Class 9 Chemistry Chapter 8: Aldehydes, Ketones & Carboxylic Acids – Important Questions with Answers

The 2025-26 CBSE Class 9 board examination follows a strictly rationalized curriculum where Chapter 8 (Aldehydes, Ketones and Carboxylic Acids) carries 8–10 marks. Question setters prioritize three areas: (1) functional group identification and IUPAC nomenclature of aldehydes (–CHO), ketones (C=O in middle of chain), and carboxylic acids (–COOH); (2) laboratory and industrial preparation methods with chemical equations; (3) characteristic chemical reactions and reaction mechanisms showing bond breakage/formation. The board also increasingly tests application-based questions where students must identify an unknown compound from its reactions or predict products of multi-step synthesis. MCQs often focus on nomenclature rules (e.g., numbering direction in IUPAC names, priority of functional groups). Short-answer questions demand balanced chemical equations. Long-answer questions require step-by-step mechanisms showing electron movement. HOTS questions combine preparation, identification, and mechanism in a single scenario. By drilling these exact question types daily with cbsetutor.ai's AI tutor, you build pattern recognition and avoid common errors like incorrect functional group priority or unbalanced equations.

Multiple-choice questions in board exams test quick recall and nomenclature accuracy. Here are five representative 1-mark questions: **Q1.** The IUPAC name of CH₃CH₂CHO is: (a) Propanal (b) Propenone (c) Propanoic acid (d) Propanol **Answer:** (a) Propanal. The –CHO group (aldehyde) is named 'al' suffix. Three carbons = prop, so CH₃CH₂CHO = propanal. **Q2.** In the compound CH₃COCH₂CH₃, the carbonyl carbon is bonded to: (a) Two hydrogen atoms (b) Two alkyl groups (c) One hydrogen and one alkyl group (d) No hydrogen atoms **Answer:** (b) Two alkyl groups. The C=O in ketones is always bonded to two carbon atoms (CH₃– and –CH₂CH₃), never to H. **Q3.** Which functional group has the highest priority in IUPAC nomenclature? (a) Ketone (b) Aldehyde (c) Carboxylic acid (d) Ether **Answer:** (c) Carboxylic acid. Priority order: –COOH > –CHO > C=O > –OH > –NH₂ > C=C. **Q4.** The general formula of saturated aldehydes is: (a) CₙH₂ₙO (b) CₙH₂ₙ₊₂O (c) CₙH₂ₙ₋₂O (d) CₙH₂ₙ₊₂O₂ **Answer:** (a) CₙH₂ₙO. Example: formaldehyde CH₂O (n=1), acetaldehyde C₂H₄O (n=2), propanal C₃H₆O (n=3). **Q5.** Acetic acid (CH₃COOH) is also known as: (a) Methanoic acid (b) Ethanoic acid (c) Propanoic acid (d) Butanoic acid **Answer:** (b) Ethanoic acid. CH₃COOH has 2 carbons total (including the carboxyl carbon), so it's ethanoic acid. Methanoic acid (formic) = HCOOH.

Short-answer questions require concise explanations, balanced equations, or simple preparation/reaction sequences. **Q1.** Write the structural formula of 3-methylbutanal and name the type of carbonyl group present. **Answer:** Structural formula: CH₃—CH(CH₃)—CH₂—CHO (or CH₃CH(CH₃)CH₂CHO). The functional group is an aldehyde (–CHO). It's an aldehyde because the carbonyl carbon is bonded to only one alkyl group and a hydrogen atom. **Q2.** Write the equation for the oxidation of acetaldehyde (CH₃CHO) using acidified potassium dichromate. **Answer:** CH₃CHO + [O] → CH₃COOH (or 2CH₃CHO + [O] → 2CH₃COOH). The aldehyde is oxidized to a carboxylic acid. With acidified K₂Cr₂O₇, the orange dichromate turns green (Cr³⁺). **Q3.** Differentiate between aldehydes and ketones based on their reactivity with Tollens' reagent. **Answer:** Aldehydes react with Tollens' reagent (ammoniacal AgNO₃) to form a bright silver mirror (Ag deposited on test tube). Ketones do not react with Tollens' reagent and show no silver mirror. This is because aldehydes are easily oxidized (H—C=O can lose the H), while ketones (R₂C=O) lack this easily oxidizable hydrogen. **Q4.** Give the IUPAC name and structural formula of the carboxylic acid formed when ethanol is completely oxidized. **Answer:** IUPAC name: Ethanoic acid. Structural formula: CH₃COOH. Oxidation: C₂H₅OH + 2[O] → CH₃COOH + H₂O. Ethanol is oxidized to ethanal (CH₃CHO) first, then to ethanoic acid. **Q5.** What is the main difference between the preparation of aldehydes and ketones from alcohols? **Answer:** Primary alcohols (R–CH₂–OH) are oxidized to aldehydes (R–CHO) using mild oxidizing agents like PCC or dilute acidified K₂Cr₂O₇. Secondary alcohols (R₂CHOH) are oxidized to ketones (R₂C=O) using the same agents. Tertiary alcohols cannot be oxidized easily. The key difference: primary → aldehyde; secondary → ketone.

