}C{\equiv}N}$ | -nitrile | cyano | $\mathrm{CH_{3}CH_{2}CN}$ propanenitrile |
| aldehyde | $\mathrm{{-}CHO}$ | -al | formyl or oxo | $\mathrm{CH_{3}CH_{2}CHO}$ propanal |
| ketone | $\mathrm{\gt C{=}O}$ | -one | oxo | $\mathrm{CH_{3}COCH_{2}CH_{3}}$ butan-2-one |
| alcohol | $\mathrm{{-}OH}$ | -ol | hydroxy | $\mathrm{CH_{3}CH(OH)CH_{3}}$ propan-2-ol |
| amine | $\mathrm{{-}NH_{2}}$ | -amine | amino | $\mathrm{CH_{3}CH_{2}NH_{2}}$ ethanamine |
| halide, nitro, ether | $\mathrm{{-}X}$, $\mathrm{{-}NO_{2}}$, $\mathrm{{-}OR}$ | — | halo, nitro, alkoxy | $\mathrm{CH_{3}OCH_{2}CH_{3}}$ methoxyethane |
📐 Rule 8.4 — Chains with a functional group or multiple bonds
- The parent chain must contain the carbon that carries the principal group and the carbons of the multiple bonds; among such chains choose the longest.
- Number the chain to give the lowest locant to the principal group, then to the multiple bonds, then to the prefixes. If a double and a triple bond would get the same locants, the double bond gets the lower number.
- Groups whose carbon is part of the chain — $\mathrm{{-}CHO}$, $\mathrm{{-}COOH}$, $\mathrm{{-}COCl}$, $\mathrm{{-}CONH_{2}}$, $\mathrm{{-}CN}$ — are always at C1, and the locant 1 is left out: butanal, butanoic acid, butanenitrile.
- The final e of -ane, -ene, -yne is dropped before a suffix starting with a vowel (propan-1-ol, butan-2-one) but kept before a consonant (propanenitrile, ethane-1,2-diol). Locants go just before the part they describe: but-2-ene, pent-3-en-1-yne.
📐 Rule 8.5 — Order of priority of functional groups
When several groups are present, the most senior one becomes the suffix and the others are prefixes:
$$\mathrm{{-}COOH}\gt\mathrm{{-}SO_{3}H}\gt\mathrm{{-}COOR}\gt\mathrm{{-}COCl}\gt\mathrm{{-}CONH_{2}}$$
$$\gt\mathrm{{-}CN}\gt\mathrm{{-}CHO}\gt\mathrm{\gt C{=}O}\gt\mathrm{{-}OH}\gt\mathrm{{-}NH_{2}}\gt\mathrm{C{=}C},\ \mathrm{C{\equiv}C}$$
The groups $\mathrm{{-}R}$, $\mathrm{C_{6}H_{5}{-}}$ (phenyl), halogens, $\mathrm{{-}NO_{2}}$ and $\mathrm{{-}OR}$ are always prefixes.
✏️ Example 2 — One functional group or one multiple bond
Name (a) $\mathrm{CH_{3}CH_{2}CH(OH)CH_{3}}$; (b) $\mathrm{(CH_{3})_{2}CHCH_{2}CHO}$; (c) $\mathrm{CH_{3}COCH_{2}CH(CH_{3})_{2}}$; (d) $\mathrm{CH_{2}{=}CHCH(CH_{3})_{2}}$; (e) $\mathrm{(CH_{3})_{2}CHCH_{2}COOH}$; (f) $\mathrm{HC{\equiv}CCH_{2}CHClCH_{3}}$.
💡 Show solution▶
(a), (b) Four-carbon chain, OH as low as possible → butan-2-ol. The aldehyde carbon is C1; methyl on C3 → 3-methylbutanal.
(c) Five-carbon chain containing $\mathrm{C{=}O}$; numbering from the left gives the ketone 2 (from the right it would be 4), methyl then at 4 → 4-methylpentan-2-one.
(d) The chain must contain both alkene carbons: 4 carbons, double bond at C1, methyl at C3 → 3-methylbut-1-ene.
(e) $\mathrm{COOH}$ carbon = C1; chain of 4; methyl on C3 → 3-methylbutanoic acid.
