Isomerism Notes for Class 11, JEE and NEET

Isomerism is part of the NCERT chapter Organic Chemistry – Some Basic Principles and Techniques (Chapter 8 in the current Class 11 Chemistry book, and Chapter 12 in older books). It is the base for naming organic compounds, for stereochemistry in Class 12, and for the “how many isomers” questions of JEE and NEET. These Isomerism Notes for Class 11 explain every type with examples, so that you can classify any pair of compounds.

Isomerism Notes for Class 11: What Is Isomerism?

Isomerism is the phenomenon in which two or more compounds have the same molecular formula but different structures or different arrangements of atoms in space. These compounds are called isomers.

Simple example: n-butane, $\mathrm{CH_{3}CH_{2}CH_{2}CH_{3}}$, and isobutane (2-methylpropane), $\mathrm{CH_{3}CH(CH_{3})CH_{3}}$, both have the molecular formula $\mathrm{C_{4}H_{10}}$, but they are different compounds with different properties.

Degree of Unsaturation

Before you count isomers, find the degree of unsaturation (the number of rings plus double bonds, with a triple bond counted as two):

$$\text{DoU} = \frac{2C + 2 + N - H - X}{2}$$

where $C$, $N$, $H$ and $X$ are the numbers of carbon, nitrogen, hydrogen and halogen atoms. For $\mathrm{C_{4}H_{8}}$, $\text{DoU} = \dfrac{2(4) + 2 - 8}{2} = 1$, so the compound has either one double bond or one ring.

India's Best Exam Preparation for Class 11th - Download Now

Isomerism Notes for Class 11

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

Classification of Isomerism in Isomerism Notes for Class 11

The Two-Branch Structure

ISOMERISM
│
├── STRUCTURAL ISOMERISM
│ (same molecular formula, different connectivity of atoms)
│ ├── Chain Isomerism
│ ├── Position Isomerism
│ ├── Functional Group Isomerism
│ ├── Metamerism
│ ├── Ring-Chain Isomerism
│ └── Tautomerism
│
└── STEREOISOMERISM
(same connectivity, different spatial arrangement)
├── Geometrical Isomerism (cis-trans)
└── Optical Isomerism (enantiomers, diastereomers)

Keep this tree diagram in your notes. Most isomerism questions in JEE and NEET fit somewhere in this structure.

Tip from the eSaral Chemistry team: When a question gives a pair of compounds, check first whether the connectivity is the same. If it differs, the pair is structural, and you then decide which of the six types applies. If it is the same, the pair is stereoisomeric. Cis–trans isomers are also called diastereomers, because they are stereoisomers that are not mirror images of each other.

Structural Isomerism in Isomerism Notes for Class 11

Chain Isomerism

The molecular formula is the same, but the carbon skeleton (the branching) is different. For $\mathrm{C_{5}H_{12}}$, there are three chain isomers:

Position Isomerism

The carbon skeleton and the functional group are the same, but the position of the functional group, the multiple bond or the substituent is different.

• $\mathrm{C_{3}H_{7}Cl}$: 1-chloropropane, $\mathrm{CH_{3}CH_{2}CH_{2}Cl}$, and 2-chloropropane, $\mathrm{CH_{3}CHClCH_{3}}$
• $\mathrm{C_{4}H_{8}}$: but-1-ene, $\mathrm{CH_{2}=CHCH_{2}CH_{3}}$, and but-2-ene, $\mathrm{CH_{3}CH=CHCH_{3}}$

Functional Group Isomerism

The molecular formula is the same, but the functional groups are different.

Metamerism

Metamers have the same functional group, but different alkyl groups on the two sides of it. It occurs in ethers, ketones, esters and secondary amines. For example, ethoxyethane, $\mathrm{CH_{3}CH_{2}OCH_{2}CH_{3}}$, and methoxypropane, $\mathrm{CH_{3}OCH_{2}CH_{2}CH_{3}}$, are metamers with the formula $\mathrm{C_{4}H_{10}O}$.

Ring-Chain Isomerism

One isomer has an open chain and the other has a ring. For $\mathrm{C_{3}H_{6}}$, propene, $\mathrm{CH_{2}=CHCH_{3}}$, and cyclopropane are ring-chain isomers.

Counting Structural Isomers in Isomerism Notes for Class 11

To count without missing any isomer, first find the degree of unsaturation, then list the carbon skeletons, and then place the functional group or double bond at each different position. Count a position from both ends to avoid counting the same isomer twice.

Stereoisomerism: Geometrical Isomerism in Isomerism Notes for Class 11

Conditions for geometrical (cis–trans) isomerism: there must be restricted rotation (a $\mathrm{C{=}C}$ double bond or a ring), and each carbon of the double bond must have two different groups.

