CBSE Class 12 Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 09

Multiple Choice Questions

Question 1. Which among the following compounds, contain \( -\text{CHO} \) group, benzene ring, \( -\text{OCH}_3 \) group, \( -\text{OH} \) group?
(a) Salicylic acid
(b) Salicylaldehyde
(c) Vanillin
(d) Cinnamaldehyde
Answer: (c) Vanillin
In simple words: Vanillin has an aldehyde group, a benzene ring, a methoxy group, and a hydroxyl group all in its chemical structure.

Exam Tip: Memorize the structural formulas of common aromatic compounds like vanillin, salicylic acid, and salicylaldehyde as they are frequently asked in functional group identification questions.

Question 2. Addition of water to butyne occurs in acidic medium and in the presence of \( \text{Hg}^{2+} \) ions as a catalyst. The product formed is
(a) \( \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{C}(=\text{O})-\text{H} \)
(b) \( \text{CH}_3-\text{CH}_2-\text{C}(=\text{O})-\text{CH}_3 \)
(c) \( \text{CH}_3-\text{CH}_2-\text{C}(=\text{O})-\text{OH} + \text{CO}_2 \)
(d) \( \text{CH}_3-\text{C}(=\text{O})-\text{OH} + \text{H}-\text{C}(=\text{O})-\text{H} \)
Answer: (b) \( \text{CH}_3-\text{CH}_2-\text{C}(=\text{O})-\text{CH}_3 \)
In simple words: Adding water to but-1-yne with mercury catalyst yields butan-2-one. This happens because the oxygen atom attaches to the second carbon atom through Markovnikov's addition.

Exam Tip: Remember that hydration of any terminal alkyne other than ethyne always produces a ketone, not an aldehyde, due to Markovnikov's rule of addition.

Question 3. The reaction, \( \text{RCN} + \text{SnCl}_2 + \text{HCl} \longrightarrow \text{RCH=NH} \xrightarrow{\text{H}_3\text{O}^+} \text{RCHO} \) is known as
(a) Etard reaction
(b) Haloform reaction
(c) Gattermann-Koch reaction
(d) Stephen reaction
Answer: (d) Stephen reaction
In simple words: This specific process of turning nitriles into aldehydes using tin chloride and acid is called the Stephen reduction.

Exam Tip: Name reactions are highly scoring; write the complete balanced chemical equation alongside the name to secure full marks.

Question 4. In the equation, \( \text{CH}_3\text{COCl} \xrightarrow{\text{H}_2/\text{Pd - BaSO}_4} \text{X} \), \( \text{X} \) is
(a) acetaldehyde
(b) propionaldehyde
(c) acetone
(d) acetic acid
Answer: (a) acetaldehyde
In simple words: When acetyl chloride reacts with hydrogen over a palladium-barium sulfate catalyst, it gets reduced to form acetaldehyde.

Exam Tip: This reaction is the Rosenmund reduction. Remember that \( \text{BaSO}_4 \) acts as a catalytic poison to prevent further reduction of the aldehyde to an alcohol.

Question 5. Rosenmund reduction is used for the preparation of aldehydes. The catalyst used in this reaction is
(a) \( \text{Pd}-\text{BaSO}_4 \)
(b) anhydrous \( \text{AlCl}_3 \)
(c) iron (III) oxide
(d) \( \text{HgSO}_4 \)
Answer: (a) \( \text{Pd}-\text{BaSO}_4 \)
In simple words: This reduction uses palladium supported on barium sulfate as the specific catalyst to convert acyl chlorides into aldehydes.

Exam Tip: Clearly mention the role of \( \text{BaSO}_4 \) as a poison for the palladium catalyst in Rosenmund reduction to show in-depth understanding.

Question 6. The compound formed as a result of oxidation of ethyl benzene by \( \text{KMnO}_4 \) is
(a) benzyl alcohol
(b) acetophenone
(c) benzophenone
(d) benzoic acid
Answer: (d) benzoic acid
In simple words: Strong oxidizers like potassium permanganate will convert any alkyl side chain on a benzene ring directly into a benzoic acid group.

