Functional Groups, Homologous Series, and IUPAC Nomenclature Practice Questions — JAMB UTME

Question 1Question →

Arrange the following straight-chain alkanes in order of increasing boiling point, starting with the alkane that has the lowest boiling point:

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Answer

The correct sequence from lowest to highest boiling point is Methane (

CH_4

), Ethane (

C_2H_6

), Propane (

C_3H_8

), and Butane (

C_4H_{10}

).

In the unbranched alkane homologous series, boiling point increases progressively with increasing carbon chain length and relative molecular mass due to stronger intermolecular van der Waals forces. Consequently, Methane (

CH_4

) has the lowest boiling point, followed by Ethane (

C_2H_6

), Propane (

C_3H_8

), and Butane (

C_4H_{10}

) with the highest boiling point.

Step-by-Step Solution

1

Identify the relationship between molecular size and boiling point in a homologous series

As the number of carbon atoms in straight-chain alkanes increases, the molecular mass increases.

Members of a homologous series share identical functional groups and exhibit a gradual gradation in physical properties.

2

Evaluate the strength of intermolecular forces

Van der Waals (dispersion) forces become stronger with increasing molecular surface area and electron count.

Greater intermolecular attraction requires more thermal energy to separate molecules into the gas phase.

3

Sequence the compounds from smallest to largest carbon chain

Methane (1C) < Ethane (2C) < Propane (3C) < Butane (4C)

Fewer carbon atoms correlate directly with a lower boiling point.

Key Concept

Physical Property Trends in Homologous Series

Question 2Question →

Match each homologous series of organic compounds listed on the left with its correct general molecular formula on the right.

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Answer

Alkanals match with

C_n H_{2n}O

; Alkanoic acids match with

C_n H_{2n}O_2

; Alkynes match with

C_n H_{2n-2}

; Alkanols match with

C_n H_{2n+2}O

.

Each class of organic compound is defined by its characteristic functional group and degree of unsaturation. Alkanols are saturated single-oxygen compounds (

C_n H_{2n+2}O

), Alkanals feature one carbonyl double bond (

C_n H_{2n}O

), Alkanoic acids feature a carboxyl group (

C_n H_{2n}O_2

), and Alkynes contain a triple bond reducing hydrogen content to

C_n H_{2n-2}

.

Step-by-Step Solution

1

Determine the general formula for aliphatic monohydric alkanols.

Alkanols consist of an alkyl group (

C_n H_{2n+1}

) bonded to a hydroxyl group (

-\text{OH}

), giving the molecular formula

C_n H_{2n+2}O

.

Saturated aliphatic monohydric alcohols have the maximum possible hydrogen-to-carbon ratio for oxygenated single-bonded species.

2

Determine the general formula for alkanals.

Alkanals contain a terminal carbonyl group (

-\text{CHO}

), introducing one double bond (

\text{C=O}

) which reduces the hydrogen atom count by two compared to alkanols, giving

C_n H_{2n}O

.

Each site of unsaturation (such as a

\pi

-bond) decreases the hydrogen atom count by two.

3

Determine the general formula for alkanoic acids.

Alkanoic acids contain a carboxyl group (

-\text{COOH}

), giving two oxygen atoms and one carbon-oxygen double bond, yielding

C_n H_{2n}O_2

.

Carboxylic acids are functional group isomers of esters and share the general formula

C_n H_{2n}O_2

.

4

Determine the general formula for alkynes.

Alkynes possess one carbon-carbon triple bond (two

\pi

-bonds), reducing the hydrogen count by four relative to alkanes (

C_n H_{2n+2}

), resulting in

C_n H_{2n-2}

.

A triple bond accounts for two degrees of unsaturation.

Key Concept

General Molecular Formulas of Organic Homologous Series

Estimated Time:1m 15s

Question 3Question →

What is the correct IUPAC name for the organic compound with the condensed structural formula $\text{CH}_3-\text{CH}_2-\text{CH}(\text{CH}_2\text{CH}_3)-\text{CH}_3$ ?

