Alkanals and Alkanones: Oxidation, Reduction, and Distinction Tests Practice Questions — JAMB UTME

Question 1Question →

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 2Question →

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 3Question →

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 4Question →

What is the IUPAC name of the organic compound formed by the reduction of butan-2-one using lithium tetrahydridoaluminate(III) ( $LiAlH_4$ )?

Show answer & explanation

Answer: butan-2-ol; 2-butanol; butan 2 ol; 2-Butanol; Butan-2-ol

Answer

butan-2-ol

The reduction of alkanones (ketones) with reducing agents such as

LiAlH_4

converts the carbonyl group (

C=O

) into a secondary alcohol group (

-CH(OH)-

). Since the carbonyl carbon in butan-2-one is at position 2, the resulting hydroxyl group is also at position 2, yielding the secondary alcohol butan-2-ol.

Step-by-Step Solution

1

Identify the functional group and carbon chain length of the reactant.

Butan-2-one is a 4-carbon alkanone (ketone) with the carbonyl group at carbon-2 (

CH_3-CO-CH_2-CH_3

).

Determining the reactant structure establishes the expected reduction product.

2

Apply the reduction reaction mechanism for alkanones.

Reducing agents such as

LiAlH_4

or

NaBH_4

reduce alkanones to secondary alcohols by adding hydrogen across the

C=O

double bond.

The carbonyl group (

C=O

) is converted to a secondary alcohol group (

-CH(OH)-

).

3

Name the resulting alcohol using IUPAC nomenclature.

CH_3-CH(OH)-CH_2-CH_3

is named butan-2-ol.

The hydroxyl group (

-OH

) remains on carbon-2 of the 4-carbon parent chain.

Key Concept

Reduction of alkanones to secondary alcohols

Estimated Time:1m 0s

Question 5Question →

Compound $W$ is a four-carbon carbonyl compound that yields a negative result when tested with ammoniacal silver nitrate solution. Complete reduction of compound $W$ using lithium aluminium hydride ( $\text{LiAlH}_4$ ) produces an alcohol, compound $Z$ . What is the IUPAC name of compound $Z$ ?

Show answer & explanation

Answer: butan-2-ol; 2-butanol; Butan-2-ol; 2-Butanol

Answer

The IUPAC name of compound

Z

is butan-2-ol.

Compound

W

does not react with Tollen's reagent (ammoniacal silver nitrate), which confirms it is an alkanone rather than an alkanal. The only four-carbon alkanone is butan-2-one. Reducing butan-2-one with

\text{LiAlH}_4

yields the secondary alcohol butan-2-ol.

Step-by-Step Solution

1

Determine the functional group of compound

W

based on the distinction test.

Compound

W

is an alkanone (ketone).

Alkanals (aldehydes) reduce ammoniacal silver nitrate (Tollen's reagent) to silver metal (silver mirror), whereas alkanones (ketones) give a negative result.

2

Identify the specific structure and IUPAC name of compound

W

.

Compound

W

is butan-2-one (

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

).

Since compound

W

is a four-carbon alkanone, its only structural isomer is butan-2-one.

3

Determine the product of the reduction reaction.

Compound

Z

is a secondary alcohol, butan-2-ol (

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

).

Reduction of alkanones with

\text{LiAlH}_4

converts the carbonyl group (

\text{C=O}

) into a secondary alcohol group (

\text{CH-OH}

).

Key Concept

Reduction of alkanones to secondary alcohols and distinction between alkanals and alkanones using Tollen's reagent

Question 6Question →

An unknown organic compound $Y$ forms a brick-red precipitate when warmed with Fehling's solution and also gives a yellow precipitate of triiodomethane ( $CHI_3$ ) when treated with iodine in sodium hydroxide solution. Which of the following compounds is $Y$ ?

Ethanal

Propan-2-one

Propanal

Methanal

Show answer & explanation

Answer: Ethanal

Answer

Ethanal

Ethanal (

CH_3CHO

) is an alkanal, so it reduces copper(II) ions in Fehling's solution to copper(I) oxide (

Cu_2O

), producing a brick-red precipitate. Furthermore, because its carbonyl group is bonded directly to a methyl group (

CH_3C=O

), it undergoes halogenation and cleavage in alkaline iodine to produce a yellow precipitate of triiodomethane (

CHI_3

).

