Match each chemical reaction involving a carbonyl compound in Column A with its correspond… — JAMB UTME

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

Warming ethanal ( $CH_3CHO$ ) with Fehling's solutionFormation of a brick-red precipitate of copper(I) oxide ( $Cu_2O$ )
Treating propanone ( $CH_3COCH_3$ ) with aqueous iodine and sodium hydroxide solutionFormation of a pale yellow precipitate of triiodomethane ( $CHI_3$ )
Reducing butan-2-one ( $CH_3COCH_2CH_3$ ) with lithium tetrahydridoaluminate(III) ( $LiAlH_4$ )Production of a secondary alcohol, butan-2-ol ( $CH_3CH(OH)CH_2CH_3$ )
Oxidizing propanal ( $CH_3CH_2CHO$ ) with acidified potassium dichromate(VI) ( $K_2Cr_2O_7$ )Production of propanoic acid ( $CH_3CH_2COOH$ ) with an orange to green color change

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

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.

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.

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.

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

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