ClearStepsCLEARSTEPS AI
Teacher Login →Student Login →

Aldehydes, Ketones and Carboxylic Acids: Class 12 Chemistry Practice Questions

25 original exam-pattern questions with full answers, matched to the current CBSE Class 12 paper design, including case-based questions. Attempt each question before opening the answer — or start a free 14-day trial ↓ for the full bank.

Q1Case-based4 marks

A food science student is studying the chemistry of common carbonyl compounds used in flavouring and preservation industries. She has four unlabelled bottles containing: (I) Formaldehyde (HCHO), (II) Acetaldehyde (CH₃CHO), (III) Acetone (CH₃COCH₃), and (IV) Acetic acid (CH₃COOH).

A food science student is studying the chemistry of common carbonyl compounds used in flavouring and preservation industries. She has four unlabelled bottles containing: (I) Formaldehyde (HCHO), (II) Acetaldehyde (CH₃CHO), (III) Acetone (CH₃COCH₃), and (IV) Acetic acid (CH₃COOH).

Based on this information, answer the following questions:

(a) The student wants to identify which two bottles contain aldehydes. She uses Tollens' reagent. Which bottles (I–IV) will give a positive result? Write the observation. (1 mark)

(b) Among bottles I and II, she further applies the iodoform test. Which bottle gives a positive result, and what is the observation? Write the balanced chemical equation for the reaction of bottle II with I₂/NaOH. (2 marks)

(c) The student notices that acetic acid (bottle IV) has a much higher boiling point (118°C) than acetone (bottle III, b.p. 56°C), despite having similar molar masses. Give one reason for this difference. (1 mark)

Show answer
(a) Bottles I (Formaldehyde, HCHO) and II (Acetaldehyde, CH₃CHO) will give a positive result with Tollens' reagent.

Observation: A bright silver mirror is deposited on the inner walls of the test tube (silver mirror test).

[Tollens' reagent — ammoniacal silver nitrate — oxidises aldehydes to carboxylate ions; Ag⁺ is reduced to Ag(s). Ketones and carboxylic acids do NOT react.]

(1 mark)

(b) Bottle II (Acetaldehyde, CH₃CHO) gives a positive iodoform test because it contains the CH₃CO– group (methyl ketone / methyl aldehyde pattern).

Bottle I (Formaldehyde, HCHO) does NOT give iodoform test — it lacks the CH₃CO– group.

Observation: A pale yellow precipitate of iodoform (CHI₃) with a characteristic antiseptic smell is formed from bottle II.

Balanced chemical equation:

CH₃CHO + 3I₂ + 3NaOH → CHI₃↓ + HCOONa + 3NaI + 3H₂O

(yellow ppt of CHI₃)

(2 marks — 1 mark for correct identification with observation; 1 mark for balanced equation)

(c) Due to intermolecular hydrogen bonding / dimer formation in acetic acid.

Acetic acid molecules associate strongly through O–H···O hydrogen bonds, forming dimers, which require significantly more energy to break during vaporisation. Acetone, having no O–H bond, cannot form such hydrogen bonds, so it has a much lower boiling point.

(1 mark)
Q2Case-based4 marks

A food chemist is analysing three unlabelled bottles in a quality-control lab. Each bottle contains one of the following compounds: Compound A (pentan-2-one), Compound B (pentanal), and Compound C (benzoic acid). She has access to standard reagents: Tollens' reagent, 2,4-DNP solution, sodium bicarbonate solution (NaHCO₃), iodoform reagent (I₂/NaOH), and Fehling's solution.

A food chemist is analysing three unlabelled bottles in a quality-control lab. Each bottle contains one of the following compounds: Compound A (pentan-2-one), Compound B (pentanal), and Compound C (benzoic acid). She has access to standard reagents: Tollens' reagent, 2,4-DNP solution, sodium bicarbonate solution (NaHCO₃), iodoform reagent (I₂/NaOH), and Fehling's solution.

(a) She adds NaHCO₃ solution separately to small samples of each compound. One compound produces brisk effervescence while the other two do not. Identify the compound that gives effervescence, and give a reason for this behaviour. (2 marks)

(b) She then performs the iodoform test on the remaining two compounds. Which compound gives a positive iodoform test? Write the chemical equation for the reaction, showing the organic product. (1 mark)

(c) She finally uses Tollens' reagent to distinguish between Compound A and Compound B. State the observation for each compound, and name the type of reaction each compound undergoes with Tollens' reagent. (1 mark)

Show answer
(a) Compound C (benzoic acid) gives brisk effervescence with NaHCO₃ solution.

Reason: Carboxylic acids are sufficiently acidic to react with NaHCO₃, releasing CO₂ gas (brisk effervescence). The reaction proceeds as:

C₆H₅COOH + NaHCO₃ → C₆H₅COONa + H₂O + CO₂↑

Compound A (pentan-2-one) and Compound B (pentanal) are a ketone and an aldehyde respectively; they do not possess sufficiently acidic protons to react with NaHCO₃ and hence give no effervescence.

∴ Compound C is identified as benzoic acid. (1 mark for correct identification + 1 mark for correct reason with equation)

(b) Compound A (pentan-2-one) gives a positive iodoform test.

Reason: Pentan-2-one contains the CH₃CO– group (methyl ketone), which reacts with I₂/NaOH to give a yellow precipitate of iodoform (CHI₃↓). Pentanal does not contain the CH₃CO– or CH₃CH(OH)– group and hence does not give a positive iodoform test.

Chemical equation:

CH₃COC₃H₇ + 3I₂ + 3NaOH → CI₃COC₃H₇ + 3NaI + 3H₂O
CI₃COC₃H₇ + NaOH → CHI₃↓ + C₃H₇COONa

(yellow precipitate)

∴ Pentan-2-one (Compound A) gives the positive iodoform test. (1 mark)

(c) Compound B (pentanal) gives a silver mirror with Tollens' reagent (ammoniacal silver nitrate solution); the aldehyde is oxidised to the corresponding carboxylate ion (pentanoate). This is an oxidation reaction.

Compound A (pentan-2-one) gives no silver mirror (no reaction) with Tollens' reagent, as ketones are not oxidised by mild oxidising agents under these conditions.

Observations:
• Compound B (pentanal): silver mirror deposited on the inner wall of the test tube. (Oxidation by Tollens' reagent)
• Compound A (pentan-2-one): no change / no silver mirror formed.

∴ Compound B is distinguished from Compound A by the silver mirror test. (1 mark)
Q3Case-based4 marks

A food science student is analysing two unlabelled bottles in a school laboratory. Bottle P contains ethanal (CH₃CHO) and Bottle Q contains propanone (CH₃COCH₃). The student performs a series of chemical tests and reactions to identify the contents and study their properties.

A food science student is analysing two unlabelled bottles in a school laboratory. Bottle P contains ethanal (CH₃CHO) and Bottle Q contains propanone (CH₃COCH₃). The student performs a series of chemical tests and reactions to identify the contents and study their properties.

(a) The student adds Tollens' reagent to a sample from each bottle. State the observation for each bottle and identify which bottle contains ethanal. (1 mark)

(b) The student then treats a fresh sample from Bottle P with NaOH solution (dilute) at low temperature. Name the type of reaction that occurs and write the structural formula of the organic product formed. (1 mark)

(c) The student treats the contents of Bottle P with I₂ and NaOH solution. A pale yellow precipitate with a distinctive smell is formed. (i) Name this reaction and (ii) write the IUPAC name of the yellow precipitate. (1 mark)

(d) The student wishes to reduce the compound in Bottle Q to the corresponding secondary alcohol using a reagent that does NOT reduce C=C double bonds. Name the suitable reducing agent and write the balanced chemical equation for the reaction. (1 mark)

Show answer
(a) Tollens' reagent contains ammoniacal silver nitrate solution ([Ag(NH₃)₂]⁺). Aldehydes reduce [Ag(NH₃)₂]⁺ to metallic silver (Ag), while ketones do not.

