Aldehydes and Ketones JEE Main previous year questions with solutions

5 solved JEE Main questions on Aldehydes and Ketones, free to read — no sign-in needed. The full chapter has 169 questions; sign in to attempt the remaining 164 in the exam simulator.

  1. Q1JEE Main 2026 (06 Apr, Shift 2)Chemical reactions for aldehydes and ketones
    'xx' is the product which is obtained by the hydrolysis of prop-1-yne in the presence of mercuric sulphate under dilute acidic medium at 333333 K. 'yy' is the product which is obtained by the reaction of ethane nitrile with methyl magnesium bromide in dry ether followed by hydrolysis. IUPAC name of product obtained from 'xx' and 'yy' in the presence of barium hydroxide followed by heating is :
    1. A.2-Methylpent-4-en-3-one
    2. B.4-Methylpent-3-en-2-one
    3. C.4-Methylpent-1-ene
    4. D.2-Methylpent-3-one
    Show answer & solution

    Answer: (B)

    Hydrolysis of prop-1-yne in the presence of HgSO4HgSO_4 and dilute H2SO4H_2SO_4 yields acetone. CH3CCH+H2OHg2+,H+CH3COCH3CH_3-C \equiv CH + H_2O \xrightarrow{Hg^{2+}, H^+} CH_3-CO-CH_3 Thus, xx is acetone. Reaction of ethane nitrile (CH3CNCH_3CN) with methyl magnesium bromide (CH3MgBrCH_3MgBr) followed by hydrolysis also yields acetone. CH3CN+CH3MgBrCH3C(CH3)=NMgBrH3O+CH3COCH3CH_3-C \equiv N + CH_3MgBr \rightarrow CH_3-C(CH_3)=NMgBr \xrightarrow{H_3O^+} CH_3-CO-CH_3 Thus, yy is acetone. Reaction of xx and yy (both acetone) in the presence of Ba(OH)2Ba(OH)_2 followed by heating results in an aldol condensation. 2CH3COCH3Ba(OH)2CH3C(OH)(CH3)CH2COCH32 CH_3-CO-CH_3 \xrightarrow{Ba(OH)_2} CH_3-C(OH)(CH_3)-CH_2-CO-CH_3 Heating causes dehydration to form an unsaturated ketone. CH3C(OH)(CH3)CH2COCH3ΔCH3C(CH3)=CHCOCH3CH_3-C(OH)(CH_3)-CH_2-CO-CH_3 \xrightarrow{\Delta} CH_3-C(CH_3)=CH-CO-CH_3 Numbering the principal carbon chain to give the lowest locant to the ketone group, the IUPAC name of the product is 4-methylpent-3-en-2-one. Answer: 4-Methylpent-3-en-2-one
  2. Q2JEE Main 2025 (03 Apr, Shift 1)Tests for aldehyde and ketones
    Number of molecules from below which cannot give ioddoform reaction is : Ethanol, Isopropyl alcohol, Bromoacetone, 2-Butanol, 2-Butanone, Butanal, 2-Pentanone, 3-Pentanone, Pentanal and 3-Pentanol
    1. A.5
    2. B.4
    3. C.3
    4. D.2
    Show answer & solution

    Answer: (B)

    Following will not give iodoform reaction/test. (1) Butanal (2) 2-Pentanone (3) Pentanal (4) 3-Pentanol
  3. Q3JEE Main 2023 (25 Jan, Shift 1)Method of preparation for both aldehydes and ketones
    Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R: Assertion A: Acetal/Ketal is stable in basic medium. Reason R: The high leaving tendency of alkoxide ion gives the stability to acetal/ketal in basic medium. In the light of the above statements, choose the correct answer from the options given below:
    1. A.A is true but R is false
    2. B.A is false but R is true
    3. C.Both A and R are true and R is the correct explanation of A
    4. D.Both A and R are true but R is NOT the correct explanation of A
    Show answer & solution

    Answer: (A)

    For Assertion: The acetals are unstable in the presence of aqueous solutions of acid because they undergo hydrolysis in the presence of acidic conditions to produce the parent aldehyde or ketone or alcohols. But are stable and do not undergo hydrolysis in the presence of a base or under neutral conditions. Hence assertion is correct. For reason: Alkoxide ion (RO)({RO}^{–}) is not considered a good leaving group Hence reason must be false. So, the correct option is A
  4. Q4JEE Main 2019 (09 Jan, Shift 1)Method of preparation for Aldehydes
    The highest value of the calculated spin only magnetic moment (in BMBM) among all the transition metal complexes is
    1. A.4.90BM4.90BM
    2. B.3.87BM3.87BM
    3. C.6.93BM6.93BM
    4. D.5.92BM5.92BM
    Show answer & solution

    Answer: (D)

    The maximum number of unpaired electrons for transition metals =5=5 or n=5n=5 Spin only magnetic moment =n(n+2)BM=\sqrt{n\left(n+2\right)}BM =5(5+2)=35=\sqrt{5\left(5+2\right)}=\sqrt{35} =5.92BM=5.92BM
  5. Q5JEE Main 2012 (26 May)Reactions for aldehydes
    Tollen's reagent and Fehling solutions are used to distinguish between
    1. A.acids and alcohols
    2. B.alkanes and alcohols
    3. C.ketones and aldehydes
    4. D.n-alkaens and branched alkanes
    Show answer & solution

    Answer: (C)

    All aldehydes and ketones show reaction with Tollen's reagent and Fehling solutions, but ketones do not show this reaction. Note :- Benzaldehyde do not give reaction with Fehling solution.

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Aldehydes and Ketones in JEE Main: previous year question analysis

Aldehydes and Ketones has appeared 169 times in JEE Main between 2002 and 2026, making it the 18th most-asked of 33 chapters and about 2.8% of the bank. Over the last 5 years it has averaged 17.4 questions per year.

Total PYQs
169
Years covered
2002–2026
Weightage rank
#18 of 33
Share of bank
2.8%

How many Aldehydes and Ketones questions appeared each year

Aldehydes and Ketones JEE Main question count by year
YearQuestionsRelative volume
20145
20153
20173
20184
201918
202012
202125
202216
202320
202420
202517
202614

Which Aldehydes and Ketones sub-topics are asked most

Every question in this chapter is tagged to a sub-topic, so you can see exactly where the marks sit before you revise.

  • Chemical reactions for aldehydes and ketones91 questions
  • Tests for aldehyde and ketones30 questions
  • Method of preparation for both aldehydes and ketones25 questions
  • Method of preparation for Aldehydes18 questions
  • Reactions for aldehydes4 questions
  • Method of preparation for Ketones1 questions

Question formats used in Aldehydes and Ketones

  • Single-correct MCQ150
  • Numerical / integer answer19

How Aldehydes and Ketones compares with nearby chapters

Counts are computed from AcadXL’s own JEE Main question bank, tagged chapter- and sub-topic-wise and checked against official answer keys. Sign in to attempt the 169 Aldehydes and Ketones questions with solutions.