IUPAC Nomenclature — Compounds Containing a Functional Group

One group is the suffix — the rest are prefixes

An alkane name has only a root and prefixes. The moment a functional group appears, a third part joins in: the suffix. A molecule may carry several functional groups, but an IUPAC name allows only one of them to be expressed as the suffix. Every other group drops down to a prefix and is cited in front of the root along with the alkyl and halo substituents.

Key Point (Definition): The principal characteristic group is the senior-most functional group present in the molecule. It alone is named as the suffix, it alone controls the choice of parent chain, and it alone gets first claim on the lowest locant. All remaining groups become prefixes.

The seniority order is fixed. Learn it top to bottom in this exact sequence:

$$\text{COOH} > \text{SO}_3\text{H} > \text{ester} > \text{acid halide} > \text{amide} > \text{nitrile} > \text{aldehyde} > \text{ketone} > \text{alcohol} > \text{amine} > \text{alkene / alkyne} > \text{alkane}$$

In words: carboxylic acid > sulphonic acid > ester > acid halide > amide > nitrile > aldehyde > ketone > alcohol > amine > alkene / alkyne > alkane. Acids first, then their derivatives, then the carbonyls, then the $\mathrm{-OH}$ and $\mathrm{-NH_2}$ pair, and multiple bonds last.

The groups that can never be a suffix

Some groups have no suffix form at all. However many of them a molecule carries, and however important they look, they are always cited as prefixes:

• halo — fluoro, chloro, bromo, iodo
• nitro ($\mathrm{-NO_2}$) and nitroso ($\mathrm{-NO}$)
• alkoxy ($\mathrm{-OR}$), which is how every ether is named
• alkyl (methyl, ethyl, isopropyl and the rest)
• phenyl ($\mathrm{-C_6H_5}$)

A molecule carrying only a chlorine therefore has no suffix beyond the parent alkane ending: $\mathrm{CH_3CH_2Cl}$ is chloroethane, still ending in -ane.

[JEE Main] A common one-mark trap is an option such as "propan-1-nitro" or "ethanochloride". Neither can exist.

Suffix and prefix forms, class by class

Class	Group	Suffix	Prefix when not senior
carboxylic acid	$\mathrm{-COOH}$	-oic acid	carboxy
ester	$\mathrm{-COOR}$	-yl …oate	alkoxycarbonyl
acid chloride	$\mathrm{-COCl}$	-oyl chloride	halocarbonyl
amide	$\mathrm{-CONH_2}$	-amide	carbamoyl
nitrile	$\mathrm{-CN}$	-nitrile	cyano
aldehyde	$\mathrm{-CHO}$	-al	oxo (or formyl)
ketone	$\mathrm{>C=O}$	-one	oxo
alcohol	$\mathrm{-OH}$	-ol	hydroxy
amine	$\mathrm{-NH_2}$	-amine	amino
alkene	$\mathrm{C=C}$	-ene	—
alkyne	$\mathrm{C \equiv C}$	-yne	—
ether	$\mathrm{-OR}$	none	alkoxy
haloalkane	$\mathrm{-X}$	none	halo
nitro compound	$\mathrm{-NO_2}$	none	nitro

Two rows repay a second look. A ketone and an aldehyde fall back to the same prefix, oxo, because both are a doubly bonded oxygen on a chain carbon. And an alcohol becomes hydroxy, never "hydroxyl-".

Building the name — four mechanical steps

The rules you used for alkanes still apply. Three additions handle the functional group.

Step 1 — the parent chain must contain the principal group. This overrides "longest chain". If a longer chain exists that misses the group, you may not use it. In

$$\mathrm{CH_3-CH_2-CH_2-CH(OH)-CH_3}$$

the longest chain is five carbons and it does carry the $\mathrm{-OH}$, so the parent is pentane. But if a six-carbon chain avoided the $\mathrm{-OH}$ while a four-carbon chain passed through it, the four-carbon chain would win. Length is only the tie-breaker among chains that already contain the group.

Step 2 — number for the lowest locant on the principal group. The priority ladder for numbering is:

$$\text{principal functional group} \; > \; \text{double or triple bond} \; > \; \text{substituent prefixes}$$

The functional group takes the lowest number even if that forces high numbers on everything else. Only when it is equally placed from both ends do you move down to the multiple bond, and only after that to the prefixes.

Step 3 — write the locants in the modern position. The number sits immediately before the piece of the name it belongs to: pent-2-ene, pentan-2-ol, pentan-2-one, but-3-en-2-ol. Not "2-pentanol".

Step 4 — the terminal "-e" rule. This is the one that decides more marks than any other single convention.