Bottle P (ethanal): A shiny silver mirror is deposited on the inner wall of the test tube (positive silver mirror test).
Bottle Q (propanone): No silver mirror is formed; the solution remains clear.

∴ Bottle P contains ethanal.

(b) When ethanal (CH₃CHO) is treated with dilute NaOH at low temperature, Aldol Condensation (Aldol Reaction) occurs.

Ethanal possesses α-hydrogen atoms. The dilute NaOH acts as a base, generating a carbanion (enolate), which attacks the carbonyl carbon of another molecule of ethanal.

Product: 3-Hydroxybutanal (Aldol product)

Structural formula of the product:

CH₃–CH(OH)–CH₂–CHO

[Structure: CH₃CH(OH)CH₂CHO — 3-hydroxybutanal, with –OH on C3 and –CHO terminal]

(c) When ethanal (CH₃CHO) is treated with I₂ and NaOH, a pale yellow precipitate with a distinctive antiseptic smell is formed.

(i) This reaction is called the Iodoform Reaction (Iodoform Test).

(ii) The yellow precipitate is CHI₃.
IUPAC name of CHI₃: Triiodomethane

The balanced equation is:
CH₃CHO + 3I₂ + 3NaOH → CHI₃↓ + HCOONa + 3NaI + 3H₂O

(d) The suitable reducing agent that reduces C=O but does NOT reduce C=C double bonds is NaBH₄ (Sodium borohydride).

NaBH₄ is a mild and selective reducing agent — it reduces carbonyl compounds (aldehydes and ketones) but leaves alkene (C=C) bonds intact, unlike LiAlH₄.

Balanced chemical equation:

CH₃COCH₃ + NaBH₄ → CH₃CH(OH)CH₃

(Full equation with solvent/hydrolysis step):
4 CH₃COCH₃ + NaBH₄ → (CH₃CHOH CH₃)₄B(intermediate) → 4 CH₃CH(OH)CH₃

[Simplified accepted form:]
CH₃COCH₃ + [H] → CH₃CH(OH)CH₃
(NaBH₄, ethanol/H₂O)

∴ Propanone is reduced to propan-2-ol (secondary alcohol) using NaBH₄.
Q4Case-based4 marks

A food chemist in a quality-control laboratory is given four unlabelled bottles containing: Bottle P – Ethanal, Bottle Q – Propanone, Bottle R – Benzaldehyde, Bottle S – Ethanoic acid. She performs a series of simple chemical tests to identify each bottle.

A food chemist is working in a quality-control laboratory. She is given four unlabelled bottles containing the following liquids:

• Bottle P : Ethanal (acetaldehyde)
• Bottle Q : Propanone (acetone)
• Bottle R : Benzaldehyde
• Bottle S : Ethanoic acid (acetic acid)

Using simple chemical tests and reactions available in the laboratory, answer the following questions:

(a) The chemist adds Tollens' reagent to a sample from each bottle. Which bottles give a positive result? Write the chemical equation for the reaction with ONE of the positive bottles. (2 marks)

(b) She then performs the iodoform test on the remaining samples. Which bottle gives a yellow precipitate? Name the yellow precipitate formed. (1 mark)

(c) She adds aqueous NaHCO₃ solution to a sample from Bottle S. What observation is made? Write the equation for this reaction. (1 mark)

Show answer
(a) Tollens' reagent (ammoniacal silver nitrate, [Ag(NH₃)₂]⁺) oxidises aldehydes to carboxylates, producing a silver mirror on the walls of the test tube. Ketones and carboxylic acids do NOT react with Tollens' reagent.

∴ Bottles P (Ethanal) and R (Benzaldehyde) give a positive silver mirror test.

Equation for the reaction with Bottle P (Ethanal):

CH₃CHO + 2[Ag(NH₃)₂]⁺ + 2OH⁻ → CH₃COO⁻ + 2Ag↓ + 4NH₃ + H₂O

(Silver mirror deposited on the inner wall of the test tube.)

(b) The iodoform test is positive for compounds containing the CH₃CO– group (methyl ketones, acetaldehyde, and secondary alcohols of the type CH₃CH(OH)R). Among the remaining bottles (Q – Propanone and S – Ethanoic acid):

• Propanone (CH₃COCH₃) contains the CH₃CO– group → gives a positive iodoform test.
• Ethanoic acid does NOT give the iodoform test.

∴ Bottle Q (Propanone) gives a yellow precipitate.

Name of the yellow precipitate: Iodoform (triiodomethane, CHI₃↓).

(c) Ethanoic acid (Bottle S) is a carboxylic acid. Carboxylic acids react with NaHCO₃ solution to liberate CO₂ gas (brisk effervescence), which distinguishes them from phenols and alcohols.

Observation: Brisk effervescence (bubbling) due to evolution of CO₂ gas is observed.

Equation:

CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂↑
Q5Case-based4 marks

A food-science student is analysing three unknown carbonyl compounds — P, Q and R — isolated from a flavouring mixture. Compound P gives a silver mirror with Tollens' reagent but does NOT give a red precipitate with Fehling's solution. Compound Q gives a red precipitate with Fehling's solution and also gives a yellow precipitate with alkaline I₂ solution (iodoform test). Compound R does NOT react with Tollens' reagent, does NOT give iodoform test, but gives an orange-yellow precipitate with 2,4-dinitrophenylhydrazine (2,4-DNP) reagent.

A food-science student is analysing three unknown carbonyl compounds — P, Q and R — isolated from a flavouring mixture. The following observations are recorded:

• Compound P gives a silver mirror with Tollens' reagent but does NOT give a red precipitate with Fehling's solution.
• Compound Q gives a red precipitate with Fehling's solution and also gives a yellow precipitate with alkaline I₂ solution (iodoform test).
• Compound R does NOT react with Tollens' reagent, does NOT give iodoform test, but gives an orange-yellow precipitate with 2,4-dinitrophenylhydrazine (2,4-DNP) reagent.

(a) Identify the class of compound to which P belongs. Give one example with its IUPAC name. (2 marks)
(b) Identify compound Q and justify your answer using the two test results. (1 mark)
(c) What functional group is confirmed by the 2,4-DNP test for compound R? Why does R not give the iodoform test? (1 mark)

Show answer
Part (a) [2 marks]

P gives a positive Tollens' test — confirming the presence of an aldehyde group (–CHO) — but a NEGATIVE Fehling's test.

Fehling's solution (Cu²⁺ in alkaline tartrate) oxidises only aliphatic (non-aromatic) aldehydes. Aromatic aldehydes do NOT reduce Fehling's reagent because the benzene ring withdraws electron density from the –CHO group, making it a weaker reducing agent under those mild alkaline-Cu²⁺ conditions. However, all aldehydes (including aromatic ones) reduce the ammoniacal AgNO₃ of Tollens' reagent.

∴ Compound P belongs to the class of aromatic aldehydes.

Example: Benzaldehyde (IUPAC name: benzaldehyde / phenylmethanal)

Benzaldehyde + Tollens' reagent → silver mirror (positive)
Benzaldehyde + Fehling's solution → no red precipitate (negative)

[Value points: correct class identification — aromatic aldehyde (1 mark); correct example with IUPAC name (1 mark)]

─────────────────────────────────────────
Part (b) [1 mark]

The two diagnostic results are:
(i) Positive Fehling's test → Q is an aliphatic aldehyde (has a free –CHO group).
(ii) Positive iodoform test → Q contains the CH₃C=O– (or CH₃CH(OH)–) structural unit.

The only aliphatic aldehyde that contains the CH₃CHO unit and satisfies both tests is acetaldehyde (ethanal, CH₃CHO).

∴ Compound Q is acetaldehyde (ethanal, CH₃CHO).

