Organic Chemistry Practice Questions — JAMB UTME — Page 5

Question 81Question →

Unlike aliphatic amines, amides such as ethanamide ( $CH_3CONH_2$ ) do not exhibit basic properties in aqueous solution. Which of the following best explains this neutral behavior?

The lone pair of electrons on the nitrogen atom is delocalized by resonance with the adjacent carbonyl group.

The nitrogen atom in ethanamide lacks an unshared pair of electrons.

Ethanamide dissociates completely in water to yield hydrogen ions (

H^+

).

The methyl group (

CH_3-

) withdraws electron density from the nitrogen atom through an inductive effect.

Show answer & explanation

Answer: The lone pair of electrons on the nitrogen atom is delocalized by resonance with the adjacent carbonyl group.

Answer

The lone pair of electrons on the nitrogen atom is delocalized by resonance with the adjacent carbonyl group.

The neutral character of amides is due to resonance stabilization. The non-bonding lone pair of electrons on the nitrogen atom overlaps with the

\pi

-system of the adjacent carbonyl group (

C=O

). This delocalization significantly decreases the electron density on nitrogen, preventing it from accepting a proton (

H^+

) and acting as a base.

Step-by-Step Solution

1

Identify the functional group and structural elements of ethanamide (

CH_3CONH_2

).

Ethanamide contains an amino group (

-NH_2

) directly attached to a carbonyl group (

-C=O

).

Understanding the adjacent arrangement of the carbonyl carbon and nitrogen is essential for evaluating electronic effects.

2

Analyze the availability of the nitrogen lone pair for protonation.

The nitrogen atom has a lone pair of electrons, but it interacts with the

\pi

-orbital of the carbonyl group, forming a resonance structure:

CH_3-C(O^-)=NH_2^+

.

Basicity depends on the availability of a lone pair to accept a proton (

H^+

). Delocalization drastically reduces lone pair availability.

3

Conclude the acid-base nature of the molecule.

Because the lone pair is delocalized, ethanamide cannot readily act as a proton acceptor, rendering it neutral in aqueous solution.

This explains the contrast between basic amines (localized lone pair) and neutral amides (delocalized lone pair).

Key Concept

Neutrality of amides due to resonance delocalization of the nitrogen lone pair into the adjacent carbonyl group

Estimated Time:45s

Question 82Question →

An organic compound $X$ with the molecular formula $\text{C}_4\text{H}_8\text{O}_2$ reacts with aqueous sodium hydroxide upon heating to yield ethanol and a salt $Y$ . What is the IUPAC name of compound $Y$ ?

Sodium ethanoate

Ethanoic acid

Sodium propanoate

Sodium methanoate

Show answer & explanation

Answer: Sodium ethanoate

Answer

Sodium ethanoate

Alkaline hydrolysis of the four-carbon ester ethyl ethanoate (

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

) with sodium hydroxide breaks the ester linkage to produce ethanol (

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

) and the sodium salt of ethanoic acid, which is sodium ethanoate (

\text{CH}_3\text{COONa}

).

Step-by-Step Solution

1

Identify the structure of the ester from its molecular formula and hydrolysis alcohol product.

The ester has molecular formula

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

and produces ethanol (

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

). The alkyl group attached to oxygen contains 2 carbons, leaving 2 carbons for the acyl group. Thus, the ester is ethyl ethanoate (

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

).

Esters have the general formula

\text{RCOOR}'

, where

\text{R}'\text{OH}

is the alcohol component.

2

Determine the products of alkaline hydrolysis.

Heating ethyl ethanoate with aqueous sodium hydroxide (

\text{NaOH}

) cleaves the ester link to yield ethanol and sodium ethanoate (

\text{CH}_3\text{COONa}

).

Alkaline hydrolysis (saponification) yields the alcohol and the alkali metal salt of the alkanoic acid.

Key Concept

Alkaline Hydrolysis (Saponification) of Esters

Question 83Question →

Primary aliphatic amines react with nitrous acid ( $HNO_2$ ) at room temperature to yield an alcohol along with rapid effervescence. Which gas is evolved during this reaction?