Key Point: Drop the terminal -e of the parent hydride when the suffix that follows begins with a vowel. Keep it when the suffix begins with a consonant. A multiplying prefix (di, tri, tetra) placed before the suffix begins with a consonant, so it brings the -e back.

Work through the four cases:

Parent	Suffix	Begins with	Result
propane	-ol	vowel o	propan-1-ol (not propane-1-ol)
propane	-amine	vowel a	propan-1-amine
propane	-nitrile	consonant n	propanenitrile
propane	-diol	consonant d	propane-1,2-diol

The -e vanishes in propan-1-ol and returns in propane-1,2-diol, and it is the letter that starts the suffix, nothing else, that decides. The same rule gives propanal (drop, a is a vowel), propanone (drop, o), propanamide (drop, a), propanoic acid (drop, then -oic), butanedioic acid (keep, d) and butanedinitrile (keep, d).

When may a locant be left out? Only when the position is not in doubt. Methanol, ethanol, propanone and ethanal need no number because there is only one place the group can sit. As soon as a second position becomes possible the number is compulsory: propan-1-ol and propan-2-ol are different compounds, and so are pentan-2-one and pentan-3-one.

Alcohols — the suffix -ol

Replace the terminal -e of the alkane by -ol, and give the carbon bearing the $\mathrm{-OH}$ the lowest possible number.

Compound	Formula	IUPAC name	Common name
1 carbon	$\mathrm{CH_3OH}$	methanol	methyl alcohol
2 carbons	$\mathrm{CH_3CH_2OH}$	ethanol	ethyl alcohol
3 carbons, end	$\mathrm{CH_3CH_2CH_2OH}$	propan-1-ol	n-propyl alcohol
3 carbons, middle	$\mathrm{(CH_3)_2CHOH}$	propan-2-ol	isopropyl alcohol
4 carbons, branched	$\mathrm{(CH_3)_3COH}$	2-methylpropan-2-ol	tert-butyl alcohol

Take $\mathrm{(CH_3)_3COH}$ carefully. The longest chain through the $\mathrm{-OH}$ carbon is three carbons, propane, with the $\mathrm{-OH}$ on C-2 and the third methyl hanging off C-2 as a substituent: 2-methylpropan-2-ol. Counting four carbons is the standard slip — the molecule does contain four, but they do not lie in one chain.

Two and three hydroxyl groups

When the same group appears more than once, use di, tri and so on, cite every locant, and remember that the multiplying prefix restores the -e.

Formula	IUPAC name	Common name
$\mathrm{HOCH_2-CH_2OH}$	ethane-1,2-diol	ethylene glycol
$\mathrm{CH_3-CH(OH)-CH_2OH}$	propane-1,2-diol	propylene glycol
$\mathrm{HOCH_2-CH_2-CH_2OH}$	propane-1,3-diol	trimethylene glycol
$\mathrm{HOCH_2-CH(OH)-CH_2OH}$	propane-1,2,3-triol	glycerol
$\mathrm{CH_3-CH(OH)-CH(OH)-CH_3}$	butane-2,3-diol	—

Both hydroxyls have to be numbered; "propanediol" alone does not say whether the second $\mathrm{-OH}$ is on C-2 or C-3.

Alcohols with a double bond

The alcohol is senior to the alkene, so the $\mathrm{-OH}$ claims the lower number and the double bond takes what is left.

• $\mathrm{CH_2=CH-CH_2OH}$: numbering from the $\mathrm{-OH}$ end puts oxygen on C-1 and the double bond between C-2 and C-3. Name: prop-2-en-1-ol.
• $\mathrm{CH_2=CH-CH(OH)-CH_3}$: the $\mathrm{-OH}$ is one carbon from the methyl end, so number from there. $\mathrm{-OH}$ on C-2, double bond from C-3. Name: but-3-en-2-ol.

Watch the ending: -ol starts with a vowel, so the -e of "ene" is dropped too — prop-2-en-1-ol, never "prop-2-ene-1-ol".

Aldehydes — the suffix -al

Replace the terminal -e of the alkane by -al. The $\mathrm{-CHO}$ carbon can only ever be at the end of a chain, because that carbon already carries a hydrogen and a doubly bonded oxygen and has just one bond left for carbon.

Key Point: In a chain aldehyde the $\mathrm{-CHO}$ carbon is always C-1, so no locant is written for the -al. Write butanal, never "butan-1-al".

Formula	IUPAC name	Common name
$\mathrm{HCHO}$	methanal	formaldehyde
$\mathrm{CH_3CHO}$	ethanal	acetaldehyde
$\mathrm{CH_3CH_2CHO}$	propanal	propionaldehyde
$\mathrm{CH_3CH_2CH_2CHO}$	butanal	butyraldehyde
$\mathrm{(CH_3)_2CHCHO}$	2-methylpropanal	isobutyraldehyde

Formaldehyde, acetaldehyde and the rest of that column are common names only — worth knowing because reagent bottles use them, but never the answer when the question asks for the IUPAC name.