Justification: The –CHO group reduces Cu²⁺ to Cu₂O (red ppt) in Fehling's test. The CH₃CO– group undergoes iodoform reaction with alkaline I₂ to give a yellow precipitate of CHI₃ (iodoform).

[Value point: correct identification as ethanal/acetaldehyde with both justifications (1 mark)]

─────────────────────────────────────────
Part (c) [1 mark]

The 2,4-DNP reagent (Brady's reagent) gives an orange-yellow precipitate (2,4-dinitrophenylhydrazone) with any compound containing a carbonyl group (C=O), i.e., aldehydes AND ketones. Since R does not give Tollens' test, it is NOT an aldehyde.

∴ The 2,4-DNP test confirms the presence of a ketone carbonyl group (C=O) in compound R.

R does NOT give the iodoform test because it lacks the CH₃CO– (methyl ketone) structural unit. The iodoform test is positive only when a CH₃CO– group (or CH₃CH(OH)– group) is present; a ketone without the methyl group adjacent to C=O (e.g., pentan-3-one, cyclohexanone) gives a negative iodoform test.

[Value points: carbonyl/ketone group confirmed by 2,4-DNP (½ mark); correct reason for negative iodoform test — absence of CH₃CO– unit (½ mark)]
Q6Case-based4 marks

A food chemistry student is analysing two unlabelled carbonyl compounds, P and Q, isolated from a fruit extract. Observation (i): Both P and Q give an orange-yellow precipitate with 2,4-DNP reagent. Observation (ii): Only P gives a silver mirror with Tollens' reagent; Q does not. Observation (iii): P gives a yellow precipitate with I₂/NaOH. Observation (iv): P treated with NaBH₄ then acidic workup gives product R; R decolourises acidic KMnO₄.

A food chemistry student is analysing two unlabelled carbonyl compounds, P and Q, isolated from a fruit extract. The following observations are recorded:

(i) Both P and Q give an orange-yellow precipitate with 2,4-dinitrophenylhydrazine (2,4-DNP) reagent.
(ii) Only P gives a silver mirror with Tollens' reagent; Q does not.
(iii) P gives a yellow precipitate with I₂/NaOH solution.
(iv) On treatment of P with NaBH₄ followed by acidic workup, a product R is obtained. R decolourises acidic KMnO₄.

(a) Identify compounds P and Q, giving one reason for each identification. (2 marks)
(b) Write the IUPAC name of compound R and state what structural feature of R causes it to decolourise acidic KMnO₄. (1 mark)
(c) The student finds that Q does NOT undergo aldol condensation with dilute NaOH. Identify ONE structural requirement that Q lacks, and name ONE compound that WOULD undergo aldol condensation under the same conditions. (1 mark)

Show answer
MARKING SCHEME

(a) Identification of P and Q — 2 marks (1 mark each)

Compound P:
P is CH₃CHO (ethanal / acetaldehyde).
• Observation (i): 2,4-DNP positive → carbonyl compound (aldehyde or ketone). ✓
• Observation (ii): Tollens' reagent positive (silver mirror) → P is an aldehyde. ✓
• Observation (iii): Iodoform test positive (yellow ppt of CHI₃↓) → P has the CH₃CO– group, i.e. P is CH₃CHO (ethanal).
(Award 1 mark for correct identification AND a valid reason based on the observations.)

Compound Q:
Q is a ketone with NO α-hydrogen (e.g. benzophenone, C₆H₅COC₆H₅, or any acceptable ketone lacking α-H).
• Observation (i): 2,4-DNP positive → carbonyl compound. ✓
• Observation (ii): Tollens' reagent negative → Q is NOT an aldehyde; Q is a ketone. ✓
• Observation (iv — inferred): Q does not undergo aldol (stated in part c) → Q has no α-hydrogen.
(Award 1 mark for correct identification as a ketone lacking α-H AND a valid reason. Benzophenone is the expected answer; any ketone with no α-H is also accepted.)

(b) IUPAC name of R and structural feature — 1 mark

NaBH₄ reduces the C=O of CH₃CHO to give the corresponding alcohol:

CH₃CHO →(NaBH₄, then H₃O⁺)→ CH₃CH₂OH

∴ R is ethanol (IUPAC name: ethanol).

Structural feature: R is a primary alcohol (–CH₂OH group). Primary (and secondary) alcohols are oxidised by acidic KMnO₄ because the C–H bond on the carbon bearing –OH can be broken by the strong oxidant, leading to decolourisation of the purple KMnO₄ solution.
(Award 1 mark for correct IUPAC name ethanol AND the identification of –OH / primary alcohol as the feature responsible.)

(c) Structural requirement and example — 1 mark

Q lacks α-hydrogen atoms (i.e. there is no C–H bond on the carbon adjacent to the carbonyl group).
Aldol condensation requires at least one α-hydrogen for enolate/enol formation in the first step.

A compound that WOULD undergo aldol condensation: ethanal (CH₃CHO) / propanal (CH₃CH₂CHO) / acetone (CH₃COCH₃) — any aldehyde or ketone possessing α-hydrogen is acceptable.
(Award 1 mark for stating 'absence of α-hydrogen' as the missing requirement AND naming any one suitable compound.)

VALUE-POINT SUMMARY
• P correctly identified as CH₃CHO (ethanal) with reason from observations — ½ + ½
• Q correctly identified as a ketone with no α-H, with reason — ½ + ½
• R named as ethanol AND primary alcohol / –OH feature stated — ½ + ½
• α-hydrogen stated as missing requirement AND one valid example given — ½ + ½
∴ Total = 4 marks
Q7Case-based4 marks

A food-science student is studying the chemistry of common kitchen ingredients. She observes the following:

• Vinegar contains ethanoic acid (acetic acid).
• Vanilla extract contains vanillin (4-hydroxy-3-methoxybenzaldehyde), an aldehyde.
• Glucose syrup, when tested in the laboratory, gives a brick-red precipitate with Fehling's solution.

Based on the above context, answer the following sub-parts.

A food-science student is studying the chemistry of common kitchen ingredients. She observes the following:

• Vinegar contains ethanoic acid (acetic acid).
• Vanilla extract contains vanillin (4-hydroxy-3-methoxybenzaldehyde), an aldehyde.
• Glucose syrup, when tested in the laboratory, gives a brick-red precipitate with Fehling's solution.

Based on the above context, answer the following:

(a) The student treats a sample of vanillin with Fehling's solution. What observation will she make? Give one reason for your answer.

(b) She then treats separate samples of ethanoic acid and vanillin each with sodium hydrogencarbonate (NaHCO₃) solution. What different observations will she make for the two compounds? Name the gas evolved, if any.

(c) Write the balanced chemical equation for the reaction of ethanoic acid with NaHCO₃.

(d) The student also tests ethanal (present in trace amounts in vinegar) with Tollens' reagent. Write the observation and identify the type of reaction taking place.

Show answer
(a) Vanillin will give a brick-red precipitate with Fehling's solution.

Reason: Vanillin contains an aldehyde (–CHO) functional group. Aldehydes are reducing agents that reduce Cu²⁺ ions (present in Fehling's solution as cupric tartrate complex) to Cu₂O, which appears as a brick-red precipitate.

[Award 1 mark: observation (½) + reason (½)]

(b) Ethanoic acid reacts with NaHCO₃ solution with brisk effervescence — CO₂ gas is evolved. Vanillin (an aldehyde) does NOT react with NaHCO₃ and no effervescence is observed.

Gas evolved: Carbon dioxide (CO₂)

Explanation: Carboxylic acids are sufficiently acidic (pKa ≈ 4.75) to liberate CO₂ from NaHCO₃. Aldehydes are not acidic enough to do so.

