Imagine you have two friends standing on opposite sides of a door. If the door is open, they can walk around and swap places easily — there's no real difference between who is on the left and who is on the right. But if the door is locked shut, they are stuck. One is permanently on the left side, the other on the right. That locked door creates two distinct arrangements: Friend A on the left, Friend B on the right versus Friend A on the right, Friend B on the left.
That locked door is the key idea behind geometrical isomerism.
In chemistry, molecules are three-dimensional. Atoms connected by a single bond can rotate freely — like an open door. But a double bond (or a ring structure) locks the atoms in place. If you have two different groups attached to each carbon of a double bond, you get two distinct spatial arrangements that cannot interconvert without breaking the bond. These are geometrical isomers (also called cis-trans or E-Z isomers).
The Precise Conditions
For a molecule to show geometrical isomerism, it must satisfy two conditions simultaneously:
Condition 1: There must be a restricted rotation around a bond — typically a carbon-carbon double bond (C=C) or a ring structure.
Condition 2: Each of the two atoms (or groups) involved in that restricted rotation must have two different substituents attached to it.
Let's unpack each.
Condition 1: Restricted Rotation
A single bond (C−C) allows free rotation — the atoms spin around the bond axis like a wheel. So no geometrical isomers exist there. A double bond (C=C) has a pi (π) bond that locks the molecule flat. Rotation would break the π bond, which requires a lot of energy (about 250–270 kJ/mol). At room temperature, this rotation simply does not happen.
Rings (like cyclopropane, cyclobutane, etc.) also restrict rotation because the ring is a closed loop — atoms cannot rotate past each other without breaking the ring.
Condition 2: Two Different Substituents on Each End
This is the "different groups" rule. Look at each carbon of the double bond (or each ring carbon involved). If both carbons have two different groups attached, geometrical isomers exist. If even one carbon has two identical groups, there is only one possible arrangement.
Watch out
A common mistake: students check only one carbon. Both carbons must have two different substituents. If one carbon has two identical groups (like two hydrogens), the molecule is identical in both arrangements — no isomerism.
How to Check: A Step-by-Step Method
Take any molecule with a double bond. Follow these steps:
Identify the double bond (or ring). Mark the two carbon atoms involved.
List the two groups attached to the first carbon. Are they different from each other? If yes, proceed. If no → no geometrical isomerism.
List the two groups attached to the second carbon. Are they different from each other? If yes → geometrical isomerism exists. If no → no geometrical isomerism.
Tip
If the two groups on a carbon are identical, the molecule is symmetric about that carbon. Flipping the other side gives the same molecule — no isomers.
Examples to Cement the Idea
Example 1: But-2-ene (CH3CH=CHCH3)
Carbon 1 of the double bond: attached to CH3 and H → different✓
Carbon 2 of the double bond: attached to CH3 and H → different✓
Result: Two geometrical isomers exist — cis (both methyl groups on the same side) and trans (methyl groups on opposite sides).
cis-But-2-ene, with both methyl groups on the same side of the C=C double bondtrans-But-2-ene, with the two methyl groups on opposite sides of the C=C double bond
Example 2: 1,2-Dichloroethene (ClCH=CHCl)
Carbon 1: attached to Cl and H → different ✓
Carbon 2: attached to Cl and H → different ✓
Result: cis and trans isomers exist.
Example 3: 1,1-Dichloroethene (Cl2C=CH2)
Carbon 1: attached to Cl and Cl → identical✗ …
Why this formula?
Geometrical Isomerism: Why the Conditions Hold
Geometrical isomerism (also called cis-trans or E-Z isomerism) arises when atoms or groups are arranged differently in space around a rigid part of a molecule — typically a double bond or a ring. The key is that rotation is restricted, so the spatial positions become fixed and distinct.
Let’s break down why the conditions are what they are.
1. The Core Requirement: Restricted Rotation
For two molecules to be geometrical isomers, they must have the same connectivity but different spatial arrangement due to a barrier to rotation.
Double bonds (C=C): The π-bond locks the two carbons in place — rotation requires breaking the π-bond (energy ~250 kJ/mol), so it doesn’t happen at room temperature.
Rings (e.g., cycloalkanes): The ring structure physically prevents free rotation about C–C single bonds within the ring.
Why this matters: Without restricted rotation, the molecule would freely interconvert between arrangements — no distinct isomers exist.
2. Condition 1: Two Different Groups on Each Carbon (for C=C)
Consider a general alkene:
C=C
Each carbon must have two different substituents (not counting the other carbon of the double bond).
Why?
If one carbon has two identical groups (e.g., both H), then swapping the groups on that carbon produces the same molecule — no isomerism.
Example:
1,2-dichloroethene (ClHC=CHCl): Each carbon has H and Cl (different) → geometrical isomers exist.
1,1-dichloroethene (Cl2C=CH2): One carbon has two Cl (identical) → no geometrical isomers.
Formal condition:
For a C=C bond, geometrical isomerism is possible iff each doubly bonded carbon bears two different substituents.
3. Condition 2: For Rings — Similar Logic
In a ring (e.g., cyclopropane, cyclohexane), the ring itself restricts rotation. Here, geometrical isomerism occurs when two substituents on different ring carbons can be on the same side (cis) or opposite sides (trans).