• But-2-ene, $\mathrm{CH_{3}CH=CHCH_{3}}$, shows cis–trans isomerism.
• But-1-ene, $\mathrm{CH_{2}=CHCH_{2}CH_{3}}$, does not, because one carbon of the double bond has two identical hydrogen atoms.
• 1,2-Dichloroethene shows it, but 1,1-dichloroethene does not.

When the groups on the two carbons are all different, name the isomers as E or Z by the priority of the groups. If the higher-priority groups are on the same side, the isomer is Z, and if they are on opposite sides, it is E.

For a polyene with $n$ double bonds, each with different groups at its ends, the number of geometrical isomers is $2^n$.

Stereoisomerism: Optical Isomerism in Isomerism Notes for Class 11

A carbon bonded to four different groups is a chiral carbon (asymmetric carbon or stereocentre), and it is marked with an asterisk.

Example: lactic acid, $\mathrm{CH_{3}CH(OH)COOH}$, has one chiral carbon (bonded to $\mathrm{CH_{3}}$, $\mathrm{OH}$, $\mathrm{H}$ and $\mathrm{COOH}$), so it has two enantiomers.

Number of Optical Isomers

For example, tartaric acid has $n = 2$ in a symmetrical molecule, so it has $2^{1} + 2^{0} = 3$ stereoisomers: two enantiomers and one meso form.

Tautomerism in Isomerism Notes for Class 11

Tautomers are isomers that interconvert rapidly and exist in a dynamic equilibrium. The most important case is keto–enol tautomerism.

• Ethanal: the keto form is $\mathrm{CH_{3}CHO}$, and the enol form is $\mathrm{CH_{2}=CHOH}$.
• Tautomerism needs an α-hydrogen (a hydrogen on the carbon next to the carbonyl group). A compound without an α-hydrogen, such as benzaldehyde, does not show keto–enol tautomerism.
• The keto form is more stable in most simple compounds.

Key Points on Isomerism in Isomerism Notes for Class 11

Solved Examples from Isomerism Notes for Class 11

Example 1 (counting isomers): Write the structural isomers of $\mathrm{C_{4}H_{10}O}$.

There are 7: four alcohols (butan-1-ol, butan-2-ol, 2-methylpropan-1-ol and 2-methylpropan-2-ol) and three ethers (methoxypropane, ethoxyethane and 2-methoxypropane).

Example 2 (degree of unsaturation): How many structural isomers does $\mathrm{C_{4}H_{8}}$ have, and how many stereoisomers in all?

$\text{DoU} = \dfrac{2(4) + 2 - 8}{2} = 1$. The structural isomers are but-1-ene, but-2-ene, 2-methylpropene, cyclobutane and methylcyclopropane, which makes 5. But-2-ene also has cis and trans forms, so there are 6 isomers in all.

Example 3 (geometrical isomerism): Which of but-1-ene, but-2-ene and 1,1-dichloroethene show geometrical isomerism?

Only but-2-ene does. In but-1-ene and in 1,1-dichloroethene, one carbon of the double bond carries two identical groups.

Example 4 (chiral carbons): Find the number of chiral carbons in butan-2-ol, $\mathrm{CH_{3}CH(OH)CH_{2}CH_{3}}$, and in 2,3-dibromobutane.

Butan-2-ol has 1 chiral carbon (C-2), so it has 2 enantiomers. 2,3-Dibromobutane has 2 chiral carbons and is symmetrical, so it has $2^{1} + 2^{0} = 3$ stereoisomers: two enantiomers and one meso form.

Example 5 (maximum optical isomers): Open-chain glucose has 4 different chiral carbons. How many stereoisomers are possible?

$2^n = 2^4 = 16$ stereoisomers.

Example 6 (tautomerism): Which of propanone and benzaldehyde shows keto–enol tautomerism?

Propanone does, because it has α-hydrogens. Benzaldehyde has no α-hydrogen, so it does not.

Common Mistakes in Isomerism Notes for Class 11 Problems

How to Study Isomerism Notes for Class 11

1. Learn the classification first, and write one example of each type from memory. Revise the shapes of carbon compounds in eSaral’s chemical bonding notes for Class 11.
2. Practise the counting questions (the isomers of $\mathrm{C_{4}H_{10}O}$ and $\mathrm{C_{4}H_{8}}$) with the degree of unsaturation.
3. Drill the conditions for geometrical and optical isomerism on 10 structures every day for a week.
4. Revise with eSaral’s Class 11 Chemistry notes, and solve the isomerism questions from eSaral’s NEET chapter-wise PYQs.