Functional groups and IUPAC naming: the twelve groups in one table, naming in five steps on one molecule, and the isomers of one formula counted

A functional group is the part of an organic molecule that reacts: the C=C of an alkene, the O–H of an alcohol, the C=O of a ketone. Twelve of them cover a Grade 12 course, and each has a suffix or prefix in its IUPAC name. Naming is five steps: find the principal group, find the longest chain that contains it, number from the end that gives the group the lowest number, name and number the branches alphabetically, assemble. Below, the twelve groups are in one table, one molecule is named with every step shown, the numbering trap is named, and the isomers of one small formula are counted.

The twelve functional groups in one table

Every structure on this page is a condensed formula: the backbone carbons joined by dashes, a branch in brackets after the carbon it hangs from. CH₃–CH(CH₃)–CH₃ is a three-carbon chain with a methyl group on the middle carbon.

Group	What to look for	Suffix or prefix	One example, named
Alkane	only C–C single bonds	-ane	CH₃–CH₂–CH₃, propane
Alkene	a C=C double bond	-ene	CH₂=CH–CH₃, propene
Alkyne	a C≡C triple bond	-yne	HC≡CH, ethyne
Alkyl halide	F, Cl, Br or I on a carbon	fluoro-, chloro-, bromo-, iodo-	CH₃Cl, chloromethane
Alcohol	–OH on a carbon	-ol	CH₃–CH₂–OH, ethanol
Ether	C–O–C	alkoxy- (or "ether" in the common name)	CH₃–O–CH₃, methoxymethane (dimethyl ether)
Aldehyde	–CHO, the C=O at the end of a chain	-al	CH₃–CHO, ethanal
Ketone	C=O between two carbons	-one	CH₃–CO–CH₃, propanone
Carboxylic acid	–COOH	-oic acid	CH₃–COOH, ethanoic acid
Ester	–COO– between two carbon groups	-yl -oate	CH₃–COO–CH₂–CH₃, ethyl ethanoate
Amine	–NH₂ (or N bonded to carbons)	-amine	CH₃–NH₂, methanamine
Amide	–CONH₂	-amide	CH₃–CONH₂, ethanamide

Four of these groups can hydrogen-bond with one another through an O–H or N–H: alcohols, carboxylic acids, amines and amides. That is why ethanol boils far above an ether or alkane of similar mass; the intermolecular forces page works that comparison.

In SCH4U this table is the first idea of the course, Organic functional groups and IUPAC nomenclature, and everything in the organic strand is built on it. The SCH4U course map shows the order.

IUPAC nomenclature: how to name organic compounds in five steps

1. Find the principal functional group. When a molecule carries more than one, the one higher in the seniority order gives the suffix and the others become prefixes. The order this course uses, which follows the IUPAC seniority of classes: carboxylic acid > ester > amide > aldehyde > ketone > alcohol > amine > alkene and alkyne. Halides and ethers never take a suffix; they are always prefixes.
2. Find the longest carbon chain that contains the principal group's carbon. Its length gives the stem: meth- 1, eth- 2, prop- 3, but- 4, pent- 5, hex- 6.
3. Number the chain from the end that gives the principal group the lowest number. An aldehyde or carboxylic acid carbon is carbon 1 by construction, so it needs no number in the name.
4. Name and number every branch (methyl, ethyl, chloro and so on), and list them alphabetically. Use di- and tri- for repeats, and ignore those prefixes when you alphabetize: ethyl comes before dimethyl.
5. Assemble: branches, then the stem, then the suffix with its number. Then check by rebuilding the molecular formula from the name.

On spelling: current IUPAC recommendations put the number right before the suffix (pentan-2-ol); many North American textbooks still write it first (2-pentanol). Both name the same molecule. Use the one your course uses.

Worked example: one alcohol, every step

Name CH₃–CH(CH₃)–CH₂–CH(OH)–CH₃.