(f) Five carbons including $\mathrm{C{\equiv}C}$; the triple bond (the "principal" feature here, since halogens are only prefixes) gets locant 1, chloro then 4 → 4-chloropent-1-yne.
✏️ Example 3 — Several groups and multiple bonds
Name (a) $\mathrm{HOCH_{2}CH_{2}COOH}$; (b) $\mathrm{CH_{3}COCH_{2}CHO}$; (c) $\mathrm{H_{2}NCH_{2}CH_{2}OH}$; (d) $\mathrm{CH_{2}{=}CHCH_{2}OH}$; (e) $\mathrm{CH_{3}CH(OH)CH_{2}CN}$; (f) $\mathrm{HC{\equiv}C{-}CH{=}CH{-}CH_{3}}$; (g) $\mathrm{CH_{3}CH_{2}COOCH_{3}}$.
💡 Show solution▶
(a), (b) $\mathrm{COOH}\gt\mathrm{OH}$: acid is the suffix, C1 = $\mathrm{COOH}$, OH on C3 → 3-hydroxypropanoic acid. $\mathrm{CHO}\gt$ ketone: C1 = $\mathrm{CHO}$, ketone O on C3 as "oxo" → 3-oxobutanal.
(c), (d) $\mathrm{OH}\gt\mathrm{NH_{2}}$: 2-aminoethan-1-ol (2-aminoethanol). OH outranks the double bond and takes C1: prop-2-en-1-ol.
(e) Nitrile outranks OH. The nitrile carbon is C1 of a four-carbon chain; OH on C3 → 3-hydroxybutanenitrile.
(f) From the left: yne 1, ene 3 → $\{1,3\}$; from the right: ene 2, yne 4 → $\{2,4\}$. Lowest set $\{1,3\}$ → pent-3-en-1-yne.
(g) An ester $\mathrm{RCOOR'}$ is named "alkyl (from $\mathrm{R'}$) + alkanoate (from $\mathrm{RCOO}$)": the acid part has 3 carbons → methyl propanoate.
📐 Rule 8.6 — Naming substituted benzenes
- A single substituent is a prefix to "benzene": chlorobenzene, nitrobenzene, ethylbenzene. Some names are retained: toluene (methylbenzene), phenol ($\mathrm{C_{6}H_{5}OH}$), aniline (benzenamine, $\mathrm{C_{6}H_{5}NH_{2}}$), benzaldehyde, benzoic acid, anisole (methoxybenzene).
- Two substituents: locants 1,2-, 1,3- or 1,4-, also called ortho (o-), meta (m-) and para (p-).
- Number the ring to give the lowest set of locants. When a retained name is the parent (toluene, phenol, benzoic acid…), the carbon carrying that group is C1; otherwise, if there is still a choice, the substituent first in alphabetical order gets C1.
- When the ring is itself a substituent, $\mathrm{C_{6}H_{5}{-}}$ is called phenyl.
✏️ Example 4 — Substituted benzenes
Give IUPAC names (and o/m/p names where they apply): (a) a benzene ring with Cl at two carbons separated by one ring carbon; (b) toluene with $\mathrm{NO_{2}}$ opposite the methyl group; (c) phenol with Cl on the neighbouring carbon; (d) benzoic acid with $\mathrm{NO_{2}}$ at the meta position; (e) phenol with $\mathrm{NO_{2}}$ at both ortho positions and the para position (picric acid); (f) a benzene ring with ethyl and methyl groups on neighbouring carbons.
💡 Show solution▶
(a), (b) 1,3-dichlorobenzene (m-dichlorobenzene). The methyl carbon of toluene is C1: 4-nitrotoluene (p-nitrotoluene), systematically 1-methyl-4-nitrobenzene.
(c), (d) OH carbon = C1: 2-chlorophenol (o-chlorophenol). COOH carbon = C1: 3-nitrobenzoic acid (m-nitrobenzoic acid).
(e) OH at C1, $\mathrm{NO_{2}}$ at 2, 4 and 6 → 2,4,6-trinitrophenol.
