Structural isomerism

Isomers are two or more compounds that have the same molecular formula but different structural formulae. Because the atoms are arranged differently, isomers are genuinely different compounds: they can have different melting points, boiling points, densities, and sometimes different chemical reactivity, even though a quick glance at their formula suggests they should be identical. When the difference lies in the carbon skeleton itself — a straight (unbranched) chain versus a branched chain — the phenomenon is called chain isomerism, one type of the broader category structural isomerism. At ICSE Class 10 the treatment stays descriptive: you compare two given structures, verify they share a molecular formula, and identify that the carbon connectivity differs. The anchor example is butane and isobutane (2-methylpropane), both $C_4H_{10}$:

• n-Butane (straight chain): $CH_3-CH_2-CH_2-CH_3$
• Isobutane (branched chain): $CH_3-CH(CH_3)-CH_3$

Both satisfy the alkane general formula

where: $n$ is the number of carbon atoms in the molecule, and $C_nH_{2n+2}$ is the formula every open-chain (acyclic) alkane must obey regardless of how its carbons are branched. Both butane and isobutane have $n=4$, so both land on $C_4H_{10}$ — but n-butane's four carbons form one continuous chain, while isobutane has a central carbon bonded to three other carbons (a branch point), so the connectivity is not the same molecule redrawn. Structural isomerism is not limited to chain branching. A second common pattern at this level is position isomerism, where the skeleton is identical but a double bond (or other feature) sits at a different position along the same chain — for example but-1-ene and but-2-ene, both $C_4H_8$, differing only in where the C=C bond falls.

Isomerism has no single grounded formula to derive — it is a rule of chemical bookkeeping, and its logic is worth stating precisely because it explains why two structures with an identical count of every element can still behave as different substances. Carbon almost always forms exactly four bonds, and a given set of atoms can be connected around those four bonds in more than one valid way while still satisfying every atom's normal valency. Step 1. Fix the molecular formula — the total count of each element's atoms — as the starting constraint; for $C_4H_{10}$ this means exactly 4 carbon atoms and exactly 10 hydrogen atoms, no more and no fewer, must appear in the finished structure. Step 2. Ask how those 4 carbon atoms can be joined to each other using single bonds while leaving every carbon with exactly four bonds total (some to carbon, the rest filled by hydrogen). There are exactly two ways to link 4 carbons in an open chain: one continuous line of four (n-butane) or three carbons hanging off a central fourth carbon (isobutane) — a branched tree instead of a line. Step 3. Once the carbon skeleton is fixed, hydrogens fill every remaining bond automatically, and in both cases the count works out to 10 hydrogens, because each structure still has the same total number of carbon-carbon and carbon-hydrogen bonds overall — branching moves bonds around, it does not create or destroy them. Step 4. Because the two skeletons are genuinely different arrangements of the same four carbons, the molecules have different shapes: n-butane is a long zig-zag, isobutane is more compact and star-shaped. Shape affects how strongly neighbouring molecules attract each other (weak intermolecular forces), which is why isomers can have different physical properties such as boiling point, even while their molecular formula — and hence their molecular mass — stays identical. This is the reasoning that justifies calling butane and isobutane different compounds rather than the same compound drawn twice.

Question: A student is given two structures, $CH_3-CH_2-CH_2-CH_3$ (n-butane) and $CH_3-CH(CH_3)-CH_3$ (isobutane), and asked to prove they are isomers, then state one physical consequence of their different structures.

Step 1. Count atoms in n-butane. It is a straight chain of 4 carbons: the two end carbons each carry 3 hydrogens (since each is bonded to only one other carbon, leaving 3 bonds free for H), and the two middle carbons each carry 2 hydrogens (bonded to two carbons, leaving 2 bonds free). Total H $= 3+2+2+3 = 10$. Molecular formula: $C_4H_{10}$.

Step 2. Count atoms in isobutane. The central carbon is bonded to 3 other carbons, using 3 of its 4 bonds, so it carries only 1 hydrogen. Each of the 3 surrounding carbons is bonded to just the central carbon, so each carries 3 hydrogens ($CH_3$ groups). Total H $= 1 + 3+3+3 = 10$. Molecular formula: $C_4H_{10}$.

Step 3. Compare the two results: both give $C_4H_{10}$, confirming the molecular formula is identical, but the structural formula is not — n-butane is an unbranched chain of 4 carbons, isobutane has a branch at the central carbon. Since same molecular formula plus different structural formula is exactly the definition of isomerism, n-butane and isobutane are structural (chain) isomers.

Step 4. State the physical consequence: because their shapes differ (long chain versus compact branched shape), the two molecules experience different strengths of intermolecular attraction, so they have slightly different boiling points despite sharing the same molecular formula and the same molecular mass — direct evidence that structure, not formula alone, governs a compound's properties.

Competitive stretch

The same recognition skill extends to alkenes. Given $CH_2=CH-CH_2-CH_3$ (but-1-ene) and $CH_3-CH=CH-CH_3$ (but-2-ene): both count out to 4 carbons and 8 hydrogens (verify: 2+1+2+3 = 8, and 3+1+1+3 = 8), so both are $C_4H_8$, matching $C_nH_{2n}$ for $n=4$. The chain length is identical in both — the only difference is where the C=C double bond sits (between carbons 1–2, versus between carbons 2–3). Same molecular formula, different structural formula (this time by bond position rather than branching), so but-1-ene and but-2-ene are also isomers of each other.

• Believing isomers are "the same compound, just drawn differently" — isomers are genuinely different compounds and can differ in boiling point, density, and chemical behaviour; they merely share one molecular formula.
• Assuming any two compounds with the same number of carbon atoms are automatically isomers — the entire molecular formula (every element's atom count, including hydrogen) must match exactly, not just the carbon count.
• Miscounting hydrogens on a branched carbon — forgetting that a carbon bonded to three other carbons has only one bond left for hydrogen is the single most common arithmetic slip in this topic.
• Trying to memorise formal isomerism categories (chain, position, functional) as a rigid syllabus list — at ICSE Class 10 this is a descriptive recognition skill using simple worked pairs, not a formal classification exercise.

Ask the tutor: Want to practise with a few more structure pairs, or see how branching changes a molecule's actual 3D shape? Ask the AI tutor to work through more isomer examples with you.