Step-by-Step Solution

1

Analyze the Fehling's solution reaction.

Fehling's test distinguishes alkanals (aldehydes) from alkanones (ketones). A positive test (brick-red

Cu_2O

precipitate) indicates

Y

must be an alkanal.

Alkanals are easily oxidized to alkanoic acids, whereas alkanones resist mild oxidation.

2

Analyze the triiodomethane (iodoform) reaction.

A positive triiodomethane test (yellow

CHI_3

precipitate) requires a methyl carbonyl group (

CH_3C=O

) or a secondary alcohol structure (

CH_3CH(OH)-

).

Iodine in aqueous alkali oxidizes and iodinates compounds containing the methyl carbonyl structural unit.

3

Combine the structural requirements.

The compound must be both an alkanal (

-CHO

) and contain a methyl carbonyl group (

CH_3C=O

). Ethanal (

CH_3CHO

) is the only alkanal that possesses a

CH_3C=O

group.

Other alkanals like methanal (

HCHO

) or propanal (

CH_3CH_2CHO

) lack the

CH_3C=O

structural unit.

Key Concept

Distinction tests for carbonyl compounds: Fehling's test and Iodoform test

Estimated Time:1m 0s

Question 7Question →

An organic compound $Z$ with the molecular formula $C_5H_{10}O$ gives a positive orange precipitate when treated with 2,4-dinitrophenylhydrazine. When warmed with acidified potassium heptaoxodichromate(VI) ( $K_2Cr_2O_7$ ), compound $Z$ is readily oxidized to a carboxylic acid containing five carbon atoms. Upon reduction with lithium tetrahydridoaluminate(III) ( $LiAlH_4$ ), compound $Z$ yields a primary alcohol that exhibits optical activity (chirality). What is the IUPAC name of compound $Z$ ?

2-methylbutanal

pentan-2-one

pentanal

3-methylbutanal

Show answer & explanation

Answer: 2-methylbutanal

Answer

2-methylbutanal

2-methylbutanal has the molecular formula

C_5H_{10}O

and contains an alkanal functional group. It reacts with 2,4-dinitrophenylhydrazine to form a precipitate. Because it is an alkanal, it undergoes oxidation with acidified potassium heptaoxodichromate(VI) to produce 2-methylbutanoic acid (a 5-carbon carboxylic acid). Upon reduction with lithium tetrahydridoaluminate(III), it forms 2-methylbutan-1-ol, which has an asymmetric carbon atom at position 2 bonded to four different substituent groups (

-H

,

-CH_3

,

-CH_2CH_3

,

-CH_2OH

), rendering the molecule chiral and optically active.

Step-by-Step Solution

1

Identify the functional group class from the 2,4-DNPH test.

A positive test with 2,4-dinitrophenylhydrazine confirms that compound

Z

is a carbonyl compound (either an alkanal or an alkanone).

Both alkanals and alkanones form colored hydrazone precipitates with 2,4-DNPH.

2

Distinguish between an alkanal and an alkanone using oxidation behavior.

Compound

Z

readily oxidizes with acidified

K_2Cr_2O_7

to a carboxylic acid with 5 carbons, proving

Z

is an alkanal (pentanal isomer).

Alkanals are easily oxidized to carboxylic acids with the same number of carbon atoms, whereas alkanones resist oxidation under mild conditions.

3

Analyze the reduction product for chirality.

Reduction of an alkanal with

LiAlH_4

yields a primary alcohol. Among 5-carbon alkanals (pentanal, 2-methylbutanal, 3-methylbutanal, 2,2-dimethylpropanal), only 2-methylbutanal reduces to 2-methylbutan-1-ol,

CH_3CH_2CH(CH_3)CH_2OH

.

In 2-methylbutan-1-ol, C-2 is bonded to four distinct groups:

-H

,

-CH_3

,

-CH_2CH_3

, and

-CH_2OH

, making it a chiral molecule capable of optical isomerism.

Key Concept

Distinction between alkanals and alkanones via oxidation, reduction of carbonyls to alcohols, and optical isomerism in branched primary alcohols.