Nitrogen gas (

N_2

)

Nitrogen(IV) oxide (

NO_2

)

Ammonia gas (

NH_3

)

Hydrogen gas (

H_2

)

Show answer & explanation

Answer: Nitrogen gas ( $N_2$ )

Answer

Nitrogen gas (

N_2

)

Primary aliphatic amines react with cold nitrous acid (

HNO_2

) to form an unstable diazonium compound that decomposes rapidly at room temperature. This decomposition produces an alcohol, water, and liberates nitrogen gas (

N_2

), which is observed as effervescence.

Step-by-Step Solution

1

Identify the functional group and reagent

The reaction involves a primary aliphatic amine (

R-NH_2

) and nitrous acid (

HNO_2

).

Nitrous acid is generated in situ from sodium trioxonitrate(III) (

NaNO_2

) and dilute hydrochloric acid (

HCl

).

2

Write the chemical reaction equation

R-NH_2 + HNO_2 \rightarrow R-OH + N_2 \uparrow + H_2O

The aliphatic diazonium salt formed as an intermediate is highly unstable and decomposes immediately at room temperature.

3

Determine the gas released

The effervescence observed is due to the release of nitrogen gas (

N_2

).

Deamination of aliphatic primary amines yields nitrogen gas as the characteristic gaseous product.

Key Concept

Reaction of primary aliphatic amines with nitrous acid to yield alcohols and nitrogen gas

Estimated Time:1m 0s

Question 84Question →

An organic compound $P$ with the molecular formula $\text{C}_4\text{H}_8\text{O}_2$ reacts with aqueous sodium hydroxide upon heating to produce a sodium salt $Q$ and an alkanol $R$ . Complete oxidation of alkanol $R$ with acidified potassium dichromate(VI) yields an alkanoic acid identical to the acid produced when salt $Q$ is acidified with dilute hydrochloric acid. What is the IUPAC name of compound $P$ ?

Methyl propanoate

Ethyl ethanoate

Propyl methanoate

Butanoic acid

Show answer & explanation

Answer: Ethyl ethanoate

Answer

Ethyl ethanoate

Ethyl ethanoate has the molecular formula

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

. Alkaline hydrolysis of ethyl ethanoate with sodium hydroxide yields sodium ethanoate (salt

Q

) and ethanol (alkanol

R

). Acidification of sodium ethanoate yields ethanoic acid (2 carbons). Complete oxidation of ethanol with acidified potassium dichromate(VI) also yields ethanoic acid (2 carbons). Because both pathways yield the exact same acid (ethanoic acid), ethyl ethanoate satisfies all conditions.

Step-by-Step Solution

1

Determine the functional group of compound P.

Compound P (

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

) reacts with

\text{NaOH}

to yield a salt and an alkanol, identifying

P

as an ester with general formula

\text{R}^1\text{COOR}^2

.

Alkaline hydrolysis (saponification) of esters produces a carboxylate salt and an alkanol.

2

Analyze the carbon distribution from the reaction products.

Acidifying salt

Q

(

\text{R}^1\text{COONa}

) gives alkanoic acid

\text{R}^1\text{COOH}

(containing

n_1 + 1

carbon atoms). Oxidation of primary alkanol

R

(

\text{R}^2\text{OH}

) gives alkanoic acid

\text{R}'\text{COOH}

(containing

n_2

carbon atoms).

Primary alkanols undergo complete oxidation with acidified

\text{K}_2\text{Cr}_2\text{O}_7

to produce alkanoic acids with the same number of carbon atoms as the alkanol.

3

Equate the carbon counts of the two acids formed.

Since both processes yield the identical acid, the acid must have 2 carbon atoms (ethanoic acid). Thus,

n_1 + 1 = 2 \implies n_1 = 1

(methyl group

\text{CH}_3-

) and

n_2 = 2

(ethyl group

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

).

The total number of carbon atoms in ester

P

is 4 (

1 + 1 + 2 = 4

). Dividing 4 total carbons equally between the acyl and alkoxy portions yields ethanoic acid derivative and ethanol derivative.

4

Deduce the structure and IUPAC name of ester P.

Ester

P

is

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

, which has the IUPAC name ethyl ethanoate.

The IUPAC name of an ester consists of the alkyl group attached to the oxygen followed by the alkanoate chain.

Key Concept

Ester Saponification and Oxidation of Alkanols

Question 85Question →

An organic compound $X$ with the molecular formula $C_3H_9N$ reacts with cold nitrous acid ( $HNO_2$ ) at $0-5^\circ\text{C}$ to produce a yellow, oily liquid without the evolution of nitrogen gas. Which of the following is the IUPAC name of compound $X$ ?