3-methylpentane

2-ethylbutane

3-ethylbutane

2-methylpentane

Show answer & explanation

Answer: 3-methylpentane

Answer

3-methylpentane

The correct IUPAC name is determined by finding the longest continuous chain of carbon atoms. Expanding the condensed formula

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

shows that the longest continuous chain contains 5 carbons (pentane). The remaining branch is a methyl group at carbon-3, making the systematic name 3-methylpentane.

Step-by-Step Solution

1

Identify the longest continuous carbon chain in the structure.

Expanding

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

shows a continuous chain of 5 carbon atoms (pentane).

IUPAC nomenclature rules dictate that the parent chain must be the longest continuous chain of carbon atoms.

2

Identify substituents and number the parent chain.

A methyl group (

-\text{CH}_3

) is attached at carbon-3.

Numbering the 5-carbon chain from either end places the substituent at position 3.

3

Combine the substituent locant, substituent name, and parent alkane name.

The correct IUPAC name is 3-methylpentane.

Combining the locant (3), substituent (methyl), and parent (pentane) yields 3-methylpentane.

Key Concept

Identification of the longest continuous carbon chain in IUPAC organic nomenclature

Estimated Time:45s

Question 4Question →

Match each organic functional group class on the left with its characteristic IUPAC naming suffix on the right.

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Answer

Alkanol (Alcohol) matches with '-ol', Alkanal (Aldehyde) matches with '-al', Alkanone (Ketone) matches with '-one', and Alkanoic acid (Carboxylic acid) matches with '-oic acid'.

Each organic functional group family corresponds to a specific IUPAC naming suffix: Alkanols take '-ol', Alkanals take '-al', Alkanones take '-one', and Alkanoic acids take '-oic acid'.

Step-by-Step Solution

1

Identify the primary functional group present in each class of organic compound.

Alkanols possess hydroxyl groups, alkanals possess aldehyde groups, alkanones possess ketone carbonyl groups, and alkanoic acids possess carboxyl groups.

The characteristic functional group determines both the homologous series and the IUPAC suffix rules.

2

Assign the standard IUPAC nomenclature suffix corresponding to each functional group.

Alkanol

\rightarrow

'-ol', Alkanal

\rightarrow

'-al', Alkanone

\rightarrow

'-one', Alkanoic acid

\rightarrow

'-oic acid'.

Standard IUPAC rules modify the parent alkane name by replacing the terminal '-e' with the designated functional group suffix.

Key Concept

IUPAC Suffixes for Functional Groups

Question 5Question →

In a homologous series of alkanoic acids, physical properties such as boiling point change systematically with increasing molecular size. Arrange the following straight-chain alkanoic acids in order of increasing boiling point, starting with the compound that has the lowest boiling point.

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Answer

The correct order of straight-chain alkanoic acids from lowest to highest boiling point is: Methanoic acid (

\text{HCOOH}

), Ethanoic acid (

\text{CH}_3\text{COOH}

), Propanoic acid (

\text{CH}_3\text{CH}_2\text{COOH}

), and Butanoic acid (

\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}

).

In any homologous series of organic compounds, physical properties such as boiling point increase with increasing relative molecular mass and carbon chain length. Methanoic acid has 1 carbon, Ethanoic acid has 2, Propanoic acid has 3, and Butanoic acid has 4. Therefore, the boiling point increases steadily from methanoic acid to butanoic acid.

Step-by-Step Solution

1

Identify the homologous series and structural difference among the compounds.

All four compounds belong to the alkanoic acid homologous series, differing consecutively by a

-\text{CH}_2-

(methylene) unit.

Members of a homologous series share similar chemical properties but show a gradual gradation in physical properties.

2

Determine how molecular mass and chain length affect the boiling point.

As the number of carbon atoms in the chain increases, the relative molecular mass increases and the surface area for intermolecular contact expands.

Greater molecular mass and contact surface area lead to stronger London dispersion (van der Waals) forces, requiring more thermal energy to boil.

3

Sequence the compounds by increasing carbon chain length and molar mass.

Methanoic acid (

1\text{ C}

) < Ethanoic acid (

2\text{ C}

) < Propanoic acid (

3\text{ C}

) < Butanoic acid (

4\text{ C}

).

