Organic Chemistry Practice Questions — JAMB UTME — Page 2

What is the hybridization state and geometry of the central carbon atom in a molecule of methane ( $CH_4$ )?

sp^3

hybridization with tetrahedral geometry and bond angles of

109.5^\circ

sp^2

hybridization with trigonal planar geometry and bond angles of

120^\circ

sp

hybridization with linear geometry and bond angles of

180^\circ

sp^3

hybridization with square planar geometry and bond angles of

90^\circ

Show answer & explanation

Answer: $sp^3$ hybridization with tetrahedral geometry and bond angles of $109.5^\circ$

Answer

sp^3

hybridization with tetrahedral geometry and bond angles of

109.5^\circ

In methane (

CH_4

), the central carbon atom forms four single covalent

\sigma

bonds with four hydrogen atoms. To achieve equivalent bonding, one

2s

and three

2p

orbitals hybridize to form four

sp^3

hybrid orbitals directed toward the corners of a regular tetrahedron, giving bond angles of

109.5^\circ

.

Step-by-Step Solution

1

Determine the valence electron configuration and bonding of the central carbon atom.

Carbon has 4 valence electrons and forms 4 single

\sigma

(sigma) bonds with four hydrogen atoms in methane (

CH_4

).

The number of single bonds and lone pairs determines the steric number of the central atom.

2

Determine the hybridization state based on the steric number.

Steric number = 4 (four

\sigma

bonds, zero lone pairs), requiring four equivalent hybrid orbitals formed by combining one

2s

and three

2p

atomic orbitals (

sp^3

).

Mixing one

s

orbital and three

p

orbitals yields four equivalent

sp^3

hybrid orbitals.

3

Determine the spatial geometry and ideal bond angle.

According to VSEPR theory, four bonding pairs position themselves as far apart as possible in 3D space, forming a tetrahedral geometry with bond angles of

109.5^\circ

.

The tetrahedral arrangement minimizes electrostatic repulsion among the four bonding electron pairs.

Key Concept

Tetrahedral Carbon, Bonding, and Hybridization

Estimated Time:45s

Question 22Question →

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.

Drag items to arrange them in the correct order

Show answer & explanation

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 23Question →

A student subjects separate samples of methyl propanoate to two different laboratory reactions:

Reaction 1: Refluxing with dilute $\text{HCl}(aq)$
Reaction 2: Refluxing with aqueous $\text{NaOH}(aq)$

Which of the following correctly compares the reversibility of these reactions and the nature of the organic products formed?

Reaction 1 is reversible yielding propanoic acid and methanol, whereas Reaction 2 is irreversible yielding sodium propanoate and methanol.

Both Reaction 1 and Reaction 2 are fully reversible equilibrium processes yielding propanoic acid and methanol.

Reaction 1 is irreversible yielding sodium propanoate, whereas Reaction 2 is reversible yielding propanoic acid.

Both Reaction 1 and Reaction 2 are irreversible single-step neutralizations yielding sodium propanoate and methanol.

Show answer & explanation

Answer: Reaction 1 is reversible yielding propanoic acid and methanol, whereas Reaction 2 is irreversible yielding sodium propanoate and methanol.

Answer

Reaction 1 is reversible yielding propanoic acid and methanol, whereas Reaction 2 is irreversible yielding sodium propanoate and methanol.

The statement specifying that Reaction 1 is reversible yielding propanoic acid and methanol, while Reaction 2 is irreversible yielding sodium propanoate and methanol, is correct. Acid-catalyzed hydrolysis of an ester is an equilibrium process that forms the parent alkanoic acid and alkanol. In contrast, alkaline hydrolysis (saponification) uses hydroxide ions to deprotonate the acid as it forms, producing an unreactive alkanoate salt (sodium propanoate) which prevents the reverse reaction, making the process quantitative and irreversible.

