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 180 questions; sign in to attempt the remaining 175 in the exam simulator.

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1. Q1JEE Main 2026 (06 Apr, Shift 2)Chemical reactions for aldehydes and ketones
   '$x$' is the product which is obtained by the hydrolysis of prop-1-yne in the presence of mercuric sulphate under dilute acidic medium at $333$ K. '$y$' 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 '$x$' and '$y$' 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
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   Answer: (B)

   Hydrolysis of prop-1-yne in the presence of $HgSO_4$ and dilute $H_2SO_4$ yields acetone. $CH_3-C \equiv CH + H_2O \xrightarrow{Hg^{2+}, H^+} CH_3-CO-CH_3$ Thus, $x$ is acetone. Reaction of ethane nitrile ($CH_3CN$) with methyl magnesium bromide ($CH_3MgBr$) followed by hydrolysis also yields acetone. $CH_3-C \equiv N + CH_3MgBr \rightarrow CH_3-C(CH_3)=NMgBr \xrightarrow{H_3O^+} CH_3-CO-CH_3$ Thus, $y$ is acetone. Reaction of $x$ and $y$ (both acetone) in the presence of $Ba(OH)_2$ followed by heating results in an aldol condensation. $2 CH_3-CO-CH_3 \xrightarrow{Ba(OH)_2} CH_3-C(OH)(CH_3)-CH_2-CO-CH_3$ Heating causes dehydration to form an unsaturated ketone. $CH_3-C(OH)(CH_3)-CH_2-CO-CH_3 \xrightarrow{\Delta} CH_3-C(CH_3)=CH-CO-CH_3$ 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
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   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
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   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 $({RO}^{–})$ is not considered a good leaving group Hence reason must be false. So, the correct option is A
4. Q4JEE 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
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   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.
5. Q5JEE Main 2026 (02 Apr, Shift 2)Chemical reactions for aldehydes and ketones
   An organic compound "$x$" where molar ratio of C, O and H are equal, on treatment with $50\%$ KOH under reflux followed by acidification produced "$y$". The most likely structure of "$y$" is: [Molar mass of '$x$' is $58$ g mol$^{-1}$]
   1. A.$CH_2 = CH - \overset{O}{\underset{\|}{C}} - OH$
   2. B.$CH_3 - CH = CH - CH = O$
   3. C.$O = C - OH, \, CH_2 - OH$ (with C-C bond)
   4. D.$CH_3 - \overset{O}{\underset{\|}{C}} - OH$
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   Answer: (C)

   Given the molar ratio of C, O, and H in compound $x$ is equal, the empirical formula is $CHO$. Empirical mass $= 12 + 1 + 16 = 29$ g mol$^{-1}$. Since the molar mass of $x$ is $58$ g mol$^{-1}$, the molecular formula is $(CHO)_2$ or $C_2H_2O_2$. The structure of $x$ is glyoxal, $OHC-CHO$. Treatment of glyoxal with $50\%$ KOH under reflux results in an intramolecular Cannizzaro reaction. One aldehyde group is oxidized to a carboxylate ion, and the other is reduced to a primary alcohol. $OHC-CHO \xrightarrow{50\% \text{ KOH}} HOCH_2-COO^{-}K^{+}$ Subsequent acidification yields glycolic acid ($y$). $HOCH_2-COO^{-}K^{+} \xrightarrow{H^{+}} HOCH_2-COOH$ The structure of $y$ is $HO-CH_2-COOH$. Answer: $O = C - OH, \, CH_2 - OH$ (with C-C bond)