JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 07

Multiple Choice Questions (MCQs) for JEE Chemistry: Aldehydes Ketones and Carboxylic Acids

Review structured MCQ sets for JEE Chemistry Aldehydes Ketones and Carboxylic Acids. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Practice Aldehydes Ketones and Carboxylic Acids MCQs for JEE Chemistry

Access the complete set of multiple-choice questions for Aldehydes Ketones and Carboxylic Acids below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: Which of the following on treatment with 50% aqueous NaOH yields the corresponding alcohol and acid?
(a) \( \ce{C6H5CHO} \)
(b) \( \ce{CH3CH2CH2CHO} \)
(c) \( \ce{CH3COCH3} \)
(d) \( \ce{CH3CHO} \)
Show Answer & Explanation

Answer: (a) \( \ce{C6H5CHO} \)

Explanation:
1. Only benzaldehyde has no alpha hydrogen and undergoes the Cannizzaro reaction.
2. Others have alpha-H and give aldol reactions.

Teacher's Note:
1. Cannizzaro: both an alcohol and an acid salt form.

Question 2: Ethanol heated with acetic acid in the presence of a few drops of conc. sulphuric acid gives the smell of:
(a) oil of wintergreen
(b) oil of mustard
(c) an ester (fruity smell)
(d) oil of bitter almonds
Show Answer & Explanation

Answer: (c) an ester (fruity smell)

Explanation:
1. Fischer esterification gives ethyl acetate, which has a fruity smell.

Teacher's Note:
1. Esters are responsible for fruit flavours.

Question 3: A carbonyl compound gives a pink colour with Schiff's reagent and a yellow precipitate when boiled with iodine and caustic alkali. It also gives a red precipitate with Fehling's solution. It is likely to be:
(a) formaldehyde
(b) propionaldehyde
(c) acetaldehyde
(d) crotonaldehyde
Show Answer & Explanation

Answer: (c) acetaldehyde

Explanation:
1. Schiff's and Fehling's tests show an aldehyde, and iodoform shows a \( \ce{CH3CO-} \) unit.
2. Acetaldehyde fulfils all three.

Teacher's Note:
1. Acetaldehyde is the only aldehyde that gives iodoform.

Question 4: A carbonyl compound gives a positive iodoform test but does not reduce Tollens' reagent or Fehling's solution. It forms a cyanohydrin whose hydrolysis gives a hydroxy acid with a methyl side chain. The compound is:
(a) acetaldehyde
(b) propionaldehyde
(c) acetone
(d) crotonaldehyde
Show Answer & Explanation

Answer: (c) acetone

Explanation:
1. No reduction of Tollens' shows it is a ketone, and iodoform shows a methyl ketone.
2. Acetone cyanohydrin hydrolyses to 2-hydroxy-2-methylpropanoic acid, which has methyl side chains.

Teacher's Note:
1. Methyl ketone plus no silver mirror points to acetone.

Question 5: Which of the following statements is wrong?
(a) all methyl ketones give a positive iodoform test
(b) acetaldehyde is the only aldehyde that gives the iodoform test
(c) all secondary alcohols give a positive iodoform test
(d) any alcohol that can be oxidised to an acetyl group gives a positive iodoform test
Show Answer & Explanation

Answer: (c) all secondary alcohols give a positive iodoform test

Explanation:
1. Only secondary alcohols of the type \( \ce{CH3CH(OH)R} \) are positive.
2. Others, such as pentan-3-ol, are negative.

Teacher's Note:
1. The structural requirement is the \( \ce{CH3CH(OH)-} \) group.

Question 6: In the sequence acetone \( \ce{->[I2/alkali] A ->[Ag, heat] B ->[dil. H2SO4, Hg2+] C} \), A, B and C are:
(a) \( \ce{CHI3} \), \( \ce{H2C=CH2} \), \( \ce{CH3CH2OH} \)
(b) \( \ce{CHI3} \), \( \ce{HC#CH} \), \( \ce{CH3CHO} \)
(c) \( \ce{CHI3} \), \( \ce{CH3C#CH} \), \( \ce{CH3COCH3} \)
(d) \( \ce{CHI3} \), \( \ce{CH3CH(OH)CH3} \), \( \ce{CH3COCH3} \)
Show Answer & Explanation

Answer: (b) \( \ce{CHI3} \), \( \ce{HC#CH} \), \( \ce{CH3CHO} \)

Explanation:
1. Acetone with iodine and alkali gives iodoform (A).
2. Silver removes iodine from iodoform to give ethyne (B).
3. Hydration of ethyne with Hg2+ gives acetaldehyde (C).

