Structural Isomerism and Stereoisomerism Practice Questions — JAMB UTME

Question 1Question →

Propanal and propanone are functional group isomers because they share the same molecular formula, $C_3H_6O$ , but belong to different homologous series with distinct functional groups.

True

False

Show answer & explanation

Answer: True

Answer

The statement is TRUE. Propanal and propanone share the same molecular formula (

C_3H_6O

) but possess different functional groups (aldehyde vs. ketone), which exemplifies functional group isomerism.

The statement is correct because propanal and propanone both have the molecular formula

C_3H_6O

but contain different functional groups (an aldehyde group and a ketone group, respectively), meeting the exact definition of functional group isomerism.

Step-by-Step Solution

1

Determine the molecular formula of propanal and propanone.

Propanal (

CH_3CH_2CHO

) has 3 carbon atoms, 6 hydrogen atoms, and 1 oxygen atom (

C_3H_6O

). Propanone (

CH_3COCH_3

) also has 3 carbon atoms, 6 hydrogen atoms, and 1 oxygen atom (

C_3H_6O

).

Isomers must share the exact same molecular formula.

2

Identify the functional groups present in both compounds.

Propanal contains an aldehyde group (

-\text{CHO}

), whereas propanone contains a ketone group (

-\text{CO}-

).

Structural isomers containing different functional groups belong to different homologous series.

3

Evaluate the relationship against the definition of functional group isomerism.

Since both compounds share the molecular formula

C_3H_6O

but differ in functional groups, the statement is true.

This directly satisfies the criteria for functional group isomerism.

Key Concept

Functional group isomerism is a form of structural isomerism where compounds have the same molecular formula but different functional groups.

Question 2Question →

For an organic compound to exhibit optical isomerism, it must possess a chiral carbon atom bonded to four different groups or atoms. Which of the following compounds possesses a chiral carbon atom and exhibits optical activity?

2-chlorobutane

1-chlorobutane

2-chloropropane

2-methylpropan-2-ol

Show answer & explanation

Answer: 2-chlorobutane

Answer

2-chlorobutane is the correct answer because its second carbon atom is attached to four distinct groups, forming a chiral center that exhibits optical isomerism.

The correct answer, 2-chlorobutane, has a central carbon atom (C-2) bonded to four completely distinct substituents: a hydrogen atom, a chlorine atom, a methyl group, and an ethyl group. Because all four attached groups are unique, the molecule lacks internal symmetry and displays optical isomerism (enantiomerism).

Step-by-Step Solution

1

Recall the condition for optical isomerism

A molecule must contain at least one chiral (asymmetric) carbon atom, which is a carbon atom bonded to four completely different atoms or groups.

Chirality prevents the mirror image of the molecule from being superimposable on the original structure.

2

Analyze the carbon environments for each compound

For 2-chlorobutane:

\text{CH}_3-\text{C}^* \text{H(Cl)}-\text{CH}_2\text{CH}_3

. The starred carbon (

\text{C}^*

) is attached to

-\text{H}

,

-\text{Cl}

,

-\text{CH}_3

, and

-\text{CH}_2\text{CH}_3

.

All four groups attached to carbon-2 are unique, creating a chiral center.

3

Check the remaining options for symmetry

1-chlorobutane has

-\text{CH}_2-

groups; 2-chloropropane has two identical

-\text{CH}_3

groups on C-2; 2-methylpropan-2-ol has three identical

-\text{CH}_3

groups on C-2.

Molecules containing identical groups attached to the same carbon atom are achiral and optically inactive.

Key Concept

Optical Isomerism and Chirality

Question 3Question →

Which of the following open-chain isomeric alkenes with the molecular formula $C_5H_{10}$ exhibits geometric (cis-trans) isomerism?

Pent-2-ene

Pent-1-ene

2-Methylbut-2-ene

3-Methylbut-1-ene

Show answer & explanation

Answer: Pent-2-ene

Answer

Pent-2-ene is the correct answer because neither of its double-bonded carbon atoms carries two identical attached groups.