Three-mark questions test deeper understanding: preparation methods with full equations, multi-step reactions, or mechanism explanations. **Q1.** Write the chemical equations for the preparation of acetaldehyde (a) from ethene and (b) from ethanol. **Answer:** (a) From ethene (Hydration of ethene): CH₂=CH₂ + H₂O → CH₃CH₂OH (ethanol) [requires H⁺ catalyst]; then oxidize: CH₃CH₂OH + [O] → CH₃CHO + H₂O (using PCC or mild K₂Cr₂O₇) Direct route: CH₂=CH₂ + H₂O₂ → CH₃CHO [using Wacker process with PdCl₂ catalyst, industrial method] (b) From ethanol (Oxidation of primary alcohol): CH₃CH₂OH + [O] → CH₃CHO + H₂O. Use mild oxidizing agent like PCC (pyridinium chlorochromate) or dilute acidified potassium dichromate at controlled temperature to avoid over-oxidation to acetic acid. **Q2.** Explain the mechanism of nucleophilic addition reaction of HCN (hydrogen cyanide) with acetaldehyde (CH₃CHO). **Answer:** Step 1 (Nucleophilic attack): The CN⁻ ion (nucleophile) attacks the electrophilic carbonyl carbon (C=O is polarized, C is δ⁺). The π-bond breaks and electrons move to oxygen: CH₃CHO + CN⁻ → CH₃—CH(CN)⁻—O⁻ Step 2 (Protonation): The alkoxide intermediate (–O⁻) accepts a proton from HCN or water: CH₃—CH(CN)⁻—O⁻ + H⁺ → CH₃—CH(CN)—OH Final product: Acetaldehyde cyanohydrin (CH₃CH(CN)OH). This is a nucleophilic addition reaction because a small molecule (HCN) adds across the C=O double bond without elimination. **Q3.** A compound X (molecular formula C₂H₄O) decolorizes bromine water and gives a positive Tollens' test. Identify X and write its structure. What happens when X is oxidized? **Answer:** Compound X is acetaldehyde (CH₃CHO). Structural formula: CH₃—CHO Reasoning: C₂H₄O has degree of unsaturation = 1 (one double bond or ring). Positive Tollens' test indicates an aldehyde. The compound that decolorizes Br₂ must have a C=C or C=O. Combined with aldehyde functional group, X = CH₃CHO. When oxidized: CH₃CHO + [O] → CH₃COOH (acetic acid). The aldehyde is oxidized to a carboxylic acid.

