The big idea: Organic molecules are sorted into classes by their functional group — a specific atom or group of atoms that gives the molecule its characteristic chemistry.
The long carbon chain (the skeleton) is fairly unreactive; the functional group is where the reactions happen. So if you can spot the group, you know how the molecule behaves and what class it belongs to.
The –OH (hydroxyl) group on an alcohol: name ends in -ol.
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Key terms: - Functional group — the reactive atom/group (e.g. –OH, –COOH, C=C) that defines the class. - Homologous series — a family of molecules with the same functional group and general formula (e.g. all alcohols). - Saturated — only single C–C bonds (alkanes). Unsaturated — has a C=C (or C≡C) bond (alkenes).
For SL you must identify these classes from a structure and match each to its functional group. Each skeletal vertex (corner or end) is a carbon; the hydrogen atoms are not drawn.
| Class | Functional group | Prefix / suffix | Example |
|---|---|---|---|
| Alkane | C–C and C–H only (saturated) | -ane | propane, C3H8 |
| Alkene | C=C double bond | -ene | propene, C3H6 |
| Alcohol | –OH (hydroxyl) | -ol | propan-1-ol |
| Halogenoalkane | –F / –Cl / –Br / –I | fluoro/chloro/bromo/iodo- | 2-chloropropane |
| Aldehyde | –CHO (carbonyl at the end) | -al | propanal |
| Ketone | C=O (carbonyl in the middle) | -one | propan-2-one |
| Carboxylic acid | –COOH (carboxyl) | -oic acid | propanoic acid |
Aldehyde vs ketone — the carbonyl position: Both contain a carbonyl (C=O), so students mix them up.
- Aldehyde (–CHO): the C=O is at the end of the chain → ends in -al. - Ketone (C=O): the C=O is in the middle of the chain → ends in -one.
Position of the carbonyl is the whole difference.
The C=O carbonyl is in the middle of the chain → a ketone (-one).
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The –COOH (carboxyl) group of a carboxylic acid: name ends in -oic acid.
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Every systematic (IUPAC) name is built from three pieces: a stem (how many carbons), a suffix (the functional group), and a locant (a number saying where the group is on the chain).
How to build the name
- Stem — count the carbons in the longest chain and use meth/eth/prop/but/pent/hex.
- Suffix — add the ending for the functional group (-ane, -ene, -ol, -al, -one, -oic acid).
- Locant — number the chain to give the functional group the lowest number, then put that number before the suffix.
- Prefix — add any substituent (e.g. chloro-, methyl-) at the front, with its own locant.
| Carbons in chain | Stem | Alkane name |
|---|---|---|
| 1 | meth- | methane |
| 2 | eth- | ethane |
| 3 | prop- | propane |
| 4 | but- | butane |
| 5 | pent- | pentane |
| 6 | hex- | hexane |
Worked example — naming an alcohol
Name the alcohol with a three-carbon chain and an –OH group on the middle carbon.
Solution
- Stem: three carbons → prop.
- Suffix: the –OH group makes it an alcohol → -ol.
- Locant: the –OH is on carbon 2, so the name is propan-2-ol.
Final answer
propan-2-ol.
A halogen (here Cl) replaces an H in a halogenoalkane: prefix chloro-.
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Worked example — naming a halogenoalkane
Name the molecule that is a three-carbon alkane with a chlorine atom on the middle carbon.
Solution
- Stem + suffix: three carbons, all single bonds → propane.
- Prefix: a chlorine substituent → chloro-.
- Locant: the Cl is on carbon 2 → 2-chloropropane.
Final answer
2-chloropropane.
How this is tested: This is one of the most reliable organic-chemistry marks.
- Paper 1A (MCQ): 'identify the functional group(s) present' in a displayed structure. - Paper 2: 'state the functional group' (1 mark) and 'deduce the IUPAC name' (1 mark) from a structure.
For a name, the markers want all three pieces correct: the right stem, the right suffix, and the locant in the right place.
Common traps: - Confusing an aldehyde (–CHO, carbonyl at the end) with a ketone (C=O in the middle). - Forgetting the locant — propan-2-ol scores; 'propanol' alone may not. - Naming a –COOH as an alcohol because you spotted the O–H within it — it is a carboxylic acid (-oic acid).
IB-style question — identify the groups (a)
(a) A molecule has the skeletal structure shown, with a –COOH group at one end and an –OH group on the third carbon. Identify the two functional groups present and name the class associated with each. [2]
How to score the marks
- Group 1: the –COOH group is a carboxyl group → the molecule is a carboxylic acid.
- Group 2: the –OH group is a hydroxyl group → it also contains an alcohol function.
Final answer
Carboxyl (–COOH, carboxylic acid) and hydroxyl (–OH, alcohol).
This molecule has TWO functional groups: a carboxyl (–COOH) and a hydroxyl (–OH).
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IB-style question — deduce the IUPAC name (b)
(b) Deduce the systematic (IUPAC) name of the unsaturated molecule shown: a four-carbon chain with a C=C double bond between carbons 2 and 3. [1]
How to score the mark
- Stem: four carbons → but-.
- Suffix + locant: a C=C double bond → -ene; it starts at carbon 2, so the locant is 2.
- Assemble stem + locant + suffix: but-2-ene.
Final answer
but-2-ene.
A C=C double bond makes it an alkene: name ends in -ene.
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