Since C-1 is fixed, every substituent number in an aldehyde is measured from the $\mathrm{-CHO}$ carbon:

• $\mathrm{CH_3CH_2CH(CH_3)CHO}$: C-1 is the CHO, C-2 bears the methyl, so 2-methylbutanal.
• $\mathrm{CH_3CH(OH)CH_2CHO}$: C-1 is the CHO, C-3 bears the $\mathrm{-OH}$, so 3-hydroxybutanal.

When the CHO sits on a ring: -carbaldehyde

A ring carbon cannot become C-1 of a chain, so the $\mathrm{-CHO}$ is treated as a group attached to the ring and the suffix -carbaldehyde is used. The carbon of the $\mathrm{-CHO}$ is not counted in the ring name.

Formula	IUPAC name	Common name
cyclohexane ring with $\mathrm{-CHO}$	cyclohexanecarbaldehyde	—
cyclopentane ring with $\mathrm{-CHO}$	cyclopentanecarbaldehyde	—
$\mathrm{C_6H_5CHO}$	benzenecarbaldehyde	benzaldehyde

Two aldehyde groups on a chain use -dial, with the -e kept: $\mathrm{OHC-CH_2-CHO}$ is propanedial.

Ketones — the suffix -one

Replace the terminal -e by -one. Unlike the aldehyde carbon, a ketone carbonyl sits inside the chain, so its position genuinely varies and a locant is required wherever more than one position is possible.

Formula	IUPAC name	Common name
$\mathrm{CH_3COCH_3}$	propanone	acetone
$\mathrm{CH_3COCH_2CH_3}$	butan-2-one	ethyl methyl ketone
$\mathrm{CH_3COCH_2CH_2CH_3}$	pentan-2-one	methyl propyl ketone
$\mathrm{CH_3CH_2COCH_2CH_3}$	pentan-3-one	diethyl ketone
cyclohexane ring with $\mathrm{C=O}$	cyclohexanone	—

Propanone needs no number. A three-carbon ketone can only have its carbonyl on the middle carbon; putting it on C-1 would make an aldehyde instead. Pentan-2-one does need one, because pentan-3-one is a real and different compound.

For $\mathrm{CH_3-CO-CH_2-CH(CH_3)-CH_3}$ the chain through the carbonyl is five carbons. Numbering from the left puts the $\mathrm{C=O}$ on C-2; from the right it would land on C-4. Lowest wins, so the methyl branch falls on C-4 and the name is 4-methylpentan-2-one.

Carboxylic acids — the suffix -oic acid

Drop the terminal -e and add -oic acid. Like the aldehyde, the $\mathrm{-COOH}$ carbon is necessarily terminal and is always C-1, so it carries no locant.

Formula	IUPAC name	Common name
$\mathrm{HCOOH}$	methanoic acid	formic acid
$\mathrm{CH_3COOH}$	ethanoic acid	acetic acid
$\mathrm{CH_3CH_2COOH}$	propanoic acid	propionic acid
$\mathrm{CH_3CH_2CH_2COOH}$	butanoic acid	butyric acid
$\mathrm{(CH_3)_2CHCOOH}$	2-methylpropanoic acid	isobutyric acid

The carbon of $\mathrm{-COOH}$ is counted in the chain: ethanoic acid has two carbons, the methyl and the acid carbon.

Two acid groups: -dioic acid

Both $\mathrm{-COOH}$ carbons are counted, they occupy the two ends of the chain, and the -e is kept because d is a consonant.

Formula	IUPAC name	Common name
$\mathrm{HOOC-COOH}$	ethanedioic acid	oxalic acid
$\mathrm{HOOC-CH_2-COOH}$	propanedioic acid	malonic acid
$\mathrm{HOOC-CH_2CH_2-COOH}$	butanedioic acid	succinic acid

Oxalic acid is ethanedioic acid, not "methanedioic acid". Two $\mathrm{-COOH}$ groups joined directly to each other give a chain of two carbons.

Ring acids use -carboxylic acid, for the reason aldehydes use -carbaldehyde: $\mathrm{C_6H_5COOH}$ is benzenecarboxylic acid, retained as benzoic acid.

Amines — the suffix -amine

Drop the terminal -e and add -amine. The parent chain is the longest carbon chain attached to the nitrogen.

Amines are classified by how many carbons the nitrogen carries: primary $\mathrm{R-NH_2}$ (one carbon on N), secondary $\mathrm{R_2NH}$ (two), tertiary $\mathrm{R_3N}$ (three).