• Step 1, the principal group. The only group is the O–H, so this is an alcohol and the suffix is -ol.
• Step 2, the longest chain. Count the backbone through the carbon that carries the OH: CH₃, CH, CH₂, CH(OH), CH₃ = 5 carbons, so the stem is pent- and the parent is pentanol. Starting from the bracketed CH₃ instead also gives 5, so the chain length does not change; there is no six-carbon path.
• Step 3, the numbering. From the left, the OH sits on carbon 4. From the right, it sits on carbon 2. The principal group gets the lower number, so number from the right: C1 = CH₃, C2 = CH(OH), C3 = CH₂, C4 = CH(CH₃), C5 = CH₃.
• Step 4, the branches. One methyl group, on carbon 4: 4-methyl.
• Step 5, assemble. 4-methylpentan-2-ol, written 4-methyl-2-pentanol in the older style.
• The formula check. Carbons: 5 in the chain + 1 in the methyl = 6. Hydrogens, carbon by carbon from the left: 3 + 1 + 3 (the methyl) + 2 + 1 + 3 = 13 on carbon, plus 1 on the O = 14. One oxygen. C₆H₁₄O, the same count you get by reading the structure atom by atom, so the name describes this molecule and no other.

The common wrong line. Numbering from the left so the branch gets the low number: 2-methylpentan-4-ol. It feels right because "lowest numbers" is the rule you remember from alkanes, but the principal group outranks a branch. Number for the OH first; the branch takes whatever number is left. The second wrong line is counting the branch into the chain and calling the molecule a hexanol. There is no straight run of six carbons; follow the dashes, and the bracketed carbon is a side branch.

Two more, briefly: an aldehyde and an ester

An aldehyde. Name CH₃–CH₂–CH(CH₃)–CH₂–CHO.

• The principal group is the CHO, so the suffix is -al and the CHO carbon is C1.
• Number from the CHO end: C1 = CHO, C2 = CH₂, C3 = CH(CH₃), C4 = CH₂, C5 = CH₃. Five carbons: pentanal.
• One methyl on C3: 3-methylpentanal. Formula: C₆H₁₂O. No number is written for the -al, because it can only be on carbon 1.

An ester. Name CH₃–CH₂–COO–CH₃.

• An ester is named in two words: the alkyl group from the alcohol part first, then the acid part with -oate.
• The acid part is CH₃–CH₂–COO–, three carbons counting the C=O carbon: propanoate.
• The alcohol part is the CH₃ on the single-bonded oxygen: methyl.
• Methyl propanoate, C₄H₈O₂.

The common wrong line for esters. Naming the halves the wrong way round. Propyl methanoate, HCOO–CH₂–CH₂–CH₃, is a different molecule with the same formula C₄H₈O₂. Find the C=O first: the carbon chain that contains it is the -oate half, and whatever sits on the other oxygen is the alkyl half.

Isomers: same formula, different molecule

Isomers share a molecular formula and differ as molecules. Structural isomers differ in how the atoms are connected. Stereoisomers have the same connections arranged differently in space; the simplest case is the cis and trans forms of a molecule with a C=C, which cannot rotate. Stereoisomers are the course's third idea, Structural and stereoisomers; here the job is counting structural isomers.

Worked count: C₄H₁₀O. First decide what the formula allows. C₄H₁₀ is a saturated alkane, and adding an oxygen adds no hydrogens, so there are no rings and no double bonds. A saturated molecule with one oxygen is either an alcohol or an ether. Count each.

Alcohols, an O–H on a four-carbon skeleton. The skeleton is either a straight chain of four or a chain of three with a methyl on the middle carbon.

• Straight chain, OH on an end carbon: butan-1-ol
• Straight chain, OH on an inner carbon: butan-2-ol
• Branched chain, OH on an end carbon: 2-methylpropan-1-ol
• Branched chain, OH on the middle carbon: 2-methylpropan-2-ol

Ethers, an O between two carbon groups whose carbons add up to four.