Geometric (cis-trans) isomerism requires restricted rotation around a carbon-carbon double bond (

C=C

) along with two different substituents attached to each of the double-bonded carbon atoms. In pent-2-ene (

CH_3-CH=CH-CH_2-CH_3

), C2 is bonded to

-H

and

-CH_3

, while C3 is bonded to

-H

and

-CH_2CH_3

. Because neither carbon carries identical groups, pent-2-ene can form distinct cis and trans stereoisomers.

Step-by-Step Solution

1

Identify the structural condition required for geometric (cis-trans) isomerism in alkenes.

For an alkene

R_1R_2C=CR_3R_4

to exhibit geometric isomerism,

R_1 \neq R_2

and

R_3 \neq R_4

must hold for both carbon atoms involved in the double bond.

If either carbon atom in the double bond is bonded to two identical atoms or groups, rotating structural representation does not create distinct non-superimposable stereoisomers.

2

Analyze the structural formula of Pent-1-ene.

Pent-1-ene is

CH_2=CH-CH_2-CH_2-CH_3

. C1 is attached to two hydrogen atoms (

-H

and

-H

).

Since C1 has identical hydrogen atoms, it cannot exhibit cis-trans isomerism.

3

Analyze the structural formula of Pent-2-ene.

Pent-2-ene is

CH_3-CH=CH-CH_2-CH_3

. C2 is bonded to

-H

and

-CH_3

. C3 is bonded to

-H

and

-CH_2CH_3

.

Both double-bonded carbons have two distinct attached groups, permitting the existence of cis-pent-2-ene and trans-pent-2-ene.

4

Analyze the structural formulas of 2-Methylbut-2-ene and 3-Methylbut-1-ene.

2-Methylbut-2-ene has two methyl groups on C2 (

(CH_3)_2C=CH-CH_3

), and 3-Methylbut-1-ene has two hydrogen atoms on C1 (

CH_2=CH-CH(CH_3)_2

).

Both compounds violate the condition of having distinct groups on each doubly bonded carbon atom.

Key Concept

Geometric (cis-trans) Isomerism in Alkenes

Estimated Time:1m 30s

Question 4Question →

Match each pair of organic compounds on the left with its corresponding type of isomerism on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Hexan-2-one and Hexan-3-one exhibit Positional isomerism; Pentane and 2,2-Dimethylpropane exhibit Chain isomerism; Ethoxyethane and Butan-1-ol exhibit Functional group isomerism; (+)-Lactic acid and (-)-Lactic acid exhibit Optical isomerism.

Hexan-2-one and Hexan-3-one differ only in the locant of the carbonyl group along an unchanged six-carbon backbone (positional isomerism). Pentane and 2,2-dimethylpropane differ in the branching of their carbon skeletons (chain isomerism). Ethoxyethane and Butan-1-ol share the formula C4H10O but contain different functional groups (functional group isomerism). (+)-Lactic acid and (-)-Lactic acid are optical enantiomers due to an asymmetric chiral carbon center.

Step-by-Step Solution

1

Examine Hexan-2-one and Hexan-3-one

Both share the molecular formula

C_6H_{12}O

and contain the carbonyl (

C=O

) functional group. In Hexan-2-one, the carbonyl carbon is at C-2, whereas in Hexan-3-one, it is at C-3.

Molecules with identical functional groups located at different positions on the carbon chain are positional isomers.

2

Examine Pentane and 2,2-Dimethylpropane

Both share the formula

C_5H_{12}

. Pentane is a straight 5-carbon chain (

CH_3-CH_2-CH_2-CH_2-CH_3

), whereas 2,2-Dimethylpropane consists of a 3-carbon chain with two methyl branches,

C(CH_3)_4

.

Molecules with the same molecular formula but different carbon chain structures are chain isomers.

3

Examine Ethoxyethane and Butan-1-ol

Both share the molecular formula

C_4H_{10}O

. Ethoxyethane (

C_2H_5-O-C_2H_5

) is an ether, while Butan-1-ol (

C_4H_9OH

) is a primary alkanol.

Molecules possessing the same molecular formula but belonging to different homologous series with distinct functional groups are functional group isomers.

4

Examine (+)-Lactic acid and (-)-Lactic acid

Lactic acid (

2

-hydroxypropanoic acid) features a central carbon atom bonded to four distinct groups:

-H

,

-OH

,

-CH_3

, and

-COOH

. This chiral center generates two non-superimposable mirror-image forms.

