Organic Chemistry Practice Questions — JAMB UTME

Question 1Question →

During the industrial synthesis of a synthetic fiber, hexanedioic acid reacts with hexane-1,6-diamine with the elimination of water molecules to form amide linkages. Which polymer is formed from this process?

Nylon-6,6

Terylene

Polyethene

Poly(methyl methacrylate)

Show answer & explanation

Answer: Nylon-6,6

Answer

Nylon-6,6 is the polyamide formed by the condensation polymerization of hexanedioic acid and hexane-1,6-diamine.

The reaction between a dicarboxylic acid containing 6 carbon atoms (hexanedioic acid) and a diamine containing 6 carbon atoms (hexane-1,6-diamine) yields a synthetic polyamide known as Nylon-6,6 via condensation polymerization with the elimination of water.

Step-by-Step Solution

1

Identify the functional groups of the monomers provided in the prompt.

Hexanedioic acid contains dicarboxylic acid groups (

-COOH

) and hexane-1,6-diamine contains diamine groups (

-NH_2

).

The reaction between carboxylic acid and amine functional groups forms peptide/amide bonds (

-CONH-

).

2

Determine the type of polymerization and the resulting macromolecule class.

The reaction involves the loss of water molecules, which defines condensation polymerization forming a polyamide.

Condensation polymerization links bifunctional monomers together with the elimination of small molecules such as

H_2O

.

3

Match the specific monomers to the corresponding polymer name.

Hexanedioic acid (6 carbon atoms) and hexane-1,6-diamine (6 carbon atoms) produce Nylon-6,6.

The numbers '6,6' in Nylon-6,6 indicate that both the diamine and dicarboxylic acid monomers contain 6 carbon atoms each.

Key Concept

Condensation Polymerization and Synthesis of Polyamides (Nylon-6,6)

Question 2Question →

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 3Question →

Consider the unsaturated hydrocarbon 2-methylbut-1-en-3-yne, which has the condensed structural formula $CH_2=C(CH_3)C\equiv CH$ . What is the total number of sigma ( $\sigma$ ) bonds present in one molecule of this compound?

Show answer & explanation

Answer: 10; 10 sigma bonds; 10 bonds

Answer

The total number of sigma (

\sigma

) bonds present in one molecule of 2-methylbut-1-en-3-yne is 10.

In 2-methylbut-1-en-3-yne (

CH_2=C(CH_3)C\equiv CH

), there are 6 carbon-hydrogen single bonds (2 from C1, 3 from the methyl group, and 1 from C4) and 4 carbon-carbon sigma bonds (1 from the C1=C2 double bond, 1 connecting C2 to the methyl carbon, 1 connecting C2 to C3, and 1 from the C3\equiv C4 triple bond). Summing these gives 10 sigma bonds in total.

Step-by-Step Solution

1

Expand the condensed structural formula to identify all individual carbon-hydrogen (C-H) bonds.

The

=CH_2

group contains 2 C-H

\sigma

bonds, the methyl group (

-CH_3

) contains 3 C-H

\sigma

bonds, and the terminal alkynyl group (

\equiv CH

) contains 1 C-H

\sigma

bond, giving a total of 6 C-H

\sigma

bonds.

Every single bond between carbon and hydrogen is a single sigma bond.

2

Identify all carbon-carbon (C-C) sigma bonds in the backbone and side chain.

The

C=C

double bond contributes 1 C-C

\sigma

bond, the single bond to the methyl branch contributes 1 C-C

\sigma

bond, the C2-C3 single bond contributes 1 C-C

\sigma

bond, and the

C\equiv C

triple bond contributes 1 C-C

\sigma

bond, giving a total of 4 C-C

\sigma

bonds.

Multiple bonds (double or triple) contain exactly one sigma bond each, with the remaining bonds being pi (

\pi

) bonds.

3

Sum the total number of C-H and C-C sigma bonds.