• One carbon and three (straight): CH₃–O–CH₂–CH₂–CH₃, methoxypropane
• One carbon and three (branched): CH₃–O–CH(CH₃)–CH₃, 2-methoxypropane
• Two and two: CH₃–CH₂–O–CH₂–CH₃, ethoxyethane (diethyl ether)

Total: 4 + 3 = 7 structural isomers, across two functional groups.

The common wrong line. Counting butan-1-ol and butan-4-ol as two. They are one molecule numbered from opposite ends, and butan-4-ol is not a valid name. The other is stopping at four because "the formula looks like an alcohol". The formula says nothing about the group; the ethers are three of the seven.

Functional groups and naming on one screen

• The twelve groups: alkane -ane, alkene -ene, alkyne -yne, alkyl halide halo-, alcohol -ol, ether alkoxy-, aldehyde -al, ketone -one, carboxylic acid -oic acid, ester -yl -oate, amine -amine, amide -amide.
• The five steps: principal group, longest chain containing it, number for the group's lowest number, branches alphabetically, assemble.
• The lowest-number rule: the principal group outranks every branch. Aldehyde and acid carbons are C1.
• The check: rebuild the formula from the name.
• Counting isomers: decide what the formula allows (rings, double bonds, which groups), then list skeletons and positions group by group, and strike out any that are the same molecule numbered backwards.

Name the three below before you read the answers. Attempting first, then checking, is chapter 3 of the free playbook.

Try one cold

(a) Name CH₃–CH(CH₃)–CH₂–CH₂–COOH.

(b) Name CH₃–CO–CH₂–CH₂–CH₃.

(c) How many structural isomers has C₃H₈O, and what are they?

Check

• (a) A carboxylic acid, so the COOH carbon is C1. Chain: COOH, CH₂, CH₂, CH, CH₃ = 5 carbons, pentanoic acid. The methyl is on C4. 4-methylpentanoic acid, C₆H₁₂O₂.
• (b) A ketone in a five-carbon chain. The C=O is on carbon 2 from the left and carbon 4 from the right, so number from the left: pentan-2-one (2-pentanone), C₅H₁₀O.
• (c) Three: propan-1-ol, propan-2-ol and methoxyethane (CH₃–O–CH₂–CH₃). The formula is saturated, so two alcohols and one ether.

If (a) came out as 2-methylpentanoic acid, you numbered from the wrong end; the acid carbon is always C1. If (c) came out as two, you forgot the ether.

Where this sits in the course

This is idea 1 of 24 in SCH4U, Organic functional groups and IUPAC nomenclature, the opening of the organic strand. It leads to Hydrocarbons: alkanes, alkenes, alkynes and their reactions, Structural and stereoisomers, Organic reaction types: addition, elimination, substitution, esterification and Polymers: addition and condensation polymerisation. Every one of those asks you to read and write names fluently, so this idea is worth securing first.

The condensed formulas are Lewis structures with the lone pairs left off; the Lewis structures page draws them in full. This strand is mostly vocabulary with one procedure, the naming, while the later calculation strands are mostly procedure. How to study chemistry shows why the two kinds are studied differently. Every idea of the three science courses has a page like this one.

How to practise it

Name ten structures, then draw ten from their names, with the groups mixed rather than one group at a time. Rohrer and Taylor (2007) found that shuffled practice felt harder and produced much better scores a week later, because a mixed set makes you identify the group before you name it, which is exactly what a test asks.

Check every name by rebuilding the formula, and mark the attempt honestly before you look. Roediger and Karpicke (2006) found that taking a test on material helped long-term retention more than restudying it, and the rebuild is a test you can run on yourself; active recall has the marking rules. For the twelve suffixes, make one card per group with the structure on one side and the suffix and an example on the other; how to write a flashcard keeps the answer off the front.