Question 8Question →

An organic compound $Z$ with the molecular formula $C_5H_{10}O$ gives a negative result with Tollen's reagent and does not form a yellow precipitate when warmed with iodine in sodium hydroxide solution. Upon reduction with lithium tetrahydridoaluminate(III) ( $LiAlH_4$ ), compound $Z$ yields a secondary alkanol. Which of the following is the correct IUPAC name of compound $Z$ ?

pentan-3-one

pentan-2-one

pentanal

3-methylbutan-2-one

Show answer & explanation

Answer: pentan-3-one

Answer

pentan-3-one

Pentan-3-one is an alkanone with the structure

CH_3CH_2COCH_2CH_3

. Because it is a ketone, it cannot be oxidized by mild oxidizing agents like Tollen's reagent. Furthermore, because its carbonyl carbon is bonded to two ethyl groups rather than a methyl group, it gives a negative triiodomethane (iodoform) test. Reduction of pentan-3-one using

LiAlH_4

yields pentan-3-ol, which is a secondary alkanol.

Step-by-Step Solution

1

Analyze the functional group class using Tollen's reagent test.

Compound

Z

gives a negative Tollen's test, confirming it is an alkanone (ketone) rather than an alkanal (aldehyde).

Alkanals are easily oxidized to alkanoic acids and reduce Tollen's reagent to metallic silver, whereas alkanones resist mild oxidation.

2

Evaluate the iodoform (triiodomethane) test requirement.

The absence of a yellow precipitate (

CHI_3

) rules out compounds containing a methyl ketone (

CH_3CO-

) group.

Only methyl ketones (

R-COCH_3

) or ethanal (

CH_3CHO

) undergo triiodomethane formation with iodine in sodium hydroxide.

3

Examine the reduction reaction and identify the specific isomer.

Among the

C_5H_{10}O

ketone isomers (pentan-2-one, pentan-3-one, and 3-methylbutan-2-one), only pentan-3-one lacks a methyl ketone group and reduces with

LiAlH_4

to give pentan-3-ol (a secondary alkanol).

Pentan-3-one has the structural formula

CH_3CH_2COCH_2CH_3

, fulfilling both qualitative test observations and structural reduction requirements.

Key Concept

Distinction tests for carbonyl compounds (Tollen's test and iodoform test) and reduction outcomes of alkanones

Estimated Time:2m 0s

Question 9Question →

Match each chemical reaction or test involving carbonyl compounds in Column A with its corresponding characteristic visual outcome in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Warming ethanal with Fehling's solution yields a brick-red precipitate; warming propanone with Tollen's reagent yields no visible reaction; warming propanal with acidified

KMnO_4

causes decolorization of the purple solution; and warming propanone with iodine in sodium hydroxide solution produces a pale yellow precipitate.

Alkanals (ethanal and propanal) are reducing agents due to the carbonyl hydrogen atom; thus, ethanal reduces Fehling's solution to a brick-red copper(I) oxide precipitate, and propanal reduces purple acidified

KMnO_4

to a colorless

Mn^{2+}

solution. Alkanones (propanone) lack this hydrogen atom, so propanone shows no reaction with Tollen's reagent. However, because propanone has a

CH_3CO-

group, it responds to the triiodomethane test by forming a pale yellow precipitate.

Step-by-Step Solution

1

Analyze the oxidation reactions of alkanals.

Ethanal reduces Fehling's solution to form brick-red

Cu_2O

, while propanal reduces acidified

KMnO_4

, turning the purple solution colorless.

Alkanals possess a hydrogen atom bonded to the carbonyl carbon, enabling easy oxidation by mild and strong oxidizing agents.

2

Analyze the oxidation behavior of alkanones.

Propanone yields no visible reaction with Tollen's reagent.

Alkanones lack a hydrogen atom on the carbonyl carbon and are resistant to oxidation by mild oxidizing agents.

3

Identify the triiodomethane (iodoform) test reaction.

Propanone forms a pale yellow precipitate of triiodomethane (

CHI_3

).

Propanone contains the

CH_3C=O

group required for a positive triiodomethane test.

Key Concept

Distinction tests and oxidation properties of alkanals and alkanones

Estimated Time:1m 0s

Question 10Question →

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 11Question →

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 12Question →

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

Alkanals and Alkanones: Oxidation, Reduction, and Distinction Tests

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