[Award 1 mark: correct different observations for both (½ + ½)]

(c) Balanced equation for ethanoic acid with NaHCO₃:

CH₃COOH(aq) + NaHCO₃(aq) → CH₃COONa(aq) + H₂O(l) + CO₂(g)↑

[Award 1 mark: balanced equation with correct products — CO₂↑ must be shown]

(d) Observation: Ethanal reduces Tollens' reagent (ammoniacal silver nitrate solution) — a shiny silver mirror is deposited on the inner walls of the test tube (silver mirror test).

Type of reaction: Oxidation (ethanal is oxidised to ethanoic acid / acetate ion; Ag⁺ is reduced to Ag metal).

Reaction:
CH₃CHO(aq) + 2[Ag(NH₃)₂]⁺(aq) + 2OH⁻(aq) → CH₃COO⁻(aq) + 2Ag(s)↓ + 4NH₃(aq) + H₂O(l)

[Award 1 mark: observation (½) + type of reaction (½)]
Q8Case-based4 marks

A food-science student is testing four carbonyl-containing samples in the laboratory. She records the following observations:

• Sample P gives a silver mirror with Tollens' reagent and also gives a yellow precipitate with alkaline I₂ solution.
• Sample Q gives a silver mirror with Tollens' reagent but does NOT give a yellow precipitate with alkaline I₂ solution.
• Sample R does NOT give a silver mirror with Tollens' reagent but DOES give a yellow precipitate with alkaline I₂ solution.
• Sample S does NOT give a silver mirror with Tollens' reagent and does NOT give a yellow precipitate with alkaline I₂ solution.

A food-science student is testing four carbonyl-containing samples in the laboratory. She records the following observations:

• Sample P gives a silver mirror with Tollens' reagent and also gives a yellow precipitate with alkaline I₂ solution.
• Sample Q gives a silver mirror with Tollens' reagent but does NOT give a yellow precipitate with alkaline I₂ solution.
• Sample R does NOT give a silver mirror with Tollens' reagent but DOES give a yellow precipitate with alkaline I₂ solution.
• Sample S does NOT give a silver mirror with Tollens' reagent and does NOT give a yellow precipitate with alkaline I₂ solution.

Based on the observations, answer the following:
(a) Identify the functional group(s) present in Sample P. Name one compound that could be Sample P.
(b) Why does Sample Q not respond to the iodoform test even though it is an aldehyde? Name one compound that could be Sample Q.
(c) Write the balanced chemical equation for the reaction of Sample R with alkaline I₂ solution.
(d) The student finds that Sample S reacts with NaHCO₃ solution to give brisk effervescence. Identify the class of compound Sample S belongs to and state why it does not respond to either test.

Show answer
MARKING SCHEME (4 marks)

(a) [1 mark]

Sample P responds to BOTH Tollens' reagent (→ silver mirror, confirming an aldehyde / –CHO group) AND the iodoform test (→ yellow CHI₃↓, confirming a CH₃CO– or CH₃CH(OH)– group).

∴ Sample P contains an aldehyde group (–CHO) AND the CH₃CO– / CH₃CH(OH)– structural unit.

One compound that could be Sample P: Acetaldehyde (Ethanal, CH₃CHO) — it possesses the –CHO group (→ Tollens' positive) and the CH₃C=O unit (→ iodoform positive).

(b) [1 mark]

The iodoform test is positive only for compounds having the CH₃CO– group (methyl ketones or acetaldehyde) or the CH₃CH(OH)– group. Sample Q is an aldehyde that lacks a CH₃ group directly attached to the carbonyl carbon; therefore it does NOT possess the CH₃CO– unit required for iodoform reaction, and the test is negative.

One compound that could be Sample Q: Benzaldehyde (C₆H₅CHO) — it is an aldehyde (→ Tollens' positive) but has no CH₃CO– group (→ iodoform negative).

(c) [1 mark]

Sample R responds to the iodoform test but NOT to Tollens' reagent → it is a methyl ketone (CH₃COR, where R ≠ H). A representative example is Acetone (Propanone, CH₃COCH₃).

Balanced equation for the iodoform reaction of acetone with alkaline I₂:

CH₃COCH₃ + 3I₂ + 3NaOH → CHI₃↓ + CH₃COONa + 3NaI + 3H₂O

(Yellow precipitate of iodoform, CHI₃↓, is formed.)

(d) [1 mark]

Sample S gives brisk effervescence with NaHCO₃ solution (→ CO₂ gas evolved), which is the characteristic test for a carboxylic acid (–COOH group). Carboxylic acids are sufficiently acidic to liberate CO₂ from NaHCO₃.

∴ Sample S belongs to the class of Carboxylic Acids.

It does NOT respond to Tollens' test: carboxylic acids do not contain a free –CHO group, so they cannot reduce Ag⁺ to Ag metal.
It does NOT respond to the iodoform test: carboxylic acids (e.g., acetic acid, CH₃COOH) do not undergo the iodoform reaction because the carbonyl carbon bears an –OH group (not directly part of the CH₃CO– unit that undergoes iodination under alkaline conditions in the iodoform sequence).

Note: If a student names CH₃COOH (acetic acid) as Sample P instead of Sample S they should be re-read against the NaHCO₃ clue which locks Sample S as a carboxylic acid. Accept any correct carboxylic acid as the identity of Sample S.
Q9MCQ1 mark

Which of the following compounds gives a positive iodoform test?

Show answer
(B) Acetophenone

Explanation: The iodoform test is positive for compounds containing the CH₃CO– group (methyl ketones) or the CH₃CH(OH)– group. Acetophenone (C₆H₅COCH₃) possesses the CH₃CO– group and therefore reacts with I₂ in the presence of NaOH to give a yellow precipitate of iodoform (CHI₃↓). Benzaldehyde lacks the CH₃CO– group, benzophenone has no methyl group attached to the carbonyl carbon, and propanal (CH₃CH₂CHO) also lacks the CH₃CO– grouping — none of these three gives the iodoform test.
Q10MCQ1 mark

Which of the following compounds gives a positive iodoform test?
(A) Benzaldehyde
(B) Acetophenone
(C) Formaldehyde
(D) Diethyl ketone

Show answer
(B) Acetophenone

Explanation: The iodoform test is positive for compounds containing the CH₃CO– group (methyl ketones) or CH₃CH(OH)– group. Acetophenone (C₆H₅COCH₃) contains the CH₃CO– group and therefore gives a yellow precipitate of CHI₃↓ (iodoform) with I₂/NaOH. Benzaldehyde lacks this group; formaldehyde (HCHO) also lacks a methyl group adjacent to C=O; diethyl ketone (C₂H₅COC₂H₅) has no methyl group directly attached to the carbonyl carbon.
Q11Short Answer1 mark

Assertion (A): Formaldehyde (HCHO) does not give iodoform test.
Reason (R): Iodoform test is given by compounds containing the CH₃CO– group or compounds that are oxidised to give the CH₃CO– group.

Show answer
Option (A): Both A and R are true, and R is the correct explanation of A.

Explanation: The iodoform test is positive only for compounds containing the CH₃CO– group (methyl ketones, acetaldehyde) or secondary alcohols of the type CH₃CH(OH)R that can be oxidised to give this group. HCHO lacks the CH₃CO– unit entirely — it has only a H–C=O group — so it does not give the iodoform test. R correctly and directly explains why A is true.
Q12Short Answer2 marks

Arrange the following carbonyl compounds in increasing order of their reactivity towards nucleophilic addition reactions:

CH₃CHO, HCHO, C₆H₅COCH₃

Show answer
Increasing order of reactivity towards nucleophilic addition:

C₆H₅COCH₃ < CH₃CHO < HCHO

Reason: Reactivity towards nucleophilic addition decreases with increasing steric hindrance around the carbonyl carbon and increasing electron density on it (due to electron-donating groups). HCHO has no alkyl/aryl substituents, making the carbonyl carbon most electrophilic and least hindered. CH₃CHO has one electron-donating methyl group, reducing electrophilicity. C₆H₅COCH₃ has both a bulky phenyl group (which also delocalises the lone pair into the ring, reducing the positive character of the carbonyl carbon) and a methyl group, making it least reactive.
Q13Short Answer2 marks

Arrange the following carbonyl compounds in increasing order of their reactivity towards nucleophilic addition reaction. Give reason.