6 \text{ (C-H } \sigma\text{ bonds)} + 4 \text{ (C-C } \sigma\text{ bonds)} = 10 \text{ total } \sigma \text{ bonds}

.

Adding all localized

\sigma

bonds yields the total count for the molecule.

Key Concept

Determination of sigma (\sigma) and pi (\pi) bond counts in complex open-chain hydrocarbons

Question 4Question →

Which of the following describes the type of orbital overlap that forms the carbon-carbon ( $\text{C}-\text{C}$ ) single bond in an ethane ( $C_2H_6$ ) molecule?

Head-on overlap of two

sp^3

hybrid orbitals

Sideways overlap of two unhybridized

p

orbitals

Head-on overlap of two

sp^2

hybrid orbitals

Sideways overlap of two

sp^3

hybrid orbitals

Show answer & explanation

Answer: Head-on overlap of two $sp^3$ hybrid orbitals

Answer

Head-on overlap of two

sp^3

hybrid orbitals

The carbon atoms in ethane (

C_2H_6

) are each bonded to four other atoms, giving them a tetrahedral arrangement and

sp^3

hybridization. The carbon-carbon single bond is a sigma (

\sigma

) bond formed by the direct head-on overlap of one

sp^3

hybrid orbital from each carbon atom.

Step-by-Step Solution

1

Determine the hybridization state of carbon atoms in ethane (

C_2H_6

)

Each carbon atom forms 4 single sigma (

\sigma

) bonds, corresponding to

sp^3

hybridization with tetrahedral geometry.

Saturated hydrocarbons (alkanes) undergo

sp^3

hybridization to accommodate four equivalent single bonds.

2

Identify the mode of overlap for the carbon-carbon single bond

The single bond between the two carbon atoms is a sigma (

\sigma

) bond.

Sigma bonds are always formed by direct end-to-end (head-on) overlap of atomic or hybrid orbitals along the bond axis.

Key Concept

Orbital overlap and hybridization in tetrahedral carbon (

sp^3

sigma bonding)

Question 5Question →

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

Drag items to arrange them in the correct order

Show answer & explanation

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 6Question →

Which of the following reagents reacts with propanal to produce a silver mirror, but shows no reaction with propanone?

Ammoniacal silver nitrate solution

Fehling's solution

Acidified potassium dichromate(VI) solution

Lithium tetrahydridoaluminate(III) in dry ether

Show answer & explanation

Answer: Ammoniacal silver nitrate solution

Answer

Ammoniacal silver nitrate solution

Ammoniacal silver nitrate solution (Tollen's reagent) acts as a mild oxidizing agent. It oxidizes propanal to propanoate ions while

Ag^+

ions are reduced to elemental silver, depositing as a silver mirror on the container walls. Propanone resists oxidation by mild oxidizing agents and yields no silver deposit.

Step-by-Step Solution

1

Identify the functional groups of the given compounds

Propanal (

CH_3CH_2CHO

) is an alkanal (aldehyde), whereas propanone (

CH_3COCH_3

) is an alkanone (ketone).

Alkanals possess a terminal carbonyl group with a hydrogen atom that is easily oxidized, while alkanones lack this hydrogen atom.

2

Determine the specific reagent that yields a silver mirror

Tollen's reagent (ammoniacal silver nitrate solution,

[Ag(NH_3)_2]^+

) is reduced by alkanals to form metallic silver (

Ag(s)

), which deposits on the glass wall as a silver mirror.

Alkanones cannot be easily oxidized by mild oxidizing agents like Tollen's reagent under standard conditions.

Key Concept

Distinction between alkanals and alkanones using Tollen's reagent

Question 7Question →

Consider the organic compound 3-methylbut-1-yne, which has the condensed structural formula $HC\equiv C-CH(CH_3)_2$ . How many carbon atoms in a single molecule of this compound are $sp^3$ hybridized?