Teacher's Note:
1. Iodoform plus silver gives acetylene.

Question 7: In the sequence \( \ce{RCOCl + H2 ->[Pd-BaSO4] A ->[HCN] B ->[H3O+] C} \), A, B and C are:
(a) \( \ce{RCHO} \), \( \ce{RCH(OH)CN} \), \( \ce{RCH(OH)CH2NH2} \)
(b) \( \ce{RCHO} \), \( \ce{RCH(OH)CN} \), \( \ce{RCH(OH)COOH} \)
(c) \( \ce{RCHO} \), \( \ce{RCOCH2CN} \), \( \ce{RCOCH2COOH} \)
(d) \( \ce{RCHO} \), \( \ce{RCH2CN} \), \( \ce{RCH2COOH} \)
Show Answer & Explanation

Answer: (b) \( \ce{RCHO} \), \( \ce{RCH(OH)CN} \), \( \ce{RCH(OH)COOH} \)

Explanation:
1. Rosenmund reduction gives the aldehyde.
2. HCN adds to give the cyanohydrin, and hydrolysis converts the -CN group into -COOH, giving the alpha-hydroxy acid.

Teacher's Note:
1. Cyanohydrin hydrolysis gives alpha-hydroxy acids.

Question 8: In the sequence \( \ce{CH3C#CH ->[dil. H2SO4, Hg2+] A ->[CHCl3] B} \) and A with \( \ce{I2/NaOH} \) gives C. A, B and C are:
(a) \( \ce{CH3CH2CHO} \), \( \ce{CH3CH2CH2Cl} \), \( \ce{CHI3} \)
(b) \( \ce{CH3COCH3} \), \( \ce{CCl3-CO-CH3} \), \( \ce{CHI3} \)
(c) \( \ce{CH3COCH3} \), \( \ce{CCl3-C(OH)(CH3)2} \), \( \ce{CHI3} \)
(d) \( \ce{CH3CH2CHO} \), \( \ce{CCl3-CH(OH)-CH2CH3} \), \( \ce{CHI3} \)
Show Answer & Explanation

Answer: (c) \( \ce{CH3COCH3} \), \( \ce{CCl3-C(OH)(CH3)2} \), \( \ce{CHI3} \)

Explanation:
1. Markovnikov hydration of propyne gives acetone (A).
2. Acetone adds chloroform in base to give chloretone, \( \ce{CCl3-C(OH)(CH3)2} \) (B).
3. With iodine and alkali it gives iodoform (C).

Teacher's Note:
1. Chloroform adds across C=O of acetone.

Question 9: Match: (i) benzophenone to diphenylmethane, (ii) benzaldehyde to 1-phenylethanol, (iii) cyclohexanone to cyclohexanol, (iv) phenyl benzoate to benzaldehyde with reagents (a) \( \ce{LiAlH4} \), (b) DIBAL-H, (c) Zn(Hg)/conc. HCl, (d) \( \ce{CH3MgBr} \).
(a) (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b)
(b) (i)-(c), (ii)-(d), (iii)-(b), (iv)-(a)
(c) (i)-(d), (ii)-(c), (iii)-(a), (iv)-(b)
(d) (i)-(d), (ii)-(a), (iii)-(c), (iv)-(b)
Show Answer & Explanation

Answer: (a) (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b)

Explanation:
1. Clemmensen reduction (\( \ce{Zn(Hg)/HCl} \)) turns C=O into CH2.
2. A Grignard reagent adds a methyl group to the aldehyde.
3. LiAlH4 reduces the ketone to the alcohol.
4. DIBAL-H partially reduces the ester to the aldehyde.

Teacher's Note:
1. Match the transformation to the reagent type.

Question 10: Methylamine reacts with acetyl chloride and forms:
(a) \( \ce{CH3NH2} \)
(b) \( \ce{CH3NHNa} \)
(c) \( \ce{CH3NHCOCH3} \)
(d) \( \ce{(CH3)2NCOCH3} \)
Show Answer & Explanation

Answer: (c) \( \ce{CH3NHCOCH3} \)

Explanation:
1. The amine nitrogen attacks the acyl chloride to form N-methylacetamide.

Teacher's Note:
1. This is N-acylation.