Step-by-Step Solution

1

Analyze Reaction 1 (Acid Hydrolysis)

Methyl propanoate reacts with water in the presence of

\text{H}^+

catalyst to form propanoic acid (

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

) and methanol (

\text{CH}_3\text{OH}

).

Acid hydrolysis of an ester is the reverse of esterification and reaches dynamic equilibrium (it is reversible).

2

Analyze Reaction 2 (Alkaline Hydrolysis / Saponification)

Methyl propanoate reacts with

\text{OH}^-

ions to form the propanoate anion (

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

) as sodium propanoate salt and methanol (

\text{CH}_3\text{OH}

).

The base reacts un-reversibly with the carboxylic acid product to form a carboxylate salt, pulling the equilibrium completely to the right and making the overall saponification irreversible.

3

Compare both reactions

Reaction 1 is reversible (producing acid + alcohol), while Reaction 2 is irreversible (producing salt + alcohol).

Differentiates reversible acid-catalyzed ester hydrolysis from irreversible base-promoted ester hydrolysis.

Key Concept

Difference between acid hydrolysis (reversible) and alkaline hydrolysis/saponification (irreversible) of esters.

Question 24Question →

In a propyne molecule ( $CH_3-C\equiv CH$ ), which hybrid orbitals overlap head-on to form the carbon-carbon single bond between the methyl carbon atom and the adjacent triple-bonded carbon atom?

sp^3

and

sp

hybrid orbitals

sp^3

and

sp^2

hybrid orbitals

sp^2

and

sp

hybrid orbitals

sp^3

and

sp^3

hybrid orbitals

Show answer & explanation

Answer: $sp^3$ and $sp$ hybrid orbitals

Answer

The carbon-carbon single bond in propyne is formed by the head-on overlap of an

sp^3

hybrid orbital from the methyl carbon atom and an

sp

hybrid orbital from the adjacent acetylenic carbon atom.

In propyne (

CH_3-C\equiv CH

), the methyl carbon atom is attached to four atoms via single bonds, giving it a tetrahedral arrangement and

sp^3

hybridization. The central carbon atom is involved in a triple bond and one single bond, giving it a linear arrangement and

sp

hybridization. Therefore, the single bond connecting these two carbon atoms is a

\sigma

bond formed by the head-on overlap of an

sp^3

hybrid orbital from the methyl carbon and an

sp

hybrid orbital from the central acetylenic carbon.

Step-by-Step Solution

1

Determine the hybridization state of the methyl carbon atom (

CH_3-

).

The methyl carbon atom forms 4

\sigma

bonds (3 with H atoms, 1 with C), giving it 4 electron domains and a tetrahedral geometry with

sp^3

hybridization.

Four equivalent bonding electron domains around a carbon atom require four

sp^3

hybrid orbitals.

2

Determine the hybridization state of the adjacent triple-bonded carbon atom (

-C\equiv

).

This carbon atom forms 2

\sigma

bonds (1 with methyl C, 1 with terminal C) and 2

\pi

bonds, giving it 2 electron domains and a linear geometry with

sp

hybridization.

Two linear bonding domains around a carbon atom require two

sp

hybrid orbitals.

3

Identify the orbitals participating in the

\sigma

single bond between these two carbon atoms.

The

\sigma

single bond is formed by the end-to-end (head-on) overlap of one

sp^3

hybrid orbital from the methyl carbon and one

sp

hybrid orbital from the acetylenic carbon.

Single bonds (

\sigma

bonds) between hybridized carbon atoms result from direct axial overlap of their respective hybrid orbitals.

Key Concept

Orbital Overlap and Carbon Hybridization in Alkynes

Question 25Question →

Match each chemical process involving alkanoic acids, esters, or fats on the left with its correct chemical description on the right.

Click a left item, then click its matching right item

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Answer

Esterification pairs with 'Reversible reaction between an alkanoic acid and an alkanol forming an ester and water'. Saponification pairs with 'Alkaline hydrolysis of fats yielding soap and glycerol'. Hydrogenation of oils pairs with 'Addition of hydrogen gas across C=C double bonds to convert liquid oils into solid fats'.