Stereoisomers that rotate plane-polarized light in opposite directions due to molecular chirality are optical isomers.

Key Concept

Types of Structural Isomerism and Stereoisomerism

Question 5Question →

But-1-ene and but-2-ene share the molecular formula $C_4H_8$ but differ in the location of their carbon-carbon double bond. Which type of structural isomerism do these two compounds exhibit?

Chain isomerism

Positional isomerism

Functional group isomerism

Geometric isomerism

Show answer & explanation

Answer: Positional isomerism

Answer

Positional isomerism

Positional isomerism occurs when compounds with the same carbon skeleton and the same functional group differ only in the location of that functional group on the chain. But-1-ene and but-2-ene both have a straight four-carbon chain, but the double bond starts at position 1 and position 2 respectively.

Step-by-Step Solution

1

Examine the structures of both molecules.

Both but-1-ene (

CH_2=CH-CH_2-CH_3

) and but-2-ene (

CH_3-CH=CH-CH_3

) have a straight four-carbon chain and an alkene functional group.

Comparing carbon chain structure and functional group identity helps determine isomer type.

2

Identify the structural difference between the two molecules.

The double bond is located between carbon-1 and carbon-2 in but-1-ene, but between carbon-2 and carbon-3 in but-2-ene.

Compounds with the same carbon framework that differ only in the location of the functional group are classified as positional isomers.

Key Concept

Positional Isomerism in Alkenes

Question 6Question →

How many acyclic structural isomers exist for the haloalkane with the molecular formula $C_4H_9Cl$ ?

2

3

4

5

Show answer & explanation

Answer: 4

Answer

There are 4 acyclic structural isomers for

C_4H_9Cl

.

The molecular formula

C_4H_9Cl

allows for both chain isomerism (butane vs. methylpropane carbon backbones) and positional isomerism (placement of the chlorine atom). This results in four distinct IUPAC structures: 1-chlorobutane, 2-chlorobutane, 1-chloro-2-methylpropane, and 2-chloro-2-methylpropane.

Step-by-Step Solution

1

Identify structural isomers based on the straight-chain butyl skeleton (

CH_3-CH_2-CH_2-CH_3

).

Two positional isomers are obtained: 1-chlorobutane (

CH_3CH_2CH_2CH_2Cl

) and 2-chlorobutane (

CH_3CH_2CH(Cl)CH_3

).

Varying the position of the chlorine atom on a four-carbon unbranched chain yields two distinct structures.

2

Identify structural isomers based on the branched methylpropane skeleton (

(CH_3)_2CH-CH_3

).

Two chain/positional isomers are obtained: 1-chloro-2-methylpropane (

(CH_3)_2CHCH_2Cl

) and 2-chloro-2-methylpropane (

(CH_3)_3CCl

).

Attaching the chlorine atom to a primary carbon versus the tertiary carbon of the methylpropane backbone yields two additional unique isomers.

3

Sum the distinct structural isomers found.

Total number of isomers =

2 + 2 = 4

.

All possible constitutional connectivities for

C_4H_9Cl

have been evaluated without double counting.

Key Concept

Structural Isomerism in Haloalkanes

Estimated Time:1m 15s

Question 7Question →

A molecule possessing two identical chiral carbon atoms, such as $2,3$ -dichlorobutane, yields a total of four optically active stereoisomers.

True

False

Show answer & explanation

Answer: False

Answer

The statement is false. A molecule with two identical chiral carbon atoms generates three stereoisomers in total: one pair of optically active enantiomers and one optically inactive meso compound.

The statement is false because

2,3

-dichlorobutane contains two identical chiral carbon atoms. Its internal plane of symmetry in the meso configuration causes internal compensation of optical rotation, yielding only two optically active enantiomers and one optically inactive meso form (three stereoisomers in total).

Step-by-Step Solution

1

Identify the chiral carbon atoms and their bonded groups in

2,3

-dichlorobutane.

Carbon-2 and Carbon-3 are both chiral centers, each attached to four groups:

-H

,

-Cl

,

-CH_3

, and

-CH(Cl)CH_3

. Because both carbons have identical sets of substituents, the chiral centers are identical.

Recognizing identical chiral centers is essential for identifying internal planes of symmetry.

2

Evaluate theoretical stereoisomer combinations using the

2^n

rule.