(f) Locants are $\{1,2\}$ either way; alphabetical order gives ethyl C1 → 1-ethyl-2-methylbenzene (also called 2-ethyltoluene).
Structural Isomerism
📘 Definition 8.6 — Isomers and structural isomerism
Isomers are different compounds with the same molecular formula. Structural isomers have their atoms joined in different orders. Four kinds are distinguished:
- Chain isomerism — different carbon skeletons: pentane, 2-methylbutane and 2,2-dimethylpropane ($\mathrm{C_{5}H_{12}}$).
- Position isomerism — the same skeleton, with a substituent, functional group or multiple bond at a different position: propan-1-ol and propan-2-ol; but-1-ene and but-2-ene.
- Functional isomerism — different functional groups: propanal and propanone ($\mathrm{C_{3}H_{6}O}$); ethanol and methoxymethane ($\mathrm{C_{2}H_{6}O}$).
- Metamerism — different alkyl groups on the two sides of the same functional group: ethoxyethane and 1-methoxypropane ($\mathrm{C_{4}H_{10}O}$); pentan-3-one and pentan-2-one.
Isomers are genuinely different substances. Butane boils at $-0.5\ ^{\circ}\mathrm{C}$ and 2-methylpropane at $-12\ ^{\circ}\mathrm{C}$: branching makes a molecule more compact, with less surface for intermolecular attraction. Ethanol (an alcohol, with hydrogen bonding) and methoxymethane (an ether, without) differ by more than $100\ ^{\circ}\mathrm{C}$ in boiling point.
🧭 Problem-solving strategy — Listing all the structural isomers of a formula
- Work out which classes the formula allows: $\mathrm{C}_{n}\mathrm{H}_{2n+2}\mathrm{O}$ → alcohols and ethers; open-chain $\mathrm{C}_{n}\mathrm{H}_{2n}\mathrm{O}$ → aldehydes and ketones; $\mathrm{C}_{n}\mathrm{H}_{2n}$ → alkenes (and also ring compounds, cycloalkanes).
- List the carbon skeletons systematically: the straight chain, then chains one carbon shorter with one branch, then two branches…
- On each skeleton, attach the group (or put the multiple bond) at every position that is not equivalent by symmetry to one already used.
- Name every structure: two drawings with the same IUPAC name are the same compound.
✏️ Example 5 — Recognising and counting structural isomers
(a) State the type of isomerism shown by each pair: (i) butan-1-ol and butan-2-ol; (ii) butanal and butan-2-one; (iii) 1-methoxypropane and ethoxyethane; (iv) pentane and 2,2-dimethylpropane; (v) but-1-ene and 2-methylpropene; (vi) ethanoic acid and methyl methanoate. (b) Write and name all the structural isomers of $\mathrm{C_{4}H_{9}Cl}$. (c) How many alcohols and how many ethers have the formula $\mathrm{C_{4}H_{10}O}$?
💡 Show solution▶
(a) (i) position (OH moved along the same chain); (ii) functional (aldehyde vs ketone); (iii) metamerism (methyl + propyl against ethyl + ethyl on the ether O); (iv) chain; (v) chain (straight vs branched skeleton); (vi) functional (carboxylic acid vs ester, both $\mathrm{C_{2}H_{4}O_{2}}$).
(b) skeletons Four carbons give two skeletons: butane and 2-methylpropane.
(b) positions Butane has two distinct positions (end and inner carbon): 1-chlorobutane, 2-chlorobutane. 2-Methylpropane also has two (a $\mathrm{CH_{3}}$ or the central CH): 1-chloro-2-methylpropane, 2-chloro-2-methylpropane. Total 4.
(c) Alcohols: replace Cl by OH in (b) → butan-1-ol, butan-2-ol, 2-methylpropan-1-ol, 2-methylpropan-2-ol: 4. Ethers: split the four carbons over the O as 1 + 3 or 2 + 2 → 1-methoxypropane, 2-methoxypropane, ethoxyethane: 3. Total 7 structural isomers.
Check $\mathrm{C_{4}H_{9}Cl}$, $\mathrm{C_{4}H_{9}OH}$ and $\mathrm{C_{4}H_{9}{-}}$ all have 4 isomers: there are exactly four butyl groups (butyl, sec-butyl, isobutyl, tert-butyl).