This counts carbons on the nitrogen, not on the carbon bearing the nitrogen — the opposite of how alcohols and haloalkanes are classified, and mixing the two up is a routine error.

Formula	IUPAC name	Class
$\mathrm{CH_3NH_2}$	methanamine	primary
$\mathrm{CH_3CH_2NH_2}$	ethanamine	primary
$\mathrm{CH_3CH_2CH_2NH_2}$	propan-1-amine	primary
$\mathrm{(CH_3)_2CHNH_2}$	propan-2-amine	primary
$\mathrm{H_2N-CH_2CH_2CH_2CH_2-NH_2}$	butane-1,4-diamine	primary (two groups)

Substituents on the nitrogen: the italic N

For a secondary or tertiary amine, pick the longest chain on nitrogen as the parent. Every other group on the nitrogen is cited as a prefix, and its "locant" is the italic letter N instead of a number, because it sits on nitrogen and not on a numbered carbon.

Formula	IUPAC name	Class
$\mathrm{CH_3-NH-CH_3}$	N-methylmethanamine	secondary
$\mathrm{CH_3-NH-CH_2CH_3}$	N-methylethanamine	secondary
$\mathrm{CH_3CH_2-NH-CH_2CH_3}$	N-ethylethanamine	secondary
$\mathrm{(CH_3)_3N}$	N,N-dimethylmethanamine	tertiary
$\mathrm{CH_3CH_2-N(CH_3)_2}$	N,N-dimethylethanamine	tertiary

Take $\mathrm{CH_3-NH-CH_2CH_3}$. Nitrogen carries a methyl and an ethyl; ethyl is longer, so the parent is ethanamine and the methyl is cited as N-methyl, giving N-methylethanamine. Reversing the choice and writing "N-ethylmethanamine" is wrong — the longer chain must be the parent.

Two identical groups on nitrogen need N repeated once for each: N,N-dimethyl, not "N-dimethyl".

$\mathrm{C_6H_5NH_2}$ is benzenamine, universally retained as aniline.

Ethers — always an alkoxy prefix

An ether has no suffix of its own. It is named as a hydrocarbon carrying an alkoxy group, exactly like a haloalkane is named as a hydrocarbon carrying a halogen.

Key Point: In $\mathrm{R-O-R'}$, the larger alkyl group becomes the parent chain and the smaller one becomes the $\mathrm{-OR}$ prefix: methoxy ($\mathrm{-OCH_3}$), ethoxy ($\mathrm{-OC_2H_5}$), propoxy ($\mathrm{-OC_3H_7}$).

Formula	IUPAC name	Common name
$\mathrm{CH_3-O-CH_3}$	methoxymethane	dimethyl ether
$\mathrm{CH_3-O-CH_2CH_3}$	methoxyethane	ethyl methyl ether
$\mathrm{CH_3CH_2-O-CH_2CH_3}$	ethoxyethane	diethyl ether
$\mathrm{CH_3-O-CH_2CH_2CH_3}$	1-methoxypropane	methyl n-propyl ether
$\mathrm{CH_3-O-CH(CH_3)_2}$	2-methoxypropane	isopropyl methyl ether
$\mathrm{C_6H_5-O-CH_3}$	methoxybenzene	anisole

The last pair is the one to fix in memory. Both $\mathrm{CH_3OCH_2CH_2CH_3}$ and $\mathrm{CH_3OCH(CH_3)_2}$ have the formula $\mathrm{C_4H_{10}O}$ and both are "methoxypropane", separated only by the locant: 1-methoxypropane has the oxygen on an end carbon, 2-methoxypropane on the middle one. Methoxymethane and methoxyethane carry no locant because their parent chains offer only one position.

Haloalkanes — halo prefix, lowest locant, alphabetical order

Fluoro, chloro, bromo and iodo are prefixes on the alkane root. Nothing else changes: the ending is still -ane.

Formula	IUPAC name	Common name
$\mathrm{CH_3Cl}$	chloromethane	methyl chloride
$\mathrm{CH_2Cl_2}$	dichloromethane	methylene chloride
$\mathrm{CHCl_3}$	trichloromethane	chloroform
$\mathrm{(CH_3)_2CHCl}$	2-chloropropane	isopropyl chloride
$\mathrm{(CH_3)_3CBr}$	2-bromo-2-methylpropane	tert-butyl bromide

Two different halogens are cited in alphabetical order, which runs bromo, chloro, fluoro, iodo — not the order of the group in the periodic table.

$\mathrm{BrCH_2-CH_2Cl}$: the chain is ethane, and numbering either way gives the locant set $\{1,2\}$, so the sets tie. The tie is broken by giving the lower number to the substituent cited first alphabetically, which is bromo. Name: 1-bromo-2-chloroethane.