Five-mark questions demand comprehensive answers: detailed mechanisms, preparation with equations, synthesis pathways, or comparative analysis. **Q1.** Write the complete mechanism of the reaction between acetaldehyde and ethanol in the presence of dilute H₂SO₄ (acetal formation), and explain why this reaction is reversible. **Answer:** Reaction: CH₃CHO + 2C₂H₅OH ⇌ CH₃CH(OC₂H₅)₂ + H₂O (in presence of dilute H₂SO₄) Mechanism: Step 1 (Nucleophilic addition of first ethanol): The carbonyl oxygen is protonated by H⁺: CH₃—C(=O)H + H⁺ → CH₃—C(⁺)H—OH. This makes the carbonyl carbon more electrophilic. Ethanol (nucleophile) attacks: CH₃—C(⁺)H—OH + OC₂H₅H → CH₃—CH(OH)—OC₂H₅ + H⁺. This is a hemiacetal intermediate. Step 2 (Dehydration): The –OH group of hemiacetal is protonated: CH₃—CH(OH)—OC₂H₅ + H⁺ → CH₃—C(⁺)(OC₂H₅)—OH. Water leaves, forming a carbocation: CH₃—C(⁺)(OC₂H₅) + H₂O. Step 3 (Nucleophilic addition of second ethanol): Another ethanol molecule attacks the carbocation: CH₃—C(⁺)(OC₂H₅) + OC₂H₅H → CH₃—CH(OC₂H₅)₂ + H⁺. Product is the acetal. Why reversible: Acetals form and hydrolyze reversibly in acidic conditions. The presence of water and acid allows the reverse reaction: acetals are hydrolyzed back to aldehyde and alcohol. Excess alcohol shifts equilibrium right (acetal formation); excess water shifts it left (acetal hydrolysis). In neutral or slightly basic conditions, acetals are stable. **Q2.** Compare the preparation of formaldehyde (HCHO) and acetaldehyde (CH₃CHO) from their respective parent alcohols. Write equations and explain why aldehydes must be removed immediately in some preparations. **Answer:** Formaldehyde preparation: CH₃OH + [O] → HCHO + H₂O (using copper catalyst at 300°C or K₂Cr₂O₇). Methanol is oxidized directly to formaldehyde. Acetaldehyde preparation: C₂H₅OH + [O] → CH₃CHO + H₂O (using PCC, or dilute acidified K₂Cr₂O₇ at controlled temp). Key difference: Formaldehyde is produced industrially via catalytic oxidation of methanol (copper or silver catalyst), a single-step process. Acetaldehyde is usually prepared from ethanol via oxidation but is less stable and over-oxidizes easily to acetic acid. Why aldehydes must be removed immediately: Aldehydes are easily oxidized to carboxylic acids. If the reaction mixture remains under oxidizing conditions, the product aldehyde will continue to be oxidized: RCHO + [O] → RCOOH. To prevent this, aldehydes are distilled off immediately as they form (in situ removal). For acetaldehyde, using a mild oxidizing agent (like PCC) and carefully controlling temperature helps minimize over-oxidation. **Q3.** Write the equations and explain the mechanism of the reaction between acetic acid (CH₃COOH) and ethanol in the presence of concentrated H₂SO₄ (esterification). Why is concentrated H₂SO₄ used instead of dilute? **Answer:** Reaction: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O (concentrated H₂SO₄ acts as catalyst and dehydrating agent) Mechanism: Step 1 (Protonation of carboxylic acid): The carboxyl oxygen is protonated by H⁺ from H₂SO₄: CH₃—COOH + H⁺ → CH₃—C(=OH)⁺—OH. This activates the carbonyl carbon for nucleophilic attack. Step 2 (Nucleophilic attack by alcohol): Ethanol (weak nucleophile) attacks the electrophilic carbonyl carbon: CH₃—C(=OH)⁺—OH + OC₂H₅H → CH₃—C(OH)(OC₂H₅)—OH⁺. An intermediate forms. Step 3 (Water elimination): Water is eliminated from the intermediate: CH₃—C(OH)(OC₂H₅)—OH⁺ → CH₃—C(=OC₂H₅)⁺ + H₂O. This is facilitated by H⁺. Step 4 (Deprotonation): Deprotonation gives the ester: CH₃—C(=OC₂H₅)⁺ + H₂O → CH₃COOC₂H₅ + H⁺. Why concentrated H₂SO₄ is essential: 1. It acts as a catalyst, providing H⁺ ions to protonate the carboxylic acid and activate the carbonyl. 2. It acts as a dehydrating agent, absorbing water produced during the reaction. By removing water, the equilibrium shifts right (Le Chatelier's principle), driving the reaction to completion and increasing yield. Dilute H₂SO₄ would not effectively remove water and the reaction would remain as an equilibrium with low ester formation. 3. Concentrated H₂SO₄ also maintains the reaction temperature, favoring esterification.

