Molecules whose mirror image is non superimposable over them are known as chiral. Which of the following molec

NCERT Exemplar · Q22

Q. Molecules whose mirror image is non superimposable over them are known as chiral. Which of the following molecules is chiral in nature?

(i) 2-Bromobutane

(ii) 1-Bromobutane

(iii) 2-Bromopropane

(iv) 2-Bromopropan-2-ol

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Concept understanding — Geometrical Isomerism Conditions

Geometrical Isomerism: The Intuition

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:

C=C

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 ( $\pi$ ) bond that locks the molecule flat. Rotation would break the $\pi$ 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 ( $CH_3CH=CHCH_3$ )

Carbon 1 of the double bond: attached to $CH_3$ and $H$ → different ✓
Carbon 2 of the double bond: attached to $CH_3$ 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 bond

trans-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 ( $Cl_2C=CH_2$ )

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:

\text{C} = \text{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 ( $\text{ClHC=CHCl}$ ): Each carbon has H and Cl (different) → geometrical isomers exist.
1,1-dichloroethene ( $\text{Cl}_2\text{C=CH}_2$ ): 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). …

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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: $\mathrm{CH_3-CHBr-CH_2-CH_3}$

Number the carbons:

C1: $\mathrm{CH_3-}$ (three H's, one C — not a chiral centre)
C2: $\mathrm{-CHBr-}$ $- CHBr -$ — this carbon is bonded to:
- a hydrogen (H)
- a bromine (Br)
- a methyl group ( $\mathrm{CH_3-}$ )
- an ethyl group ( $\mathrm{-CH_2CH_3}$ )

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: $\mathrm{CH_2Br-CH_2-CH_2-CH_3}$

C1: $\mathrm{-CH_2Br}$ — 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 $\mathrm{CH_2}$ group). No chiral centre.

The molecule is achiral. …

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This is a classic trap in stereochemistry. Let's break down the common mistakes students make when tackling this exact problem, and how to avoid each.

✗ Mistake 1: Confusing "chiral" with "having a chiral centre"

Many students think: If a molecule has a chiral carbon, it must be chiral.

But that's not always true — a molecule can have chiral centres and still be achiral if it has a plane of symmetry (meso compound).

✓ How to avoid:

Always check for internal symmetry (plane of symmetry) before concluding chirality.
A chiral centre is a necessary but not sufficient condition for chirality.

✗ Mistake 2: Forgetting to check for symmetry in the whole molecule

Students often look only at one carbon and ignore the rest of the molecule.

For example, in 2-Bromopropan-2-ol (iv), the central carbon has four different groups? Let's check:

Carbon: attached to –Br, –OH, –CH₃, and –CH₃ → Two identical methyl groups → not a chiral centre.

✓ How to avoid:

Draw the full structure.
Check every substituent on the carbon in question — if any two are identical, it's not a chiral centre.

✗ Mistake 3: Assuming all halogenated alkanes are chiral

Just because a molecule has a bromine atom doesn't make it chiral.

Example: 1-Bromobutane (ii) has Br at the end — the carbon with Br is attached to two H atoms → achiral.

✓ How to avoid:

A carbon must have four different groups to be a chiral centre.
Count groups carefully: –H counts as a group!

✗ Mistake 4: Misidentifying the chiral centre in 2-Bromobutane

2-Bromobutane (i) has the structure:

CH₃–CHBr–CH₂–CH₃

The carbon with Br is attached to:

–H
–Br
–CH₃
–CH₂CH₃

All four are different → chiral centre exists.

No plane of symmetry → molecule is chiral.

✓ How to avoid:

Write the full condensed formula.
List the four groups explicitly.
Check for symmetry in the whole molecule.

✗ Mistake 5: Overlooking that 2-Bromopropane is symmetric

2-Bromopropane (iii):

CH₃–CHBr–CH₃

The central carbon is attached to:

–H
–Br
–CH₃
–CH₃

Two identical methyl groups → not a chiral centre. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
$K[Cr(H_2O)_2(C_2O_4)_2]$
›Reveal solutionSolution
Geometrical isomerism requires at least two different possible spatial arrangements of ligands. $M(AA)_2B_2$ (complex I) has cis/trans forms; $M(AA)_3$ (II) and $MA_5B$ (III) each have only one possible arrangement, so they do not show geometrical isomerism.