$\mathrm{CH_3-CHBr-CH_2Cl}$: numbering from the $\mathrm{CH_2Cl}$ end gives $\{1,2\}$; from the methyl end it gives $\{2,3\}$. The sets do not tie, so first point of difference settles it before alphabetical order is consulted. Name: 2-bromo-1-chloropropane.

[JEE/NEET] Alphabetical order decides only the order of citation and, when locant sets tie, which substituent gets the lower number. It never overrides a genuinely lower locant set.

Nitro compounds — the nitro prefix

$\mathrm{-NO_2}$ has no suffix form at all. It is always nitro-.

Formula	IUPAC name
$\mathrm{CH_3NO_2}$	nitromethane
$\mathrm{CH_3CH_2NO_2}$	nitroethane
$\mathrm{CH_3CH_2CH_2NO_2}$	1-nitropropane
$\mathrm{(CH_3)_2CHNO_2}$	2-nitropropane
$\mathrm{(CH_3)_2CH-CH_2NO_2}$	2-methyl-1-nitropropane
$\mathrm{C_6H_5NO_2}$	nitrobenzene

In 2-methyl-1-nitropropane the chain is three carbons, the nitro sits on C-1 and the methyl on C-2, and m precedes n, so methyl is cited first.

Nitriles — the suffix -nitrile

The carbon of the $\mathrm{-C \equiv N}$ group is counted as part of the chain, and it is C-1. That single fact produces most of the mistakes in this class.

Key Point: $\mathrm{CH_3CN}$ has two carbons in its chain, so it is ethanenitrile, not "methanenitrile". The nitrile carbon is C-1 and the methyl is C-2.

Because -nitrile begins with the consonant n, the terminal -e is kept: ethane + nitrile = ethanenitrile.

Formula	IUPAC name	Common name
$\mathrm{CH_3CN}$	ethanenitrile	methyl cyanide, acetonitrile
$\mathrm{CH_3CH_2CN}$	propanenitrile	ethyl cyanide
$\mathrm{CH_3CH_2CH_2CN}$	butanenitrile	n-propyl cyanide
$\mathrm{(CH_3)_2CH-CH_2CN}$	3-methylbutanenitrile	isobutyl cyanide
$\mathrm{NC-CH_2CH_2-CN}$	butanedinitrile	succinonitrile

Check 3-methylbutanenitrile atom by atom: C-1 is the nitrile carbon, C-2 is the $\mathrm{CH_2}$, C-3 is the $\mathrm{CH}$ carrying the methyl branch, C-4 is the terminal $\mathrm{CH_3}$. Four carbons in the chain, methyl on C-3.

On a ring the same problem arises as with $\mathrm{-CHO}$, and the same solution is used: the suffix -carbonitrile, with the nitrile carbon outside the ring count. A cyclohexane ring bearing $\mathrm{-CN}$ is cyclohexanecarbonitrile, and $\mathrm{C_6H_5CN}$ is benzenecarbonitrile, retained as benzonitrile. When a nitrile is not the senior group it becomes the prefix cyano.

Esters, acid chlorides and amides — in outline

These three are all derivatives of a carboxylic acid, and all three outrank the aldehyde and the ketone.

Esters, $\mathrm{R-COO-R'}$. The name comes in two words. The alkyl group attached to oxygen is named first as a separate word; the acid part loses "-ic acid" and takes -oate.

Formula	IUPAC name	Common name
$\mathrm{HCOOCH_3}$	methyl methanoate	methyl formate
$\mathrm{CH_3COOCH_3}$	methyl ethanoate	methyl acetate
$\mathrm{CH_3COOCH_2CH_3}$	ethyl ethanoate	ethyl acetate
$\mathrm{CH_3CH_2COOCH_3}$	methyl propanoate	methyl propionate

In $\mathrm{CH_3COOC_2H_5}$ the ethyl sits on oxygen, so ethyl is the first word, and the two-carbon acid part gives ethanoate.

Acid chlorides, $\mathrm{R-COCl}$. Replace "-oic acid" of the parent acid by -oyl chloride.

Formula	IUPAC name	Common name
$\mathrm{CH_3COCl}$	ethanoyl chloride	acetyl chloride
$\mathrm{CH_3CH_2COCl}$	propanoyl chloride	propionyl chloride
$\mathrm{C_6H_5COCl}$	benzenecarbonyl chloride	benzoyl chloride

Amides, $\mathrm{R-CONH_2}$. Replace the terminal -e by -amide; the a is a vowel so the -e goes.

Formula	IUPAC name	Common name
$\mathrm{HCONH_2}$	methanamide	formamide
$\mathrm{CH_3CONH_2}$	ethanamide	acetamide
$\mathrm{CH_3CH_2CONH_2}$	propanamide	propionamide
$\mathrm{CH_3CONHCH_3}$	N-methylethanamide	N-methylacetamide

Substituents on the amide nitrogen use the italic N as in amines.

