Isomerism

Isomerism is the phenomenon in which two or more compounds have an identical molecular formula but different structures, and hence different physical and/or chemical behaviour. The defining test is simple to state and easy to apply: fix the molecular formula, then ask whether the atoms are actually joined together (and arranged in space) the same way.

where: the subscripts $a, b, c, \ldots$ are the atom counts of each element — identical on both sides by definition of "isomer" — while "structure" stands for either the bonding sequence (connectivity) or the 3-D spatial arrangement, at least one of which must differ. Isomerism splits into two branches, and Class 11 develops only the first in detail.

A. Structural (constitutional) isomerism — the bonding sequence itself differs. Four named sub-types, distinguished by exactly what about the connectivity changes:

• Chain isomerism — the carbon skeleton's branching pattern differs, functional group unchanged. Butane $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3$ vs 2-methylpropane $(\text{CH}_3)_3\text{CH}$, both $\text{C}_4\text{H}_{10}$.
• Position isomerism — the skeleton and functional group are both the same; only where the group (or a multiple bond) sits on that skeleton changes. Propan-1-ol vs propan-2-ol, both $\text{C}_3\text{H}_8\text{O}$.
• Functional-group isomerism — the isomers belong to entirely different compound families. Ethanol (alcohol) vs dimethyl ether (ether), both $\text{C}_2\text{H}_6\text{O}$.
• Metamerism — alkyl groups are distributed differently on either side of the same divalent atom/group (–O–, –N<, –COO–, and similar), with both isomers staying in the same class. Diethyl ether vs methyl propyl ether, both $\text{C}_4\text{H}_{10}\text{O}$, both ethers.

B. Stereoisomerism — the bonding sequence is identical; only the 3-D spatial arrangement differs. It includes geometrical (cis-trans, from restricted rotation around a double bond) and optical (from chirality) isomerism. NCERT Class 11 asks only that you know this category exists and roughly how it differs conceptually from structural isomerism — full R/S and cis-trans work is a Class 12 topic and is deliberately out of scope here.

The four-way classification is not an arbitrary list to memorise — it falls out of asking a fixed sequence of questions about two isomers, one at a time, and each question isolates exactly one structural feature.

Step 1. First confirm they really are isomers at all: write out both molecular formulas and check every atom count matches. If they do not match, the pair is simply two different compounds, not isomers, and the classification exercise does not even begin.

Step 2. Compare the carbon skeleton — the pattern of carbon-to-carbon bonds, ignoring for a moment what functional group sits where. If the skeleton itself is branched differently (a straight chain versus a branched one, for instance), that difference alone is enough to call it chain isomerism, regardless of what happens next.

Step 3. If the skeleton is identical, look at the functional group. Ask two things in order: is it the same kind of group in both molecules, and if so, does it sit at the same position on the skeleton? Same group, different position, same skeleton gives position isomerism. Different kind of group entirely — different compound family altogether — gives functional-group isomerism.

Step 4. A special case of "different kind of group" needs its own rule: if both compounds are of the same class (both ethers, both esters, both secondary amines) but the alkyl groups are shuffled around a shared divalent linking atom or group, that shuffle is metamerism, not plain functional-group isomerism — the family stays fixed, only the alkyl split changes. This is why the sequence in Step 3 must be followed carefully: check same-class-different-alkyl-split before concluding functional-group isomerism.

Step 5. Only after all of structure, group and position are found identical do you reach for stereoisomerism — same connectivity throughout, different arrangement in 3-D space. Class 11 stops here: you are expected to recognise that this fifth possibility exists, not to draw or name specific geometrical or optical isomers.

Identify the type of isomerism between $\text{CH}_3{-}\text{CH}_2{-}\text{CH}_2{-}\text{CH}_2{-}\text{OH}$ (butan-1-ol) and $\text{CH}_3{-}\text{CH}(\text{OH}){-}\text{CH}_2{-}\text{CH}_3$ (butan-2-ol).

Step 1. Check the molecular formulas. Both structures are $\text{C}_4\text{H}_{10}\text{O}$ — four carbons, ten hydrogens, one oxygen, on both sides — so they are confirmed isomers of each other before any further comparison is needed.

Step 2. Compare the carbon skeleton. Both molecules have the same unbranched four-carbon chain (a straight butane backbone), so the skeleton itself is identical. This rules out chain isomerism immediately.

Step 3. Compare the functional group and its position. Both carry the same functional group, –OH (an alcohol). In butan-1-ol the –OH sits on C1, the end carbon; in butan-2-ol it sits on C2, an internal carbon. The group is the same kind, but its position on the chain differs.

Step 4. Conclude. Identical skeleton, identical functional group, different position of that group: this pair shows position isomerism.

Competitive stretch

Now classify $\text{CH}_3{-}\text{COOH}$ (ethanoic acid) against $\text{H}{-}\text{COO}{-}\text{CH}_3$ (methyl methanoate), both $\text{C}_2\text{H}_4\text{O}_2$.

Step 1. Ethanoic acid carries a carboxylic acid group, –COOH. Methyl methanoate carries an ester linkage, –COO–, and belongs to an entirely different family of compounds. Because the two molecules are of different classes despite sharing a formula, this is functional-group isomerism, not metamerism — metamerism would require both compounds to already be esters, only differing in how the alkyl groups are split across the –COO– linkage (methyl propanoate vs ethyl ethanoate, both $\text{C}_4\text{H}_8\text{O}_2$ and both esters, is the metameric pair to contrast against). Answer: functional-group isomerism.

• Jumping straight to "position isomerism" without first checking the skeleton — a different branching pattern always overrides a superficial resemblance in functional-group placement; skeleton comes before position.
• Calling every same-class pair with different alkyl groups "metamerism" without checking that the divalent linking group (–O–, –COO–, etc.) is actually shared — metamerism has this specific structural requirement, it is not just "same family, different formula split."
• Treating functional-group isomers as chemically similar because the molecular formula matches — an alcohol and an ether with the same formula behave completely differently in reactions; always identify the actual functional group present.
• Attempting to assign cis-trans or R/S labels at Class 11 level — the syllabus only requires knowing stereoisomerism exists as a category, not classifying specific stereoisomers.

Ask the tutor: want a mixed set of unlabelled isomer pairs to classify one by one using the Step 1-5 decision order above, or a short preview of how geometrical and optical isomerism will build on this foundation next year?