Exam Tip: No matter how long the alkyl side chain is, as long as it has at least one benzylic hydrogen, vigorous oxidation with alkaline \( \text{KMnO}_4 \) always yields benzoic acid.

Question 7. Which of the following is most reactive in nucleophilic addition reactions?
(a) \( \text{HCHO} \)
(b) \( \text{CH}_3\text{CHO} \)
(c) \( \text{CH}_3\text{COCH}_3 \)
(d) \( \text{CH}_3\text{COC}_2\text{H}_5 \)
Answer: (a) \( \text{HCHO} \)
In simple words: Formaldehyde is the most reactive because it has no bulky alkyl groups to block the attacking nucleophile and no alkyl groups to reduce the positive charge on the carbonyl carbon.

Exam Tip: Reactivity towards nucleophilic addition decreases with an increase in steric hindrance and the +I effect of alkyl groups; thus, aldehydes are always more reactive than ketones.

Question 8. Which of the following compounds will give butanone on oxidation with alkaline \( \text{KMnO}_4 \) solution?
(a) Butan-1-ol
(b) Butan-2-ol
(c) Both (a) and (b)
(d) None of the options
Answer: (b) Butan-2-ol
In simple words: Butan-2-ol is a secondary alcohol, and oxidising a secondary alcohol yields a ketone like butanone.

Exam Tip: Primary alcohols oxidize to aldehydes and then to carboxylic acids, while secondary alcohols stop at ketones under normal oxidizing conditions.

Question 9. The reagent which does not react with both, acetone and benzaldehyde?
(a) Sodium hydrogen sulphite
(b) Phenyl hydrazine
(c) Fehling's solution
(d) Grignard reagent
Answer: (c) Fehling's solution
In simple words: Fehling's solution does not react with ketones like acetone, nor does it react with aromatic aldehydes like benzaldehyde.

Exam Tip: Note that aliphatic aldehydes reduce Fehling's solution, but aromatic aldehydes (such as benzaldehyde) and ketones do not, making this a useful distinguishing test.

Question 10. A compound (A) with molecular formula \( \text{C}_5\text{H}_{10}\text{O} \), forms a phenyl hydrazone and gives negative Tollen's and iodoform tests. The compound on reduction gives n-pentane. The compound (A) is
(a) pentan-3-one
(b) pentanal
(c) pentanol
(d) pentan-2-one
Answer: (a) pentan-3-one
In simple words: Pentan-3-one is a ketone, so it fails Tollen's test. It also lacks a methyl ketone group, meaning it cannot undergo the iodoform reaction.

Exam Tip: Use a systematic elimination process for such identification questions: Tollen's test rules out aldehydes, while the iodoform test rules out methyl ketones.

Question 11. When \( \text{C}_6\text{H}_5\text{COOCOCH}_3 \) is treated with \( \text{H}_2\text{O} \), the product obtained is
(a) benzoic acid and ethanol
(b) benzoic acid and ethanoic acid
(c) acetic acid and phenol
(d) benzoic anhydride and methanol
Answer: (b) benzoic acid and ethanoic acid
In simple words: Hydrolyzing this mixed acid anhydride splits it into its parent carboxylic acids, which are benzoic acid and acetic (ethanoic) acid.

Exam Tip: Hydrolysis of any acid anhydride always gives the corresponding carboxylic acids by cleaving the \( \text{C}-\text{O}-\text{C} \) linkage.

Question 12. Consider the following reaction.
\[ \begin{matrix} \text{H} \\ \text{ } \end{matrix} \text{C=O} + \begin{matrix} \text{H} \\ \text{ } \end{matrix} \text{C=O} + \text{Conc. KOH} \xrightarrow{\Delta} \text{A} + \text{B} \]
Identify A and B from the given options.
(a) A-Methanol, B-Potassium formate
(b) A-Ethanol, B-Potassium formate
(c) A-Methanal, B-Ethanol
(d) A-Methanol, B-Potassium acetate
Answer: (a) A-Methanol, B-Potassium formate
In simple words: Formaldehyde undergoes the Cannizzaro reaction with concentrated base, where one molecule is reduced to methanol and the other is oxidized to potassium formate.