Propan-1-amine

N

-methylethanamine

Propan-2-amine

N,N

-dimethylmethanamine

Show answer & explanation

Answer: $N$ -methylethanamine

Answer

The correct compound is

N

-methylethanamine because secondary aliphatic amines react with nitrous acid (

HNO_2

) at low temperatures to produce insoluble, yellow oily

N

-nitrosamines without liberating nitrogen gas.

Secondary aliphatic amines such as

N

-methylethanamine react with cold nitrous acid (

HNO_2

) to form

N

-nitrosamines. These compounds are insoluble in water and appear as yellow oily liquids. Because no aliphatic diazonium intermediate breaks down to release gas, no nitrogen gas effervescence is observed.

Step-by-Step Solution

1

Classify the given structural isomers of

C_3H_9N

by amine degree.

Propan-1-amine and propan-2-amine are primary (

1^\circ

) amines;

N

-methylethanamine is a secondary (

2^\circ

) amine;

N,N

-dimethylmethanamine is a tertiary (

3^\circ

) amine.

Amine classification determines the distinct reaction pathway and observable products with nitrous acid.

2

Analyze the reaction behavior of each amine class with nitrous acid (

HNO_2

) at

0-5^\circ\text{C}

.

Primary amines evolve

N_2

gas and form alcohols; secondary amines form yellow oily

N

-nitrosamines without gas evolution; tertiary amines form soluble nitrite salts.

Nitrous acid is used as a qualitative reagent to distinguish between

1^\circ

,

2^\circ

, and

3^\circ

amines.

3

Match the observation (yellow oily liquid, no nitrogen gas) to the correct compound.

The observation corresponds uniquely to a secondary amine, which is

N

-methylethanamine.

Only secondary amines undergo nitrosation at the nitrogen atom to form neutral, oily

N

-nitrosamine layers.

Key Concept

Distinction tests for primary, secondary, and tertiary amines using nitrous acid (

HNO_2

)

Question 86Question →

When propanal ( $CH_3CH_2CHO$ ) is warmed with Fehling's solution, a brick-red precipitate is formed. What is the chemical formula of this precipitate, and what organic product is formed from the reaction?

Cu_2O

and propanoic acid

CuO

and propan-1-ol

Cu_2O

and propanone

Cu(OH)_2

and propanoic acid

Show answer & explanation

Answer: $Cu_2O$ and propanoic acid

Answer

The precipitate formed is copper(I) oxide (

Cu_2O

) and the organic oxidation product is propanoic acid.

Fehling's solution contains alkaline solution of copper(II) sulfate complexed with tartrate ions. When heated with an alkanal like propanal, the aldehyde group is oxidized to a carboxylic acid (propanoic acid), while

Cu^{2+}

is reduced to insoluble brick-red copper(I) oxide (

Cu_2O

).

Step-by-Step Solution

1

Identify the functional group and reaction type

Propanal is an alkanal containing the terminal carbonyl group (

-CHO

). Warming with Fehling's solution causes a redox distinction test.

Alkanals act as reducing agents and readily undergo oxidation, whereas Fehling's solution contains complexed

Cu^{2+}

ions.

2

Determine the inorganic reduction product (precipitate)

The deep blue

Cu^{2+}

ions are reduced to copper(I) oxide (

Cu_2O

), which precipitates as a insoluble brick-red solid.

The reduction half-reaction is

2Cu^{2+} + 2OH^- + 2e^- → Cu_2O + H_2O

.

3

Determine the organic oxidation product

Propanal (

CH_3CH_2CHO

) is oxidized to propanoic acid (

CH_3CH_2COOH

).

Oxidation of an alkanal inserts an oxygen atom into the

C-H

bond of the aldehyde group without altering the carbon backbone length.

Key Concept

Fehling's distinction test for alkanals and their oxidation products

Question 87Question →

Propanoic acid is heated under reflux with ethanol in the presence of concentrated tetraoxosulfate(VI) acid to produce a sweet-smelling liquid ester. This ester is separated and subsequently boiled under reflux with aqueous potassium hydroxide until saponification is complete. Which pair of organic products is isolated from the saponification mixture?

