Isomerism

Isomer questions hand Extended candidates a molecular formula and two instructions (draw, then name), and the mark scheme rejects anything that is secretly the same molecule drawn twice. That trap, the bent-chain duplicate, catches more candidates than the chemistry does. The content is one definition and a short catalogue of C4 structures.

The definition

Structural isomers are compounds with the same molecular formula but different structural formulae.

Two marking points live in that sentence: same molecular formula and different structural formulae. The atoms are identical in number and kind; they are connected differently. Because the connectivity differs, isomers are genuinely different compounds with different names and different boiling points (branched isomers boil lower than straight chains, a detail that occasionally earns a mark).

The C4H10 pair

Butane’s molecular formula, C4H10, fits the alkane general formula CnH2n+2 (general formulae are catalogued under homologous series). Two structures exist:

Isomer	Structure	Description
butane	CH3−CH2−CH2−CH3	straight chain of four carbons
methylpropane	(CH3)3CH	three-carbon chain with a CH3 branch on the middle carbon

That is the complete list. A “third isomer” of C4H10 is always butane wearing a disguise: the chain drawn with a corner in it. The test: count the longest continuous carbon chain. If it is four, the molecule is butane no matter how the page bends it.

The C4H8 set

C4H8 fits CnH2n, so these are alkenes, and the double bond adds a second kind of isomerism, positional isomerism:

• but-1-ene: CH2=CH−CH2−CH3, double bond between carbons 1 and 2
• but-2-ene: CH3−CH=CH−CH3, double bond between carbons 2 and 3
• methylpropene: CH2=C(CH3)−CH3, branched

Number the chain from the end nearer the double bond, and quote the lower-numbered carbon of the pair: the double bond in CH3−CH2−CH=CH2 starts at carbon 1 (counting from the right), so it is but-1-ene, not but-3-ene.

Alcohols show positional isomerism too: propan-1-ol (OH on the end carbon) and propan-2-ol (OH on the middle carbon) are the standard pair, relevant when alcohols questions go Extended.

How to find all the isomers, reliably

1. Draw the longest possible straight chain first.
2. Shorten the chain by one carbon and attach the spare as a branch; try each distinct position.
3. For alkenes and alcohols, slide the functional group along the chain, position by position.
4. Audit for duplicates: name every structure. Two drawings that earn the same name are the same isomer.

Step 4 is the one that saves marks. Naming is the duplicate detector: methylpropane can only be drawn one way once you try to name your “new” structure.

Worked exam question

(a) Define structural isomers. [2] (b) Draw the displayed formulae of the two isomers of C4H10 and name each one. [3] (c) Explain why CH3CH2CH2CH3 and the same chain drawn in an L-shape are not isomers of each other. [1]

Model answer, mark by mark:

• (a) M1: same molecular formula. M2: different structural formulae. (Both clauses required.)
• (b) M3: butane drawn with all bonds: four carbons in a chain, ten hydrogens each on its own bond. M4: methylpropane drawn fully: propane chain, CH3 branch on C2, all hydrogens shown. M5: both names correct and matched to the right structure.
• (c) M6: both drawings have the same structural formula. The longest chain is four carbons in each, so they are the same compound (butane); a bend on paper does not change the bonding.

The mistakes that cost marks

1. The bent-chain “isomer”. An L-shaped four-carbon chain is still butane. Count the longest chain before claiming a new structure.
2. Half a definition. “Same formula, different structure” without molecular and structural leaves the examiner to guess which formulae you mean. Use both technical terms.
3. Branches on the end carbon. A CH3 “branch” on carbon 1 just extends the chain. Methylpropane’s branch sits on the middle carbon. Anywhere else recreates butane.
4. Hydrogens lost in displayed formulae. Every isomer of C4H10 needs exactly ten drawn hydrogens. Count them against the molecular formula before moving on.
5. Numbering from the wrong end. But-3-ene does not exist; numbered from the nearer end it is but-1-ene.

How examiners want it phrased

Student wording	Mark-scheme wording
”Same atoms, different shape"	"Same molecular formula but different structural formulae"
"The double bond is in another place"	"But-1-ene and but-2-ene are positional isomers; the C=C is between different carbon atoms"
"It’s just bent, so it’s different"	"The longest carbon chain is unchanged, so the structural formula and the compound are the same"
"Branched ones are lighter"	"Branched isomers have lower boiling points than their straight-chain isomers”

Isomerism closes the Supplement strand of organic chemistry, and it rewards method over memory: chain, shorten, slide, name. Ten minutes of supervised drawing practice catches the duplicate habit for good, exactly the sort of fix we make in a free trial lesson.

Test yourself

Three Supplement questions. Answer from memory, then click to check.

Q1 (3 marks). Name the three structural isomers of C4H8 that contain a C=C double bond.

Show answer

• but-1-ene [1] • but-2-ene [1] • methylpropene [1]

Q2 (2 marks). Explain why propan-1-ol and propan-2-ol are structural isomers of each other.

Show answer

• they have the same molecular formula, C3H7OH (C3H8O) [1] • but different structural formulae: the −OH group is on the end carbon in propan-1-ol and on the middle carbon in propan-2-ol [1]

Q3 (2 marks). A student names the compound CH3CH2CH=CH2 “but-3-ene”. Give the correct name and explain the student’s error.

Show answer

• but-1-ene [1] • the chain must be numbered from the end nearer the double bond, and the lower-numbered carbon of the C=C is quoted [1]