Which group wins — a decision table

Every row below is a compound carrying two functional groups. Read across: the senior one becomes the suffix, the junior one becomes the prefix from the table in the first block.

Compound	Groups present	Senior group	Junior becomes	IUPAC name
$\mathrm{HOCH_2CH_2CHO}$	alcohol, aldehyde	aldehyde	hydroxy	3-hydroxypropanal
$\mathrm{HOCH_2CH_2CH_2COCH_3}$	alcohol, ketone	ketone	hydroxy	5-hydroxypentan-2-one
$\mathrm{HOCH_2CH_2COOH}$	alcohol, acid	acid	hydroxy	3-hydroxypropanoic acid
$\mathrm{CH_3COCH_2COOH}$	ketone, acid	acid	oxo	3-oxobutanoic acid
$\mathrm{OHC-CH_2CH_2CH_2-COOH}$	aldehyde, acid	acid	oxo	5-oxopentanoic acid
$\mathrm{H_2NCH_2CH_2OH}$	amine, alcohol	alcohol	amino	2-aminoethan-1-ol
$\mathrm{H_2NCH_2CH_2CHO}$	amine, aldehyde	aldehyde	amino	3-aminopropanal
$\mathrm{NC-CH_2CH_2-CHO}$	nitrile, aldehyde	nitrile	oxo	4-oxobutanenitrile
$\mathrm{CH_3COCH_2CONH_2}$	ketone, amide	amide	oxo	3-oxobutanamide
$\mathrm{CH_2=CH-CH_2OH}$	alkene, alcohol	alcohol	(stays a suffix)	prop-2-en-1-ol
$\mathrm{ClCH_2CH_2CH_2OH}$	halogen, alcohol	alcohol	chloro	3-chloropropan-1-ol
$\mathrm{O_2NCH_2CH_2COOH}$	nitro, acid	acid	nitro	3-nitropropanoic acid
$\mathrm{CH_3OCH_2CH_2CHO}$	ether, aldehyde	aldehyde	methoxy	3-methoxypropanal

Three habits fall out of this table.

One. A halogen, a nitro group or an alkoxy group never enters the seniority contest at all. In 3-chloropropan-1-ol the alcohol is the only candidate for the suffix even though chlorine looks the more reactive group.

Two. An aldehyde or ketone demoted to a prefix becomes oxo, and the oxo carbon is still counted in the chain. In 5-oxopentanoic acid C-1 is the $\mathrm{-COOH}$ carbon and C-5 is the $\mathrm{-CHO}$ carbon.

Three. The senior group also decides the direction of numbering. In $\mathrm{HOCH_2CH_2CH_2COCH_3}$ the ketone must get the lower number, so numbering starts from the methyl end: C-2 the carbonyl, C-5 the $\mathrm{CH_2OH}$. Numbering from the alcohol end instead puts C-1 on the $\mathrm{CH_2OH}$ and the carbonyl on C-4, giving "1-hydroxypentan-4-one" — the right suffix on the wrong locants. Reverse the seniority as well and the result is "4-oxopentan-1-ol". Both are wrong.

A checklist to run on every name

1. List every functional group present.
2. Rank them; the senior one is the suffix, all others are prefixes.
3. Choose the longest chain that contains the senior group.
4. Number so the senior group gets the lowest locant; break remaining ties with the multiple bond, then first point of difference, then alphabetical order.
5. Assemble: prefixes in alphabetical order, then root, then the multiple-bond suffix, then the functional-group suffix.
6. Apply the -e rule, and use commas between numbers and hyphens between numbers and letters, with no spaces except in "…oic acid", "…oyl chloride" and the two-word ester name.

Worked problems

Question 1: A branched alcohol where the chain is shorter than the carbon count

Give the IUPAC name of $\mathrm{(CH_3)_2C(OH)-CH_2-CH_3}$.

Answer:

The molecule has five carbons in total, so my first instinct is "pentan-". I check whether five carbons actually lie in one chain. The central carbon carries two methyls, one $\mathrm{-OH}$ and one ethyl. Any chain through it can pick up only one of the two methyls, so the longest chain is methyl + central carbon + $\mathrm{CH_2}$ + $\mathrm{CH_3}$, which is four carbons. Root: but-.

The $\mathrm{-OH}$ is on the second carbon of that chain from the methyl end, so it is C-2. The leftover methyl is also on C-2.

Ans: 2-methylbutan-2-ol

Watch out: Total carbon count is not chain length. Counting five here and writing "pentan-2-ol" is the commonest version of this error.