Potassium propanoate and ethanol

Potassium ethanoate and propan-1-ol

Propanoic acid and ethanol

Ethanoic acid and propan-1-ol

Show answer & explanation

Answer: Potassium propanoate and ethanol

Answer

The isolated products of the alkaline hydrolysis are potassium propanoate and ethanol.

In the initial esterification reaction, propanoic acid and ethanol react in the presence of concentrated tetraoxosulfate(VI) acid catalyst to form ethyl propanoate (

\text{C}_2\text{H}_5\text{COOCH}_2\text{CH}_3

). When ethyl propanoate is subsequently hydrolyzed under basic conditions with aqueous potassium hydroxide (saponification), the carbonyl-oxygen ester linkage is irreversibly cleaved. This forms the potassium salt of the carboxylic acid (potassium propanoate) and regenerates the alcohol (ethanol).

Step-by-Step Solution

1

Identify the ester formed in the esterification step

Propanoic acid (

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

) + Ethanol (

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

)

\xrightarrow{\text{conc. H}_2\text{SO}_4}

Ethyl propanoate (

\text{C}_2\text{H}_5\text{COOCH}_2\text{CH}_3

) + Water (

\text{H}_2\text{O}

)

Esterification combines the alkanoic acid acyl group (

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

) with the alkoxy group (

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

) of the alkanol.

2

Determine the saponification reaction of ethyl propanoate with potassium hydroxide

Ethyl propanoate (

\text{C}_2\text{H}_5\text{COOCH}_2\text{CH}_3

) +

\text{KOH(aq)} \rightarrow

Potassium propanoate (

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

) + Ethanol (

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

)

Alkaline hydrolysis cleaves the ester bond to produce the potassium salt of the alkanoic acid and the free alkanol.

Key Concept

Alkaline Hydrolysis (Saponification) of Esters

Estimated Time:2m 0s

Question 88Question →

Match each chemical reaction or process involving alkanoic acid derivatives on the left with its corresponding principal product on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The reaction of ethanoic acid and ethanol (esterification) pairs with Ethyl ethanoate and water. The alkaline hydrolysis of oil (saponification) pairs with Sodium salt of fatty acid (soap) and glycerol. The addition of hydrogen to liquid oil (catalytic hydrogenation) pairs with Solid saturated fat (margarine).

Esterification of ethanoic acid and ethanol yields ethyl ethanoate and water. Saponification of vegetable oils with aqueous sodium hydroxide yields fatty acid sodium salts (soap) and glycerol. Catalytic hydrogenation of unsaturated vegetable oils yields solid saturated fats (margarine).

Step-by-Step Solution

1

Identify the products of esterification

Ethanoic acid reacts with ethanol in the presence of concentrated tetraoxosulfate(VI) acid to produce ethyl ethanoate and water.

The hydroxyl group of the acid combines with hydrogen from the alkanol to form water, linking the remaining fragments into an ester.

2

Identify the products of saponification

Triglycerides react with boiling aqueous sodium hydroxide to yield sodium alkanoates (soap) and propane-1,2,3-triol (glycerol).

Alkaline cleavage of ester linkages in fats yields carboxylate salts and frees the triol backbone.

3

Identify the products of catalytic hydrogenation

Unsaturated fatty acid chains in liquid vegetable oils undergo addition of hydrogen to form saturated chains, hardening the oil into solid fat.

Reducing double bonds increases the melting point, transforming liquid oils into margarine.

Key Concept

Reactions and Industrial Products of Alkanoic Acids, Esters, Fats, and Oils

Question 89Question →

Match each organic reaction involving an amine or amide on the left with its corresponding principal product on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Reduction of propanamide with

LiAlH_4

pairs with propylamine; Reaction of propanamide with

Br_2/KOH

pairs with ethylamine; Alkaline hydrolysis of propanamide pairs with sodium propanoate and ammonia; Acylation of methylamine with ethanoyl chloride pairs with

N

-methylethanamide.

Each reaction pair is determined by its specific mechanistic pathway:

LiAlH_4

reduces the carbonyl group to methylene (retaining carbon count to form propylamine);

Br_2/KOH

undergoes Hofmann degradation to lose the carbonyl carbon (forming ethylamine); basic hydrolysis cleaves the

C-N

bond (yielding sodium propanoate and ammonia); and acylation of methylamine with ethanoyl chloride produces the substituted amide (

N

-methylethanamide).