Question 11: The intermediates formed during the reaction of \( \ce{RCONH2} \) with \( \ce{Br2} \) and KOH are:
(a) \( \ce{RCONHBr} \) and \( \ce{RNCO} \)
(b) \( \ce{RNHCOBr} \) and \( \ce{RNCO} \)
(c) \( \ce{RNH-Br} \) and \( \ce{RCONHBr} \)
(d) \( \ce{RCONBr2} \)
Show Answer & Explanation

Answer: (a) \( \ce{RCONHBr} \) and \( \ce{RNCO} \)

Explanation:
1. The amide is N-brominated to give RCONHBr, and then the rearrangement gives an isocyanate, RNCO.
2. Hydrolysis gives the amine.

Teacher's Note:
1. Isocyanate formation is the key step of Hofmann rearrangement.

Question 12: \( \ce{CH3CH2Cl ->[NaCN] X ->[Ni/H2] Y ->[acetic anhydride] Z} \). Z is:
(a) \( \ce{CH3CH2CH2NHCOCH3} \)
(b) \( \ce{CH3CH2CH2NH2} \)
(c) \( \ce{CH3CH2CH2CONHCH3} \)
(d) \( \ce{CH3CH2CH2CONHCOCH3} \)
Show Answer & Explanation

Answer: (a) \( \ce{CH3CH2CH2NHCOCH3} \)

Explanation:
1. NaCN gives propanenitrile (X), hydrogenation gives propylamine (Y).
2. Acetylation of the amine gives N-propylacetamide (Z).

Teacher's Note:
1. The nitrile adds one carbon.

Question 13: Butanenitrile is hydrolysed with \( \ce{H3O+} \) to give A, which reacts with \( \ce{NH3} \) on heating to give B. B is:
(a) butanoic acid
(b) butanal
(c) butyl isocyanide
(d) butanamide
Show Answer & Explanation

Answer: (d) butanamide

Explanation:
1. Acid hydrolysis of the nitrile gives butanoic acid (A).
2. The ammonium salt on heating loses water to give butanamide (B).

Teacher's Note:
1. Amide forms by dehydration of the ammonium salt.

Question 14: Sodium ethoxide reacted with ethanoyl chloride. The compound produced is:
(a) ethyl chloride
(b) ethyl ethanoate
(c) diethyl ether
(d) 2-butanone
Show Answer & Explanation

Answer: (b) ethyl ethanoate

Explanation:
1. The ethoxide ion replaces chloride at the acyl carbon to give the ester ethyl ethanoate.

Teacher's Note:
1. This is nucleophilic acyl substitution.

Question 15: Trichloroacetaldehyde was subjected to Cannizzaro's reaction using NaOH. The mixture of products contains sodium trichloroacetate and another compound. The other compound is:
(a) 2,2,2-trichloropropanol
(b) chloroform
(c) 2,2,2-trichloroethanol
(d) trichloromethanol
Show Answer & Explanation

Answer: (c) 2,2,2-trichloroethanol

Explanation:
1. Chloral has no alpha-hydrogen, so Cannizzaro disproportionation occurs.
2. One molecule is oxidised to the acid salt and the other is reduced to \( \ce{CCl3CH2OH} \).

Teacher's Note:
1. Reduction product of an aldehyde is the primary alcohol.

Question 16: Acids do not give the characteristic reactions of the C=O group because of:
(a) dimerisation
(b) resonance
(c) cyclic structures
(d) the attached alkyl radical
Show Answer & Explanation

Answer: (b) resonance

Explanation:
1. In a carboxylic acid, the lone pair on the -OH oxygen is delocalised into the C=O group.
2. So the carbonyl carbon is much less electrophilic than in aldehydes and ketones.

Teacher's Note:
1. Resonance reduces the positive charge on carbon.

Question 17: When 23 g of sodium reacts with methyl alcohol, the gas liberated is:
(a) half a mole of hydrogen
(b) one mole of hydrogen
(c) one mole of oxygen
(d) none
Show Answer & Explanation

Answer: (a) half a mole of hydrogen

Explanation:
1. 23 g of sodium is one mole.
2. \( \ce{2CH3OH + 2Na -> 2CH3ONa + H2} \), so one mole of Na gives half a mole of \( \ce{H2} \).

Teacher's Note:
1. Use the balanced equation.

Question 18: Which of the following compounds are capable of reacting with Tollens' reagent?
(a) formaldehyde
(b) formic acid
(c) acetaldehyde
(d) all of the above
Show Answer & Explanation

Answer: (d) all of the above

Explanation:
1. Formaldehyde and acetaldehyde are aldehydes.
2. Formic acid also has an aldehydic -CHO group and is oxidised.

Teacher's Note:
1. Formic acid is a reducing acid.

