Structural isomerism

Isomers are compounds with identical molecular formula but different structures, and therefore usually different physical and chemical properties. Structural isomerism covers every case where that difference is a matter of connectivity — which atom is bonded to which — with no need to think about the molecule's shape in space. ISC Class 11 sorts structural isomerism into four named types, each defined by what specifically differs between the two structures.

• Chain isomerism. The carbon skeleton itself is arranged differently — straight versus branched — while everything else about the molecule (functional group, if any) matches. $\text{C}_4\text{H}_{10}$ gives n-butane (straight chain) and isobutane (branched); the molecules are hydrocarbons here, but chain isomerism happens in any homologous series, since it is purely a skeleton effect.
• Position isomerism. The skeleton and the functional group are identical; only the location of that group (or of a double/triple bond) along the chain differs. Propan-1-ol and propan-2-ol are both $\text{C}_3\text{H}_8\text{O}$, both alcohols on an identical three-carbon chain, differing only in whether $-\text{OH}$ sits on carbon 1 or carbon 2.
• Functional-group isomerism. The formula is shared but the molecules belong to entirely different chemical families because they carry different functional groups. Ethanol ($\text{CH}_3\text{CH}_2\text{OH}$) and dimethyl ether ($\text{CH}_3\text{OCH}_3$) are both $\text{C}_2\text{H}_6\text{O}$ — one an alcohol, the other an ether — and their properties diverge sharply as a result.
• Metamerism. Both molecules share the same functional group, from the same homologous series, but split their carbon atoms unevenly across the two sides of that group. Metamerism only arises for functional groups sandwiched between two alkyl chains — ethers, ketones, esters, secondary amines. Methyl propyl ether ($\text{CH}_3-\text{O}-\text{C}_3\text{H}_7$, a 1+3 carbon split) and diethyl ether ($\text{C}_2\text{H}_5-\text{O}-\text{C}_2\text{H}_5$, a 2+2 split) are metamers of $\text{C}_4\text{H}_{10}\text{O}$.

A fifth category, stereoisomerism (geometrical and optical isomerism), also produces the same molecular formula from different molecules, but the difference there is spatial — the connectivity is identical and only the 3-D arrangement of atoms varies. ISC Class 11 flags stereoisomerism as a separate, later topic; full R/S or D/L nomenclature is out of scope here. Before drawing candidate structures, it helps to know how much unsaturation (rings or multiple bonds) a formula even permits — that constrains which functional groups are physically possible for a given formula, narrowing the search.

where: $\text{DoU}$ is the degree of unsaturation (a whole number counting rings plus $\pi$ bonds), $C$ is the number of carbon atoms, and $H$ is the number of hydrogen atoms in the molecular formula (oxygen and other divalent-or-less heteroatoms do not enter the count). A $\text{DoU}$ of $0$ means a fully saturated, acyclic molecule — only chain, position, functional-group (saturated families), or metameric isomers are possible; a $\text{DoU}$ of $1$ opens the door to one ring or one double bond (such as a $\text{C}=\text{O}$), which is exactly what separates an alcohol/ether pair from an aldehyde/ketone pair at the same formula.

Structural isomerism has no single algebraic formula the way projectile motion or acid-base titration does — it is a classification scheme, so the practical skill is a repeatable procedure for deciding which of the four types (if any) relates two given structures. Two structures are isomers at all only if they share the same molecular formula (same $C$, $H$, and heteroatom counts) — check that first, since sharing a name-family alone is not enough. Once the formula matches, ask three questions in order.

Step 1: Is the functional group the same in both structures? If no, and both are fully valid molecules, it is functional-group isomerism — stop there, since a change of functional group family overrides every other consideration.

Step 2: If the functional group is the same, is the carbon skeleton (the connectivity pattern ignoring the functional group) identical? If the skeleton differs — straight versus branched — while the functional group and its relative attachment pattern stay analogous, it is chain isomerism.

Step 3: If both functional group and skeleton match, does only the position of the functional group along that skeleton differ? If yes, it is position isomerism. But if the functional group sits between two alkyl chains (an ether oxygen, a ketone carbonyl, an ester linkage) and what differs is how many carbons fall on each side of it rather than a position along one continuous chain, it is metamerism instead — the two are easy to conflate, and the distinguishing question is always "is there a group with two attachment arms whose carbon count is being redistributed?"

Question: For the molecular formula $\text{C}_4\text{H}_8\text{O}$, find its degree of unsaturation, then identify a pair of oxygen-containing isomers with different functional groups and name the isomerism relating them.

Step 1. Compute the degree of unsaturation using equation (1) with $C = 4$, $H = 8$: $\text{DoU} = \dfrac{2(4) + 2 - 8}{2} = \dfrac{2}{2} = 1$. One degree of unsaturation means the molecule contains exactly one ring or one $\pi$ bond — with a single oxygen present, a $\text{C}=\text{O}$ (carbonyl) is the most direct possibility, though a saturated ring with an ether oxygen is also consistent with $\text{DoU} = 1$.

Step 2. Build a carbonyl-based structure: butanal, $\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}$, an aldehyde with the $\text{C}=\text{O}$ terminal.

Step 3. Build a second carbonyl-based structure with the same formula but the $\text{C}=\text{O}$ internal instead of terminal: butan-2-one, $\text{CH}_3\text{COCH}_2\text{CH}_3$, a ketone.

Step 4. Compare functional groups. Butanal is an aldehyde ($-\text{CHO}$), butan-2-one is a ketone ($\text{C}=\text{O}$ flanked by two carbons) — different families sharing $\text{C}_4\text{H}_8\text{O}$ and the same $\text{DoU} = 1$.

Answer: butanal and butan-2-one are functional-group isomers of $\text{C}_4\text{H}_8\text{O}$; both correctly carry one degree of unsaturation from the carbonyl double bond.

Competitive stretch

Push the same formula further: does $\text{C}_4\text{H}_8\text{O}$ also permit position isomers within the ketone family alone? Only one internal position exists for a $\text{C}=\text{O}$ on a four-carbon chain (butan-2-one), since placing it at carbon 1 would make it an aldehyde, not a ketone — so no ketone-ketone position pair exists here. But among the aldehydes, 2-methylpropanal, $(\text{CH}_3)_2\text{CHCHO}$, has the same $-\text{CHO}$ group as butanal on a branched rather than straight skeleton, making butanal and 2-methylpropanal chain isomers of each other, both distinct again from butan-2-one.

• Mixing up metamerism with position isomerism — metamerism concerns carbon distribution around a two-armed functional group (an ether oxygen, a ketone carbonyl); position isomerism concerns where a group sits along one continuous chain. Ask whether the group has two attachment arms before deciding.
• Assuming a matching molecular formula alone proves isomerism — it is necessary but not sufficient; the structures must also be genuinely distinct, valid molecules, not the same structure drawn twice.
• Forgetting that a nonzero degree of unsaturation does not force a carbonyl specifically — a ring (like an ether-bridged cyclic structure) can equally supply $\text{DoU} = 1$, so equation (1) narrows possibilities without picking a unique structure.
• Treating stereoisomers as structural isomers — geometrical and optical isomers have identical connectivity and differ only in spatial arrangement, a separate category this syllabus does not require you to draw or name in full here.

Ask the tutor: try classifying the isomerism relating a few molecular formulas you pick yourself, and check your reasoning step by step together.