Propanone, Ethanal, 2-Methylpropanal, Methanal

Show answer
Increasing order of reactivity towards nucleophilic addition:

Propanone < 2-Methylpropanal < Ethanal < Methanal

Reason: Reactivity towards nucleophilic addition depends on two factors — steric hindrance and electronic effects. Greater the steric hindrance (bulkier the substituents around the carbonyl carbon), lower is the reactivity. Also, electron-donating alkyl groups decrease the electrophilicity of the carbonyl carbon, further reducing reactivity.

Methanal (HCHO) has no alkyl group, hence least steric hindrance and maximum positive character on carbonyl carbon — most reactive.
Ethanal (CH₃CHO) has one methyl group — less reactive than methanal.
2-Methylpropanal [(CH₃)₂CHCHO] has a branched alkyl group — more steric hindrance than ethanal.
Propanone (CH₃COCH₃) has two methyl groups on either side of the carbonyl carbon — maximum steric hindrance and greatest electron donation — least reactive.

∴ Increasing order: Propanone < 2-Methylpropanal < Ethanal < Methanal
Q14Short Answer2 marks

Arrange the following carbonyl compounds in increasing order of their reactivity towards nucleophilic addition reactions. Give a reason for the order:
Formaldehyde (HCHO), Acetaldehyde (CH₃CHO), Acetone (CH₃COCH₃), Benzaldehyde (C₆H₅CHO)

Show answer
Increasing order of reactivity towards nucleophilic addition:

CH₃COCH₃ < C₆H₅CHO < CH₃CHO < HCHO

(1 mark for correct order)

Reason: Reactivity towards nucleophilic addition depends on two factors — steric effect and electronic effect. A greater positive charge density on the carbonyl carbon and lesser steric hindrance increases reactivity.

• HCHO is most reactive — no alkyl/aryl substituent, so maximum positive charge on carbonyl carbon and minimum steric hindrance.
• CH₃CHO is next — one electron-donating methyl group reduces positive charge slightly and provides some steric hindrance.
• C₆H₅CHO (benzaldehyde) is less reactive than CH₃CHO — the phenyl ring delocalises the lone pair into the ring through resonance, reducing the electrophilicity of the carbonyl carbon, and also provides greater steric hindrance.
• CH₃COCH₃ is least reactive — two electron-donating methyl groups reduce positive charge on carbonyl carbon the most and provide greater steric crowding around the carbon.

(1 mark for correct reason)

∴ Increasing order: CH₃COCH₃ < C₆H₅CHO < CH₃CHO < HCHO
Q15Short Answer3 marks

Answer the following:
(i) Give ONE reason why formic acid (HCOOH) is a stronger acid than acetic acid (CH₃COOH).
(ii) Arrange the following carbonyl compounds in decreasing order of their reactivity towards nucleophilic addition of HCN, giving reason:
HCHO, CH₃COCH₃, CH₃CHO
(iii) Write the reaction involved in Rosenmund reduction.

Show answer
(i) Acetic acid has an electron-donating methyl (−CH₃) group attached to the carboxyl group. Due to the +I (inductive) effect of the −CH₃ group, the electron density on the O−H bond increases, making it harder to release H⁺. In formic acid, the carboxyl group is attached only to hydrogen, which has no electron-donating effect. Thus, HCOOH releases H⁺ more readily and is a stronger acid than CH₃COOH.

(Award 1 mark for: +I effect of −CH₃ / no electron-donating substituent in HCOOH → HCOOH is stronger acid)

(ii) Decreasing order of reactivity towards nucleophilic addition of HCN:

HCHO > CH₃CHO > CH₃COCH₃

Reason: Reactivity towards nucleophilic addition depends on (a) the electrophilicity of the carbonyl carbon (electronic factor) and (b) the steric accessibility of the carbonyl carbon (steric factor).

• HCHO has no alkyl group — no steric hindrance and no +I effect, so the carbonyl carbon is most electrophilic and most accessible → most reactive.
• CH₃CHO has one electron-donating methyl group — partial decrease in electrophilicity and slight steric hindrance → less reactive than HCHO.
• CH₃COCH₃ has two methyl groups — greater +I effect reduces electrophilicity further and greater steric hindrance around the carbonyl carbon → least reactive.

(Award 1 mark for correct order with valid reason — steric/electronic)

(iii) Rosenmund Reduction:

An acyl chloride (acid chloride) is reduced to the corresponding aldehyde by hydrogen gas in the presence of palladium supported on barium sulphate (Pd–BaSO₄) as catalyst. The catalyst is partially poisoned (with quinoline–S or BaSO₄) to prevent over-reduction to alcohol.

Reaction:

R−COCl + H₂ →(Pd−BaSO₄ / quinoline−S)→ R−CHO + HCl

Example:

CH₃COCl + H₂ →(Pd−BaSO₄)→ CH₃CHO + HCl

(Award 1 mark for: correct reactant acid chloride + H₂, correct conditions Pd–BaSO₄ over arrow, correct product aldehyde + HCl — equation must be balanced)
Q16Short Answer3 marks

Write the reaction involved in the following and name it:

CH₃CHO + CH₃CHO → CH₃CH(OH)CH₂CHO (dil. NaOH catalyst)

Show answer
CH₃CHO + CH₃CHO →(dil. NaOH) CH₃CH(OH)CH₂CHO

This reaction is known as Aldol Condensation (Aldol Addition).
Q17Short Answer3 marks

An organic compound (A) with molecular formula C₃H₆O gives a positive 2,4-DNP test but does not reduce Tollens' reagent. Compound (A) on reaction with I₂/NaOH gives a yellow precipitate (B). On reduction of (A) with NaBH₄, compound (C) is obtained.

(i) Identify compounds (A), (B) and (C).
(ii) Write the balanced chemical equation for the reaction that gives compound (B).
(iii) State why compound (A) does not reduce Tollens' reagent.

Show answer
IDENTIFICATION OF COMPOUNDS — (1 mark)

Compound (A) gives a positive 2,4-DNP test → contains a C=O group (aldehyde or ketone).
Compound (A) does NOT reduce Tollens' reagent → it is a KETONE, not an aldehyde.
Molecular formula C₃H₆O with a ketone group → (A) is propan-2-one (acetone), CH₃COCH₃.

Compound (B) = yellow precipitate formed with I₂/NaOH → CHI₃ (iodoform).

Compound (C) = product of NaBH₄ reduction of a ketone → secondary alcohol = propan-2-ol, CH₃CH(OH)CH₃.

∴ (A) = CH₃COCH₃ (propan-2-one / acetone)
(B) = CHI₃ (iodoform, yellow precipitate)
(C) = CH₃CH(OH)CH₃ (propan-2-ol)

──────────────────────────────────────────
BALANCED EQUATION FOR IODOFORM REACTION — (1 mark)

CH₃COCH₃ + 3I₂ + 3NaOH → CH₃COONa + CHI₃↓ + 3NaI + 2H₂O

(Yellow precipitate CHI₃↓ confirms positive iodoform test.)

──────────────────────────────────────────
REASON — (1 mark)

Tollens' reagent (ammoniacal silver nitrate) oxidises only aldehydes — compounds that have a hydrogen atom directly attached to the carbonyl carbon (–CHO group). Compound (A), propan-2-one, is a ketone; it has no such α-hydrogen on the carbonyl carbon and cannot be further oxidised under mild conditions. Hence it does not reduce Tollens' reagent and gives no silver mirror.
Q18Short Answer3 marks

Answer the following questions:
(a) Arrange the following compounds in increasing order of their reactivity towards nucleophilic addition reaction:
(i) HCHO (ii) CH₃CHO (iii) C₆H₅COCH₃
(b) Why is the boiling point of carboxylic acids much higher than that of alcohols of comparable molecular mass?
(c) Complete the following reaction:
CH₃CH₂CHO + HCN →

Show answer
(a) Reactivity towards nucleophilic addition depends on two factors: (i) the magnitude of the partial positive charge on the carbonyl carbon (electronic factor) and (ii) the steric hindrance around the carbonyl carbon.

