Chapter 8: Aldehydes, Ketones and Carboxylic Acids Application Questions | chemistry Class 12 CBSE

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Chapter 8: Aldehydes, Ketones and Carboxylic Acids Application Questions | chemistry Class 12 CBSE | ByteNotes.in

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A student needs to distinguish between propanal and propanone using simple chemical tests. Suggest two tests with observations.

Tollens' test: add ammoniacal silver nitrate to each compound and warm; propanal (an aldehyde) produces a silver mirror on the inner walls of the test tube; propanone (a ketone) gives no reaction.
Fehling's test: heat each compound with Fehling's solution (A + B mixed); propanal produces a brick-red/reddish-brown precipitate of Cu₂O; propanone gives no precipitate as ketones are not oxidised by Fehling's reagent.
Iodoform test (confirmatory for propanone): treat with I₂/NaOH solution; propanone (a methyl ketone) gives a yellow precipitate of iodoform (CHI₃); propanal does not give this test.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

An organic compound X gives a silver mirror with Tollens' reagent, does not give a red precipitate with Fehling's reagent, and on vigorous oxidation gives benzoic acid. Identify X and explain.

Since X gives a positive Tollens' test, it is an aldehyde; since it does not reduce Fehling's reagent, it must be an aromatic aldehyde (Fehling's test is negative for aromatic aldehydes).
Since vigorous oxidation of X gives benzoic acid, X must be benzaldehyde (C₆H₅CHO), where the –CHO group is oxidised to –COOH on the benzene ring.
Benzaldehyde is an aromatic aldehyde that reduces Tollens' reagent (Ag⁺ → Ag mirror) but not Fehling's reagent, and its oxidation product is benzoic acid, confirming the identification.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 3

Applied Concept

With reference to the esterification reaction, explain why the percentage yield of ethyl ethanoate is improved by using excess ethanol or by removing water as it forms.

Esterification is a reversible reaction: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O, and the system reaches equilibrium with a mixture of all four species.
By Le Chatelier's principle, using excess of one reactant (cheaper ethanol) shifts equilibrium to the right, increasing ester yield; alternatively, removing water (by molecular sieves, distillation, or a dehydrating agent) also shifts equilibrium to the right.
Removing the ester as it forms (by distillation of the lower-boiling ester) similarly drives the reaction forward; the mineral acid catalyst (H₂SO₄) only speeds equilibrium attainment without shifting position.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 4

Applied Concept

A chemist wants to prepare benzaldehyde from toluene in the laboratory without using the chlorination route. Suggest two named reactions that can be used, with conditions.

Etard reaction: treat toluene with chromyl chloride (CrO₂Cl₂) in carbon disulphide (CS₂) solvent; the methyl group is oxidised to a chromium complex intermediate, which upon hydrolysis (H₃O⁺) gives benzaldehyde.
CrO₃ method: treat toluene with chromic oxide (CrO₃) in acetic anhydride at 273–283 K; toluene is converted to benzylidene diacetate, which on acid hydrolysis gives benzaldehyde.
Both methods stop the oxidation at the aldehyde stage; strong oxidising agents (like KMnO₄) would over-oxidise the methyl group all the way to benzoic acid.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

In a food chemistry lab, a scientist needs to identify whether a natural extract contains a methyl ketone (like acetophenone). Describe the test with expected observations.

Iodoform test: add the extract to a solution of iodine (I₂) in sodium hydroxide (NaOH) and warm; a yellow precipitate of iodoform (CHI₃) with its characteristic antiseptic smell confirms the presence of a CH₃CO– group (methyl ketone) or a CH₃CH(OH)– group.
The chemistry involves three successive α-iodinations of the methyl group to form a triiodomethyl ketone intermediate, followed by base-catalysed cleavage to give CHI₃ and the sodium salt of a carboxylic acid with one fewer carbon.
Acetophenone (C₆H₅COCH₃) would give a positive iodoform test since it has a CH₃CO– group; the yellow precipitate is confirmatory, and a 2,4-DNP test can additionally confirm the carbonyl group.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 6

Applied Concept

With reference to the structure of the carboxyl group, explain why the carboxylic carbon is less electrophilic than a simple aldehyde carbonyl carbon.

In carboxylic acids, the carbon of the carboxyl group is attached to an –OH group whose oxygen lone pairs participate in resonance with the C=O bond, donating electron density to the carbonyl carbon.
This resonance creates a partial double bond character in the C–OH bond and reduces the partial positive charge on the carboxyl carbon, making it less electrophilic than the carbonyl carbon of an aldehyde.
As a result, carboxylic acids do not undergo typical nucleophilic addition reactions with HCN, NaHSO₃, or alcohols (which aldehydes readily undergo); instead, they react via substitution at the carbonyl carbon.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 7

Applied Concept

A student carries out an Aldol condensation between two molecules of ethanal. Write the structural formula of the initial aldol product and the final condensation product.