Stereoisomerism: Geometrical and Optical Isomers
📘 Definition 8.7 — Stereoisomerism
Stereoisomers have the same structural formula (the same atoms joined in the same order) but a different arrangement of their atoms in space.
- Geometrical (cis–trans) isomerism arises because the groups on a $\mathrm{C{=}C}$ bond cannot rotate (the π bond would break). If similar groups are on the same side of the double bond the isomer is cis; if on opposite sides, trans.
- Optical isomerism arises when a molecule is not superimposable on its mirror image — it is chiral, like a left and a right hand. The two mirror-image forms are enantiomers. They have identical melting points, boiling points and densities, but rotate the plane of plane-polarised light by equal amounts in opposite directions (dextrorotatory, (+); laevorotatory, (−)). An equimolar mixture of the two, a racemic mixture, is optically inactive.
📐 Rule 8.7 — When do stereoisomers exist?
- Geometrical isomers exist for $\mathrm{abC{=}Cxy}$ only if $\mathrm{a}\ne\mathrm{b}$ and $\mathrm{x}\ne\mathrm{y}$: each carbon of the double bond must carry two different groups. But-2-ene ($\mathrm{CH_{3}CH{=}CHCH_{3}}$) has cis and trans forms; but-1-ene ($\mathrm{CH_{2}{=}CHCH_{2}CH_{3}}$) and 2-methylpropene do not.
- A carbon atom bonded to four different groups is an asymmetric (chiral) carbon, often marked with *. A molecule with exactly one such carbon is always chiral and exists as a pair of enantiomers: butan-2-ol, $\mathrm{CH_{3}{-}\overset{*}{C}H(OH){-}CH_{2}CH_{3}}$ (groups H, OH, $\mathrm{CH_{3}}$, $\mathrm{C_{2}H_{5}}$).
✏️ Example 6 — Spotting stereoisomers
(a) Which of these show geometrical isomerism: but-1-ene, but-2-ene, 2-methylbut-2-ene, pent-2-ene, 1,2-dichloroethene, 1,1-dichloroethene? (b) Which of these contain a chiral carbon: propan-2-ol, butan-2-ol, 2-chlorobutane, 3-methylpentane, 2-methylbutan-1-ol, lactic acid $\mathrm{CH_{3}CH(OH)COOH}$?
💡 Show solution▶
(a) yes but-2-ene ($\mathrm{CH_{3}}$, H on each carbon), pent-2-ene ($\mathrm{CH_{3}}$/H and $\mathrm{C_{2}H_{5}}$/H), 1,2-dichloroethene ($\mathrm{ClCH{=}CHCl}$: Cl/H on each carbon).
(a) no but-1-ene ($\mathrm{CH_{2}{=}}$: two H on C1), 2-methylbut-2-ene ($\mathrm{(CH_{3})_{2}C{=}}$: two $\mathrm{CH_{3}}$ on C2), 1,1-dichloroethene ($\mathrm{Cl_{2}C{=}CH_{2}}$).
(b) yes butan-2-ol (C2: H, OH, $\mathrm{CH_{3}}$, $\mathrm{C_{2}H_{5}}$); 2-chlorobutane (C2: H, Cl, $\mathrm{CH_{3}}$, $\mathrm{C_{2}H_{5}}$); 2-methylbutan-1-ol (C2: H, $\mathrm{CH_{3}}$, $\mathrm{C_{2}H_{5}}$, $\mathrm{CH_{2}OH}$); lactic acid (C2: H, OH, $\mathrm{CH_{3}}$, $\mathrm{COOH}$).
(b) no propan-2-ol (C2 carries two $\mathrm{CH_{3}}$); 3-methylpentane (C3 carries two $\mathrm{C_{2}H_{5}}$).
🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — counting stereoisomers
Not required by the CBSE syllabus. JEE Main and NEET list stereoisomerism; JEE Advanced lists enantiomers, diastereomers and meso forms of compounds with up to two asymmetric centres (without R/S and E/Z labels).