Exam Tip: Formaldehyde lacks alpha-hydrogen atoms, making it undergo self-oxidation and reduction (disproportionation) when heated with a concentrated alkali.

Question 13. Which of the following does not give aldol condensation reaction?
(a) \( \text{CH}_3-\text{CHO} \)
(b) \( \text{C}_6\text{H}_{11}-\text{CHO} \) (Cyclohexanecarbaldehyde)
(c) \( \text{C}_6\text{H}_5-\text{CHO} \) (Benzaldehyde)
(d) \( \text{CH}_3\text{COCH}_3 \)
Answer: (c) \( \text{C}_6\text{H}_5-\text{CHO} \) (Benzaldehyde)
In simple words: Benzaldehyde lacks any alpha-hydrogen atoms, which means it cannot participate in the aldol condensation reaction.

Exam Tip: For any carbonyl compound to undergo aldol condensation, it must possess at least one alpha-hydrogen atom adjacent to the carbonyl group.

Question 14. A carbonyl compound X undergoes the reactions given in the table below.

Which of the following could compound X be?
(a) \( \text{CH}_3-\text{CH}_2-\text{CHO} \)
(b) \( \text{CH}_3-\text{CO}-\text{CH}_3 \)
(c) \( \text{CH}_3-\text{CHO} \)
(d) \( \text{H}-\text{CHO} \)
Answer: (c) \( \text{CH}_3-\text{CHO} \)
In simple words: Acetaldehyde gives a positive Tollens' test because it is an aldehyde, a positive iodoform test because of its methyl group next to carbonyl, and undergoes aldol condensation due to its alpha-hydrogens.

Exam Tip: Acetaldehyde is the only aldehyde that gives a positive iodoform test because it contains the required active methyl group attached directly to the carbonyl carbon.

Question 15. Which of the following compounds gives an oxime with hydroxylamine?
(a) \( \text{CH}_3\text{COCH}_3 \)
(b) \( \text{CH}_3\text{COOH} \)
(c) \( (\text{CH}_3\text{CO})_2\text{O} \)
(d) \( \text{CH}_3\text{COCl} \)
Answer: (a) \( \text{CH}_3\text{COCH}_3 \)
In simple words: Acetone is a ketone, and it undergoes nucleophilic addition with hydroxylamine to form acetoxime.

Exam Tip: Only aldehydes and ketones form oximes with hydroxylamine; carboxylic acids, anhydrides, and acid chlorides do not undergo this nucleophilic addition reaction.

Question 16. What is the correct IUPAC name of the given compound?
\[ \begin{matrix} \text{CH}_3 \\ | \\ \text{CH}_3-\text{C}-\text{CH}_2\text{CH}_3 \\ | \\ \text{COOH} \end{matrix} \]
(a) 2, 2-dimethylbutanoic acid
(b) 2-carboxyl-2-methylbutane
(c) 2-ethyl-2-methylpropanoic acid
(d) 3-methylbutane carboxylic acid
Answer: (a) 2, 2-dimethylbutanoic acid
In simple words: By numbering from the carboxylic acid carbon as 1, the longest continuous carbon chain has 4 carbons (butanoic acid), with two methyl groups on carbon 2.

Exam Tip: Always begin numbering the longest carbon chain from the carboxylic acid carbon, as it receives the highest priority in IUPAC nomenclature.

Question 17. The acid formed when propyl magnesium bromide is treated with \( \text{CO}_2 \) followed by acid hydrolysis.
(a) \( \text{C}_3\text{H}_7\text{COOH} \)
(b) \( \text{C}_2\text{H}_5\text{COOH} \)
(c) \( \text{CH}_3\text{COOH} \)
(d) \( \text{C}_3\text{H}_7\text{OH} \)
Answer: (a) \( \text{C}_3\text{H}_7\text{COOH} \)
In simple words: Propyl magnesium bromide reacts with carbon dioxide to add one carbon, forming a salt that yields butanoic acid after hydrolysis.