Esterification is defined as the reversible condensation between an alkanoic acid and an alkanol. Saponification is the base-catalyzed hydrolysis of esters/fats to produce soap and glycerol. Hydrogenation reduces unsaturation in vegetable oils by adding hydrogen across carbon-carbon double bonds.

Step-by-Step Solution

1

Analyze Esterification

Esterification combines an alkanoic acid and an alkanol in a reversible equilibrium reaction to form an ester and water.

This matches the second description.

2

Analyze Saponification

Saponification breaks down triacylglycerols (fats/oils) using sodium or potassium hydroxide, yielding glycerol and salts of fatty acids (soap).

This matches the first description.

3

Analyze Hydrogenation of oils

Hydrogenation saturates the double bonds in unsaturated vegetable oils using a nickel catalyst to turn liquid oils into solid margarine.

This matches the third description.

Key Concept

Reactions and industrial processes of alkanoic acids, esters, fats, and oils

Question 26Question →

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 27Question →

During the esterification reaction between ethanoic acid ( $\text{CH}_3\text{COOH}$ ) and ethanol ( $\text{C}_2\text{H}_5\text{OH}$ ), concentrated tetraoxosulfate(VI) acid ( $\text{H}_2\text{SO}_4$ ) is added to the reaction mixture. Which of the following statements correctly accounts for the dual function of concentrated $\text{H}_2\text{SO}_4$ in maximizing the equilibrium yield of ethyl ethanoate?

It functions as a catalyst to increase the reaction rate and as a dehydrating agent to absorb water, shifting the equilibrium position to the right.

It acts as a strong inorganic acid reactant that undergoes irreversible ionic neutralization with ethanol to produce ethyl ethanoate complete to completion.

It serves solely as a catalyst, which selectively increases the rate of the forward reaction and permanently increases the equilibrium constant (

K_c

).

It acts primarily as an oxidizing agent that converts ethanol into ethanoic acid before esterification begins.

Show answer & explanation

Answer: It functions as a catalyst to increase the reaction rate and as a dehydrating agent to absorb water, shifting the equilibrium position to the right.

Answer

Concentrated tetraoxosulfate(VI) acid acts as both a catalyst to speed up the reaction rate and a dehydrating agent to remove water, shifting the reversible equilibrium toward the formation of ethyl ethanoate.

The correct response accurately identifies both functions of concentrated tetraoxosulfate(VI) acid in esterification: it acts as a catalyst to increase the rate at which equilibrium is reached and as a strong dehydrating agent that removes water from the system, driving the reversible equilibrium to the right to maximize ester production according to Le Chatelier's principle.

Step-by-Step Solution

1

Analyze the chemical equation for esterification

\text{CH}_3\text{COOH}(l) + \text{C}_2\text{H}_5\text{OH}(l) \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5(l) + \text{H}_2\text{O}(l)

Esterification is a reversible organic reaction between an alkanoic acid and an alkanol to form an ester and water.

2

Identify the catalytic role of concentrated

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

Provides

\text{H}^+

ions to protonate the carbonyl oxygen, accelerating both forward and backward rates.

Lowering activation energy allows the system to reach equilibrium faster.

3

Identify the dehydrating role and apply Le Chatelier's principle

Concentrated

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

absorbs water (

\text{H}_2\text{O}

), decreasing product concentration.

Removing a product from a reversible system shifts the equilibrium position to the right, increasing the yield of the ester.

Key Concept

Reversibility of Esterification and Dual Role of Concentrated H2SO4

Question 28Question →

Match each chemical process or reagent involving alkenes and alkynes in Column A with its corresponding chemical observation or primary product in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairing links calcium carbide hydrolysis to ethyne gas preparation, propene with bromine in carbon tetrachloride to reddish-brown decolorization, but-1-yne with ammoniacal silver nitrate to white precipitate formation, and but-2-yne with ammoniacal silver nitrate to no visible reaction.