Methanoic acid has the lowest boiling point (~

101^\circ\text{C}

) and Butanoic acid has the highest (~

163^\circ\text{C}

).

Key Concept

Gradation of physical properties in a homologous series

Question 6Question →

What is the correct IUPAC name for the branched alkane with the structural formula $\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}_3$ ?

2-methylbutane

3-methylbutane

2-ethylpropane

pentane

Show answer & explanation

Answer: 2-methylbutane

Answer

The correct IUPAC name for the compound is 2-methylbutane.

The longest continuous carbon chain consists of 4 carbon atoms (butane). Numbering from the end closest to the branch assigns the methyl group to position 2, giving the correct IUPAC name 2-methylbutane.

Step-by-Step Solution

1

Identify the longest continuous carbon chain.

The longest chain has 4 carbon atoms, which corresponds to the parent alkane butane.

IUPAC rules mandate that the parent structure is named according to the longest continuous chain of carbon atoms.

2

Number the parent carbon chain to give substituents the lowest locant numbers.

Numbering from left to right assigns the methyl group to carbon-2 (

C

-2), whereas right to left gives carbon-3 (

C

-3). Thus, left-to-right numbering is correct.

Substituents must receive the lowest possible numerical locants.

3

Assemble the complete IUPAC name.

The substituent prefix '2-methyl' combined with the parent 'butane' gives 2-methylbutane.

The full name consists of the substituent position, substituent name, and parent alkane name.

Key Concept

IUPAC rules for branched alkanes: identifying the longest continuous carbon chain and assigning the lowest position number to alkyl substituents.

Estimated Time:45s

Question 7Question →

Arrange the following organic functional groups in decreasing order of priority (highest priority first) for selection as the principal functional group suffix when naming polyfunctional compounds according to IUPAC nomenclature rules:

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Answer

The correct sequence in decreasing order of IUPAC principal functional group priority is: Carboxylic acid (

-\text{COOH}

), Ester (

-\text{COOR}

), Aldehyde (

-\text{CHO}

), Alcohol (

-\text{OH}

), and Amine (

-\text{NH}_2

).

According to official IUPAC nomenclature seniority rules for principal functional groups: Carboxylic acids (

-\text{COOH}

) take top priority, followed by acid derivatives like Esters (

-\text{COOR}

), then Aldehydes (

-\text{CHO}

), Alcohols (

-\text{OH}

), and finally Amines (

-\text{NH}_2

).

Step-by-Step Solution

1

Identify the highest-priority functional group among carboxylic derivatives and oxygen/nitrogen species.

Carboxylic acids (

-\text{COOH}

) occupy the top hierarchy level among organic functional groups.

IUPAC nomenclature assigns the highest principal suffix priority to carboxylic acid functional groups over esters, carbonyls, alcohols, and amines.

2

Compare carboxylic acid derivatives to carbonyl and hydroxyl groups.

Ester (

-\text{COOR}

) ranks higher than Aldehyde (

-\text{CHO}

).

Carboxylic acid derivatives (esters, acyl halides, amides) take priority over aldehydes and ketones.

3

Determine priority among oxygen- and nitrogen-containing groups.

Aldehyde (

-\text{CHO}

) > Alcohol (

-\text{OH}

) > Amine (

-\text{NH}_2

).

Carbonyl groups rank higher than hydroxyl groups, which in turn rank higher than amino groups.

Key Concept

IUPAC Functional Group Seniority Hierarchy for Polyfunctional Nomenclature

Question 8Question →

Match each condensed organic structural formula listed on the left with its correct functional group classification on the right.

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Answer

\text{CH}_3\text{COCH}_3

matches with Alkanone;

\text{CH}_3\text{CH}_2\text{CHO}

matches with Alkanal;

\text{CH}_3\text{CH}_2\text{COOH}

matches with Alkanoic acid;

\text{HCOOCH}_3

matches with Alkanoate (Ester).