Question 2: An aldehyde with a branch

Give the IUPAC name of $\mathrm{CH_3-CH_2-CH(CH_3)-CHO}$.

Answer:

The $\mathrm{-CHO}$ carbon has to be C-1, so I have no choice about direction. Counting from it: C-1 is the CHO, C-2 is the $\mathrm{CH}$ carrying the methyl, C-3 is the $\mathrm{CH_2}$, C-4 is the terminal $\mathrm{CH_3}$. Four carbons: root but-.

Suffix -al, no locant needed for it. Methyl prefix on C-2.

Ans: 2-methylbutanal

Watch out: Never write "butan-1-al". A chain aldehyde carries no locant on the -al.

Question 3: A secondary amine

Give the IUPAC name of $\mathrm{CH_3-NH-CH_2-CH_3}$.

Answer:

Nitrogen carries two carbon groups, a methyl and an ethyl, so this is a secondary amine.

The parent must be the longer chain on nitrogen: ethyl, giving ethanamine. The methyl is left over and sits on nitrogen, so it is cited as N-methyl.

Ans: N-methylethanamine

Watch out: "N-ethylmethanamine" uses the shorter chain as parent and is wrong. The N is a locant, not a substituent name, so it stays italic and is never replaced by a number.

Question 4: A branched ether

Give the IUPAC name of $\mathrm{(CH_3)_2CH-O-CH_3}$.

Answer:

An ether has no suffix. The larger group becomes the parent chain and the smaller becomes alkoxy.

The larger group is the three-carbon isopropyl, so the parent is propane. The smaller group is $\mathrm{-OCH_3}$, methoxy.

The oxygen is attached to the middle carbon of the propane chain, C-2.

Ans: 2-methoxypropane

Watch out: "Methoxypropane" without a number does not distinguish this from 1-methoxypropane, $\mathrm{CH_3OCH_2CH_2CH_3}$. Both are $\mathrm{C_4H_{10}O}$.

Question 5: Two different halogens

Give the IUPAC name of $\mathrm{CH_3-CHBr-CH_2Cl}$.

Answer:

No functional group has a suffix here, so the ending is plain -ane. Three carbons: propane.

Numbering from the $\mathrm{CH_2Cl}$ end: chloro on C-1, bromo on C-2, set $\{1,2\}$. From the methyl end: bromo on C-2, chloro on C-3, set $\{2,3\}$. First point of difference is 1 against 2, so $\{1,2\}$ wins.

Citation is alphabetical, so bromo before chloro.

Ans: 2-bromo-1-chloropropane

Watch out: Alphabetical order fixes the order of writing, not the numbering. Here chloro genuinely has the lower locant even though bromo is written first.

Question 6: A nitrile

Give the IUPAC name of $\mathrm{(CH_3)_2CH-CH_2-CN}$.

Answer:

The nitrile carbon counts as part of the chain and is C-1. From it: C-2 is the $\mathrm{CH_2}$, C-3 is the $\mathrm{CH}$ with the methyl, C-4 is a $\mathrm{CH_3}$. Four chain carbons, root but-.

The suffix -nitrile starts with n, a consonant, so butane keeps its -e: butanenitrile. Methyl prefix on C-3.

Ans: 3-methylbutanenitrile

Watch out: Forgetting to count the nitrile carbon gives "2-methylpropanenitrile", which is a different compound altogether.

Question 7: An alcohol and a ketone together

Give the IUPAC name of $\mathrm{HO-CH_2CH_2CH_2-CO-CH_3}$.

Answer:

Two groups: an alcohol and a ketone. The ketone is senior, so -one is the suffix and the alcohol drops to the prefix hydroxy.

The chain has five carbons and contains both groups, so the root is pent-.

The ketone must get the lowest locant. Numbering from the methyl end puts the carbonyl on C-2; from the $\mathrm{-OH}$ end it would be C-4. So I number from the methyl: C-1 methyl, C-2 carbonyl, C-3, C-4, C-5 the $\mathrm{CH_2OH}$.

Pentane loses its -e before -one.

Ans: 5-hydroxypentan-2-one

Watch out: "2-oxopentan-5-ol" gets the seniority backwards: it keeps the correct numbering but names the ketone as a prefix and the alcohol as the suffix. The alcohol never beats the ketone.

Question 8: An aldehyde and an acid together

Give the IUPAC name of $\mathrm{OHC-CH_2-CH_2-CH_2-COOH}$.

Answer:

Carboxylic acid is the most senior group of all, so it takes the suffix and the aldehyde becomes the prefix oxo.

The $\mathrm{-COOH}$ carbon is C-1. Counting on: C-2, C-3 and C-4 are the three $\mathrm{CH_2}$ groups and C-5 is the $\mathrm{-CHO}$ carbon. Five carbons, so pentanoic acid with an oxo on C-5.