HCHO has no alkyl group attached — maximum partial positive charge on carbonyl carbon and no steric hindrance.
CH₃CHO has one electron-donating methyl group — partial positive charge decreases slightly and steric hindrance is moderate.
C₆H₅COCH₃ (methyl phenyl ketone / acetophenone) has a bulky phenyl group and a methyl group — both reduce the partial positive charge (conjugation with ring further delocalises the charge) and create the maximum steric hindrance.

∴ Increasing order of reactivity:
C₆H₅COCH₃ < CH₃CHO < HCHO

(b) Due to extensive intermolecular hydrogen bonding, carboxylic acids exist as dimers even in the vapour phase. Each molecule forms two O–H···O hydrogen bonds with its neighbour, making the effective molecular mass nearly double. In alcohols, hydrogen bonding exists but no dimer formation occurs in the vapour phase. Therefore, a much larger amount of energy is required to break the association in carboxylic acids, resulting in a much higher boiling point than that of alcohols of comparable molecular mass.

(c) CH₃CH₂CHO + HCN →

Product: CH₃CH₂CH(OH)CN
(2-hydroxybutanenitrile / propionaldehyde cyanohydrin)

The reaction proceeds by nucleophilic addition of CN⁻ to the carbonyl carbon, followed by protonation:

CH₃CH₂CHO + HCN → CH₃CH₂–CH(OH)–CN

(This is a cyanohydrin; the product has a new chiral centre and is obtained as a racemic mixture.)
Q19Short Answer3 marks

Write the products formed in each of the following reactions:

(i) CH₃CHO + NH₂OH →

(ii) C₆H₅CHO + KOH (conc.) →

(iii) CH₃COCH₃ + LiAlH₄ → (followed by H₃O⁺)

Show answer
(i) Reaction of ethanal with hydroxylamine (Oxime formation):

CH₃CHO + NH₂OH → CH₃CH=N−OH + H₂O

The product is acetaldoxime (ethanaloxime). The reaction is a nucleophilic addition–elimination (condensation) reaction where the lone pair on nitrogen attacks the carbonyl carbon, followed by loss of water.

[1 mark — correct product CH₃CH=N−OH with H₂O]

(ii) Reaction of benzaldehyde with concentrated KOH (Cannizzaro Reaction):

Benzaldehyde has no α-hydrogen. In the presence of concentrated KOH, it undergoes Cannizzaro reaction — a disproportionation in which one molecule is oxidised to the carboxylate salt and another is reduced to the alcohol.

2C₆H₅CHO + KOH (conc.) → C₆H₅COOK + C₆H₅CH₂OH

Products: Potassium benzoate (C₆H₅COOK) and benzyl alcohol (C₆H₅CH₂OH).

[1 mark — both products correctly identified with balanced equation]

(iii) Reduction of propanone with LiAlH₄ followed by H₃O⁺ work-up:

LiAlH₄ reduces ketones to secondary alcohols. The hydride ion (H⁻) acts as a nucleophile and attacks the carbonyl carbon.

CH₃COCH₃ →(i) LiAlH₄ / dry ether (ii) H₃O⁺→ CH₃CH(OH)CH₃

Product: Propan-2-ol (isopropanol), a secondary alcohol.

[1 mark — correct product propan-2-ol / CH₃CH(OH)CH₃]

∴ Summary of products:
(i) CH₃CH=N−OH (acetaldoxime)
(ii) C₆H₅COOK + C₆H₅CH₂OH (potassium benzoate + benzyl alcohol)
(iii) CH₃CH(OH)CH₃ (propan-2-ol)
Q20Short Answer3 marks

An organic compound (A) with molecular formula C₈H₈O forms an orange-yellow precipitate with 2,4-dinitrophenylhydrazine, does NOT reduce Fehling's solution, and gives a positive iodoform test. On vigorous oxidation with KMnO₄/H⁺, it gives benzoic acid.
(i) Identify compound (A) and write its IUPAC name.
(ii) Write the chemical equation for the reaction of (A) with 2,4-dinitrophenylhydrazine.
(iii) Write the chemical equation for the reaction of (A) with I₂/NaOH (iodoform test) and name the yellow precipitate formed.

Show answer
(i) Identification of compound (A): [1 mark]

Clues analysed:
• Forms 2,4-DNP derivative → contains C=O group (aldehyde or ketone).
• Does NOT reduce Fehling's solution → NOT an aldehyde; compound is a KETONE.
• Positive iodoform test → contains the CH₃CO– group (methyl ketone).
• Vigorous oxidation gives benzoic acid → contains a phenyl (C₆H₅–) group attached to carbon.
• Molecular formula C₈H₈O → C₆H₅–CO–CH₃ (MW = 120 g mol⁻¹ ✓).

∴ Compound (A) is Acetophenone (Methyl phenyl ketone).
IUPAC name: 1-Phenylethan-1-one

(ii) Reaction of (A) with 2,4-Dinitrophenylhydrazine (2,4-DNP): [1 mark]

C₆H₅COCH₃ + (NO₂)₂C₆H₃NHNH₂ → C₆H₅C(=NNH–C₆H₃(NO₂)₂)CH₃ + H₂O

[Acetophenone reacts with 2,4-dinitrophenylhydrazine to form an orange-yellow 2,4-dinitrophenylhydrazone precipitate and water. This confirms the presence of the carbonyl (C=O) group.]

(iii) Reaction of (A) with I₂/NaOH (Iodoform test): [1 mark]

C₆H₅COCH₃ + 3I₂ + 3NaOH → C₆H₅COONa + CHI₃↓ + 3NaI + 2H₂O

The yellow precipitate formed is Iodoform (CHI₃).

[Acetophenone gives a positive iodoform test because it contains the CH₃CO– group. The methyl group is successively iodinated and then the CI₃ group is cleaved by NaOH to give CHI₃ (iodoform), which appears as a yellow precipitate with a characteristic antiseptic smell.]
Q21Case-based4 marks

A food-testing laboratory receives three unlabelled bottles, each containing one of the following compounds: ethanal (CH₃CHO), propan-2-one (CH₃COCH₃), and propanoic acid (CH₃CH₂COOH). Observations from four chemical tests are tabulated above.

A food-testing laboratory receives three unlabelled bottles, each containing one of the following compounds: ethanal (CH₃CHO), propan-2-one (CH₃COCH₃), and propanoic acid (CH₃CH₂COOH). A trainee chemist performs a series of chemical tests to identify each bottle. Study the observations recorded in the table below and answer the questions that follow.

| Test | Bottle P | Bottle Q | Bottle R |
|---|---|---|---|
| Litmus test | No change | No change | Turns blue litmus red |
| Tollens' reagent (warm) | Silver mirror formed | No silver mirror | No silver mirror |
| Iodoform test (I₂/NaOH, warm) | Yellow ppt. formed | Yellow ppt. formed | No yellow ppt. |
| 2,4-DNP reagent | Orange-yellow ppt. | Orange-yellow ppt. | No ppt. |

(a) Identify compounds P, Q, and R. Give one reason to justify the identification of P over Q. (2 marks)
(b) Write the balanced chemical equation for the reaction of compound Q with I₂ in the presence of NaOH, showing the organic product(s) formed. (1 mark)
(c) The trainee suggests that compound R could also be distinguished from P and Q by reacting with Na₂CO₃ solution. Is this suggestion correct? Justify your answer by writing the relevant chemical equation. (1 mark)

Show answer
(a) Identification of P, Q, and R:

Bottle R turns blue litmus red → it is a carboxylic acid → R is propanoic acid (CH₃CH₂COOH).