The theoretical maximum without symmetry considerations is

2^2 = 4

configurations:

(2R, 3R)

,

(2S, 3S)

,

(2R, 3S)

, and

(2S, 3R)

.

The formula

2^n

determines the upper limit of stereoisomers for

n

chiral centers.

3

Examine internal symmetry and optical activity of the configurations.

The

(2R, 3R)

and

(2S, 3S)

forms are non-superimposable mirror images (enantiomers) and are optically active. However, the

(2R, 3S)

and

(2S, 3R)

structures are identical to each other due to an internal plane of symmetry. This single achiral meso compound is optically inactive via internal compensation.

Internal symmetry cancels optical rotation, leaving only two optically active stereoisomers and one optically inactive meso stereoisomer (3 total).

Key Concept

Meso compounds and internal optical compensation in molecules with identical chiral centers.

Question 8Question →

Match each type of organic isomerism on the left with its corresponding example pair of compounds on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Chain isomerism pairs with butane and

2

-methylpropane; Positional isomerism pairs with drop-in pair butan-

1

-ol and butan-

2

-ol; Functional group isomerism pairs with ethanoic acid and methyl methanoate; Geometric isomerism pairs with

cis

-but-

2

-ene and

trans

-but-

2

-ene.

Each type of isomerism is matched to its definitive structural characteristic: chain isomerism involves skeleton branching changes (butane and

2

-methylpropane), positional isomerism involves relocations of the same functional group (butan-

1

-ol and butan-

2

-ol), functional group isomerism involves different organic families sharing a formula (ethanoic acid and methyl methanoate), and geometric isomerism involves spatial orientations across a double bond (

cis

-but-

2

-ene and

trans

-but-

2

-ene).

Step-by-Step Solution

1

Analyze the carbon skeletons of butane and

2

-methylpropane.

Both have formula

C_4H_{10}

, but one is unbranched while the other is branched, defining chain isomerism.

Chain isomers possess identical molecular formulas but different carbon chain arrangements.

2

Examine the functional group positions in butan-

1

-ol and butan-

2

-ol.

The

-\text{OH}

group is on carbon-

1

in butan-

1

-ol and carbon-

2

in butan-

2

-ol, defining positional isomerism.

Positional isomers have the same functional group located on different carbon atoms along the same parent chain.

3

Compare the functional groups of ethanoic acid and methyl methanoate.

Ethanoic acid is a carboxylic acid (

CH_3COOH

) while methyl methanoate is an ester (

HCOOCH_3

). Both have the formula

C_2H_4O_2

, defining functional group isomerism.

Functional group isomers share a molecular formula but belong to different organic homologous families.

4

Evaluate the spatial arrangement in

cis

-but-

2

-ene and

trans

-but-

2

-ene.

The double bond restricts rotation, placing methyl groups on the same side (

cis

) or opposite sides (

trans

), defining geometric isomerism.

Geometric isomerism is a stereoisomerism type arising from restricted double-bond rotation.

Key Concept

Classification of Structural Isomerism and Stereoisomerism

Question 9Question →

How many of the acyclic structural isomers with the molecular formula $C_4H_8Cl_2$ possess at least one chiral carbon atom?

3

2

4

5

Show answer & explanation

Answer: 3

Answer

3 acyclic structural isomers of

C_4H_8Cl_2

possess at least one chiral carbon atom.

The correct response is 3. A chiral carbon atom is defined as a carbon atom bonded to four different atoms or functional groups. Among the nine constitutional acyclic isomers of dichlorobutane (

C_4H_8Cl_2

), only 1,2-dichlorobutane (at carbon-2), 1,3-dichlorobutane (at carbon-2), and 2,3-dichlorobutane (at carbon-2 and carbon-3) possess carbon atoms meeting this criterion.

Step-by-Step Solution

1

Identify all acyclic structural (constitutional) isomers of

C_4H_8Cl_2

.

There are 9 total acyclic structural isomers: 1,1-dichlorobutane, 1,2-dichlorobutane, 1,3-dichlorobutane, 1,4-dichlorobutane, 2,2-dichlorobutane, 2,3-dichlorobutane, 1,1-dichloro-2-methylpropane, 1,2-dichloro-2-methylpropane, and 1,3-dichloro-2-methylpropane.