Show answer & explanation

Answer: 3; three; 3 carbon atoms; 3 carbons

Answer

There are 3

sp^3

hybridized carbon atoms in one molecule of 3-methylbut-1-yne.

In 3-methylbut-1-yne (

HC\equiv C-CH(CH_3)_2

), there are 5 total carbon atoms. The two terminal/alkyne carbons (

C_1

and

C_2

) participate in a triple bond, giving them 2

\sigma

bonds each and an

sp

hybridization state. The central methine carbon (

C_3

) is bonded to four distinct atoms (

C_2

,

H

, and two methyl carbons) via single

\sigma

bonds, making it

sp^3

hybridized. The two methyl group carbons (

C_4

and

C_5

) are each single-bonded to three hydrogen atoms and

C_3

, making them

sp^3

hybridized as well. Thus, exactly 3 carbon atoms are

sp^3

hybridized.

Step-by-Step Solution

1

Expand the condensed structural formula to identify every carbon atom and its bonding environment.

The expanded structure is

H-C_1 \equiv C_2 - C_3H(C_4H_3)(C_5H_3)

, containing a total of 5 carbon atoms.

Expanding the formula clarifies the number of single (

\sigma

) and multiple bonds connected to each carbon atom.

2

Determine the hybridization state of the triply bonded carbon atoms (

C_1

and

C_2

).

C_1

and

C_2

are each involved in one triple bond and one single bond, forming 2

\sigma

bonds and 2

\pi

bonds. Thus, both

C_1

and

C_2

are

sp

hybridized.

A carbon atom with 2 steric domains (linear geometry) uses

sp

hybrid orbitals.

3

Determine the hybridization state of the methine carbon atom (

C_3

) and the two methyl carbon atoms (

C_4

and

C_5

).

C_3

forms four single

\sigma

bonds (one to

C_2

, one to

H

, and two to methyl carbons).

C_4

and

C_5

each form four single

\sigma

bonds (one to

C_3

and three to

H

). Therefore,

C_3

,

C_4

, and

C_5

are all

sp^3

hybridized.

A carbon atom bonded to 4 separate atoms via single

\sigma

bonds has 4 steric domains (tetrahedral geometry) and undergoes

sp^3

hybridization.

4

Count the total number of

sp^3

hybridized carbon atoms.

3 carbon atoms (

C_3

,

C_4

, and

C_5

) are

sp^3

hybridized.

Combining the results from steps 2 and 3 gives 2

sp

carbons and 3

sp^3

carbons.

Key Concept

Identification of carbon hybridization states (

sp^3, sp^2, sp

) in aliphatic molecules

Estimated Time:1m 30s

Question 8Question →

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

Click a left item, then click its matching right item

Show answer & explanation

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 9Question →

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 10Question →

What volume of oxygen at STP is required for the complete combustion of $10\text{ dm}^3$ of propane gas ( $C_3H_8$ ) measured at the same temperature and pressure?

10\text{ dm}^3

30\text{ dm}^3

50\text{ dm}^3

22.4\text{ dm}^3

Show answer & explanation

Answer: $50\text{ dm}^3$

Answer

The volume of oxygen required at STP for the complete combustion of

10\text{ dm}^3

of propane is

50\text{ dm}^3

.

According to the balanced chemical equation

C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)

,

1\text{ dm}^3

of propane requires

5\text{ dm}^3

of oxygen for complete combustion at the same temperature and pressure. Therefore,

10\text{ dm}^3

of propane requires

10 \times 5 = 50\text{ dm}^3

of oxygen.

Step-by-Step Solution

1

Write and balance the chemical equation for the complete combustion of propane gas.

C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)

A balanced chemical equation provides the correct mole and volume ratios of reactants and products.

2

Apply Gay-Lussac's Law of Combining Volumes for gases at the same temperature and pressure.

1\text{ volume of } C_3H_8 \text{ reacts with } 5\text{ volumes of } O_2

The volume ratio of gaseous reactants equals the ratio of their stoichiometric coefficients.

