Chemistry · Lesson

Constitutional isomers and stereoisomers

Classify a pair of structures in a consistent order: composition, connectivity, then specified spatial arrangement. Enumerate a few small formulas without counting rotated drawings or conformations as new constitutions.

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By the end, you can

Distinguish constitutional isomers, configurational stereoisomers and conformations.
Apply the simple E/Z condition using the two groups on each alkene carbon.
Check bounded small-formula counts against equivalent atom relabelings.
Keep protonation and resonance out of an isomer count.

Compare composition before comparing drawings

Begin with molecular formula and charge. If those differ, the examples are not isomers under the same molecular formula: CH₃NH₂ and CH₃NH₃⁺ illustrate a protonation change. Next compare atom connectivity and bond orders while allowing equivalent atoms to be relabeled. Different drawings of the same graph are not automatically different constitutional isomers.

If formula matches and constitution differs, the pair are constitutional isomers. Ethanol and dimethyl ether are one example; propanal and propanone another. If constitution matches, compare the specified spatial arrangement. Enantiomers are non-superposable mirror partners; other distinct stereoisomers are diastereomers. A conformation produced by a selected single-bond rotation remains the same constitution. These categories require enough information: an unmarked flat drawing cannot establish an unspecified configuration.

An E/Z comparison needs two different groups at each end

For a selected simple alkene, configurational E/Z alternatives require each double-bond carbon to have two distinguishable attached groups. Compare the higher-priority group at each end using the Cahn–Ingold–Prelog rules. Z puts those higher-priority groups on the same side of the double-bond reference; E puts them on opposite sides.

But-2-ene has a methyl and a hydrogen at each end, so it has E and Z forms. But-1-ene has two hydrogens at the terminal double-bond carbon and has no E/Z pair. A page rotation preserves the configuration. Interconverting an E/Z pair requires a process that changes the double-bond configuration; it is not ordinary free rotation about the depicted C=C bond. Mechanism, irradiation, temperature and other conditions govern that process.

State what the enumeration allows

For the counts below, consider ordinary neutral closed-shell connected structures of the specified elemental formulas, with no isotopic substitutions. Count equivalent atom relabelings only once, and do not count ordinary ring or single-bond conformers as separate constitutions. C₄H₁₀ has two: butane and 2-methylpropane.

C₄H₈ has five constitutional examples in this bounded scope: but-1-ene, but-2-ene, 2-methylpropene, cyclobutane and methylcyclopropane. Its DBE=1 allows either one ring or one double bond. Only the but-2-ene constitution adds an E/Z partner, giving six specified configurational species for this list.

Separate constitution counts from specified configurations

Formula	Constitutional classes	Specified configurational species	What adds a partner?
C₄H₁₀	2	2	Neither constitution has the selected configurational pair
C₄H₈	5	6	But-2-ene has E and Z forms
C₄H₁₀O	7	8	Butan-2-ol has R and S forms

Exact bounded counts for ordinary neutral connected closed-shell formulas, excluding isotopic substitution and ordinary conformer counts. Distinct drawings related by equivalent atom relabeling count once. The configurational column includes E/Z for but-2-ene and R/S for butan-2-ol; it is not a universal all-organic enumeration method.

A formula can allow both alcohols and ethers

C₄H₁₀O has DBE=0 in the ordinary neutral scope. Its seven constitutions are four alcohols— butan-1-ol, butan-2-ol, 2-methylpropan-1-ol and 2-methylpropan-2-ol—and three ethers: 1-methoxypropane, 2-methoxypropane and ethoxyethane. Butan-2-ol has two enantiomers, so specifying configurations gives eight species for these constitutions.

That count does not make every four-carbon oxygen structure equally stable, accessible or useful in a given environment. It is a graph-and-configuration count under stated assumptions, not a synthesis catalogue. Resonance contributors are descriptions of one electronic entity and add no isomers. Proton transfer changes composition or the location of nuclei and must be classified on its own terms rather than slipped into a resonance or conformer count.

Worked examples

Remove equivalent carbon relabelings

Why do straight-chain and branched C₄H₁₀ give two constitutional classes?

DBE=0; the four-carbon hydrocarbon skeleton is connected, acyclic and single bonded in this scope.
One graph has carbon degrees 1,2,2,1: the butane path.
The other has degrees 3,1,1,1: the 2-methylpropane branch.
Renumbering, rotating or bending either graph does not create another constitution.

Two constitutional classes, rather than one class for each possible drawing orientation.

Expand only the constitution with an E/Z pair

Starting from the five listed C₄H₈ constitutions, count their specified configurations.

But-1-ene has two identical H at one double-bond end; 2-methylpropene also fails the two-different-groups condition.
But-2-ene has E and Z forms.
Cyclobutane and methylcyclopropane each contributes one configurational species in this unsubstituted/monosubstituted list.
The total is 1+2+1+1+1=6.

Five constitutional classes and six configurational species under the stated scope.

Keep a contributor pair separate from an isomer pair

Classify ethanol/dimethyl ether, acetate contributor A/B, and anti/gauche butane.

Ethanol/dimethyl ether: same formula, different O connectivity.
Acetate A/B: same labeled nuclei and electron total, different electron-bookkeeping contributors.
Anti/gauche butane: same connectivity, different single-bond conformations.

Constitutional isomers; resonance representations of one ion; conformations of one constitution, respectively.

Try it yourself

Does CH₂=CHCH₂CH₃ have an E/Z pair at that double bond?
Show the answer and reasoning
No.
The CH₂ carbon has two identical hydrogens.
In the stated C₄H₁₀O enumeration, how many constitutional classes are ethers?
Show the answer and reasoning
Three.
They are 1-methoxypropane, 2-methoxypropane and ethoxyethane.
How are E-but-2-ene and Z-but-2-ene related?
Show the answer and reasoning
They are configurational diastereomers.
They share constitution but have distinct non-mirror-partner double-bond configurations.
Does viewing the same configured molecule from the opposite side of the page create a new isomer?
Show the answer and reasoning
No.
A change in viewpoint preserves the molecular identity; the drawing must retain its conventions consistently.

What to carry forward

Compare formula and charge, then connectivity, then configuration.
Equivalent atom labels and page orientations do not add isomers.
Check the E/Z condition at both double-bond ends.
Every enumeration needs a declared chemical and counting scope.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.