Both P and Q give a positive iodoform test and a positive 2,4-DNP test, so both contain the CH₃CO– group. However, only P gives a silver mirror with Tollens' reagent → P is an aldehyde → P is ethanal (CH₃CHO).

Q gives no silver mirror → Q is a ketone → Q is propan-2-one (CH₃COCH₃).

∴ P = ethanal (CH₃CHO), Q = propan-2-one (CH₃COCH₃), R = propanoic acid (CH₃CH₂COOH).

Justification for P over Q:
Due to the presence of an aldehydic hydrogen (–CHO group), ethanal (P) is oxidised by Tollens' reagent (ammoniacal AgNO₃) to form a silver mirror, whereas propan-2-one (Q), being a ketone, lacks the –CHO group and does not reduce Tollens' reagent. This distinguishes P from Q.

(b) Reaction of compound Q (propan-2-one) with I₂/NaOH (iodoform reaction):

CH₃COCH₃ + 3I₂ + 3NaOH → CH₃COO⁻Na⁺ + CHI₃↓ + 3NaI + 2H₂O

(Balanced equation: iodoform CHI₃ yellow precipitate is the organic product; sodium propanoate/sodium acetate is formed as the other organic product.)

∴ The organic products are iodoform (CHI₃), a yellow precipitate, and sodium ethanoate (CH₃COONa).

(c) The suggestion is correct.

Propanoic acid (R) reacts with Na₂CO₃ solution to produce brisk effervescence of CO₂ gas, which is not given by ethanal (P) or propan-2-one (Q).

Relevant equation:

2CH₃CH₂COOH + Na₂CO₃ → 2CH₃CH₂COONa + H₂O + CO₂↑

Neither ethanal nor propan-2-one reacts with Na₂CO₃ (they are not acidic enough to liberate CO₂). ∴ The suggestion is correct and compound R is confirmed as propanoic acid.
Q22Case-based4 marks

Riya is working in a school chemistry laboratory with four unlabelled bottles containing ethanal (CH₃CHO), propanone (CH₃COCH₃), benzaldehyde (C₆H₅CHO), and methanoic acid (HCOOH). She needs to identify each compound using simple chemical tests and also predict which compound has the highest boiling point.

Riya is working in a school chemistry laboratory and is given four unlabelled bottles containing the following compounds: ethanal, propanone, benzaldehyde, and methanoic acid. She is asked to identify each compound using simple chemical tests and also to predict which compound will give the highest boiling point among these four. Help Riya by answering the following:

(a) Which compound among the four will give a positive Tollens' test but NOT a positive iodoform test? Name the compound and write the chemical equation for the Tollens' test reaction. (2 marks)

(b) Which compound will give a positive iodoform test? Name it and state the observation. (1 mark)

(c) Which compound will have the highest boiling point among the four, and why? (1 mark)

Show answer
(a) Benzaldehyde (C₆H₅CHO) gives a positive Tollens' test but does NOT give a positive iodoform test.

Reason: Benzaldehyde is an aldehyde (has –CHO group) so it reduces Tollens' reagent (ammoniacal silver nitrate), but it does NOT contain the CH₃CO– group required for the iodoform test.

Tollens' test reaction:

C₆H₅CHO + 2[Ag(NH₃)₂]⁺ + 2OH⁻ → C₆H₅COO⁻ + 2Ag(s)↓ + 4NH₃ + H₂O

Observation: A shiny silver mirror (or grey-black precipitate of silver) is deposited on the inner walls of the test tube.

(b) Ethanal (CH₃CHO) gives a positive iodoform test.

Reason: Ethanal contains the CH₃CO– group (methyl ketone/methyl carbonyl group), which is the structural requirement for the iodoform test.

Observation: A yellow precipitate of iodoform (CHI₃) is formed with a characteristic antiseptic smell.

CH₃CHO + 3I₂ + 3NaOH → CHI₃↓ (yellow) + HCOONa + 3NaI + 3H₂O

(c) Methanoic acid (HCOOH) has the highest boiling point among the four compounds.

Due to extensive intermolecular hydrogen bonding / dimer formation between carboxylic acid molecules, a large amount of energy is required to overcome these strong associations, resulting in the highest boiling point.
Q23Case-based4 marks

A pharmaceutical chemist is analysing two unlabelled bottles — Bottle X contains benzaldehyde (C₆H₅CHO) and Bottle Y contains acetophenone (C₆H₅COCH₃). Both are colourless liquids with aromatic odours. She carries out a series of chemical tests and also explores the nucleophilic addition behaviour of each compound.

A pharmaceutical chemist is analysing two unlabelled bottles — Bottle X contains benzaldehyde (C₆H₅CHO) and Bottle Y contains acetophenone (C₆H₅COCH₃). Both are colourless liquids with aromatic odours. She carries out a series of chemical tests and also explores the nucleophilic addition behaviour of each compound.

(a) She treats both compounds with 2,4-dinitrophenylhydrazine (2,4-DNP) reagent. What does she observe for each bottle, and what does this test confirm? (1 mark)

(b) She then adds Tollens' reagent to separate samples of X and Y and warms gently. Describe the observation for each, and identify which bottle is benzaldehyde. (1 mark)

(c) She wishes to convert benzaldehyde (Bottle X) into benzyl alcohol (C₆H₅CH₂OH). Name the reagent she should use and write the balanced chemical equation for this conversion. (1 mark)

(d) Comparing the nucleophilic addition reactivity of benzaldehyde and acetophenone, which compound is MORE reactive and give ONE reason based on electronic or steric factors. (1 mark)

Show answer
(a) 2,4-DNP Test:
Both Bottle X (benzaldehyde) and Bottle Y (acetophenone) give an orange-yellow precipitate with 2,4-DNP reagent.
This confirms the presence of the carbonyl group (C=O) in both compounds — the test is positive for both aldehydes and ketones.

(b) Tollens' Reagent Test:
Bottle X (benzaldehyde): A bright silver mirror is deposited on the inner wall of the test tube (silver mirror formation).
Bottle Y (acetophenone): No silver mirror is formed; the solution remains colourless.
∴ Bottle X is benzaldehyde — aldehydes are oxidised by Tollens' reagent; ketones are not.

(c) Conversion of benzaldehyde to benzyl alcohol:
Reagent: NaBH₄ (sodium borohydride) / LiAlH₄ (lithium aluminium hydride)

C₆H₅CHO + [H] → C₆H₅CH₂OH
NaBH₄
(or LiAlH₄)

Balanced equation using NaBH₄:
4 C₆H₅CHO + NaBH₄ → 4 C₆H₅CH₂OH + NaBO₂

(or simplified form accepted by CBSE:)
C₆H₅CHO + 2[H] → C₆H₅CH₂OH

(d) Benzaldehyde (C₆H₅CHO) is MORE reactive towards nucleophilic addition than acetophenone (C₆H₅COCH₃).
Reason: Acetophenone has two groups attached to the carbonyl carbon — a phenyl group and a methyl group — which create greater steric hindrance, making it harder for the nucleophile to attack. In addition, the methyl group has a +I (electron-donating inductive) effect, which increases electron density on the carbonyl carbon and reduces the partial positive charge (δ+), making it less electrophilic. Benzaldehyde has only a hydrogen atom in place of the methyl group, so it is less hindered and more electrophilic.
∴ Benzaldehyde undergoes nucleophilic addition more readily than acetophenone.
Q24Case-based4 marks

A food quality inspector visits a small-scale vinegar manufacturing unit. She explains to the workers that vinegar is essentially a dilute aqueous solution of ethanoic acid (acetic acid), produced by the oxidation of ethanol. She further demonstrates that ethanoic acid, despite having a carbonyl group like aldehydes and ketones, behaves very differently in chemical tests.