3

Multiply the given volume of propane by the volume ratio coefficient.

\text{Volume of } O_2 = 10\text{ dm}^3 \times 5 = 50\text{ dm}^3

Calculating

10 \times 5

gives the total volume of oxygen required.

Key Concept

Combustion Stoichiometry and Gay-Lussac's Law of Combining Volumes

Question 11Question →

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 12Question →

A triglyceride derived from a single saturated alkanoic acid undergoes complete saponification with excess aqueous $\text{NaOH}$ , yielding glycerol and a sodium soap salt. When the isolated soap salt is acidified with excess dilute $\text{HCl}$ , a pure saturated alkanoic acid $X$ is liberated. Neutralization of $10.24\text{ g}$ of acid $X$ requires exactly $40.0\text{ cm}^3$ of a $1.00\text{ mol dm}^{-3}$ $\text{NaOH}$ solution. What is the correct IUPAC name of the alkanoic acid $X$ ? (Atomic masses: $\text{H} = 1$ , $\text{C} = 12$ , $\text{O} = 16$ )

Hexadecanoic acid

Tetradecanoic acid

Pentadecanoic acid

Octadecanoic acid

Show answer & explanation

Answer: Hexadecanoic acid

Answer

Hexadecanoic acid

The acid

X

reacts with

\text{NaOH}

in a 1:1 stoichiometry.

0.0400\text{ mol}

of

\text{NaOH}

neutralizes

0.0400\text{ mol}

of

X

, giving a molar mass of

256\text{ g mol}^{-1}

. Setting the general formula for saturated alkanoic acids

\text{C}_n\text{H}_{2n}\text{O}_2

equal to

256

yields

14n + 32 = 256

, which solves to

n = 16

. The 16-carbon saturated carboxylic acid is Hexadecanoic acid.

Step-by-Step Solution

1

Calculate the number of moles of NaOH used in the neutralization reaction

\text{Moles of NaOH} = \text{Concentration} \times \text{Volume in dm}^3 = 1.00\text{ mol dm}^{-3} \times \frac{40.0}{1000}\text{ dm}^3 = 0.0400\text{ mol}

Neutralization uses volume and molarity to determine mole quantity.

2

Determine the molar mass of the monocarboxylic acid X

Since monocarboxylic acid reacts with NaOH in a 1:1 mole ratio (

\text{R-COOH} + \text{NaOH} \rightarrow \text{R-COONa} + \text{H}_2\text{O}

), moles of

X = 0.0400\text{ mol}

. Thus,

\text{Molar mass of } X = \frac{10.24\text{ g}}{0.0400\text{ mol}} = 256\text{ g mol}^{-1}

.

Molar mass is the mass divided by the amount in moles.

3

Use the general formula for a saturated alkanoic acid to find the number of carbon atoms (n)

The general formula for a saturated monocarboxylic acid is

\text{C}_n\text{H}_{2n}\text{O}_2

. Molar mass

= 12n + 2n + 32 = 14n + 32 = 256 \implies 14n = 224 \implies n = 16

.

Determining

n

gives the total number of carbon atoms in the IUPAC parent chain.

4

Assign the official IUPAC name for a 16-carbon saturated alkanoic acid

\text{C}_{16}\text{H}_{32}\text{O}_2

is named Hexadecanoic acid.

The IUPAC suffix for a 16-carbon alkanoic acid is hexadecanoic acid.

Key Concept

Determination of alkanoic acid stoichiometry from saponification and neutralization data

Estimated Time:3m 0s

Question 13Question →

A $45.0\text{ g}$ sample of impure glucose containing $80.0\%$ pure glucose ( $C_6H_{12}O_6$ ) by mass undergoes complete fermentation in the presence of zymase enzyme at suitable conditions. What is the volume of carbon dioxide gas, in $\text{dm}^3$ , released at standard temperature and pressure (STP)?