Structural isomers must be systematically generated based on butane and methylpropane carbon skeletons.

2

Examine each structural isomer for the presence of a chiral (asymmetric) carbon atom.

A chiral carbon must be bonded to four completely different atoms or groups.

Chirality requires tetrahedral asymmetry around a carbon atom.

3

Determine which specific isomers contain chiral carbons.

1. 1,2-dichlorobutane: Carbon-2 is bonded to

-H

,

-Cl

,

-CH_2Cl

, and

-CH_2CH_3

(Chiral).
2. 1,3-dichlorobutane: Carbon-2 is bonded to

-H

,

-Cl

,

-CH_3

, and

-CH_2CH_2Cl

(Chiral).
3. 2,3-dichlorobutane: Carbon-2 and Carbon-3 are both bonded to

-H

,

-Cl

,

-CH_3

, and

-CH(Cl)CH_3

(Chiral).

All other structural isomers have identical groups attached to each carbon (e.g., two

-Cl

atoms on C-1 in 1,1-dichlorobutane, or two methyl groups on C-2 in methylpropane derivatives).

Key Concept

Chirality and Structural Isomerism in Haloalkanes

Question 10Question →

Which of the following structural isomers of hexane, $C_6H_{14}$ , contains exactly one tertiary carbon atom?

2-Methylpentane

2,3-Dimethylbutane

2,2-Dimethylbutane

Hexane

Show answer & explanation

Answer: 2-Methylpentane

Answer

2-Methylpentane contains exactly one tertiary carbon atom.

A tertiary carbon atom is defined as a carbon atom bonded to three other carbon atoms. In 2-methylpentane, the carbon atom at position 2 is bonded to C-1, C-3, and the branch methyl group, making it the only tertiary carbon atom in the molecule.

Step-by-Step Solution

1

Define a tertiary carbon atom

A tertiary carbon atom (

3^\circ

) is directly bonded to three other carbon atoms.

Carbon classification depends on the number of attached alkyl/carbon groups.

2

Analyze the carbon skeleton of 2-Methylpentane

In

CH_3-CH(CH_3)-CH_2-CH_2-CH_3

, C-2 is bonded to C-1, C-3, and the methyl group.

This structural arrangement yields exactly one tertiary carbon atom.

3

Evaluate the remaining options

2,3-Dimethylbutane has two tertiary carbons, 2,2-Dimethylbutane has one quaternary carbon (zero tertiary), and unbranched hexane has zero tertiary carbons.

Only 2-Methylpentane satisfies the condition of having exactly one tertiary carbon atom.

Key Concept

Classification of carbon atoms in structural isomers

Question 11Question →

But-2-ene exhibits geometric (cis-trans) isomerism because each carbon atom involved in the double bond is bonded to two non-identical groups, whereas but-1-ene does not exhibit geometric isomerism.

True

False

Show answer & explanation

Answer: True

Answer

The statement is true because geometric isomerism requires restricted rotation about the

C=C

bond together with two distinct groups attached to each double-bonded carbon atom—a condition satisfied by but-2-ene but not by but-1-ene.

The statement correctly describes the structural rule for geometric isomerism in alkenes. But-2-ene meets the condition because both double-bonded carbons carry two non-identical groups (

-H

and

-CH_3

), whereas but-1-ene fails the condition because its terminal carbon carries two identical hydrogen atoms.

Step-by-Step Solution

1

Identify the structural requirements for geometric (cis-trans) isomerism.

Geometric isomerism in alkenes requires a rigid

C=C

double bond where each of the two unsaturated carbon atoms is attached to two non-identical substituents.

If either carbon atom of the double bond carries two identical groups, swapping those groups produces an identical molecule rather than a distinct stereoisomer.

2

Analyze the substituent groups attached to the double-bonded carbons in but-2-ene (

CH_3-CH=CH-CH_3

).

Carbon-2 is attached to

-H

and

-CH_3

, and Carbon-3 is also attached to

-H

and

-CH_3

.

Since both double-bonded carbons have two different groups attached, but-2-ene exists as two stereoisomers: cis-but-2-ene and trans-but-2-ene.

3

Analyze the substituent groups attached to the double-bonded carbons in but-1-ene (

CH_2=CH-CH_2-CH_3

).

