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1.8.3Unit 1. Structure, Bonding and Introduction to Organic Chemistry

Nomenclature and Classification — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions

5 min read 6 parts Lesson 29 of 36 Unit quiz

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Part 1 of 6Naming Rules

1.8.3 Nomenclature and Classification

Systematic names describe the structure of a molecule exactly, so any chemist can draw it.

Stem
  • Count the carbons in the longest continuous chain: meth-, eth-, prop-, but-, pent-, hex- for one to six carbons.
Suffix or ending
  • Add -ane for alkanes, -ene for alkenes and -ol for alcohols, showing the main functional group.
Side groups
  • Alkyl branches such as methyl and ethyl, and halogens such as chloro and bromo, are prefixes.
Locants
  • Number the chain from the end that gives the lowest numbers to the functional group, then to the substituents.

Part 2 of 6Writing the Name

1
Find the longest carbon chain that contains the functional group and use its stem.
2
Number the carbons so that the functional group gets the lowest possible number.
3
List side groups in alphabetical order before the stem, each with its position number.
4
Use di-, tri- or tetra- for repeated groups, and repeat the number for each one, as in 2,2-dimethyl.
5
Use commas between numbers and hyphens between numbers and letters, for example 2-bromo-2-methylpropane.

Part 3 of 6Classification

Classification
Primary
  • The carbon attached to the functional group is bonded to only one other carbon, as in propan-1-ol and 1-bromobutane.
Secondary
  • The carbon holding the functional group is bonded to two other carbons, as in propan-2-ol and 2-bromopropane.
Tertiary
  • The carbon holding the functional group is bonded to three other carbons, as in 2-methylpropan-2-ol.
Why it matters
  • Primary, secondary and tertiary compounds react differently, for example in oxidation of alcohols and in substitution of halogenoalkanes.
NameClass
Butan-1-olPrimary
Butan-2-olSecondary
2-Methylpropan-2-olTertiary
2-Bromo-2-methylpropaneTertiary

Part 4 of 6Practice Patterns

Practice Patterns
Branched alkane
  • CH₃CH(CH₃)CH₂CH₃ has a four-carbon chain with a methyl on carbon 2, so it is 2-methylbutane.
Alkene position
  • CH₃CH=CHCH₃ is but-2-ene because the double bond starts at carbon 2.
Halogenoalkane
  • CH₃CHBrCH₃ is 2-bromopropane, which is a secondary halogenoalkane.
Two identical groups
  • (CH₃)₃CCl is 2-chloro-2-methylpropane, a tertiary halogenoalkane.

Part 5 of 6Worked Example

Question Name (CH₃)₂CHCH₂OH and say whether it is a primary, secondary or tertiary alcohol.

  1. The longest chain containing the carbon bearing –OH has three carbons, so the stem is propane.
  2. Number from the end nearest –OH–OH is on carbon 1 and a methyl group is on carbon 2.
  3. The carbon with –OH is bonded to only one other carbon.

Answer 2-methylpropan-1-ol, a primary alcohol.

Part 6 of 6Exam Checklist & Tips

1.8.3 – Key Points for the Exam

  • IUPAC The international body whose rules give each organic compound a unique systematic name.
  • Locant The number showing the position of a group or bond on a carbon chain.
  • Primary A carbon bonded to one other carbon that carries the functional group.
  • Secondary A carbon bonded to two other carbons that carries the functional group.
  • Tertiary A carbon bonded to three other carbons that carries the functional group.

Common Mistakes

  • Choosing a chain that is not the longest one because it was drawn that way.
  • Numbering from the wrong end so the functional group does not get the lowest number.
  • Placing substituents in the order of their numbers instead of alphabetical order.
  • Classifying by the number of carbons in the molecule instead of the carbon bonded to the functional group.

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