Each condensed formula corresponds directly to its functional group family:

\text{CH}_3\text{COCH}_3

is propanone (an alkanone),

\text{CH}_3\text{CH}_2\text{CHO}

is propanal (an alkanal),

\text{CH}_3\text{CH}_2\text{COOH}

is propanoic acid (an alkanoic acid), and

\text{HCOOCH}_3

is methyl methanoate (an ester / alkanoate).

Step-by-Step Solution

1

Identify the characteristic functional group in each condensed formula.

\text{CH}_3\text{COCH}_3

has a non-terminal carbonyl

\text{-CO-}

;

\text{CH}_3\text{CH}_2\text{CHO}

has a formyl

\text{-CHO}

;

\text{CH}_3\text{CH}_2\text{COOH}

has a carboxyl

\text{-COOH}

;

\text{HCOOCH}_3

has an ester linkage

\text{-COO-}

.

Functional groups determine the class of organic compounds.

2

Match each identified group to its IUPAC homologous series classification.

Non-terminal carbonyl (

\text{-CO-}

) = Alkanone; Formyl (

\text{-CHO}

) = Alkanal; Carboxyl (

\text{-COOH}

) = Alkanoic acid; Ester linkage (

\text{-COO-}

) = Alkanoate (Ester).

IUPAC nomenclature categorizes compounds based on these specific functional group structures.

Key Concept

Classification of Organic Compounds by Functional Groups

Question 9Question →

Match each characteristic functional group structure listed on the left with its corresponding organic class name on the right.

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Answer

The correct pairings are:

-\text{NH}_2

matches with Primary amine,

-\text{CONH}_2

matches with Amide,

-\text{O}-

matches with Ether, and

-\text{COOR}

matches with Ester.

Each organic compound class is identified by its specific functional group arrangement:

-\text{NH}_2

specifies primary amines,

-\text{CONH}_2

specifies primary amides, an oxygen atom bridging two carbon groups (

-\text{O}-

) specifies ethers, and

-\text{COOR}

specifies esters.

Step-by-Step Solution

1

Identify the amino functional group

-\text{NH}_2

is recognized as an amino group attached to a carbon chain.

Compounds containing the

-\text{NH}_2

group attached directly to an alkyl carbon belong to the class of primary amines.

2

Identify the carboxamide functional group

-\text{CONH}_2

contains a carbonyl bonded to an amino group.

This structural unit defines the primary amide family.

3

Identify the ether linkage

-\text{O}-

represents a single oxygen atom bridging two carbon atoms.

An oxygen atom connected to two alkyl or aryl groups (

\text{R-O-R'}

) characterizes an ether.

4

Identify the ester group

-\text{COOR}

contains a carbonyl carbon attached to an alkoxy group.

This group is derived from an alkanoic acid and an alkanol, forming an ester.

Key Concept

Identification of characteristic functional groups in organic compounds

Question 10Question →

Match each organic functional group class on the left with its corresponding characteristic IUPAC suffix on the right.

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Answer

Carboxylic acid matches '-oic acid', Acid amide matches '-amide', Acyl chloride matches '-oyl chloride', and Ester matches '-oate'.

Each organic class is correctly matched to its IUPAC principal suffix: carboxylic acids use '-oic acid', amides use '-amide', acyl chlorides use '-oyl chloride', and esters use '-oate'.

Step-by-Step Solution

1

Identify the characteristic functional group structure for each carbonyl derivative class.

Carboxylic acids have

-\text{COOH}

, amides have

-\text{CONH}_2

, acyl chlorides have

-\text{COCl}

, and esters have

-\text{COOR}

.

Functional groups define the chemical family and dictate IUPAC naming rules.

2

Pair each class with its designated IUPAC principal suffix.

-\text{COOH} \rightarrow \text{-oic acid}

,

-\text{CONH}_2 \rightarrow \text{-amide}

,

-\text{COCl} \rightarrow \text{-oyl chloride}

, and

-\text{COOR} \rightarrow \text{-oate}

.

Standard IUPAC rules assign specific characteristic suffixes to represent the main functional group in systematic names.

