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8873 / 2027
Polymers and Organic Chemistry

Lesson 2 of 12

Organic isomers

Systematic drawing prevents repeated structures and missing isomers.

A-Level 8873, revised syllabus (2026-2027)

  • I can describe constitutional isomerism.Syllabus

    Syllabus 8873, 9(c). Describe constitutional isomerism.

  • I can explain cis-trans isomerism in alkenes.Syllabus

    Syllabus 8873, 9(d). Explain cis-trans isomerism in alkenes.

  • I can deduce isomers from a known molecular formula.Syllabus

    Syllabus 8873, 9(e). Deduce isomers from a known molecular formula.

Change connectivity or fixed spatial arrangement

Systematic drawing prevents repeated structures and missing isomers.

Constitutional (structural) isomers have the same molecular formula, but their atoms are joined differently. You can change the carbon chain, move the functional group to a different position, or use a different functional group that fits the formula. Turning a chain round or drawing it the other way does not make a new isomer.

Worked exampleFind every alcohol with four carbons

Deduce the constitutional alcohol isomers with molecular formula C4H10O. Restrict the task to alcohols.

Try it first. Then show the worked stepsHide the worked steps
  1. Use the straight four-carbon skeleton. OH can be on carbon 1 or carbon 2: butan-1-ol, CH3CH2CH2CH2OH; and butan-2-ol, CH3CH(OH)CH2CH3. Positions 3 and 4 duplicate 2 and 1 after reversing the chain.
  2. Use the branched three-carbon skeleton with a methyl branch. OH can be on an end carbon: 2-methylpropan-1-ol, (CH3)2CHCH2OH.
  3. Or put OH on the central carbon: 2-methylpropan-2-ol, (CH3)3COH.
  4. Check each carbon has four bonds and each structure has C4H10O. These are the four alcohol constitutional isomers; the restriction matters because other functional-group classes can also share this formula.
Answer

Four alcohol isomers: two primary, one secondary and one tertiary.

Worked exampleSearch across functional-group families

A compound has formula C3H6O2 and is known to be a saturated, acyclic carboxylic acid or ester with no other functional group. Deduce all possibilities and decide which could yield methanol on alkaline hydrolysis.

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  1. For the acid family, one carbon belongs to COOH. The remaining two form an ethyl group: CH3CH2COOH, propanoic acid. With only three carbons there is no different branched acid skeleton.
  2. For an ester, count the carbonyl carbon as part of the acid-derived fragment. Split the total of three carbons between the acid-derived and alcohol-derived fragments. The alcohol fragment must contain at least one carbon.
  3. A 2 + 1 split gives CH3COOCH3, methyl ethanoate. A 1 + 2 split gives HCOOCH2CH3, ethyl methanoate. Methanoate has H directly attached to its carbonyl carbon; do not insert an extra CH3.
  4. All three structures have 3 C, 6 H and 2 O. Reversing a drawing does not create another structure, but swapping the two ester fragments does change connectivity.
  5. Methyl ethanoate gives methanol and ethanoate on alkaline hydrolysis. Ethyl methanoate gives ethanol and methanoate; propanoic acid undergoes neutralisation instead.
Answer

The specified family restriction gives three constitutional isomers. Methanol production selects methyl ethanoate. The general formula alone does not identify the functional group.

Cis-trans isomerism keeps the same connections but changes the arrangement in space around a C=C bond. A double bond is one sigma bond and one pi bond. Rotating it would break the sideways overlap that forms the pi bond, so it cannot rotate freely. Each carbon of the double bond must carry two different groups for cis and trans forms to exist.

But-2-ene has two fixed arrangements

Each carbon in the C=C has H and CH3 attached. In cis-but-2-ene the two CH3 groups lie on the same side; in trans-but-2-ene they lie on opposite sides. Rotation about the C=C is restricted by the pi bond.

Cis and trans forms share the same molecular formula and connectivity. They are not two more constitutional isomers.

Worked exampleSeparate the two kinds of isomer count

Deduce the acyclic alkenes with formula C4H8, including cis-trans forms.

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  1. A straight skeleton gives but-1-ene, CH2=CHCH2CH3, and but-2-ene, CH3CH=CHCH3. A branched skeleton gives 2-methylpropene, CH2=C(CH3)2.
  2. But-2-ene has H and CH3 on each double-bonded carbon, so both cis and trans forms exist.
  3. But-1-ene and 2-methylpropene each contain a CH2 end of the double bond. Its two identical H substituents prevent cis-trans isomerism.
Answer

There are three constitutional alkene isomers, or four distinct alkene structures when cis- and trans-but-2-ene are counted separately. Rings are excluded by the question.

Check your understanding

Which of CHCl=CHBr and CH2=CClBr has cis-trans isomers?

  1. CH2=CClBr only, because one carbon carries two different halogens
  2. Both of them, because each has a C=C bond that cannot rotate
  3. CHCl=CHBr only, because each C=C carbon has two different groups
Show the answer

CHCl=CHBr only, because each C=C carbon has two different groups

In CHCl=CHBr the two H atoms can be on the same or opposite sides. In CH2=CClBr one carbon carries two H atoms, so swapping them gives the same molecule.