Hydrolysis of calcium carbide is a standard laboratory preparation of ethyne gas. Electrophilic addition of bromine across the double bond of propene decolorizes the reddish-brown bromine solution, demonstrating general unsaturation. Ammoniacal silver trioxonitrate(V) reacts specifically with terminal alkynes like but-1-yne because of their acidic terminal hydrogens to form a white silver alkynide precipitate, whereas internal alkynes like but-2-yne lack a terminal acidic hydrogen and show no reaction.

Step-by-Step Solution

1

Identify the preparation method for ethyne

Calcium carbide reacts with water according to

\text{CaC}_2 + 2\text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_2 + \text{Ca(OH)}_2

, yielding ethyne gas.

This is the primary laboratory method for generating ethyne.

2

Evaluate the reaction testing for general unsaturation in propene

Bromine in

\text{CCl}_4

adds across the carbon-carbon double bond of propene to form 1,2-dibromopropane.

The consumption of molecular bromine causes the reddish-brown color of the solution to discharge.

3

Distinguish between terminal and internal alkynes using ammoniacal silver trioxonitrate(V)

But-1-yne possesses a terminal

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

bond with an acidic proton that forms a white silver alkynide precipitate, whereas internal but-2-yne lacks terminal hydrogen atoms and gives no precipitate.

Ammoniacal

\text{AgNO}_3

selectively reacts only with terminal alkynes based on terminal hydrogen acidity.

Key Concept

Unsaturation testing, ethyne synthesis, and differentiation of terminal from internal alkynes

Estimated Time:1m 30s

Question 29Question →

A gaseous hydrocarbon rapidly decolourizes bromine water in tetrachloromethane, but fails to form a precipitate when bubbled through an ammoniacal solution of silver nitrate. Which of the following hydrocarbons exhibits this chemical behavior?

Propene

Propyne

Ethyne

Propane

Show answer & explanation

Answer: Propene

Answer

Propene

Propene is an alkene (

CH_3CH=CH_2

). Its carbon-carbon double bond allows it to undergo electrophilic addition with bromine water, decolourizing the reddish-brown solution. However, because it is an alkene rather than a terminal alkyne, it lacks an acidic hydrogen atom bonded to a triply-bonded carbon atom, so it does not produce a precipitate when treated with ammoniacal silver nitrate.

Step-by-Step Solution

1

Analyze the reaction with bromine water.

Rapid decolourization of bromine water confirms the presence of unsaturation (a double or triple carbon-carbon bond). Saturated hydrocarbons like propane are eliminated.

Unsaturated hydrocarbons undergo addition reactions across their double or triple bonds to absorb bromine.

2

Analyze the outcome with ammoniacal silver nitrate solution.

The absence of a precipitate indicates that the hydrocarbon is NOT a terminal alkyne (

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

).

Only terminal alkynes possess acidic acetylenic hydrogen atoms capable of being replaced by silver ions to form insoluble metallic acetylides.

3

Identify the hydrocarbon matching both criteria.

Propene (

CH_3CH=CH_2

) is an alkene. It decolourizes bromine water due to unsaturation but yields a negative test with ammoniacal silver nitrate.

Alkenes contain carbon-carbon double bonds but lack acidic acetylenic hydrogen atoms.

Key Concept

Distinguishing alkenes from terminal alkynes using unsaturation and acetylenic hydrogen reagents

Estimated Time:1m 0s

Question 30Question →

Benzene preferentially undergoes electrophilic substitution reactions rather than electrophilic addition reactions under standard conditions because addition reactions disrupt the aromatic resonance stabilization energy of the ring.

True

False

Show answer & explanation

Answer: True

Answer

The statement is True. Benzene undergoes electrophilic substitution instead of addition to preserve its aromatic resonance stabilization energy.

