Topic 2 of 12
Organic isomers
Systematic drawing prevents repeated structures and missing isomers.
A-Level 8873, revised syllabus (2026-2027)
Change connectivity or fixed spatial arrangement
Systematic drawing prevents repeated structures and missing isomers.
Constitutional (structural) isomers have the same molecular formula but different atom-to-atom connections. You can change the carbon skeleton, move a functional group to an inequivalent position, or use a different functional group compatible with the formula. Rotating or reversing the same drawn chain does not make a new constitutional isomer.
Worked example
Find every alcohol with four carbons
Deduce the constitutional alcohol isomers with molecular formula C4H10O. Restrict the task to alcohols.
- 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.
- Use the branched three-carbon skeleton with a methyl branch. OH can be on an end carbon: 2-methylpropan-1-ol, (CH3)2CHCH2OH.
- Or put OH on the central carbon: 2-methylpropan-2-ol, (CH3)3COH.
- 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.
Four alcohol isomers: two primary, one secondary and one tertiary.
Worked example
Search 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.
- 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.
- 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.
- 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.
- 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.
- Methyl ethanoate gives methanol and ethanoate on alkaline hydrolysis. Ethyl methanoate gives ethanol and methanoate; propanoic acid undergoes neutralisation instead.
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 connectivity but changes the spatial arrangement around a C=C bond. A double bond contains a sigma bond and a pi bond. Free rotation would destroy the sideways orbital overlap forming the pi bond, so the arrangement is restricted. Each double-bonded carbon must have two different substituents for cis-trans alternatives 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.
Worked example
Separate the two kinds of isomer count
Deduce the acyclic alkenes with formula C4H8, including cis-trans forms.
- 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.
- But-2-ene has H and CH3 on each double-bonded carbon, so both cis and trans forms exist.
- But-1-ene and 2-methylpropene each contain a CH2 end of the double bond. Its two identical H substituents prevent cis-trans isomerism.
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.