A molecule with $n$ chiral carbons has at most $2^{n}$ stereoisomers; with $n$ double bonds that can show cis–trans isomerism, a factor 2 per double bond. Symmetry can lower the count: 2,3-dichlorobutane has two chiral carbons but only 3 stereoisomers, a pair of enantiomers and one meso form that is its own mirror image (it has an internal mirror plane). Hepta-2,5-diene, $\mathrm{CH_{3}CH{=}CHCH_{2}CH{=}CHCH_{3}}$, has two such double bonds but only 3 geometrical isomers (cis,cis; cis,trans; trans,trans), because the molecule reads the same from both ends.Exercises
I
Alkanes and haloalkanes.
- 1 Name $\mathrm{CH_{3}CH_{2}CH(CH_{3})CH_{2}CH_{3}}$.
- 2 Name $\mathrm{(CH_{3})_{2}CHCH(CH_{3})CH_{2}CH_{3}}$.
- 3 Name $\mathrm{CH_{3}C(CH_{3})_{2}CH_{2}CH_{2}CH_{3}}$.
- 4 Name $\mathrm{CH_{3}CH_{2}C(CH_{3})(C_{2}H_{5})CH_{2}CH_{3}}$.
- 5 Name $\mathrm{CH_{2}ClCH_{2}CH(CH_{3})CH_{3}}$.
- 6 Name $\mathrm{CH_{3}CHBrCHBrCH_{2}CH_{3}}$.
- 7 Each name is wrong. Draw the structure it describes and give the correct name: (i) 2-ethylbutane; (ii) 4-methylpentane; (iii) 1,1-dimethylethane.
- 8 Write the condensed and molecular formulae of 4-ethyl-2,2-dimethylhexane.
💡 Show solutions▶
1 3-methylpentane.
2 Five-carbon chain, methyls at $\{2,3\}$ (from the left) against $\{3,4\}$: 2,3-dimethylpentane.
3 2,2-dimethylpentane.
4 Longest chain 5 (every path through an ethyl group also gives 5); C3 carries ethyl and methyl: 3-ethyl-3-methylpentane.
5 Four-carbon chain; from the Cl end $\{1,3\}$, from the other end $\{2,4\}$: 1-chloro-3-methylbutane.
6 2,3-dibromopentane.
7 (i) $\mathrm{CH_{3}CH(C_{2}H_{5})CH_{2}CH_{3}}$: longest chain 5 → 3-methylpentane. (ii) Numbered from the wrong end → 2-methylpentane. (iii) $\mathrm{(CH_{3})_{2}CHCH_{3}}$: longest chain 3 → 2-methylpropane.
8 $\mathrm{(CH_{3})_{3}CCH_{2}CH(C_{2}H_{5})CH_{2}CH_{3}}$; $\mathrm{C_{10}H_{22}}$ (6 + 2 + 2 carbons).
II
Functional groups, multiple bonds and benzene rings.
- 1 Name $\mathrm{CH_{3}CH_{2}CH_{2}COOH}$.
- 2 Name $\mathrm{CH_{3}CH(CH_{3})CH_{2}OH}$.
- 3 Name $\mathrm{CH_{3}CH_{2}COCH_{2}CH_{2}CH_{3}}$.
- 4 Name $\mathrm{CH_{3}CH{=}CHCH_{2}CH_{3}}$ and $\mathrm{CH_{2}{=}C(CH_{3})CH_{2}CH_{3}}$.
- 5 Name $\mathrm{HC{\equiv}CCH(CH_{3})_{2}}$.
- 6 Name $\mathrm{HOCH_{2}CH_{2}CH_{2}CHO}$, and say why the OH group is a prefix here.
- 7 Name $\mathrm{CH_{2}{=}CHCOOH}$ and $\mathrm{NCCH_{2}CH_{2}CN}$.
- 8 Name (i) a benzene ring carrying Cl and $\mathrm{CH_{3}}$ para to each other; (ii) $\mathrm{C_{6}H_{5}CH_{2}OH}$.
💡 Show solutions▶
1 Butanoic acid.
2 2-methylpropan-1-ol (OH carbon = C1 of a 3-carbon chain).
3 Hexan-3-one (from the left the ketone is at 3, from the right at 4).
4 Pent-2-ene; 2-methylbut-1-ene.
5 3-methylbut-1-yne.