(Relative atomic masses: $C = 12.0$ , $H = 1.0$ , $O = 16.0$ ; Molar volume of gas at STP $= 22.4\text{ dm}^3\text{mol}^{-1}$ )

Show answer & explanation

Answer: 8.96

Answer

The volume of carbon dioxide gas released at STP is

8.96\text{ dm}^3

.

The complete fermentation of glucose is represented by the equation

C_6H_{12}O_6 \xrightarrow{\text{zymase}} 2C_2H_5OH + 2CO_2

. Taking into account the

80.0\%

purity, the mass of active glucose is

0.800 \times 45.0\text{ g} = 36.0\text{ g}

, which corresponds to

\frac{36.0}{180.0} = 0.200\text{ mol}

. Because

1\text{ mol}

of glucose yields

2\text{ mol}

of

CO_2

,

0.400\text{ mol}

of

CO_2

is produced. At STP,

0.400\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 8.96\text{ dm}^3

.

Step-by-Step Solution

1

Determine the mass of pure glucose in the impure sample

36.0 g of pure glucose

Only the active pure glucose undergoes fermentation.

2

Calculate the molar mass of glucose (

C_6H_{12}O_6

)

180.0 g/mol

Needed to convert mass of reactant into molar amount.

3

Calculate the number of moles of glucose fermented

0.200 mol of glucose

Moles = Mass / Molar mass.

4

Determine moles of CO2 evolved using reaction stoichiometry

0.400 mol of CO2

Fermentation of 1 mole of hexose sugar produces 2 moles of ethanol and 2 moles of carbon dioxide.

5

Calculate the volume of CO2 gas at STP

8.96 dm^3

Volume = Moles × Molar volume at STP.

Key Concept

Fermentation Stoichiometry and Molar Yield of Alkanols

Question 14Question →

An acyclic hydrocarbon $X$ with the molecular formula $C_5H_8$ rapidly decolourises bromine water. However, when $X$ is treated with ammoniacal silver nitrate solution, no precipitate is observed. Upon complete catalytic hydrogenation in the presence of a nickel catalyst, $X$ is converted into pentane. What is the IUPAC name of hydrocarbon $X$ ?

Show answer & explanation

Answer: pent-2-yne; 2-pentyne; Pent-2-yne; 2-Pentyne

Answer

Pent-2-yne (or 2-pentyne)

Hydrocarbon

X

has the molecular formula

C_5H_8

, corresponding to two degrees of unsaturation. Complete hydrogenation to pentane confirms an unbranched five-carbon chain. Decolourisation of bromine water verifies unsaturation. Because

X

yields no precipitate with ammoniacal silver nitrate solution, it lacks acidic terminal acetylenic hydrogens (

R-C \equiv C-H

). Therefore, the triple bond must be located internally between carbon-2 and carbon-3, making the compound pent-2-yne.

Step-by-Step Solution

1

Determine the degree of unsaturation and carbon skeleton of hydrocarbon

X

.

Degree of unsaturation is 2, and the carbon skeleton is a straight 5-carbon chain.

The molecular formula

C_5H_8

corresponds to

C_nH_{2n-2}

, indicating two degrees of unsaturation (an alkyne or alkadiene). Complete catalytic hydrogenation yields pentane (

C_5H_{12}

), proving an unbranched five-carbon chain.

2

Analyze the reaction with bromine water.

Hydrocarbon

X

contains carbon-carbon multiple bonds.

Decolourisation of bromine water confirms the presence of unsaturation.

3

Evaluate the test with ammoniacal silver nitrate solution.

Hydrocarbon

X

is an internal (non-terminal) alkyne.

Terminal alkynes possess acidic acetylenic hydrogen atoms (

R-C \equiv C-H

) that react with ammoniacal silver nitrate to form a characteristic white silver acetylide precipitate. The absence of a precipitate rules out pent-1-yne and confirms that the triple bond is located internally at C-2.