Benzene's delocalized

\pi

-electron cloud makes it exceptionally stable. Electrophilic substitution allows the ring to react with electrophiles while retaining its planar, aromatic structure and resonance energy.

Step-by-Step Solution

1

Analyze the electronic structure and aromaticity of benzene.

Benzene is a cyclic, planar molecule with

6

delocalized

\pi

-electrons, rendering it exceptionally stable due to aromatic resonance energy.

Determining structural stability dictates chemical reactivity.

2

Compare the outcome of substitution versus addition reactions on the ring.

Substitution retains the unbroken

6\pi

-electron cloud (aromaticity), whereas addition destroys the cyclic delocalization, forming a non-aromatic cyclohexadiene derivative.

Preserving aromaticity lowers the activation energy and thermodynamic barrier for substitution relative to addition.

3

Evaluate the validity of the statement.

The statement accurately describes why substitution is preferred over addition.

Direct mapping between resonance stabilization energy and reaction pathway preference.

Key Concept

Aromatic Stability and Electrophilic Substitution Reactivity of Benzene

Question 31Question →

During the industrial conversion of starch into ethanol through fermentation, starch is hydrolyzed to maltose, maltose is converted to glucose, and glucose is decomposed into ethanol and carbon(IV) oxide. Which of the following represents the correct sequential order of enzymes catalyzed in this process?

Diastase

\rightarrow

Maltase

\rightarrow

Zymase

\rightarrow

Diastase

\rightarrow

Maltase

\rightarrow

Diastase

\rightarrow

Zymase

Diastase

\rightarrow

Zymase

\rightarrow

Maltase

Show answer & explanation

Answer: Diastase $\rightarrow$ Maltase $\rightarrow$ Zymase

Answer

The correct sequence of enzymes in the fermentation of starch to ethanol is diastase, followed by maltase, and finally zymase.

The production of ethanol from starch involves three sequential enzyme reactions: diastase hydrolyzes starch to maltose, maltase hydrolyzes maltose into glucose units, and zymase ferments glucose to yield ethanol and carbon(IV) oxide gas.

Step-by-Step Solution

1

Identify the enzyme that hydrolyzes starch to maltose

Starch is converted to maltose by the enzyme diastase present in malt.

Complex polysaccharides like starch require diastase for initial breakdown into disaccharides.

2

Identify the enzyme that converts maltose into glucose

Maltose is converted into glucose by the enzyme maltase.

Maltase specifically breaks down the disaccharide maltose into monosaccharide glucose units.

3

Identify the enzyme that ferments glucose into ethanol

Glucose is decomposed into ethanol and carbon(IV) oxide by the enzyme zymase.

Zymase secreted by yeast catalyzes the final fermentation step converting monosaccharides into alcohol.

Key Concept

Sequential enzyme-catalyzed reactions in starch fermentation to produce ethanol

Estimated Time:1m 0s

Question 32Question →

Match each chemical transformation or test involving alkanals and alkanones with its corresponding observation or principal product.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Oxidation of propanal with acidified

K_2Cr_2O_7

matches color change from orange to green yielding propanoic acid; Reduction of propanone using

LiAlH_4

matches formation of a secondary alcohol, propan-2-ol; Warming ethanal with Tollen's reagent matches deposition of a shiny silver mirror coating; Warming propanone with

I_2

in aqueous

NaOH

matches formation of a characteristic yellow precipitate of triiodomethane (

CHI_3

).

Each chemical process uniquely corresponds to its diagnostic observation or reaction outcome: propanal oxidizes to propanoic acid with an orange-to-green color change (

Cr^{3+}

); propanone reduces to propan-2-ol; ethanal reduces Tollen's reagent to produce a silver mirror; and propanone gives a positive iodoform test producing yellow

CHI_3

precipitate.