Exam Tip: Grignard reagents react with dry ice (\( \text{CO}_2 \)) to yield a carboxylic acid that has one more carbon atom than the original alkyl group.

Question 18. Which of the following acid is the strongest?
(a) \( \text{CH}_3\text{COOH} \)
(b) \( \text{CH}_2\text{ClCOOH} \)
(c) \( \text{CHCl}_2\text{COOH} \)
(d) \( \text{CCl}_3\text{COOH} \)
Answer: (d) \( \text{CCl}_3\text{COOH} \)
In simple words: Trichloroacetic acid has three strongly electron-withdrawing chlorine atoms that stabilize the carboxylate anion, making it highly acidic.

Exam Tip: The presence of electron-withdrawing groups (like chlorine) on the alpha carbon increases the acidic strength by dispersing the negative charge of the conjugate base.

Question 19. The carboxylic acid group does not give the usual addition and elimination reactions of aldehydes and ketones because
(a) O—H bond is more polar than C = O group
(b) COOH group gets ionised
(c) carboxylate ion gets stabilised by resonance
(d) COOH group does not contain carbonyl group
Answer: (c) carboxylate ion gets stabilised by resonance
In simple words: Resonance stabilization within the carboxyl group reduces the double-bond character of the carbonyl carbon, making it less susceptible to nucleophilic attack.

Exam Tip: Remember that resonance in the carboxyl group substantially reduces the electrophilic character of the carbonyl carbon compared to aldehydes or ketones.

Question 20. Arrange the following acids in the decreasing order of acidic strength.
\[ \text{HC}\equiv\text{C}-\text{COOH}\text{ (I)} \qquad \text{C}_6\text{H}_5\text{COOH}\text{ (II)} \qquad \text{CH}_2=\text{CH}-\text{COOH}\text{ (III)} \qquad \text{CH}_3\text{COOH}\text{ (IV)} \]
(a) I > II > III > IV
(b) I < II < III < IV
(c) I > III > II > IV
(d) I > IV > III > II
Answer: (a) I > II > III > IV
In simple words: Acidic strength increases with the electronegativity of the carbon atom attached to the carboxyl group, which decreases in the order of sp (alkynyl) > sp2 (aryl/alkenyl) > sp3 (alkyl).

Exam Tip: Compare the hybridization of the alpha carbon: higher s-character (sp > sp2 > sp3) means greater electronegativity, which strongly enhances the acid's strength.

Question 22. The product formed during Hell-Volhard-Zelinsky reaction is
(a) \( \text{R}-\text{CH}(\text{X})-\text{COOH} \)
(b) \( \text{R}-\text{CH}_2-\text{COX} \)
(c) \( \text{R}-\text{CX}_2-\text{COOH} \)
(d) \( \text{R}-\text{CH}(\text{X})-\text{CH}_2-\text{COOH} \)
Answer: (a) \( \text{R}-\text{CH}(\text{X})-\text{COOH} \)
In simple words: In this reaction, a halogen atom replaces a hydrogen atom specifically at the alpha carbon of a carboxylic acid.

Exam Tip: The HVZ reaction is specific for the halogenation of carboxylic acids at the alpha position and requires the presence of at least one alpha-hydrogen.

Question 23. Carboxylic acids do not undergo Friedel Crafts reaction because
(a) COOH group is meta directing
(b) COOH group is resonance stabilised
(c) Carboxyl group is deactivating and gets bonded to Friedel-Craft's catalyst
(d) All of the options
Answer: (c) Carboxyl group is deactivating and gets bonded to Friedel-Craft's catalyst
In simple words: The highly deactivating carboxyl group forms a complex with the anhydrous aluminium chloride catalyst, which completely deactivates the catalyst.