4

Deduce the final IUPAC name of hydrocarbon

X

.

pent-2-yne

Combining a straight 5-carbon chain with an internal triple bond between carbon-2 and carbon-3 gives pent-2-yne.

Key Concept

Distinction between terminal and non-terminal alkynes using ammoniacal silver nitrate test and carbon skeleton determination via hydrogenation

Estimated Time:2m 0s

Question 15Question →

An unknown gaseous hydrocarbon $X$ decolorizes acidified potassium tetraoxomanganate(VII) solution and produces a reddish-brown precipitate when bubbled into an ammoniacal solution of copper(I) chloride. Which of the following IUPAC structural formulas represents hydrocarbon $X$ ?

\text{CH}_3\text{CH}_2\text{C}\equiv\text{CH}

\text{CH}_3\text{C}\equiv\text{CCH}_3

\text{CH}_3\text{CH}=\text{CHCH}_3

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

Show answer & explanation

Answer: $\text{CH}_3\text{CH}_2\text{C}\equiv\text{CH}$

Answer

The correct structural formula is

\text{CH}_3\text{CH}_2\text{C}\equiv\text{CH}

(but-1-yne).

The compound

\text{CH}_3\text{CH}_2\text{C}\equiv\text{CH}

(but-1-yne) contains a carbon-carbon triple bond which decolorizes acidified

\text{KMnO}_4

via oxidation. Furthermore, because it is a terminal alkyne with a hydrogen atom directly bonded to an

sp

-hybridized carbon, it reacts with ammoniacal copper(I) chloride solution to yield a reddish-brown precipitate of copper(I) acetylide.

Step-by-Step Solution

1

Analyze the reaction with acidified potassium tetraoxomanganate(VII) (\text{KMnO}_4) solution.

The decolorization of acidified

\text{KMnO}_4

proves that hydrocarbon

X

is unsaturated (contains carbon-carbon double or triple bonds). This eliminates saturated alkanes like butane.

Unsaturated hydrocarbons undergo oxidation addition across carbon-carbon multiple bonds.

2

Analyze the reaction with ammoniacal copper(I) chloride (\text{Cu}_2\text{Cl}_2) solution.

Formation of a reddish-brown precipitate (copper(I) diallylide/acetylide) confirms the presence of a terminal alkyne carrying an acidic acetylenic hydrogen attached to an

sp

-hybridized carbon atom (

-\text{C}\equiv\text{C}-\text{H}

).

Only terminal alkynes have sufficiently acidic hydrogen atoms to be substituted by copper(I) or silver ions in ammoniacal solutions.

3

Differentiate between terminal alkynes, internal alkynes, and alkenes based on structural formulas.

\text{CH}_3\text{CH}_2\text{C}\equiv\text{CH}

is but-1-yne (a terminal alkyne) which fulfills both conditions.

\text{CH}_3\text{C}\equiv\text{CCH}_3

is an internal alkyne lacking a terminal acidic hydrogen.

Internal alkynes and alkenes fail the ammoniacal copper(I) chloride test despite being unsaturated.

Key Concept

Distinction between general unsaturation tests and terminal alkyne confirmation tests.

Estimated Time:1m 30s

Question 16Question →

An organic compound $X$ with the molecular formula $\text{C}_4\text{H}_8\text{O}_2$ does not react with sodium hydrogentrioxocarbonate(IV) solution to liberate gas. Upon refluxing $X$ with dilute sodium hydroxide solution, two organic products, $Y$ and $Z$ , are formed. Acidification of product $Z$ yields ethanoic acid, while mild oxidation of product $Y$ yields an alkanal that gives a positive Tollen's test. Which of the following correctly identifies the IUPAC name of compound $X$ and the chemical nature of its reaction with dilute sodium hydroxide?