Carbon-1 is attached to two identical hydrogen atoms (

-H

and

-H

).

The presence of two identical hydrogen atoms on Carbon-1 prevents the formation of cis-trans isomers for but-1-ene.

Key Concept

Structural Criteria for Geometric (Cis-Trans) Isomerism

Question 12Question →

Which of the following alkenes with the molecular formula $C_5H_{10}$ exhibits geometric (cis-trans) isomerism?

Pent-2-ene

Pent-1-ene

2-Methylbut-2-ene

3-Methylbut-1-ene

Show answer & explanation

Answer: Pent-2-ene

Answer

Pent-2-ene is the only isomer listed that exhibits geometric (cis-trans) isomerism because each carbon of the double bond is attached to two distinct groups.

Pent-2-ene has a double bond between carbon-2 and carbon-3. Carbon-2 is attached to a hydrogen atom and a methyl group (

CH_3

), while carbon-3 is attached to a hydrogen atom and an ethyl group (

CH_2CH_3

). Because neither carbon atom of the double bond holds two identical groups, spatial restriction gives rise to distinct cis and trans stereoisomers.

Step-by-Step Solution

1

Recall the necessary structural condition for geometric (cis-trans) isomerism in alkenes.

For a molecule to show cis-trans isomerism around a double bond

C=C

, each of the two carbon atoms in the double bond must be attached to two different atoms or groups.

If either carbon in the double bond has two identical groups attached, rotating spatial arrangements results in identical molecules.

2

Examine the connectivity of each option at the

C=C

double bond.

Pent-2-ene has

C_2

bonded to

-H

and

-CH_3

, and

C_3

bonded to

-H

and

-CH_2CH_3

. Both carbons have two different groups.

This satisfies the criteria for both cis and trans geometric arrangements.

3

Check the remaining options for duplicate attached groups on double-bonded carbons.

Pent-1-ene and 3-methylbut-1-ene both have a terminal

=CH_2

(two H atoms on

C_1

). 2-Methylbut-2-ene has two methyl groups on

C_2

.

None of these three options satisfy the non-identical substituent requirement on both double-bonded carbons.

Key Concept

Geometric Isomerism Requirements in Alkenes

Estimated Time:1m 0s

Question 13Question →

An organic compound with the molecular formula $C_4H_{10}O$ forms four isomeric alkanols. Which of these alkanol isomers contains an asymmetric carbon atom and exhibits optical isomerism?

Butan-1-ol

Butan-2-ol

2-Methylpropan-1-ol

2-Methylpropan-2-ol

Show answer & explanation

Answer: Butan-2-ol

Answer

Butan-2-ol is the only isomeric alkanol of formula

C_4H_{10}O

containing a chiral carbon atom.

Butan-2-ol has a chiral carbon at position 2 (

CH_3-CH(OH)-CH_2-CH_3

), which is bonded to four different substituent groups:

-H

,

-OH

,

-CH_3

, and

-CH_2CH_3

. The presence of this asymmetric center causes optical activity.

Step-by-Step Solution

1

Identify the requirement for optical isomerism.

A molecule exhibits optical isomerism if it contains at least one chiral (asymmetric) carbon atom—a carbon atom bonded to four different groups or atoms.

Chirality causes non-superimposable mirror images (enantiomers) capable of rotating plane-polarized light.

2

Analyze the structural formulas of the four isomeric alkanols of

C_4H_{10}O

.

1) Butan-1-ol:

CH_3-CH_2-CH_2-CH_2OH

2) Butan-2-ol:

CH_3-CH(OH)-CH_2-CH_3

3) 2-Methylpropan-1-ol:

(CH_3)_2CH-CH_2OH

4) 2-Methylpropan-2-ol:

(CH_3)_3C-OH

Writing structural formulas reveals the substituent groups attached to each carbon atom.

3

Examine carbon-2 in butan-2-ol.

Carbon-2 is attached to

-H

,

-OH

,

-CH_3

, and

-CH_2CH_3

. All four substituents are distinct.

Since carbon-2 has four different attached groups, it is an asymmetric (chiral) carbon atom.

Key Concept

Optical isomerism and chirality in alkanols

Structural Isomerism and Stereoisomerism

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