Concept and Intuition

Whether an octahedral complex shows geometrical (cis-trans) isomerism depends on its ligand-substitution pattern, not just on having a mix of ligands:

$M(AA)_2B_2$ (two bidentate symmetric chelates + two monodentate ligands): the two B ligands can be adjacent (cis) or opposite (trans) — genuine geometrical isomerism exists (and the cis form is additionally chiral).

$M(AA)_3$ (three identical bidentate chelates, e.g. $[Co(en)_3]^{3+}$ ): by symmetry there is only one way to arrange three identical chelate rings around the octahedron — no cis/trans distinction is possible. Only optical isomerism (mirror-image Δ and Λ forms) exists.

$MA_5B$ (five identical monodentate ligands + one different one): since all six octahedral positions are equivalent by symmetry when five ligands are identical, placing the lone B ligand at "any" position gives the same single structure — no geometrical isomerism is possible.

Step-by-Step Solution

I: $K[Cr(H_2O)_2(C_2O_4)_2]$ = $[Cr(C_2O_4)_2(H_2O)_2]^-$ , an $M(AA)_2B_2$ complex → cis and trans forms exist → does show geometrical isomerism. …
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AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
$Q. Which of the following exhibit cis-trans isomerism? (I) 2-Methylpent-2-ene (II) Styrene (III) 2-Chlorobut-2-ene (IV) 1-Phenylprop-1-ene The correct answer is (A) I & II only (B) III & IV only (C) I & III only (D) II & IV only$
›Reveal solutionSolution
This tests the basic criterion for geometrical (cis–trans) isomerism: each alkene carbon must bear two different substituents. Only 2-chlorobut-2-ene and 1-phenylprop-1-ene qualify.

Concept and Intuition

Geometric (cis-trans) isomerism about a C=C double bond arises only when restricted rotation is combined with each sp2 carbon of the double bond carrying two non-identical groups. If either carbon has two identical substituents, the molecule and its "other geometry" are actually the same compound — no isomerism.

Step-by-Step Solution

2-Methylpent-2-ene: $CH_3{-}C(CH_3){=}CH{-}CH_2{-}CH_3$ . The C2 (left alkene carbon) bears two methyl groups — identical substituents — so no cis-trans isomerism, regardless of what's on C3.

Styrene: $C_6H_5{-}CH{=}CH_2$ . The terminal alkene carbon ( $=CH_2$ ) carries two hydrogens — identical — so no cis-trans isomerism.

2-Chlorobut-2-ene: $CH_3{-}CCl{=}CH{-}CH_3$ . C2 carries $CH_3$ and $Cl$ (different); C3 carries $H$ and $CH_3$ (different). Both alkene carbons have two different groups → geometric isomerism exists (cis and trans forms). …
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AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
$Q. Which of the following exhibit only geometrical isomerism? (A) Diaquadioxalatochromate (III) ion (B) Dichloridobis(ethane-1, 2-diamine)platinum (IV) ion (C) Triamminetrinitrito - N cobalt (III) (D) Tris(ethane-1, 2-diamine)cobalt (III) ion$
›Reveal solutionSolution
Comparing the isomerism possibilities of each octahedral coordination-compound type ( $M(AA)_2B_2$ , $MA_3B_3$ , $M(AA)_3$ ) shows that only the $MA_3B_3$ -type complex is restricted to geometrical (fac/mer) isomerism alone. The answer is (C).

Concept and Intuition

For octahedral complexes, the type and combination of ligands determines what kinds of isomerism are possible. A complex of type $M(AA)_2B_2$ (two bidentate chelating ligands plus two monodentate ligands) can exist as cis and trans geometrical isomers; critically, the cis isomer of this type lacks any plane of symmetry and is chiral (shows optical isomerism), while the trans isomer is not chiral. A complex of type $M(AA)_3$ (three identical bidentate chelating ligands) has only one possible geometric arrangement (no cis/trans distinction exists), so it shows only optical isomerism, always as a pair of non-superimposable mirror-image (Δ/Λ) forms. A complex of type $MA_3B_3$ can arrange its ligands as facial (fac, three of one type on one triangular face) or meridional (mer, three of one type in a plane through the metal) — both of these arrangements possess a mirror plane of symmetry and are therefore achiral, so this type shows geometrical isomerism only, with no optical activity.