Ethyl ethanoate; an irreversible alkaline hydrolysis reaction

Ethyl ethanoate; a reversible esterification equilibrium reaction

Methyl propanoate; an irreversible alkaline hydrolysis reaction

Butanoic acid; a reversible acid-base neutralization reaction

Show answer & explanation

Answer: Ethyl ethanoate; an irreversible alkaline hydrolysis reaction

Answer

The IUPAC name of compound

X

is ethyl ethanoate, and its reaction with dilute sodium hydroxide is an irreversible alkaline hydrolysis reaction.

Compound

X

does not evolve carbon(IV) oxide gas with sodium hydrogentrioxocarbonate(IV), ruling out alkanoic acids and establishing that

X

is an ester. Alkaline hydrolysis of

X

yields sodium ethanoate (

Z

) and ethanol (

Y

), because acidification of

Z

produces ethanoic acid (

2

carbons) and oxidation of ethanol (

Y

) yields ethanal, an alkanal that gives a silver mirror with Tollen's reagent. Therefore,

X

is ethyl ethanoate (

\text{CH}_3\text{COOCH}_2\text{CH}_3

). Base hydrolysis of esters converts the carboxyl moiety into a resonance-stabilized carboxylate ion, preventing the reverse reaction and rendering the hydrolysis irreversible.

Step-by-Step Solution

1

Determine the functional group class of compound

X

Compound

X

is an ester.

Isomers with formula

\text{C}_4\text{H}_8\text{O}_2

can be alkanoic acids or esters. Since

X

does not react with

\text{NaHCO}_3

to evolve

\text{CO}_2

gas, it lacks the free carboxyl acid group (

-\text{COOH}

) and must be an ester.

2

Deduce the structure of the alkanoate (acid) portion of the ester

The alkanoate part is ethanoate (

\text{CH}_3\text{COO}^-

).

Alkaline hydrolysis of ester

X

produces carboxylate salt

Z

. Acidification of

Z

yields ethanoic acid (

\text{CH}_3\text{COOH}

), which contains 2 carbon atoms.

3

Deduce the structure of the alkyl (alcohol) portion of the ester

The alkyl group is ethyl (

-\text{C}_2\text{H}_5

), making

Y

ethanol (

\text{CH}_3\text{CH}_2\text{OH}

).

The total number of carbon atoms in

X

is 4. Subtracting 2 carbons from the acid part leaves 2 carbons for the alcohol

Y

(ethanol). Mild oxidation of ethanol yields ethanal (an alkanal), which reduces Tollen's reagent.

4

Identify the reaction type with dilute

\text{NaOH}

Irreversible alkaline hydrolysis (saponification).

Hydrolysis of an ester with alkali consumes hydroxyl ions (

\text{OH}^-

) to form an unreactive carboxylate anion (

\text{CH}_3\text{COO}^-

), driving the reaction to completion irreversibly.

Key Concept

Chemical differentiation between carboxylic acids and esters, ester hydrolysis kinetics, and structural deduction.

Question 17Question →

Match each chemical reaction or process involving alkanoic acids, esters, fats, or oils on the left with its corresponding chemical outcome or product description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings match: (1) Acid-catalyzed esterification of propane-1,2,3-triol with hexadecanoic acid to producing tripalmitin fat and water; (2) Alkaline hydrolysis of glyceryl tristearate with aqueous NaOH to yielding propane-1,2,3-triol and sodium octadecanoate soap; (3) Catalytic hydrogenation of glyceryl trioleate to converting an unsaturated liquid triacylglycerol into a saturated solid fat; and (4) Acid-catalyzed reflux of ethyl ethanoate with water to reversibly producing ethanoic acid and ethanol.

The correct pairings logically connect each organic process to its definitive product or reaction characteristic: esterification of hexadecanoic acid with glycerol yields tripalmitin; saponification of glyceryl tristearate using NaOH irreversibly yields glycerol and sodium octadecanoate soap; catalytic hydrogenation saturates double bonds in glyceryl trioleate turning liquid oil into solid fat; and acid hydrolysis of ethyl ethanoate is a reversible equilibrium yielding ethanoic acid and ethanol.