Key Concept

IUPAC Nomenclature Suffixes for Carbonyl Derivatives

Question 11Question →

Arrange the following hydrocarbons in order of increasing boiling point, starting with the compound that has the lowest boiling point:

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Answer

The correct order from lowest to highest boiling point is 2,2-dimethylpropane, 2-methylbutane, pentane, and hexane.

Boiling points of alkanes depend on molecular mass and surface area contact. Hexane has the highest molecular mass and largest surface area, yielding the highest boiling point. Among the structural isomers of pentane, increased branching produces a more spherical shape with smaller surface contact area, reducing intermolecular van der Waals forces. Consequently, 2,2-dimethylpropane (most branched) has the lowest boiling point, followed by 2-methylbutane, straight-chain pentane, and finally hexane.

Step-by-Step Solution

1

Compare molecular mass and carbon chain length among the given hydrocarbons.

hexane (

C_6H_{14}

) has a larger molecular mass and longer chain than the pentane isomers (

C_5H_{12}

), giving it the strongest London dispersion forces and highest boiling point.

Boiling point increases with molar mass in a homologous series due to increased electron cloud polarizability.

2

Compare the degree of branching among the structural isomers of pentane (

C_5H_{12}

).

2,2-dimethylpropane is highly branched (compact spherical shape), 2-methylbutane is moderately branched, and pentane is a straight chain.

Increased branching makes the molecule more spherical, reducing the surface area available for intermolecular van der Waals contact.

3

Rank all four compounds in order of increasing boiling point.

2,2-dimethylpropane < 2-methylbutane < pentane < hexane.

More branched isomers have lower boiling points than less branched isomers of the same molecular formula, and higher molar mass alkanes boil at higher temperatures than lower mass alkanes.

Key Concept

Effect of molecular mass and chain branching on alkane boiling points

Estimated Time:1m 0s

Question 12Question →

An organic compound has the following condensed structural formula:

\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}(\text{C}_2\text{H}_5)\text{CH}_2\text{C}\equiv\text{CH}

What is the correct IUPAC name for this compound?

4-ethyl-5-methylhex-1-yne

4-isopropylhex-1-yne

3-ethyl-2-methylhex-5-yne

3-isopropylhex-5-yne

Show answer & explanation

Answer: 4-ethyl-5-methylhex-1-yne

Answer

4-ethyl-5-methylhex-1-yne

The compound is numbered starting from the terminal alkyne carbon to give the triple bond the lowest locant (

C1

). At

C4

, there are two possible 6-carbon chains. Following IUPAC tie-breaking guidelines, the continuous chain that yields the greater number of substituents (ethyl at

C4

and methyl at

C5

) is chosen as the parent chain, resulting in 4-ethyl-5-methylhex-1-yne.

Step-by-Step Solution

1

Identify the principal functional group and assign lowest locant numbering

The alkyne triple bond (

-C\equiv C-

) is at position 1, so numbering starts from the rightmost carbon: C1 is

HC\equiv

, C2 is

-C-

, C3 is

-CH_2-

, C4 is

-CH(C_2H_5)-

.

IUPAC rules state that principal functional groups receive the lowest possible locants.

2

Determine the longest continuous carbon chain containing the triple bond

From C4, continuing straight through

-CH(CH_3)CH_3

gives 6 carbons (hex-1-yne). Going down the ethyl group

-CH_2CH_3

also gives 6 carbons (hex-1-yne).

Both potential paths yield a parent chain length of 6 carbons.

3

Apply the tie-breaking rule for chains of equal length

The chain straight through

-CH(CH_3)CH_3

has 2 substituents (ethyl at C4, methyl at C5). The path through the ethyl group has only 1 substituent (isopropyl at C4). Thus, the chain with 2 substituents is selected.

When two chains of equal length compete for parent status, IUPAC rules dictate selecting the chain with the maximum number of substituents.

4

Assemble the IUPAC name in alphabetical order

Alphabetize 'ethyl' before 'methyl' to give 4-ethyl-5-methylhex-1-yne.

Substituents are listed alphabetically regardless of their locant numbers.