Ans: 5-oxopentanoic acid

Watch out: The oxo carbon is inside the chain count. Writing "4-oxopentanoic acid" or "pentanedial" both lose the acid.

Question 9: Name to structure

Write the structure of 3-methylbutan-2-ol.

Answer:

"butan" tells me four carbons in a row. "-2-ol" puts an $\mathrm{-OH}$ on C-2. "3-methyl" puts a $\mathrm{CH_3}$ on C-3.

So: C-1 is $\mathrm{CH_3}$; C-2 is $\mathrm{CH}$ with the $\mathrm{-OH}$; C-3 is $\mathrm{CH}$ with the methyl branch; C-4 is $\mathrm{CH_3}$.

I check valencies. C-2 has bonds to C-1, C-3, the $\mathrm{-OH}$ and one H: four. C-3 has bonds to C-2, C-4, the methyl and one H: four. Good.

Ans: $\mathrm{CH_3-CH(OH)-CH(CH_3)-CH_3}$

Question 10: Name to structure with two prefixes

Write the structure of 2-chloro-3-methylbutanal.

Answer:

"butanal" means a four-carbon chain with the $\mathrm{-CHO}$ as C-1, and no locant is written for the -al because C-1 is forced.

"2-chloro" puts $\mathrm{Cl}$ on C-2. "3-methyl" puts $\mathrm{CH_3}$ on C-3. C-4 is the last $\mathrm{CH_3}$.

Valency check on C-2: bonds to C-1, C-3, $\mathrm{Cl}$ and one H, four in all.

Ans: $\mathrm{CH_3-CH(CH_3)-CHCl-CHO}$

Question 11: Name to structure with an N locant

Write the structure of N,N-dimethylethanamine.

Answer:

"ethanamine" means the parent chain on nitrogen is an ethyl group, $\mathrm{CH_3CH_2-}$.

"N,N-dimethyl" puts two methyl groups on the nitrogen, not on any carbon. That uses up the nitrogen's remaining two bonds, so no hydrogen is left on N.

Nitrogen now holds three carbons, so this is a tertiary amine, and nitrogen has exactly three bonds as it must.

Ans: $\mathrm{CH_3CH_2-N(CH_3)_2}$

Watch out: Each group on nitrogen needs its own N. "N-dimethylethanamine" with a single N is wrong, and "2,2-dimethylethanamine" would put both methyls on carbon, which describes $\mathrm{H_2N-CH_2-CH(CH_3)_2}$ — a completely different amine.

Question 12: Find the error

A student writes propane-2-ol for $\mathrm{(CH_3)_2CHOH}$. What is wrong, and what is the correct name?

Answer:

The chain and the locant are both right: three carbons with the $\mathrm{-OH}$ on the middle one.

The mistake is the -e. The suffix -ol begins with the vowel o, so the terminal -e of propane must be dropped.

Ans: propan-2-ol

Question 13: Find the error

A student writes 2-hydroxybutan-4-al for $\mathrm{CH_3-CH(OH)-CH_2-CHO}$. Find every mistake.

Answer:

There are two mistakes.

The first is the locant on the aldehyde. The $\mathrm{-CHO}$ carbon must be C-1, so it can never be numbered 4, and it never carries a locant at all.

The second follows from fixing the first. Once the $\mathrm{-CHO}$ carbon is C-1, I count on: C-2 is the $\mathrm{CH_2}$, C-3 is the $\mathrm{CH}$ carrying the $\mathrm{-OH}$, C-4 is the $\mathrm{CH_3}$. So the hydroxy is on C-3, not C-2.

Aldehyde beats alcohol in seniority, so -al stays the suffix and hydroxy stays the prefix; that part the student had right.

Ans: 3-hydroxybutanal

Question 14: Find the error

A student writes 1-hydroxy-4-chlorobutane for $\mathrm{Cl-CH_2CH_2CH_2-CH_2-OH}$. Find every mistake.

Answer:

Three mistakes here.

First, an alcohol has a suffix, so $\mathrm{-OH}$ must be expressed as -ol and not as the prefix hydroxy. Chlorine has no suffix form, so it is the only prefix.

Second, the citation order. Prefixes are always cited in alphabetical order, so chloro must be written before hydroxy. The locants themselves are the only part the student got right: the $\mathrm{-OH}$ is the principal group, so it does take C-1, and the chlorine does fall on C-4.

Third, the ending. With -ol as the suffix, butane drops its terminal -e.

Ans: 4-chlorobutan-1-ol

Watch out: "Hydroxy" is only used when something senior to the alcohol is already claiming the suffix. A halogen is never senior to anything, because it has no suffix at all.