The initial aldol product (from base-catalysed self-condensation of two molecules of ethanal/CH₃CHO) is 3-hydroxybutanal: CH₃–CH(OH)–CH₂–CHO, where one ethanal acts as nucleophile (via enolate) and the other as electrophile.
On heating, 3-hydroxybutanal undergoes dehydration (loss of water from the β-OH and α-H) to give the aldol condensation product: but-2-enal (CH₃–CH=CH–CHO), an α,β-unsaturated aldehyde.
The aldol reaction illustrates carbon–carbon bond formation using α-acidic hydrogens; the word 'aldol' is derived from 'aldehyde' and 'alcohol', reflecting the two functional groups in the initial product.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 8

Applied Concept

Arrange the following carboxylic acids in increasing order of acid strength and justify: CH₃COOH, CF₃COOH, ClCH₂COOH, BrCH₂COOH.

The acidity increases with the strength of the electron-withdrawing group at the α-carbon; stronger inductive withdrawal stabilises the carboxylate anion more, increasing acid strength (decreasing pKa).
Methyl group is electron-donating (destabilises carboxylate), making acetic acid the weakest; Br and Cl are moderately electron-withdrawing halogens, with Cl > Br in electronegativity, so ClCH₂COOH > BrCH₂COOH.
Three fluorine atoms in CF₃COOH provide the strongest inductive withdrawal (F is the most electronegative element), making trifluoroacetic acid by far the strongest. Order: CH₃COOH < BrCH₂COOH < ClCH₂COOH < CF₃COOH.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 9

Applied Concept

An organic compound C₈H₈O forms a 2,4-DNP derivative, does not reduce Tollens' or Fehling's reagent, gives a positive iodoform test, and on oxidation gives benzoic acid. Identify it.

Formation of a 2,4-DNP derivative confirms a carbonyl compound (aldehyde or ketone); failure to reduce Tollens' or Fehling's reagent confirms it is a ketone (not an aldehyde).
Positive iodoform test confirms the presence of a CH₃CO– group (methyl ketone); oxidation giving benzoic acid (C₇H₆O₂, one carboxyl on benzene ring) means one substituent on the ring was a methyl ketone.
Molecular formula C₈H₈O with one ring plus CH₃CO– accounts for the formula; the compound is acetophenone (phenyl methyl ketone, C₆H₅COCH₃), whose oxidation by H₂CrO₄ gives benzoic acid and a one-carbon fragment.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 10

Applied Concept

With reference to the bisulphite addition reaction, explain how it can be used to separate benzaldehyde from a mixture with acetophenone.

Sodium hydrogensulphite (NaHSO₃) reacts with aldehydes and some methyl ketones to form crystalline, water-soluble bisulphite addition compounds; the equilibrium lies to the right for aldehydes but largely to the left for bulky ketones.
Benzaldehyde reacts completely with NaHSO₃ to form its crystalline bisulphite adduct, which is water-soluble and can be filtered off from the non-reacting acetophenone (steric hindrance of the phenyl ring prevents complete bisulphite addition in acetophenone under normal conditions).
The pure benzaldehyde is then regenerated by treating the adduct with dilute mineral acid or dilute alkali, which reverses the addition reaction and liberates the free aldehyde.

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Chapter 8: Aldehydes, Ketones and Carboxylic Acids | Application Questions

A student needs to distinguish between propanal and propanone using simple chemical tests. Suggest two tests with observations.

An organic compound X gives a silver mirror with Tollens' reagent, does not give a red precipitate with Fehling's reagent, and on vigorous oxidation gives benzoic acid. Identify X and explain.

With reference to the esterification reaction, explain why the percentage yield of ethyl ethanoate is improved by using excess ethanol or by removing water as it forms.

A chemist wants to prepare benzaldehyde from toluene in the laboratory without using the chlorination route. Suggest two named reactions that can be used, with conditions.

In a food chemistry lab, a scientist needs to identify whether a natural extract contains a methyl ketone (like acetophenone). Describe the test with expected observations.

With reference to the structure of the carboxyl group, explain why the carboxylic carbon is less electrophilic than a simple aldehyde carbonyl carbon.

A student carries out an Aldol condensation between two molecules of ethanal. Write the structural formula of the initial aldol product and the final condensation product.

Arrange the following carboxylic acids in increasing order of acid strength and justify: CH₃COOH, CF₃COOH, ClCH₂COOH, BrCH₂COOH.

An organic compound C₈H₈O forms a 2,4-DNP derivative, does not reduce Tollens' or Fehling's reagent, gives a positive iodoform test, and on oxidation gives benzoic acid. Identify it.

With reference to the bisulphite addition reaction, explain how it can be used to separate benzaldehyde from a mixture with acetophenone.

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