6 4-hydroxybutanal: the aldehyde outranks the alcohol, so $\mathrm{CHO}$ is the suffix (C1) and OH becomes "hydroxy".
7 Prop-2-enoic acid (acrylic acid; the COOH carbon is C1, the double bond starts at C2); butanedinitrile (both nitrile carbons are chain ends: "-dinitrile", e kept before the consonant).
8 (i) 1-chloro-4-methylbenzene (4-chlorotoluene, p-chlorotoluene); (ii) phenylmethanol (benzyl alcohol): the chain is one carbon carrying OH and a phenyl group.
III
Structural isomers.
- 1 Define chain isomerism and position isomerism, with one example of each.
- 2 What type of isomers are propan-1-ol and methoxyethane?
- 3 What type of isomers are 2-methylbutane and 2,2-dimethylpropane?
- 4 What type of isomers are pentan-2-one and pentan-3-one?
- 5 What type of isomers are 1-chloropropane and 2-chloropropane?
- 6 Name all the structural isomers of $\mathrm{C_{5}H_{12}}$. Which has the lowest boiling point, and why?
- 7 Name the two open-chain carbonyl compounds $\mathrm{C_{3}H_{6}O}$. What kind of isomers are they?
- 8 How many structural isomers does $\mathrm{C_{5}H_{11}Cl}$ have? How many of them contain a chiral carbon? (Use Interactive 2 to check.)
💡 Show solutions▶
1 Chain: different carbon skeletons (butane, 2-methylpropane). Position: same skeleton, group in a different place (butan-1-ol, butan-2-ol).
2 Functional isomers ($\mathrm{C_{3}H_{8}O}$: alcohol and ether).
3 Chain isomers ($\mathrm{C_{5}H_{12}}$).
4 Metamers (methyl + propyl against ethyl + ethyl on the $\mathrm{C{=}O}$); they can also be called position isomers, since the carbonyl group moves along the same five-carbon chain.
5 Position isomers.
6 Pentane, 2-methylbutane, 2,2-dimethylpropane. 2,2-Dimethylpropane (about $10\ ^{\circ}\mathrm{C}$, against $28\ ^{\circ}\mathrm{C}$ and $36\ ^{\circ}\mathrm{C}$): the most compact, almost spherical molecule has the smallest surface for intermolecular attraction.
7 Propanal ($\mathrm{CH_{3}CH_{2}CHO}$) and propanone ($\mathrm{CH_{3}COCH_{3}}$): functional isomers.
8 8 isomers: 1-, 2- and 3-chloropentane; 1-chloro-2-methylbutane, 2-chloro-2-methylbutane, 2-chloro-3-methylbutane, 1-chloro-3-methylbutane; 1-chloro-2,2-dimethylpropane. Chiral carbon in 3 of them: 2-chloropentane, 1-chloro-2-methylbutane, 2-chloro-3-methylbutane.
IV
Stereoisomers.
- 1 Explain why but-2-ene shows geometrical isomerism but but-1-ene does not.
- 2 Does 2-methylbut-2-ene, $\mathrm{CH_{3}CH{=}C(CH_{3})_{2}}$, show geometrical isomerism?
- 3 Which of $\mathrm{CHCl{=}CHBr}$ and $\mathrm{CH_{2}{=}CBr_{2}}$ has cis and trans forms?
- 4 Mark the chiral carbon in alanine, $\mathrm{CH_{3}CH(NH_{2})COOH}$, and list its four groups.
- 5 Is 2-methylbutane chiral? Explain.
- 6 Find the alkanes with the fewest carbon atoms that contain a chiral carbon.
- 7 Applied. The two enantiomers of carvone smell of spearmint and of caraway. Could they be separated by distillation? What property of light distinguishes them?
- 8 How many open-chain isomers, including geometrical isomers, have the formula $\mathrm{C_{4}H_{8}}$? Name them.
💡 Show solutions▶
1 Rotation about $\mathrm{C{=}C}$ is blocked. In but-2-ene each double-bond carbon carries two different groups ($\mathrm{CH_{3}}$, H), so two arrangements exist. In but-1-ene C1 carries two H atoms: swapping them gives the same molecule.