A food quality inspector visits a small-scale vinegar manufacturing unit. She explains to the workers that vinegar is essentially a dilute aqueous solution of ethanoic acid (acetic acid), produced by the oxidation of ethanol. She further demonstrates that ethanoic acid, despite having a carbonyl group like aldehydes and ketones, behaves very differently in chemical tests.

Based on the above context, answer the following questions:

(a) The inspector tests three unlabelled bottles — Bottle X contains ethanol (CH₃CH₂OH), Bottle Y contains ethanal (CH₃CHO), and Bottle Z contains ethanoic acid (CH₃COOH). She performs Tollens' test on each. Which bottle(s) give a positive result? Give one reason. (2 marks)

(b) The inspector notes that ethanoic acid has a much higher boiling point (118°C) than ethanal (20°C), even though both have similar molecular masses. Give reason. (1 mark)

(c) She also observes that ethanoic acid does NOT give the characteristic nucleophilic addition reactions of the carbonyl group. Give reason. (1 mark)

Show answer
(a) Bottle Y (ethanal, CH₃CHO) gives a positive Tollens' test — a silver mirror is deposited on the inner walls of the test tube.

Bottle X (ethanol) and Bottle Z (ethanoic acid) give NO positive result.

Reason: Tollens' reagent [Ag(NH₃)₂]⁺ is a mild oxidising agent that oxidises aldehydes to carboxylate ions, reducing Ag⁺ to metallic silver (silver mirror). Alcohols and carboxylic acids are not oxidised under these conditions.

Reaction:
CH₃CHO(aq) + 2[Ag(NH₃)₂]⁺(aq) + 2OH⁻(aq) → CH₃COO⁻(aq) + 2Ag(s)↓ + 4NH₃(aq) + H₂O(l)
(Silver mirror formed)

∴ Only Bottle Y gives a positive Tollens' test.

(b) Due to intermolecular hydrogen bonding and dimer formation in ethanoic acid, a large amount of energy is required to break these strong associations in the liquid state, resulting in a much higher boiling point than ethanal.

Ethanoic acid exists as dimers in the liquid state through two strong O–H···O hydrogen bonds, whereas ethanal (an aldehyde) cannot form hydrogen bonds between its own molecules (no O–H group), and therefore has a much lower boiling point.

(c) Due to resonance, the lone pair of electrons on the oxygen of the –OH group is donated into the carbonyl C=O, giving the carboxylate/acyl group partial double-bond character. This increases the electron density on the carbonyl carbon, making it far less electrophilic and hence resistant to nucleophilic addition. Additionally, the carbonyl carbon of –COOH is already bonded to two electronegative oxygen atoms, reducing its susceptibility to nucleophilic attack.

The resonance can be represented as:
[Structure: CH₃–C(=O)–OH ↔ CH₃–C(–O⁻)=OH⁺]

∴ Nucleophilic addition to the carbonyl group of carboxylic acids does not occur under normal conditions.
Q25Case-based4 marks

A food-testing laboratory receives three unlabelled bottles — P, Q and R — each said to contain one of the following compounds: ethanal (CH₃CHO), propan-2-one (CH₃COCH₃), and methanoic acid (HCOOH). The chemist performs the following sequential tests:

Test 1: All three samples are treated with 2,4-DNP reagent. P and Q give an orange-yellow precipitate; R does not.

Test 2: P and Q are treated with Tollens' reagent. Only P gives a silver mirror; Q does not.

Test 3: R is treated with NaHCO₃ solution. Brisk effervescence is observed.

A food-testing laboratory receives three unlabelled bottles — P, Q and R — each said to contain one of the following compounds: ethanal (CH₃CHO), propan-2-one (CH₃COCH₃), and methanoic acid (HCOOH). The chemist performs the following sequential tests:

Test 1: All three samples P, Q and R are treated with 2,4-dinitrophenylhydrazine (2,4-DNP) reagent. P and Q give an orange-yellow precipitate; R does not.

Test 2: P and Q are then treated with Tollens' reagent (ammoniacal silver nitrate solution). Only P gives a silver mirror; Q does not.

Test 3: R is treated with NaHCO₃ solution. Brisk effervescence is observed.

(a) Identify compounds P, Q and R. Give one reason for your identification of R. [2 marks]

(b) The chemist further treats compound P with Tollens' reagent under basic conditions and then acidifies the product. Write the balanced chemical equation for the overall reaction of P with Tollens' reagent, identifying the organic product formed. [1 mark]

(c) Compound Q is reacted with iodine and aqueous sodium hydroxide (NaOH) solution. Name the reaction, identify the yellow precipitate formed, and state one other compound that would give a positive result in this test. [1 mark]

Show answer
MARKING SCHEME — 4 marks

(a) Identification of P, Q and R: [2 marks]

Test 1 (2,4-DNP): The 2,4-DNP reagent gives an orange-yellow precipitate with compounds containing a carbonyl group (C=O), i.e., both aldehydes and ketones. P and Q give this precipitate, so P and Q each contain a C=O group. R does not react, so R lacks a carbonyl group. Therefore R is methanoic acid (HCOOH). [½ mark]

Test 2 (Tollens' reagent): Only P gives a silver mirror. Tollens' reagent is oxidised only by aldehydes (which are themselves oxidised to the corresponding carboxylate). P is therefore ethanal (CH₃CHO) and Q is propan-2-one (CH₃COCH₃), since ketones do not reduce Tollens' reagent. [½ mark]

∴ P = Ethanal (CH₃CHO), Q = Propan-2-one (CH₃COCH₃), R = Methanoic acid (HCOOH). [½ mark]

Reason for identification of R: R gives brisk effervescence with NaHCO₃ solution. Due to its acidic nature (pKa ≈ 3.75), methanoic acid reacts with NaHCO₃ to liberate CO₂ gas — a reaction that aldehydes and ketones do not undergo. This confirms R is the carboxylic acid. [½ mark]

(b) Reaction of P (ethanal) with Tollens' reagent: [1 mark]

CH₃CHO + 2[Ag(NH₃)₂]⁺ + 2OH⁻ → CH₃COO⁻ + 2Ag(s)↓ + 4NH₃ + H₂O

On acidification: CH₃COO⁻ + H⁺ → CH₃COOH

∴ The organic product formed (after acidification) is acetic acid (ethanoic acid, CH₃COOH). A silver mirror deposit of Ag(s) is observed on the inner walls of the test tube. [1 mark — award for balanced ionic equation with Ag↓ and correct organic product; ECF applies if P misidentified in (a)]

(c) Reaction of Q (propan-2-one) with I₂/NaOH: [1 mark]

Name of reaction: Iodoform reaction (Iodoform test). [½ mark]

Yellow precipitate formed: Iodoform (CHI₃). [½ mark — both parts needed for the 1 mark; accept either order]

Overall equation for reference:
CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃↓ + CH₃COONa + 3NaI + 3H₂O

One other compound giving a positive iodoform test: Ethanol (CH₃CH₂OH) / Acetaldehyde (CH₃CHO) / any methyl ketone (CH₃COR) / any secondary alcohol of the type CH₃CH(OH)R. [Award ½ mark for any one valid example — accept acetaldehyde, ethanol, butan-2-ol, etc.]

Want unlimited practice on Aldehydes, Ketones and Carboxylic Acids?

The full ClearSteps bank has 25+ reviewed questions on this chapter alone — with step-wise solutions, difficulty levels, and progress tracking. Verify your WhatsApp number and your free trial starts right here.

CBSE · CLASS 12
🎟️14-day free trial · code PRAC auto-applied
Mobile Number
+91
OTP will be sent to your WhatsApp

⭐ Trusted by 12,000+ students across India

All chapters →Chapter-wise PYQ →Sample Papers →
Aldehydes, Ketones and Carboxylic Acids Class 12 Questions