Step-by-Step Solution

1

Analyze the reaction of propane-1,2,3-triol (glycerol) with hexadecanoic acid (palmitic acid).

Identified as esterification forming tripalmitin and water.

Hexadecanoic acid (

C_{15}H_{31}COOH

) esterifies with glycerol to form glyceryl tripalmitin (

C_{51}H_{98}O_6

), a saturated fat.

2

Analyze the alkaline hydrolysis of glyceryl tristearate using aqueous

NaOH

.

Identified as saponification yielding glycerol and sodium octadecanoate.

Base hydrolysis of fats irreversibly converts ester groups into glycerol and carboxylate salts (soap).

3

Analyze the catalytic hydrogenation of glyceryl trioleate.

Identified as hardening of oils.

Addition of

H_2

across the

C=C

double bonds of unsaturated oleic acid residues converts liquid oil into saturated solid fat.

4

Analyze the acid-catalyzed reaction of ethyl ethanoate with water.

Identified as reversible ester hydrolysis.

Acid hydrolysis of esters is reversible, establishing an equilibrium mixture of the parent alkanoic acid (ethanoic acid) and alkanol (ethanol).

Key Concept

Reactivity and Interconversion of Alkanoic Acids, Esters, Fats, and Oils

Question 18Question →

An organic compound $X$ with the molecular formula $\text{C}_4\text{H}_{10}\text{O}$ resists oxidation when treated with acidified potassium dichromate(VI) ( $\text{K}_2\text{Cr}_2\text{O}_7/\text{H}^+$ ). When compound $X$ is heated with concentrated tetraoxosulfate(VI) acid ( $\text{H}_2\text{SO}_4$ ) at $170^\circ\text{C}$ , it undergoes dehydration to produce a major organic product $Y$ . What is the IUPAC name of compound $Y$ ?

2-methylpropanal

But-2-ene

2-methylpropene

But-1-ene

Show answer & explanation

Answer: 2-methylpropene

Answer

2-methylpropene

The compound resisting oxidation must be a tertiary alkanol because the hydroxyl-bearing carbon lacks an alpha-hydrogen atom. The only tertiary alkanol with formula

\text{C}_4\text{H}_{10}\text{O}

is 2-methylpropan-2-ol. Subjecting 2-methylpropan-2-ol to intra-molecular dehydration using concentrated tetraoxosulfate(VI) acid at

170^\circ\text{C}

removes water to form 2-methylpropene as the major alkene product.

Step-by-Step Solution

1

Determine the structural class of compound X from its resistance to oxidation.

Compound X is a tertiary alkanol.

Primary and secondary alkanols are readily oxidized by acidified potassium dichromate(VI), whereas tertiary alkanols resist mild oxidation because the carbon atom bonded to the hydroxyl group (

-OH

) carries no hydrogen atoms.

2

Identify the specific isomer of formula

\text{C}_4\text{H}_{10}\text{O}

corresponding to a tertiary alkanol.

Compound X is 2-methylpropan-2-ol,

(\text{CH}_3)_3\text{C-OH}

.

Among the four structural isomers of

\text{C}_4\text{H}_{10}\text{O}

alkanols, only 2-methylpropan-2-ol is tertiary.

3

Determine the elimination product when 2-methylpropan-2-ol undergoes acid-catalyzed dehydration.

Dehydration yields 2-methylpropene,

(\text{CH}_3)_2\text{C=CH}_2

.

Heating with concentrated

\text{H}_2\text{SO}_4

at

170^\circ\text{C}

removes a molecule of water (the

-OH

group and a hydrogen atom from an adjacent methyl group), forming an alkene.

Key Concept

Classification of alkanols based on oxidation behavior and acid-catalyzed dehydration to alkenes

Estimated Time:2m 0s

Question 19Question →

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 20Question →

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

Click a left item, then click its matching right item

Show answer & explanation

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

Organic Chemistry

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