2 No: C2 carries two identical $\mathrm{CH_{3}}$ groups.
3 $\mathrm{CHCl{=}CHBr}$ (H/Cl and H/Br); $\mathrm{CH_{2}{=}CBr_{2}}$ has two identical groups on each carbon.
4 C2: H, $\mathrm{NH_{2}}$, $\mathrm{CH_{3}}$, $\mathrm{COOH}$ — four different groups.
5 No. C2 carries H, two $\mathrm{CH_{3}}$ groups and $\mathrm{C_{2}H_{5}}$; no carbon has four different groups, and the molecule is superimposable on its mirror image.
6 Seven carbons: 3-methylhexane (C3: H, $\mathrm{CH_{3}}$, $\mathrm{C_{2}H_{5}}$, $\mathrm{C_{3}H_{7}}$) and 2,3-dimethylpentane (C3: H, $\mathrm{CH_{3}}$, $\mathrm{C_{2}H_{5}}$, $\mathrm{CH(CH_{3})_{2}}$). No alkane with 6 or fewer carbons has one.
7 No: enantiomers have identical boiling points. They rotate plane-polarised light by equal angles in opposite directions.
8 Four: but-1-ene, cis-but-2-ene, trans-but-2-ene, 2-methylpropene (three structural isomers, one of which has two geometrical forms).
🎯 Exam corner
E1. NEET The IUPAC name of $\mathrm{(CH_{3})_{2}CHCH_{2}CH(CH_{3})CH_{2}CH_{3}}$ is (A) 2,4-dimethylhexane (B) 3,5-dimethylhexane (C) 2-ethyl-4-methylpentane (D) 1,1,3-trimethylpentane
💡 Show solution▶
Idea Longest chain: 6 carbons. Methyls at $\{2,4\}$ from the left, $\{3,5\}$ from the right; (C) and (D) do not use the longest chain.
Answer (A).
E2. JEE Main The correct IUPAC name of $\mathrm{CH_{2}{=}CH{-}CH_{2}{-}C{\equiv}CH}$ is (A) pent-4-en-1-yne (B) pent-1-en-4-yne (C) pent-1-yn-4-ene (D) pent-4-yn-1-ene
💡 Show solution▶
Locants From the left: ene 1, yne 4 → $\{1,4\}$. From the right: yne 1, ene 4 → $\{1,4\}$. A tie, so the double bond gets the lower number. "-ene" is always cited before "-yne".
Answer (B).
E3. JEE Main Numerical answer. How many of the structurally isomeric alcohols $\mathrm{C_{5}H_{11}OH}$ contain a chiral carbon atom?
💡 Show solution▶
List The 8 alcohols: pentan-1-ol, pentan-2-ol, pentan-3-ol, 2-methylbutan-1-ol, 2-methylbutan-2-ol, 3-methylbutan-2-ol, 3-methylbutan-1-ol, 2,2-dimethylpropan-1-ol.
Test Four different groups on one carbon: pentan-2-ol (C2: H, OH, $\mathrm{CH_{3}}$, $\mathrm{C_{3}H_{7}}$), 2-methylbutan-1-ol (C2: H, $\mathrm{CH_{3}}$, $\mathrm{C_{2}H_{5}}$, $\mathrm{CH_{2}OH}$), 3-methylbutan-2-ol (C2: H, OH, $\mathrm{CH_{3}}$, $\mathrm{CH(CH_{3})_{2}}$). Pentan-3-ol has two ethyl groups on C3; the others have no carbon with four different groups.
Answer 3.
E4. Board · 3 marks Write the IUPAC names of (i) $\mathrm{CH_{3}COCH_{2}CH_{2}CH_{3}}$; (ii) $\mathrm{(CH_{3})_{3}COH}$; (iii) a phenol with $\mathrm{NO_{2}}$ on the carbon next to the OH group.
💡 Show solution▶
(i) Pentan-2-one (1 mark).
(ii) 2-methylpropan-2-ol (1 mark).
(iii) 2-nitrophenol (o-nitrophenol) (1 mark).