Step-by-Step Solution

1

Analyze the reduction reaction of primary amides.

Reducing

CH_3CH_2CONH_2

with

LiAlH_4

reduces the

C=O

bond to a

-CH_2-

group without altering the total carbon count, yielding

CH_3CH_2CH_2NH_2

(propylamine).

Amide reduction retains the full carbon skeleton.

2

Identify the reaction of primary amides with

Br_2

and

KOH

.

This is Hofmann degradation, which removes the carbonyl carbon (

C=O

) as carbonate, reducing the carbon length by 1. Propanamide (3 carbons) yields ethylamine (2 carbons).

Hofmann degradation shortens the carbon chain by one atom.

3

Examine the basic hydrolysis of amides.

Nucleophilic attack of

OH^-

on the carbonyl carbon of propanamide cleaves the amide bond to generate propanoate anion (forming sodium propanoate with

Na^+

) and ammonia gas.

Base hydrolysis of amides yields a carboxylate salt and ammonia.

4

Examine the nucleophilic substitution between methylamine and ethanoyl chloride.

The nitrogen lone pair of methylamine attacks ethanoyl chloride, releasing

HCl

to form a secondary amide,

N

-methylethanamide (

CH_3CONHCH_3

).

Primary amines undergo acylation to form secondary amides.

Key Concept

Chemical Reactions and Interconversions of Amines and Amides

Estimated Time:1m 30s

Question 90Question →

Which of the following statements correctly distinguishes the esterification reaction between ethanoic acid and ethanol from the neutralization reaction between ethanoic acid and aqueous sodium hydroxide?

Esterification is a slow, reversible covalent reaction catalyzed by an acid, whereas neutralization is a rapid, irreversible ionic reaction.

Esterification is a rapid, irreversible ionic reaction, whereas neutralization is a slow reversible reaction that requires refluxing.

Esterification yields a alkanoic acid salt and water, whereas neutralization yields an ester and glycerol.

Concentrated tetraoxosulfate(VI) acid acts as an oxidizing agent in esterification, while sodium hydroxide acts as a dehydrating catalyst in neutralization.

Show answer & explanation

Answer: Esterification is a slow, reversible covalent reaction catalyzed by an acid, whereas neutralization is a rapid, irreversible ionic reaction.

Answer

Esterification is a slow, reversible covalent reaction catalyzed by an acid, whereas neutralization is a rapid, irreversible ionic reaction.

The statement identifying esterification as a slow, reversible covalent reaction catalyzed by an acid and neutralization as a rapid, irreversible ionic reaction is correct because organic esterification involves covalent bond rearrangement reaching dynamic equilibrium, whereas neutralization is an ionic combination forming water completely.

Step-by-Step Solution

1

Analyze the nature of esterification

Esterification takes place between an alkanoic acid and an alkanol in the presence of concentrated

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

. It is a reversible, molecular (covalent) reaction that proceeds slowly to reach equilibrium.

Covalent bond cleavage and formation require activation energy and proceed reversibly.

2

Analyze the nature of neutralization

Neutralization takes place between ethanoic acid and a strong base (

\text{NaOH}

), forming sodium ethanoate salt and water completely.

The reaction involves free ions in solution (

\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}

), making it rapid and virtually quantitative (irreversible).

3

Compare the key characteristics

Esterification is slow, reversible, and acid-catalyzed; neutralization is rapid, irreversible, and ionic.

This highlights the fundamental difference between organic ester formation and acid-base salt formation.

Key Concept

Reversibility and Kinetics of Esterification vs Neutralization

Estimated Time:1m 0s

Question 91Question →

Match each synthetic polymer or biomolecule listed in Column A with its corresponding chemical linkage and structural classification in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Nylon-6,6 pairs with polyamide linkage formed from hexanedioic acid and hexane-1,6-diamine; Terylene pairs with polyester linkage formed from benzene-1,4-dicarboxylic acid and ethane-1,2-diol; Starch pairs with glycosidic linkage formed from

\alpha

-D-glucose monomers; Protein pairs with peptide linkage formed from

\alpha

-amino acid monomers.