Step-by-Step Solution

1

Analyze the oxidation reaction of propanal

Propanal is an alkanal oxidized by acidified dichromate to propanoic acid (

CH_3CH_2COOH

) with dichromate reducing from orange

Cr_2O_7^{2-}

to green

Cr^{3+}

.

Alkanals are readily oxidized to alkanoic acids.

2

Determine the product of propanone reduction

Reducing propanone (

CH_3COCH_3

) with

LiAlH_4

adds hydrogen across the carbonyl double bond, converting the ketone into propan-2-ol (

CH_3CH(OH)CH_3

).

Reduction of alkanones yields secondary alcohols.

3

Evaluate the reaction between ethanal and Tollen's reagent

Ethanal oxidizes to ethanoic acid while reducing

Ag^+

ions in ammoniacal silver nitrate to elemental silver metal (

Ag

), depositing as a silver mirror.

Tollen's reagent specifically distinguishes alkanals from alkanones.

4

Analyze the iodoform test for propanone

Compounds containing the

CH_3C=O

group react with iodine in sodium hydroxide solution to yield triiodomethane (

CHI_3

), a yellow precipitate.

Propanone is a methyl ketone which gives a positive triiodomethane test.

Key Concept

Reactions and Distinction Tests of Alkanals and Alkanones

Question 33Question →

Which of the following chemical changes occurs during the industrial conversion of a liquid vegetable oil into solid margarine via catalytic hydrogenation?

The ester linkages are hydrolyzed to yield glycerol and free fatty acids

The degree of unsaturation decreases as carbon-carbon double bonds are converted to single bonds

The alkanoic acid chains undergo dehydration to produce gaseous alkynes

The triacylglycerol molecules are oxidized to form short-chain dicarboxylic acids

Show answer & explanation

Answer: The degree of unsaturation decreases as carbon-carbon double bonds are converted to single bonds

Answer

The degree of unsaturation decreases as carbon-carbon double bonds are converted to single bonds.

Vegetable oils are naturally liquid at room temperature due to their high degree of unsaturation (presence of multiple

C=C

double bonds in the fatty acid tails). During industrial hydrogenation, hydrogen gas (

H_2

) is reacted with the oil using a nickel catalyst at elevated temperatures. This addition reaction converts unsaturated

C=C

bonds into saturated

C-C

single bonds, thereby decreasing the degree of unsaturation, increasing the packing efficiency of the molecules, and raising the melting point to produce solid margarine.

Step-by-Step Solution

1

Identify the chemical structure of vegetable oils

Vegetable oils are liquid triacylglycerols (esters of glycerol and long-chain fatty acids) containing a high proportion of unsaturated carbon-carbon double bonds (

C=C

).

Liquid fats have lower melting points due to kinks created by double bonds in the hydrocarbon chains.

2

Determine the reaction mechanism of catalytic hydrogenation

Addition of hydrogen gas (

H_2

) in the presence of a finely divided nickel catalyst converts

C=C

double bonds into

C-C

single bonds.

Adding hydrogen saturates the hydrocarbon chains, causing them to pack more tightly and raising their melting point to solidify the oil into margarine.

Key Concept

Hardening of Oils and Hydrogenation of Unsaturated Lipids

Estimated Time:1m 0s

Question 34Question →

Arrange the following sequential steps involved in the laboratory preparation and isolation of soap (saponification) from vegetable oil in the correct chronological order from first to last:

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order of steps for the preparation and isolation of soap is: first, refluxing the vegetable oil with concentrated sodium hydroxide (saponification); second, adding concentrated sodium chloride solution (salting out); third, filtering the mixture to separate the solid soap curd; and fourth, washing the solid soap with cold distilled water to remove impurities.

Saponification begins by refluxing vegetable oil (a triglyceride ester) with concentrated sodium hydroxide, hydrolyzing the ester links to yield sodium alkanoates (soap) and glycerol. Next, concentrated sodium chloride solution is added (salting out) to precipitate the soap by decreasing its solubility. The mixture is then filtered to isolate the solid soap curd from the liquid filtrate, and finally, the solid residue is washed with cold distilled water to remove excess sodium hydroxide.