E5. NEET Which pair of compounds are metamers? (A) propan-1-ol and propan-2-ol (B) ethoxyethane and 1-methoxypropane (C) butanal and butan-2-one (D) butane and 2-methylpropane
💡 Show solution▶
Idea Metamers have the same functional group with different alkyl groups on its two sides. (A) position, (C) functional, (D) chain isomers.
Answer (B): $\mathrm{C_{2}H_{5}{-}O{-}C_{2}H_{5}}$ and $\mathrm{CH_{3}{-}O{-}C_{3}H_{7}}$, both $\mathrm{C_{4}H_{10}O}$.
E6. JEE Advanced One or more options correct. Which of these compounds are optically active (chiral)? (A) 2-chlorobutane (B) 3-chloropentane (C) 1-chloro-2-methylbutane (D) 2-chloro-2-methylbutane
💡 Show solution▶
(A), (B) 2-chlorobutane, C2: H, Cl, $\mathrm{CH_{3}}$, $\mathrm{C_{2}H_{5}}$ → chiral. 3-chloropentane, C3: H, Cl, $\mathrm{C_{2}H_{5}}$, $\mathrm{C_{2}H_{5}}$ → not chiral.
(C), (D) 1-chloro-2-methylbutane, C2: H, $\mathrm{CH_{3}}$, $\mathrm{C_{2}H_{5}}$, $\mathrm{CH_{2}Cl}$ → chiral. 2-chloro-2-methylbutane, C2: Cl, $\mathrm{CH_{3}}$, $\mathrm{CH_{3}}$, $\mathrm{C_{2}H_{5}}$ → not chiral.
Answer (A), (C).
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes
Research question. Shops, factories and customs offices still use common and trade names: isopropyl alcohol, acetone, MTBE, iso-octane, ethylene glycol. How reliably can you turn such a name into a structure and an IUPAC name — and how do chemical databases make sure that one structure has exactly one identity?
Why it matters. Ambiguous names cause real harm: medication errors from look-alike drug names are tracked by the World Health Organization, and safety data sheets, poison-centre records and international trade rules (the Harmonized System used by customs worldwide) all depend on identifying a chemical unambiguously. Databases such as PubChem assign every structure a machine-readable identifier (the InChI) generated from its connectivity — the same idea as Rule 8.3: one structure, one name.
What to do.
- Find the condensed formulae of isopropyl alcohol, acetone, MTBE (methyl tert-butyl ether, a petrol additive), iso-octane and ethylene glycol.
- Give each its IUPAC name using Rules 8.3–8.5, and check your names in Interactive 1 where possible.
- Look each compound up in PubChem and compare your name with the "IUPAC name" field. Note any differences in style (for example, locants that are left out).
- For one compound, list two structural isomers and check that their database entries really are different substances (boiling points).
Tools: Rules 8.3–8.6, Interactives 1 and 2, PubChem (pubchem.ncbi.nlm.nih.gov).
Hand in: A table (common name, condensed formula, your IUPAC name, database name, molecular formula) and a short note on one isomer pair.
Going further. Read how the InChI string is built from a structure. Why can it, unlike a common name, be generated by a computer and searched on the web?
Instructor note: propan-2-ol, $\mathrm{(CH_{3})_{2}CHOH}$; propanone (propan-2-one), $\mathrm{CH_{3}COCH_{3}}$; MTBE = 2-methoxy-2-methylpropane, $\mathrm{CH_{3}OC(CH_{3})_{3}}$; iso-octane = 2,2,4-trimethylpentane; ethylene glycol = ethane-1,2-diol, $\mathrm{HOCH_{2}CH_{2}OH}$. Databases often write "propan-2-one" and "2-methoxy-2-methylpropane". Example isomer pair: propan-2-ol ($82\ ^{\circ}\mathrm{C}$) and propan-1-ol ($97\ ^{\circ}\mathrm{C}$).
🧪 Self-check quiz
0/5 correct
Q1. The IUPAC name of $\mathrm{(CH_{3})_{3}CH}$ is
Q2. Which group has the highest priority when choosing the suffix?
Q3. Propanal and propanone are
Q4. Which compound shows geometrical (cis–trans) isomerism?
Q5. Which molecule contains a chiral carbon?