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How do we name organic compounds and represent the different ways their atoms can be arranged?

Formulae (empirical, molecular, general, structural, displayed and skeletal), homologous series and functional groups, IUPAC nomenclature, and isomerism (structural and E/Z stereoisomerism with Cahn-Ingold-Prelog priority).

An OCR H432 module 4 answer on the basics of organic chemistry: the six types of formula, homologous series and functional groups, IUPAC nomenclature, and structural and E/Z stereoisomerism using Cahn-Ingold-Prelog priority rules.

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  1. What this topic is asking
  2. Types of formula
  3. Homologous series and functional groups
  4. IUPAC nomenclature
  5. Isomerism
  6. Examples in context
  7. Try this

What this topic is asking

OCR specification point 4.1.1 wants you to use the different types of formula, explain what a homologous series and a functional group are, name organic compounds by IUPAC rules, and recognise the two families of isomerism: structural (chain, position and functional group) and stereoisomerism (E/Z, decided by Cahn-Ingold-Prelog priority). This is the language that every later organic topic uses.

Types of formula

Skeletal formulae are the workhorse of organic chemistry because they are quick to draw and make functional groups stand out. Each line end or vertex is a carbon, and enough hydrogens are assumed to give each carbon four bonds.

Homologous series and functional groups

The Module 4 series are the alkanes, alkenes, alcohols and haloalkanes. An alkyl group has the general formula CnH2n+1\text{C}_n\text{H}_{2n+1} (methyl, ethyl, propyl) and appears as a branch.

IUPAC nomenclature

To name a compound:

  • Find the longest carbon chain containing the functional group; this gives the stem (meth, eth, prop, but, pent).
  • Add the suffix for the main functional group (-ane, -ene, -ol).
  • Add prefixes for substituents (methyl, chloro) with locant numbers, chosen to give the lowest set of locants.
  • Use di, tri, tetra for repeated groups and list substituents alphabetically.

So CH3CH(CH3)CH3\text{CH}_3\text{CH(CH}_3)\text{CH}_3 is 2-methylpropane and CH3CH=CHCH3\text{CH}_3\text{CH=CHCH}_3 is but-2-ene.

Isomerism

E/Z isomerism needs two conditions: a C=C\text{C=C} double bond that restricts rotation, and two different groups on each carbon of the double bond.

Examples in context

Example 1. Why nomenclature matters in industry. A single molecular formula such as C4H10\text{C}_4\text{H}_{10} covers butane and 2-methylpropane (isobutane), which have different boiling points and uses (isobutane is a refrigerant). Systematic naming removes ambiguity so chemists worldwide describe the same molecule.

Example 2. E/Z isomerism in vision and fats. Restricted rotation about C=C\text{C=C} bonds underlies the cis-to-trans switch of retinal in the eye and the difference between cis (healthier) and trans (less healthy) fats, both direct consequences of the geometric isomerism introduced here.

Try this

Q1. State what is meant by a homologous series. [2 marks]

  • Cue. A series of compounds with the same functional group and general formula, differing by CH2\text{CH}_2, with similar chemical properties and a gradual change in physical properties.

Q2. Give the molecular formula of the alkene with empirical formula CH2\text{CH}_2 and three carbon atoms, and name it. [2 marks]

  • Cue. C3H6\text{C}_3\text{H}_6, propene.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20193 marksA hydrocarbon has the empirical formula CH2\text{CH}_2 and a relative molecular mass of 56.056.0. (a) Determine its molecular formula. (b) Draw the skeletal formula of one possible position isomer that is an alkene.
Show worked answer β†’

(a) The empirical formula mass of CH2\text{CH}_2 is 12.0+2(1.0)=14.012.0 + 2(1.0) = 14.0. Dividing 56.0/14.0=456.0 / 14.0 = 4, so the molecular formula is (CH2)4=C4H8(\text{CH}_2)_4 = \text{C}_4\text{H}_8 (1).

(b) C4H8\text{C}_4\text{H}_8 as an alkene is butene. A valid skeletal formula shows a four-carbon chain with the double bond drawn between carbons 1 and 2 (but-1-ene) or carbons 2 and 3 (but-2-ene) (1). Carbon atoms sit at the ends and bends of the lines and hydrogens on carbon are omitted (1).

Markers reward the correct molecular formula, an alkene skeletal structure with four carbons, and the skeletal convention applied correctly.

OCR 20212 marksState, with a reason, whether but-2-ene shows E/Z isomerism, and name the two stereoisomers.
Show worked answer β†’

But-2-ene does show E/Z isomerism (1), because the C=C\text{C=C} double bond restricts rotation and each carbon of the double bond carries two different groups (a CH3\text{CH}_3 and an H), the requirement for E/Z isomerism.

The two isomers are E-but-2-ene (the two methyl groups on opposite sides) and Z-but-2-ene (the two methyl groups on the same side) (1).

Markers reward the correct decision plus the restricted-rotation and different-groups reason, and the correct E and Z names.

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