Step-by-Step Solution

1

Reflux vegetable oil with aqueous sodium hydroxide solution.

Complete alkaline hydrolysis of the triester (triglyceride) into glycerol and sodium alkanoates (soap).

Base-catalyzed ester hydrolysis is required to break down the vegetable oil into fatty acid sodium salts.

2

Add concentrated sodium chloride solution (brine) to the reaction mixture.

Precipitation of sodium alkanoate (soap) as a floating curd.

Increasing the concentration of sodium ions forces the soap salt out of solution due to the common-ion effect and reduced solubility.

3

Filter the mixture using a funnel and filter paper.

Crude solid soap is retained on the filter paper while glycerol and brine pass through as filtrate.

Filtration separates the insoluble soap precipitate from the soluble glycerol byproduct.

4

Rinse the collected soap residue with cold water.

Purified soap free from unreacted sodium hydroxide and salt.

Cold water dissolves remaining surface impurities without dissolving significant amounts of the solid soap.

Key Concept

Saponification process and salting out of soap

Estimated Time:1m 30s

Question 35Question →

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:

Drag items to arrange them in the correct order

Show answer & explanation

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 36Question →

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

Click a left item, then click its matching right item

Show answer & explanation

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 37Question →

An organic liquid $X$ with the molecular formula $C_4H_8O$ forms a yellow precipitate when warmed with iodine in alkaline solution, but shows no observable reaction with Tollen's reagent. When compound $X$ is reacted with lithium tetrahydridoaluminate(III) ( $LiAlH_4$ ) in dry ether, it produces compound $Y$ . What is the IUPAC name of compound $Y$ ?

Butan-2-ol

Butan-1-ol

2-Methylpropan-2-ol

Butanoic acid

Show answer & explanation

Answer: Butan-2-ol

Answer

Butan-2-ol

The positive iodoform test shows the presence of a methyl ketone carbonyl group (

CH_3C=O

), while the negative Tollen's test confirms that compound X is a ketone (butan-2-one) rather than an aldehyde. Reducing a ketone with lithium tetrahydridoaluminate(III) (

LiAlH_4

) converts the carbonyl carbon to a secondary alcohol group, yielding butan-2-ol.

Step-by-Step Solution

1

Analyze the chemical test results to identify the functional group of compound

X

.

Compound

X

(

C_4H_8O

) gives a positive triiodomethane (iodoform) test (yellow precipitate) but a negative Tollen's test (no silver mirror). This confirms that compound

X

is a methyl ketone, specifically butan-2-one (

CH_3COCH_2CH_3

).

Alkanals reduce Tollen's reagent to silver metal, whereas alkanones do not. The positive iodoform test indicates the presence of a

CH_3C=O

carbonyl group.

2

Determine the reaction outcome when compound

X

is reduced using

LiAlH_4

.

Reduction of the ketone butan-2-one with

LiAlH_4

converts the carbonyl group (

C=O

) into a secondary alcohol group (

-CH(OH)-

).

LiAlH_4

acts as a hydride donor reducing agent, turning alkanones into secondary alkanols.

3

Deduce the IUPAC name of the resulting product compound

Y

.

Reducing

CH_3COCH_2CH_3

produces

CH_3CH(OH)CH_2CH_3

, which is named butan-2-ol.

The four-carbon chain with the hydroxyl group at carbon-2 is named butan-2-ol.

Key Concept

Distinction between alkanals and alkanones via Tollen's and Iodoform tests, and reduction of ketones to secondary alcohols.

Question 38Question →

Arrange the following organic compounds, each having a relative molecular mass of approximately $58-60\text{ g/mol}$ , in order of INCREASING boiling point (from lowest boiling point to highest boiling point).