Higher-order thinking questions integrate multiple concepts and require logical reasoning. **Case-Study Question:** A student is given four organic compounds A, B, C, and D with the same molecular formula C₃H₆O. Compound A gives a positive Tollens' test and decolorizes bromine water. Compound B gives a negative Tollens' test but decolorizes bromine water. Compound C gives a negative Tollens' test and does not decolorize bromine water but is soluble in water and turns red litmus blue. Compound D gives no reaction with Tollens' or bromine water and is insoluble in water. Identify compounds A, B, C, and D. Explain your reasoning. Write the structure of A and state what product is formed when A is oxidized. **Step-by-Step Solution:** Step 1 (Analyze molecular formula): C₃H₆O has degree of unsaturation = 1. This means one double bond (C=O or C=C) or a ring. With a single functional group, the four isomers must be: • Propanal (aldehyde) • Propanone/acetone (ketone) • Allyl alcohol (alcohol with C=C) • Cyclopropanol (alcohol with ring) Step 2 (Interpret Tollens' test): Positive Tollens' test → aldehyde. Negative Tollens' test → not an aldehyde (ketone, alcohol, ether, etc.). Compound A: Positive Tollens' + decolorizes Br₂ → This must be propanal (CH₃CH₂CHO), which has an aldehyde (Tollens' positive) AND the statement "decolorizes bromine water" likely refers to its ability to undergo nucleophilic addition reactions (though strictly, only C=C decolorizes Br₂; here, C=O carbonyl can react with Br₂ slowly). Most reliably, A = propanal. Compound B: Negative Tollens' + decolorizes Br₂ → Decolorizes Br₂ → must have C=C double bond. B = allyl alcohol (CH₂=CH—CH₂OH). This is an unsaturated alcohol. The alkene decolorizes Br₂; it does not give Tollens' test. Compound C: Negative Tollens' + does not decolorize Br₂ + soluble in water + turns red litmus blue → Turns litmus blue indicates the compound is basic or an alcohol with acidic H attached to oxygen. Wait—this is problematic. C₃H₆O alcohols are neutral. Re-read: "turns red litmus blue." Only compounds with basic properties turn red litmus blue. But C₃H₆O alcohols are neutral or very weakly acidic. Re-interpretation: If the question means "turns blue litmus red," then C has acidic properties, implying a carboxylic acid. But C₃H₆O cannot be a carboxylic acid (RCOOH would be at least C₁H₂O₂). Let's assume the question meant the compound is soluble in water and the hydrogen on the –OH makes it weakly acidic. Then C = cyclopropanol (three-membered ring with OH). It's soluble (OH group), doesn't decolorize Br₂ (no C=C), negative Tollens'. Compound D: No Tollens', no Br₂ decolorization, insoluble in water → This is propanone (CH₃COCH₃), a ketone. It has no aldehyde (no Tollens'), no C=C (no Br₂), and is less soluble in water than alcohols. Step 3 (Identify A): **A = Propanal (CH₃CH₂CHO)** Structure of A: CH₃—CH₂—CHO (or CH₃CH₂CHO) Step 4 (Oxidation of A): CH₃CH₂CHO + [O] → CH₃CH₂COOH (propanoic acid). Using mild oxidation (K₂Cr₂O₇), the aldehyde is oxidized to a carboxylic acid. **Summary:** A = Propanal (CH₃CH₂CHO) — aldehyde B = Allyl alcohol (CH₂=CHCH₂OH) — unsaturated alcohol C = Cyclopropanol (three-membered ring with OH) — cyclic alcohol D = Propanone (CH₃COCH₃) — ketone When propanal is oxidized: CH₃CH₂CHO + [O] → CH₃CH₂COOH (propanoic acid, a carboxylic acid).

CBSETUTOR.ai's adaptive AI tutor is specifically designed to help Class 9 students master Chapter 8 (Aldehydes, Ketones and Carboxylic Acids) through daily, intelligent drilling of board-level questions. **Personalized Question Generation:** The AI generates unlimited variations of nomenclature questions (e.g., "Name C₄H₈O," "Name 2-methylpropanal"), preparation equations, and mechanism problems tailored to your current skill level. If you make errors in functional group priority, the AI detects this and generates 5–10 targeted follow-up questions on IUPAC rules. **Real-Time Error Diagnosis:** When you solve a question, the AI provides instant feedback: "Your chemical equation is unbalanced. Aldehydes need exactly one H on the carbonyl carbon—not two." It explains the error conceptually, not just marking you wrong. **Spaced Repetition with Board Patterns:** The AI tracks which question types you struggle with (e.g., mechanism questions, preparation by oxidation) and resurfaces them on a scientifically-proven spaced schedule. You'll drill an acetal formation mechanism on Day 1, Day 3, Day 7, and Day 14 until it's automatic. **Step-by-Step Solution Walkthrough:** For complex 5-mark mechanism questions, the AI breaks down each step with electron-movement diagrams, formal charges, and intermediate structures. You can slow-play the mechanism or speed-play it as you gain confidence. **Mock Test Simulation:** Weekly mini-tests simulate actual board exam format: 5 MCQs (1 mark each), 3 short-answer (2 marks), 2 medium (3 marks), 1 long-answer (5 marks). The AI scores you, identifies weak topics, and creates a custom revision plan. **Concept Linking:** The AI connects Chapter 8 to Chapter 7 (Hydrocarbons) and Chapter 10 (Metals & Non-metals) so you see how aldehydes arise from alkenes and why oxidation reactions matter across chemistry. **Start a 3-day free trial at cbsetutor.ai** and experience adaptive drilling on Chapter 8 questions right now—no credit card required.