Exam Tip: Explain both points: the deactivating nature of the \( -\text{COOH} \) group and its complexation with the Lewis acid catalyst \( \text{AlCl}_3 \) to write a complete answer.

Question 24. Electrophilic substitution in benzoic acid takes place at the meta-position. Which of the following is the reason for the reaction above?
(a) The carboxyl group activates only the meta-position.
(b) The carboxyl group deactivates only the ortho and para-positions.
(c) The carboxyl group activates the meta-position more than the ortho and para-positions.
(d) The carboxyl group deactivates the meta-position less than the ortho and para-positions.
Answer: (d) The carboxyl group deactivates the meta-position less than the ortho and para-positions.
In simple words: Since the electron-withdrawing carboxyl group pulls electron density mostly from the ortho and para positions, the meta-position remains relatively more electron-rich.

Exam Tip: Keep in mind that meta-directing groups do not actually activate the meta-position; they merely deactivate it much less than the ortho and para positions.

Assertion-Reason

Directions In the following questions, an Assertion (A) is followed by a corresponding Reason (R). Use the following keys to choose the appropriate answer:
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.

Question. Assertion (A) Formaldehyde is a planar molecule.
Reason (R) It contains \( \text{sp}^2 \)-hybridised carbon atom.
Answer: (a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
In simple words: Formaldehyde has a flat, planar shape because its carbon atom forms three bonds using sp2 hybrid orbitals, creating a coplanar structure with 120-degree bond angles.

Exam Tip: Relate hybridization to molecular geometry: \( \text{sp}^2 \)-hybridized carbon always leads to a trigonal planar structure with bond angles around 120 degrees.

Question. Assertion (A) The solubility of aldehydes and ketones in water decreases with increase in the size of alkyl group.
Reason (R) Alkyl groups are electron-repelling groups.
Answer: (c) (A) is true, but (R) is false.
In simple words: Larger alkyl chains make the molecule more hydrophobic, which decreases its solubility in water, regardless of the electron-repelling nature of those alkyl groups.

Exam Tip: Remember that solubility of carbonyl compounds in water is governed by hydrogen bonding, which is hindered by the larger non-polar hydrophobic parts of the molecule.

Question. Assertion (A) Compounds containing \( -\text{CHO} \) group are easily oxidised to corresponding carboxylic acids.
Reason (R) Carboxylic acids can be reduced to alcohols by treatment with \( \text{LiAlH}_4 \).
Answer: (b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
In simple words: Aldehydes oxidize easily because of the hydrogen atom directly attached to the carbonyl carbon, which is not explained by the fact that carboxylic acids can be reduced back to alcohols using lithium aluminium hydride.

Exam Tip: Aldehydes are highly prone to oxidation compared to ketones because of the easily oxidizable C-H bond present on the carbonyl carbon.

Question. Assertion (A) Aldehydes and ketones, both react with Tollen’s reagent to form silver mirror.
Reason (R) Both aldehydes and ketones contain a carbonyl group.
Answer: (d) (A) is false, but (R) is true.
In simple words: Only aldehydes can reduce Tollen's reagent to form a silver mirror; ketones do not react with it, although both compound classes contain a carbonyl group.

Exam Tip: Tollen's reagent is a mild oxidizing agent that selectively oxidizes aldehydes to carboxylate ions, making it an excellent distinguishing test between aldehydes and ketones.

Question. Assertion (A) Pentan-2-one can be distinguished from pentan-3-one by iodoform test.
Reason (R) Pentan-2-one is a methyl ketone, while the pentan-3-one is not.
Answer: (a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
In simple words: Pentan-2-one gives a positive iodoform test because it contains a methyl ketone group, whereas pentan-3-one lacks this specific group and does not react.

Exam Tip: The iodoform test requires the presence of a methyl ketone (\( \text{CH}_3\text{CO}- \)) group or an alcohol oxidizable to a methyl ketone, yielding a bright yellow precipitate of iodoform (\( \text{CHI}_3 \)).