Key Concept

IUPAC Nomenclature of Alkynes with Branched Chains and Tie-Breaking Rules

Estimated Time:2m 0s

Question 13Question →

Arrange the following organic compounds in order of INCREASING solubility in water, starting with the least soluble compound:

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Answer

The correct sequence from least soluble to most soluble in water is: Pentane, Pentan-1-ol, Butane-1,4-diol, Ethane-1,2-diol.

Water solubility in organic compounds is determined by the balance between hydrophilic (water-attracting) functional groups and hydrophobic (water-repelling) carbon chains. Pentane is non-polar and cannot form hydrogen bonds, making it the least soluble. Pentan-1-ol has one

-\text{OH}

group but a large 5-carbon hydrophobic chain, giving it low solubility. Butane-1,4-diol contains two

-\text{OH}

groups, significantly enhancing hydrogen bonding with water molecules. Ethane-1,2-diol has two

-\text{OH}

groups and the shortest hydrophobic carbon chain (2 carbons), making it the most soluble in water.

Step-by-Step Solution

1

Analyze the functional groups present in each compound.

Pentane has no polar functional groups. Pentan-1-ol has one

-\text{OH}

group. Butane-1,4-diol has two

-\text{OH}

groups. Ethane-1,2-diol has two

-\text{OH}

groups.

Water is a polar solvent that dissolves compounds capable of forming hydrogen bonds through polar functional groups like

-\text{OH}

.

2

Compare the hydrophobic alkyl chain lengths for compounds with the same functional group capability.

Ethane-1,2-diol has a smaller non-polar carbon chain (2 carbons) compared to Butane-1,4-diol (4 carbons).

A larger non-polar alkyl region increases hydrophobic interaction, reducing solubility in water.

3

Establish the complete order of increasing solubility.

Pentane < Pentan-1-ol < Butane-1,4-diol < Ethane-1,2-diol.

Solubility increases with an increasing ratio of hydrophilic functional groups (

-\text{OH}

) to non-polar alkyl chain length.

Key Concept

Solubility of organic compounds based on functional group polarity and hydrophobic alkyl chain length

Question 14Question →

Match each polyfunctional organic compound listed on the left with its principal IUPAC functional group suffix on the right, based on IUPAC nomenclature priority rules.

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Answer

The correct pairings match

\text{HOCH}_2\text{CH}_2\text{COOH}

with -oic acid,

\text{CH}_3\text{COCH}_2\text{CHO}

with -al,

\text{CH}_3\text{CH(OH)COCH}_3

with -one, and

\text{CH}_2=\text{CHCH}_2\text{OH}

with -ol.

In IUPAC nomenclature for polyfunctional organic compounds, the principal functional group determines the ending suffix of the compound name. Carboxylic acid (

-\text{COOH}

) outranks alcohol (

-\text{OH}

), giving -oic acid; aldehyde (

-\text{CHO}

) outranks ketone (

-\text{CO}-

), giving -al; ketone (

-\text{CO}-

) outranks alcohol (

-\text{OH}

), giving -one; and alcohol (

-\text{OH}

) outranks alkene (

\text{C}=\text{C}

), giving -ol.

Step-by-Step Solution

1

Identify all functional groups present in each compound.

Compound 1 contains hydroxyl and carboxyl; Compound 2 contains ketone and aldehyde; Compound 3 contains hydroxyl and ketone; Compound 4 contains alkene and alcohol.

Identifying all present functional groups is necessary before applying priority rules.

2

Apply standard IUPAC functional group seniority order.

Priority order: Carboxylic acid (

-\text{COOH}

) > Aldehyde (

-\text{CHO}

) > Ketone (

-\text{CO}-

) > Alcohol (

-\text{OH}

) > Alkene (

\text{C}=\text{C}

).

The group with higher priority determines the principal suffix of the IUPAC name, while lower priority groups are named as prefixes.

3

Assign the principal suffix to each compound.

Compound 1 matches -oic acid, Compound 2 matches -al, Compound 3 matches -one, and Compound 4 matches -ol.

Matches each compound structure to its appropriate IUPAC ending based on priority.

Key Concept

IUPAC Functional Group Seniority Hierarchy

Functional Groups, Homologous Series, and IUPAC Nomenclature

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