Why?
The ring is a closed loop — you cannot rotate one carbon relative to another without breaking bonds.
If the two substituents are on different carbons, their relative orientation (same side / opposite sides) is fixed.
Condition:
The ring must have at least two substituents (could be same or different) on different carbons.
If both substituents are on the same carbon, swapping them doesn’t change the molecule (no isomerism).
Example:
1,2-dimethylcyclopropane: Two methyl groups on adjacent carbons → cis and trans isomers exist.
1,1-dimethylcyclopropane: Both methyls on same carbon → no geometrical isomerism.
4. The E-Z Notation (Why It’s Needed)
When the four substituents on a C=C are all different, cis-trans naming fails. The Cahn-Ingold-Prelog priority rules assign E (opposite sides) or Z (same side).
Why this works:
Priority is based on atomic number (higher = higher priority). …
Concept: Chirality requires a carbon bonded to four different substituents (a chiral centre). A molecule is chiral if it is non-superimposable on its mirror image.
Reasoning:
Check each molecule for a chiral carbon.
(i) 2-Bromobutane: CH3CHBrCH2CH3 — the second carbon is attached to H, Br, CH3, and CH2CH3 (all different).
(ii) 1-Bromobutane: CH2BrCH2CH2CH3 — no carbon with four different groups.
(iii) 2-Bromopropane: CH3CHBrCH3 — the second carbon has two identical methyl groups, so it is achiral. …
A molecule is chiral if it has a carbon atom bonded to four different groups (a chiral centre). Among the given options, only 2-Bromobutane has such a carbon, making it the chiral molecule.
Why chirality matters — and how to spot it
Chirality is a property of molecular handedness: a chiral molecule and its mirror image cannot be superimposed, like your left and right hands. For most organic molecules at the JEE/NEET level, chirality arises from a stereogenic centre — typically a carbon atom with four different substituents. If any two groups on that carbon are identical, the molecule is achiral (it has a plane of symmetry).
So the task is simple: check each molecule for a carbon with four distinct attachments.
Step-by-step analysis
1. 2-Bromobutane
Structure: CH3−CHBr−CH2−CH3
Number the carbons:
C1: CH3− (three H's, one C — not a chiral centre)
C2: −CHBr− — this carbon is bonded to:
a hydrogen (H)
a bromine (Br)
a methyl group (CH3−)
an ethyl group (−CH2CH3)
All four groups are different. Therefore, C2 is a chiral centre. The molecule exists as a pair of enantiomers.
Tip
A quick check: if the carbon is attached to four different atoms or groups (count the atoms directly attached, then look at the next sphere if needed), it's chiral. Here, H, Br, CH₃, and CH₂CH₃ are all distinct.
2. 1-Bromobutane
Structure: CH2Br−CH2−CH2−CH3
C1: −CH2Br — two hydrogens, one bromine, one carbon. Two H's are identical → not chiral.
C2, C3, C4: each has at least two identical substituents (e.g., two H's on a CH2 group). No chiral centre.
Method: Chirality Detection via Asymmetric Carbon (Stereocenter) Analysis
Concept First — Why This Works
A molecule is chiral if it has a non-superimposable mirror image. The most common cause is the presence of an asymmetric carbon (a carbon bonded to four different groups). If no such carbon exists, the molecule is usually achiral (superimposable on its mirror image).
Steps
Draw the structure of each molecule (condensed or line formula).
Identify each carbon that is bonded to four different atoms/groups.
Check for symmetry — even if a carbon has four different groups, the molecule may still be achiral if it has a plane of symmetry.
Conclude: If at least one asymmetric carbon exists and the molecule lacks a plane of symmetry, it is chiral.
Applying to the Options
(i) 2-Bromobutane
Structure: CH3−CHBr−CH2−CH3
Carbon-2: bonded to H, Br, CH3, CH2CH3 — four different groups✓
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
AHSEC Higher Secondary (HS) Final Examination 2022Set ANNUAL1 mark
Q.How many geometrical isomers will be possible for [Pt(Py)(NH3)BrCl] compound?
OR
Write the hybridization state of central atom of the following co-ordination complexes along with their magnetic properties: [Cu(NH3)4]2+ and [Ni(CO)4].
›Reveal solutionSolution
A square-planar complex with 4 different unidentate ligands (type Mabcd) has exactly 3 possible geometric (cis/trans) arrangements, one for each choice of which ligand pair sits trans to each other.
[Pt(Py)(NH3)BrCl] is a square-planar complex (Pt2+, d8, 4-coordinate) bearing 4 DIFFERENT unidentate ligands: pyridine (Py), ammine (NH3), bromido (Br), and chlorido (Cl) — this is the general type Mabcd.
In a square-planar arrangement, each ligand has one ligand directly opposite (trans) to it and two adjacent (cis) to it. With 4 different ligands, the isomers differ based on which specific pair of ligands ends up trans to each other. There are 3 distinct ways to pair up the 4 different ligands into two trans-pairs:
(Py trans to NH3) with (Br trans to Cl)
(Py trans to Br) with (NH3 trans to Cl)
(Py trans to Cl) with (NH3 trans to Br)
Each of these gives a structurally distinct (non-superimposable) square-planar isomer, so a total of 3 geometrical isomers are possible.