Question. Assertion (A) \( \alpha \)-hydrogen atoms of carbonyl compounds are acidic.
Reason (R) The strong electron releasing effect of the carbonyl group make the stabilisation of the conjugate base by the resonance.
Answer: (c) (A) is true, but (R) is false.
In simple words: The carbonyl group is highly electron-withdrawing, not electron-releasing, which increases the acidity of alpha-hydrogen atoms and stabilizes the resulting conjugate base.

Exam Tip: Always describe both the strong electron-withdrawing nature (inductive effect) of the carbonyl group and the resonance stabilization of the resulting carbanion (enolate ion) when explaining alpha-hydrogen acidity.

Question. Assertion (A) Aromatic aldehydes and formaldehyde undergo Cannizzaro reaction.
Reason (R) Aromatic aldehydes are almost as reactive as formaldehyde.
Answer: (c) (A) is true, but (R) is false.
In simple words: Both aromatic aldehydes and formaldehyde undergo the Cannizzaro reaction because they lack alpha-hydrogen atoms, but aromatic aldehydes are actually much less reactive than formaldehyde.

Exam Tip: Aromatic aldehydes are less reactive than formaldehyde toward nucleophilic attacks due to the resonance donation of electron density from the benzene ring to the carbonyl carbon.

Question. Assertion (A) A carboxylate ion (\( \text{RCOO}^- \)) is stabilised by resonance to a greater extent as compared to the acid (\( \text{RCOOH} \)).
Reason (R) The contributing structures of \( \text{RCOO}^- \) are equivalent, while those of \( \text{RCOOH} \) are not.
Answer: (a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
In simple words: The carboxylate ion is highly stable because its resonance structures are identical, meaning the negative charge is shared equally between two identical oxygen atoms, unlike the uncharged acid which has non-equivalent resonance structures involving charge separation.

Exam Tip: Draw the resonance structures of both the carboxylate ion (identical and equivalent) and the carboxylic acid (non-equivalent, involves charge separation) to clearly explain the difference in stability.

Question. Assertion (A) Acetic acid but not formic acid can be halogenated in the presence of red P and \( \text{Cl}_2 \).
Reason (R) Acetic acid is a weaker acid than formic acid.
Answer: (b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
In simple words: Acetic acid has alpha-hydrogen atoms and can undergo the Hell-Volhard-Zelinsky halogenation, whereas formic acid cannot because it lacks alpha-hydrogens, which is unrelated to their difference in acid strength.

Exam Tip: The HVZ reaction requires the presence of an alpha-hydrogen atom; formic acid (\( \text{HCOOH} \)) has no carbon atom other than the carboxyl carbon, so it lacks an alpha-hydrogen and cannot undergo this reaction.

Question. Assertion (A) Benzoic acid does not undergo Friedel-Crafts reaction.
Reason (R) Carboxyl group is deactivating and the catalyst aluminium chloride gets bonded to the carboxyl group.
Answer: (a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
In simple words: The carboxyl group in benzoic acid is strongly deactivating, and the Lewis acid catalyst aluminium chloride binds directly to it, preventing the reaction from taking place.

Exam Tip: Explain that the strongly electron-withdrawing nature of the \( -\text{COOH} \) group, along with its complexation with the Lewis acid catalyst \( \text{AlCl}_3 \), makes the benzene ring completely unreactive towards Friedel-Crafts alkylation or acylation.

Question. Assertion (A) Aromatic carboxylic acids do not undergo Friedel-Crafts reaction.
Reason (R) Carboxyl group is deactivating and the catalyst aluminium chloride gets bonded to the carboxyl group.
Answer: (a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
In simple words: The carboxyl group acts as a strong deactivating agent on the aromatic ring, and the aluminium chloride catalyst attaches to it, halting any Friedel-Crafts reaction.

Exam Tip: Be consistent in noting that aromatic carboxylic acids and nitrobenzene are generally used as solvents or are inert in Friedel-Crafts reactions due to their highly deactivated aromatic rings.