Each polymer is correctly matched to its functional linkage and monomer constituents: Nylon-6,6 is a polyamide formed from hexanedioic acid and hexane-1,6-diamine; Terylene is a polyester formed from benzene-1,4-dicarboxylic acid and ethane-1,2-diol; Starch is a polysaccharide held together by glycosidic linkages between glucose units; Protein is a natural polymer made of amino acids linked by peptide bonds.

Step-by-Step Solution

1

Identify the monomer composition and functional groups of Nylon-6,6

Nylon-6,6 contains amide linkages formed between carboxylic acid (

-\text{COOH}

) groups of hexanedioic acid and amino (

-\text{NH}_2

) groups of hexane-1,6-diamine.

Synthetic polyamides require a di-acid and a di-amine reactant.

2

Identify the monomer composition and functional groups of Terylene

Terylene contains ester linkages (

-\text{COO}-

) formed between benzene-1,4-dicarboxylic acid and ethane-1,2-diol.

Polyesters are produced by reacting a dicarboxylic acid with a dihydric alcohol (diol).

3

Determine the structural linkages present in Starch

Starch is a polysaccharide composed of

\alpha

-D-glucose monomers linked via condensation through glycosidic bonds.

Carbohydrates form ether-like condensation links known as glycosidic linkages.

4

Determine the structural linkages present in Proteins

Proteins are natural polymers made of

\alpha

-amino acids joined together by peptide linkages (

-\text{CO}-\text{NH}-

).

The reaction between the carboxyl group of one amino acid and the amino group of another forms a peptide bond.

Key Concept

Structural linkages and monomeric constituents of synthetic condensation polymers and natural biomolecules

Estimated Time:1m 30s

Question 92Question →

Which of the following carbonyl compounds yields a secondary alcohol upon reduction with lithium tetrahydridoaluminate(III) ( $LiAlH_4$ )?

Propanone

Ethanal

Butanal

Methanal

Show answer & explanation

Answer: Propanone

Answer

Propanone yields a secondary alcohol (propan-2-ol) when reduced by lithium tetrahydridoaluminate(III).

Propanone is a ketone (alkanone). When reduced with lithium tetrahydridoaluminate(III) (

LiAlH_4

), the carbonyl carbon (

C=O

) is converted to a secondary alcohol group (

-CH(OH)-

). Specifically, propanone (

CH_3COCH_3

) is reduced to propan-2-ol (

CH_3CH(OH)CH_3

), which contains a carbon atom bonded to two other carbon atoms and the hydroxyl group.

Step-by-Step Solution

1

Identify the functional groups of the given carbonyl compounds.

Propanone (

CH_3COCH_3

) is a ketone (alkanone), whereas ethanal (

CH_3CHO

), butanal (

CH_3CH_2CH_2CHO

), and methanal (

HCHO

) are aldehydes (alkanals).

Alkanals and alkanones exhibit distinct reduction behaviors based on the position of the carbonyl group (

C=O

).

2

Apply the general reduction reaction rules for alkanals and alkanones using

LiAlH_4

.

Reduction of an alkanal (

R-CHO

) produces a primary alcohol (

R-CH_2OH

). Reduction of an alkanone (

R-CO-R'

) produces a secondary alcohol (

R-CH(OH)-R'

).

The hydride ion (

H^-

) adds to the carbonyl carbon atom, converting the ketone group into a secondary hydroxyl group.

3

Determine which compound forms a secondary alcohol.

Propanone (

CH_3COCH_3

) is reduced to propan-2-ol (

CH_3CH(OH)CH_3

), which is a secondary alcohol.

Propan-2-ol has the hydroxyl-bearing carbon attached to two other carbon atoms, fitting the definition of a secondary alcohol.

Key Concept

Reduction of Alkanals and Alkanones

Question 93Question →

Match each chemical reaction involving a carbonyl compound in Column A with its corresponding chemical product or visual observation in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Warming ethanal with Fehling's solution matches the formation of a brick-red precipitate of copper(I) oxide. Treating propanone with aqueous iodine and sodium hydroxide matches the formation of a pale yellow precipitate of triiodomethane. Reducing butan-2-one with lithium tetrahydridoaluminate(III) matches the production of a secondary alcohol, butan-2-ol. Oxidizing propanal with acidified potassium dichromate(VI) matches the production of propanoic acid with an orange to green color change.