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence in order of increasing boiling point is Butane (

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

), followed by Methyl formate (

\text{HCOOCH}_3

), then Propan-1-ol (

\text{C}_3\text{H}_7\text{OH}

), and finally Ethanoic acid (

\text{CH}_3\text{COOH}

).

Boiling points depend on the relative strength of intermolecular forces when molecular masses are comparable (~60 g/mol). Butane is non-polar and exhibits only weak dispersion forces (lowest boiling point). Methyl formate is polar and exhibits dipole-dipole attractions. Propan-1-ol forms strong hydrogen bonds via its hydroxyl group. Ethanoic acid forms even stronger hydrogen bonds and stable cyclic dimers, giving it the highest boiling point.

Step-by-Step Solution

1

Identify the primary type of intermolecular force present in each compound of similar molar mass (

\approx 58-60\text{ g/mol}

).

Butane has London dispersion forces; Methyl formate has dipole-dipole forces; Propan-1-ol has hydrogen bonding; Ethanoic acid has extensive hydrogen bonding and dimer formation.

Boiling point increases as the strength of intermolecular forces holding the liquid molecules together increases.

2

Compare the compounds without hydrogen bonding capabilities (Butane vs. Methyl formate).

Butane is non-polar and exhibits only weak dispersion forces. Methyl formate has a polar carbonyl group (

\text{C=O}

) causing dipole-dipole attractions, making its boiling point higher than butane.

Permanent dipole-dipole attractions are stronger than instantaneous dispersion forces for molecules of similar size.

3

Compare the hydrogen-bonded compounds (Propan-1-ol vs. Ethanoic acid).

Propan-1-ol forms intermolecular hydrogen bonds through its single hydroxyl group. Ethanoic acid forms stronger hydrogen bonds using both its carbonyl oxygen and hydroxyl hydrogen to form stable cyclic dimers.

Dimerization in alkanoic acids effectively doubles the molecular interaction area, requiring significantly more thermal energy to break apart during vaporization.

4

Arrange the compounds in order of increasing boiling point.

Butane < Methyl formate < Propan-1-ol < Ethanoic acid.

The progression of intermolecular force strength directly dictates the trend in boiling points.

Key Concept

Intermolecular Forces and Boiling Point Trends in Alkanoic Acids, Esters, Alkanols, and Alkanes

Question 39Question →

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 40Question →

What is the bond angle between adjacent single covalent bonds formed by a tetrahedral $sp^3$ hybridized carbon atom in a saturated organic compound?

109.5^\circ

120^\circ

180^\circ

90^\circ

Show answer & explanation

Answer: $109.5^\circ$

Answer

The bond angle of a tetrahedral sp³ hybridized carbon atom is 109.5°.

In saturated organic compounds such as alkanes, the central carbon atom forms four single sigma bonds using four equivalent

sp^3

hybrid orbitals. According to VSEPR theory, four bonding electron pairs surrounding a central atom arrange themselves in a regular tetrahedral shape to minimize electrostatic repulsion, yielding a standard bond angle of

109.5^\circ

.

Step-by-Step Solution

1

Determine the hybridization state of a carbon atom forming four single sigma bonds.

Mixing one 2s orbital and three 2p orbitals gives four equivalent

sp^3

hybrid orbitals.

Saturated carbon forms four single bonds by directed valence orbital mixing.

2

Apply VSEPR theory to determine spatial orientation for four electron pairs around the central carbon.

To minimize electron pair repulsion, the four orbitals point toward the vertices of a regular tetrahedron.

Symmetrical four-coordinate electron pair repulsion produces tetrahedral spatial orientation.

3

Identify the characteristic inter-bond angle of a regular tetrahedron.

The angle between any two adjacent bonds is

109.5^\circ

(or

109^\circ 28'

).

This angle maximizes the distance between the four bonding electron pairs in three-dimensional space.

Key Concept

Tetrahedral Geometry and sp³ Hybridization Bond Angle

Estimated Time:45s

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