Question. Assertion (A) Bromination of benzoic acid gives m-bromobenzoic acid.
Reason (R) Carboxyl group increases the electron density at the meta-position.
Answer: (c) (A) is true, but (R) is false.
In simple words: Bromination happens at the meta-position because the carboxyl group deactivates ortho and para positions much more, leaving the meta-position relatively more electron-dense, rather than actually increasing its electron density.

Exam Tip: Remember that electron-withdrawing groups deactivate all ring positions, but they deactivate the meta-position less than the ortho/para positions, making the meta-position the preferred site for electrophilic attack.

Case-Study

Question 37. Read the case given below and answer the following questions.
A student conducted the following experiments on three organic compounds labelled A, B and C.
Compound A: When treated with Tollen's reagent, a silver mirror was observed. On reaction with Fehling's solution, a brick-red precipitate was formed. The compound also reacted with sodium bisulphite to produce a crystalline solid.
Compound B: On treatment with Brady's reagent (2,4-dinitrophenylhydrazine), an orange precipitate was formed. However, it did not give a positive test with Tollen's reagent or Fehling's solution. When heated with iodine in the presence of alkali, a yellow precipitate with an antiseptic odour was observed.
Compound C: On adding to a solution of sodium bicarbonate, vigorous effervescence was observed. When treated with neutral \( \text{FeCl}_3 \), a violet coloration appeared.

Question (i) Explain the following observations:
(a) Formation of a silver mirror with Tollen's reagent for compound A.
(b) Yellow precipitate with iodine and alkali for compound B.
Answer:
(a) Aldehydes have the ability to reduce Tollen's reagent (\( \text{Ag}^+ \) ions) into metallic silver, creating a shiny silver mirror.
(b) The iodoform reaction proves that a methyl ketone group (\( -\text{CH}_3\text{CO} \)) is present, producing a bright yellow solid of iodoform (\( \text{CHI}_3 \)).
In simple words: Compound A contains an aldehyde group which reduces silver ions to form a silver mirror, while Compound B has a methyl ketone group that reacts with iodine and alkali to form a yellow iodoform precipitate.

Exam Tip: Be sure to write the ionic equations or specify the active functional groups (\( -\text{CHO} \) for Tollen's and \( -\text{COCH}_3 \) for iodoform) to get full credit.

Question (ii) Write the structures of the compound 'A' and 'B' based on the tests performed.
Answer: Based on the experimental results, Compound A is identified as acetaldehyde, \( \text{CH}_3\text{CHO} \), and Compound B is identified as acetone, \( \text{CH}_3\text{COCH}_3 \).
In simple words: Compound A is acetaldehyde because it reacts with Tollen's and Fehling's solutions, while Compound B is acetone because it is a methyl ketone.

Exam Tip: When structures are asked, always draw the complete structural formula showing all atoms and bonds clearly.

Question (ii) Or. Identify the functional groups present in compounds A and B.
Answer: In Compound A, the active group is an aldehyde (\( -\text{CHO} \)), and in Compound B, it is a ketone, specifically a methyl ketone (\( -\text{CO}-\text{CH}_3 \)).
In simple words: Compound A contains an aldehyde functional group, whereas Compound B contains a methyl ketone functional group.

Exam Tip: Specify "methyl ketone" rather than just "ketone" to show that you understand why Compound B gave a positive iodoform test.

Question 38. (iii) What is the reason behind effervescence with sodium bicarbonate for compound C?
Answer: The visible bubbling is due to the release of carbon dioxide (\( \text{CO}_2 \)) gas when the acidic carboxyl group of Compound C (which is salicylic acid) reacts with sodium bicarbonate.
In simple words: Compound C is salicylic acid, and its acidic group reacts with sodium bicarbonate to release carbon dioxide gas as bubbles.

Exam Tip: Write the chemical equation: \( \text{R-COOH} + \text{NaHCO}_3 \longrightarrow \text{R-COONa} + \text{H}_2\text{O} + \text{CO}_2\uparrow \) to demonstrate the source of the effervescence.