Each carbonyl compound reacts according to its specific structural features: alkanals (ethanal, propanal) are easily oxidized by mild and strong oxidizing agents like Fehling's solution and acidified dichromate, respectively. Methyl ketones (propanone) uniquely yield yellow iodoform upon treatment with alkaline iodine solution. Ketones (butan-2-one) reduce under hydride transfer (

LiAlH_4

) to form secondary alcohols.

Step-by-Step Solution

1

Analyze the distinction test for ethanal (alkanal) using Fehling's solution

Alkanals reduce Fehling's solution containing copper(II) tartrate complex to insoluble red copper(I) oxide (

Cu_2O

).

Alkanals are easily oxidized to alkanoic acids due to the presence of the carbonyl hydrogen atom.

2

Analyze the triiodomethane (iodoform) reaction of propanone

Propanone contains the methyl carbonyl structure (

CH_3-CO-

), which reacts with

I_2/OH^-

to precipitate yellow

CHI_3

.

The iodoform test specifically identifies compounds containing a methyl group attached directly to a carbonyl carbon.

3

Determine the reduction product of the alkanone (butan-2-one)

Reduction of a ketone yields a secondary alcohol, turning

C=O

into

CH-OH

. Thus, butan-2-one gives butan-2-ol.

The carbonyl group of an alkanone has two alkyl substituents, forming a secondary alcohol carbon upon addition of hydrogen.

4

Determine the oxidation product of propanal

Oxidation of propanal adds oxygen across the C-H bond to yield propanoic acid, while reducing

Cr_2O_7^{2-}

(orange) to

Cr^{3+}

(green).

Acidified

K_2Cr_2O_7

acts as a strong oxidizing agent towards alkanals.

Key Concept

Chemical tests and redox behavior of alkanals vs. alkanones

Question 94Question →

A vegetable oil containing glyceryl tristearate is boiled with aqueous sodium hydroxide during soap manufacturing. Which of the following sets of products is formed from this saponification reaction?

Propane-1,2,3-triol and sodium stearate

Stearic acid and sodium hydroxide

Propane-1,2,3-triol and octadecan-1-ol

Octadecane and sodium trioxocarbonate(IV)

Show answer & explanation

Answer: Propane-1,2,3-triol and sodium stearate

Answer

Propane-1,2,3-triol and sodium stearate

Saponification is the alkaline hydrolysis of fats and oils (triesters of glycerol). When glyceryl tristearate is heated with aqueous sodium hydroxide, the ester bonds are irreversibly broken to yield propane-1,2,3-triol (glycerol) and sodium stearate, which is a soap.

Step-by-Step Solution

1

Identify the functional group and reactants

Glyceryl tristearate is a triacylglycerol (fat/ester) reacting with a strong alkali (

\text{NaOH}

).

Understanding the nature of the reactants helps determine the type of chemical process taking place.

2

Determine the reaction mechanism (saponification)

Alkaline hydrolysis cleaves the three ester ester bonds in the triglyceride.

Base-catalyzed hydrolysis of esters is irreversible and yields the alcohol component and the carboxylate salt.

3

Identify the resulting products

The alcohol component formed is propane-1,2,3-triol (glycerol) and the salt formed is sodium stearate (soap).

The glycerol backbone is released as propane-1,2,3-triol while the long-chain fatty acid chains form sodium carboxylate salts.

Key Concept

Saponification of Fats and Oils

Question 95Question →

What is the total number of sigma ( $\sigma$ ) bonds in a single molecule of prop-2-enal (acrolein, $\text{CH}_2=\text{CH}-\text{CHO}$ )?

Show answer & explanation

Answer: 7; seven; 7 sigma bonds; 7 sigma

Answer

7

Expanding prop-2-enal (

\text{CH}_2=\text{CH}-\text{CHO}

) reveals four

\text{C}-\text{H}

single bonds, one

\text{C}-\text{C}

single bond, one

\text{C}=\text{C}

double bond (composed of one

\sigma

and one

\pi

bond), and one

\text{C}=\text{O}

double bond (composed of one

\sigma

and one

\pi

bond). Summing all head-on orbital overlaps yields a total of 7 sigma (

\sigma

) bonds.

Step-by-Step Solution

1

Draw the expanded structural formula of prop-2-enal.