Step-by-Step Solution

(A) Diaquadioxalatochromate(III) ion, $[Cr(C_2O_4)_2(H_2O)_2]^-$ : type $M(AA)_2B_2$ . Its cis isomer is chiral (optically active) and its trans isomer is not — so this complex shows both geometrical and optical isomerism, not geometrical isomerism alone.

(B) Dichloridobis(ethane-1,2-diamine)platinum(IV) ion, $[Pt(en)_2Cl_2]^{2+}$ : also type $M(AA)_2B_2$ , with the same situation as (A) — cis form chiral, trans form not, so both geometrical and optical isomerism are present.

(C) Triamminetrinitrito–N cobalt(III), $[Co(NH_3)_3(NO_2)_3]$ : type $MA_3B_3$ . This shows fac and mer geometrical isomers. Both the fac isomer ( $C_{3v}$ symmetry, has mirror planes) and the mer isomer ( $C_s$ symmetry, has a mirror plane) are achiral. So this complex shows only geometrical isomerism, with no optical isomerism at all — this is the answer. …
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AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
$[\mathrm{Co(en)(NH_3)_2Cl_2}]\mathrm{Cl}$
›Reveal solutionSolution
Tests which octahedral complex types show geometrical (cis-trans) isomerism; the answer is (C) I, II & IV only since the tris-chelate $[Co(en)_3]^{3+}$ has no cis/trans forms.

Concept and Intuition

Geometrical isomerism (cis-trans) in octahedral complexes arises when two or more identical ligands (or ligating groups) can occupy either adjacent (cis) or opposite (trans) positions. Complexes of type $MA_4B_2$ , $MA_3B_3$ , and mixed-ligand types with two identical monodentate ligands like $M(AA)B_2C_2$ or $M(AA)_2B_2$ show this. However, a complex where all three ligand positions are filled by the same symmetric bidentate chelate, i.e. $M(AA)_3$ , has only ONE possible geometric arrangement (the chelate rings are geometrically forced into one shape) — such complexes show only optical isomerism (as non-superimposable mirror images), never geometrical isomerism.

Step-by-Step Solution

I) $[Co(en)(NH_3)_2Cl_2]^+$ : ligand set is one en (bidentate, counts as occupying 2 cis sites) + 2 NH3 + 2 Cl. The two Cl's can be cis or trans to each other → geometrical isomerism exists.

II) $[Co(NH_3)_4Cl_2]^+$ : type $MA_4B_2$ — the two Cl ligands can be cis (adjacent) or trans (opposite) → geometrical isomerism exists. …
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AP EAPCET 2021Set ap-2021-09-06-FN1 markMCQ
$C_4H_6Cl_2$
›Reveal solutionSolution
Dichlorocyclobutane ( $C_4H_6Cl_2$ ) has three substitution patterns (1,1-, 1,2-, 1,3-); only the 1,2- and 1,3- patterns show cis/trans isomerism, each contributing a cis and a trans form, giving 4 cis/trans isomers in total.

Concept and Intuition

Cyclobutane ( $C_4H_8$ ) with two hydrogens replaced by chlorine gives dichlorocyclobutane, $C_4H_6Cl_2$ . Because the ring holds the carbon skeleton rigid (no free rotation around the ring bonds the way there is in an open chain), placing two substituents on ring carbons in different relative positions can create genuine, non-interconvertible geometric (cis/trans) isomers — exactly analogous to cis/trans isomerism in cyclic compounds generally. The key is that cis/trans isomerism requires two different substituents on each of two ring carbons that are directly compared (i.e., no ring carbon carrying two identical Cl's), so a substitution pattern with both Cl atoms on the same carbon cannot show cis/trans isomerism at all.

Step-by-Step Solution

Enumerate the possible relative positions of two Cl atoms on a four-membered ring: 1,1- (geminal, same carbon), 1,2- (adjacent carbons), and 1,3- (opposite/across the ring).

1,1-dichlorocyclobutane: both Cl's are on one carbon; that carbon has no distinguishable "up/down" substituent pair to compare across the ring, so no cis/trans isomerism is possible here — it is a single compound.

1,2-dichlorocyclobutane: the two Cl's are on adjacent ring carbons, each of which also bears an H; the two Cl's can be on the same face of the ring (cis) or on opposite faces (trans) — 2 distinct isomers. …
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Q.Molecules whose mirror image is non superimposable over them are known as chiral. Which of the following molecules is chiral in nature?

Why chirality matters — and how to spot it

Step-by-step analysis

1. 2-Bromobutane

2. 1-Bromobutane

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