Question 19: Fehling's solution is:
(a) a solution of magenta dye bleached by \( \ce{SO2} \)
(b) an ammoniacal solution of \( \ce{AgNO3} \)
(c) a mixture of \( \ce{CuSO4} \) solution and a solution of NaOH and sodium potassium tartrate
(d) an alcoholic solution of 2,4-dinitrophenylhydrazine
Show Answer & Explanation

Answer: (c) a mixture of \( \ce{CuSO4} \) solution and a solution of NaOH and sodium potassium tartrate

Explanation:
1. Fehling's A is aqueous CuSO4 and Fehling's B is alkaline sodium potassium tartrate; they are mixed before use.

Teacher's Note:
1. Tartrate keeps Cu(II) in solution as a complex.

Question 20: Alkanoic acids can be made by hydrolysis of:
(a) trihaloalkanes
(b) 1,1,1-trihaloalkanes
(c) Grignard reagents
(d) ketones
Show Answer & Explanation

Answer: (b) 1,1,1-trihaloalkanes

Explanation:
1. Hydrolysis of RCX3 forms an unstable triol that loses water to give RCOOH.

Teacher's Note:
1. Three halogens on one carbon become an acid group.

Question 21: Acids have much higher boiling points than isomeric esters because:
(a) acids form dimers by H-bonding
(b) acids are volatile in steam
(c) esters are non-volatile
(d) acids can ionise to give protons in aqueous solution
Show Answer & Explanation

Answer: (a) acids form dimers by H-bonding

Explanation:
1. Carboxylic acids form hydrogen-bonded dimers while esters have no O-H group.

Teacher's Note:
1. H-bonding raises the boiling point.

Question 22: The reaction of ethyl acetate with sodium ethoxide to give acetoacetic ester is known as:
(a) Claisen condensation
(b) Clemmensen reduction
(c) Cannizzaro reaction
(d) Etard reaction
Show Answer & Explanation

Answer: (a) Claisen condensation

Explanation:
1. The ester enolate attacks another ester molecule and loses ethoxide to give a beta-keto ester.
2. This is Claisen condensation.

Teacher's Note:
1. Claisen condensation: ester + ester gives beta-keto ester.

Question 23:
Assertion (A): Acetamide on reaction with \( \ce{Br2} \)/NaOH gives \( \ce{CH3NH2} \).
Reason (R): The carbon of the amide functional group becomes part of \( \ce{Na2CO3} \).
(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.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. In Hofmann degradation the carbonyl carbon is lost as carbonate.
2. This is why the amine has one carbon less. R explains A.

Teacher's Note:
1. Carbonyl carbon leaves as \( \ce{CO3^2-} \).

Question 24:
Assertion (A): Ethyl acetoacetate forms butane-1,4-diol with \( \ce{NaBH4} \).
Reason (R): \( \ce{NaBH4} \) can reduce ketone and ester groups.
(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.
Show Answer & Explanation

Answer: (d) A is false, but R is true.

Explanation:
1. NaBH4 reduces only the ketone group, giving ethyl 3-hydroxybutanoate.
2. It does not reduce esters, so both statements are false.

Teacher's Note:
1. LiAlH4 is needed to reduce esters.

Free study material for Chemistry

Practice MCQs for JEE Chemistry Aldehydes Ketones and Carboxylic Acids

About Aldehydes Ketones and Carboxylic Acids MCQs for JEE Chemistry

Review structured objective questions for JEE Chemistry Aldehydes Ketones and Carboxylic Acids. Built according to official JEE guidelines, these MCQ sets support daily revision and core concept reinforcement.

How to Verify Your MCQ Answers

Each question includes structured solution keys mapped directly to standard JEE textbooks, helping students evaluate their reasoning and correct mistakes early in their revision.

Enhance Speed with Online MCQ Tests

Follow up your worksheet practice by attempting the interactive online Chemistry MCQ test for this chapter to evaluate your execution speed. All platform resources are free to access.

FAQs

Where can I access latest JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 07?

You can get most exhaustive JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 07 for free on StudiesToday.com. These MCQs for JEE Chemistry are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Chemistry JEE material?

Yes, our JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 07 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Chemistry MCQs help in scoring full marks in JEE exams?

By solving our JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 07, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Chemistry.

Do you provide answers and explanations for JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 07?

Yes, Chemistry MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Chemistry JEE MCQs online?

Yes, you can also access online interactive tests for JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 07 on StudiesToday.com as they provide instant answers and score to help you track your progress in Chemistry.