The expanded formula showing all individual atoms and bonds is

\text{H}_2\text{C}=\text{CH}-\text{C}(=\text{O})\text{H}

.

Expanding the structural formula ensures that all implicit single bonds, double bonds, and hydrogen attachments are explicitly visible for counting.

2

Count all carbon-hydrogen (

\text{C}-\text{H}

) single sigma bonds.

There are 2

\text{C}-\text{H}

bonds on the terminal alkene carbon, 1

\text{C}-\text{H}

bond on the central alkene carbon, and 1

\text{C}-\text{H}

bond on the aldehyde carbon, giving a total of 4

\text{C}-\text{H}

\sigma

bonds.

Every single covalent bond formed with hydrogen involves head-on

sp^2-s

orbital overlap and constitutes one

\sigma

bond.

3

Count the sigma bonds among the carbon-carbon and carbon-oxygen links.

The

\text{C}=\text{C}

double bond contains 1

\sigma

bond, the

\text{C}-\text{C}

single bond contains 1

\sigma

bond, and the

\text{C}=\text{O}

double bond contains 1

\sigma

bond, giving 3 heavy-atom

\sigma

bonds.

Every covalent bond—whether single, double, or triple—contains exactly one

\sigma

bond resulting from axial head-on orbital overlap.

4

Sum the total number of sigma bonds.

4\text{ (C-H \sigma\ bonds)} + 3\text{ (C-C and C-O \sigma\ bonds)} = 7\text{ \sigma\ bonds}

.

Adding all localized head-on orbital overlaps gives the total count of sigma bonds in the molecule.

Key Concept

Determination of sigma (

\sigma

) and pi (

\pi

) bond counts in organic structures

Question 96Question →

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

Click a left item, then click its matching right item

Show answer & explanation

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 97Question →

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

Click a left item, then click its matching right item

Show answer & explanation

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 98Question →

Match each carbon species or bond descriptor on the left with its corresponding hybridization state, geometric configuration, or orbital overlap mode on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The central carbon in methane matches

sp^3

hybridization with tetrahedral geometry (

109.5^\circ

); the

\pi

bond component matches sideways overlap of unhybridized

p

orbitals; the central carbon in carbon dioxide matches

sp

hybridization with linear geometry (

180^\circ

); and the carbon in ethene matches

sp^2

hybridization with trigonal planar geometry (

120^\circ

).

Each carbon atom's hybridization and spatial arrangement depend directly on its steric number (number of

\sigma

bonds). Methane features four

\sigma

bonds (

sp^3

,

109.5^\circ

tetrahedral). Ethene features three

\sigma

bonds per carbon (

sp^2

,

120^\circ

trigonal planar). Carbon dioxide features two

\sigma

bonds (

sp

,

180^\circ

linear).

\Pi

bonds are characterized by the sideways overlap of unhybridized

2p

atomic orbitals.

Step-by-Step Solution

1

Determine the steric number and geometry of the carbon in methane (

\text{CH}_4

).

Four single

\sigma

bonds give a steric number of 4, which dictates

sp^3

hybridization and a tetrahedral angle of

109.5^\circ

.

Mixing one

s

and three

p

orbitals forms four equivalent

sp^3

hybrid orbitals pointing to tetrahedral corners.

2

Identify how a carbon-carbon

\pi

bond is formed.

It forms via lateral/sideways overlap of parallel, unhybridized

p

orbitals above and below the internuclear axis.

Head-on overlap forms

\sigma

bonds, whereas parallel side-by-side overlap creates

\pi

electron clouds.

3

Analyze the steric environment around the carbon in carbon dioxide (

\text{CO}_2

).

The carbon forms two

\sigma

bonds (one to each oxygen atom) and two

\pi

bonds, yielding a steric number of 2, corresponding to

sp

hybridization and

180^\circ

linear geometry.

Two hybrid orbitals position themselves as far apart as possible at

180^\circ

.

4

Determine the hybridization and bond angles of carbon in ethene (

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

).

Each carbon atom forms three

\sigma

bonds (steric number 3), requiring

sp^2

hybridization with a trigonal planar shape and

120^\circ

bond angles.

Three hybrid orbitals lie in a single plane separated by

120^\circ

.

Key Concept

Carbon Hybridization, Geometry, and Orbital Overlap

Question 99Question →

Arrange the following hydrocarbons 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 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 100Question →

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

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