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JEE Main Previous Year Questions (2025): Aldehydes Ketones and Carboxylic Acids
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Page 1 A : 1 Q ' absorbs 2 moles of hydrogen and ' X ' upon oxidation wit h gives The total number of bonds present in the compound ' ' is __________. JEE Main 2025 (Online) 23rd January Evening Shift Ans 7 2 : Solution: K M n O 4 | H + s X ' ------------------------------------------------------------------------------------------------- I : 2 Q n the Claisen-Schmidt reaction to prepare, dibenzalacetone from 5.3 g of benzaldehyde, a total of 3.51 g of product was obtained. The percentage yield in this reaction was _______ %. JEE Main 2025 (Online) 29th January Evening Shift Ans: 60 Page 2 A : 1 Q ' absorbs 2 moles of hydrogen and ' X ' upon oxidation wit h gives The total number of bonds present in the compound ' ' is __________. JEE Main 2025 (Online) 23rd January Evening Shift Ans 7 2 : Solution: K M n O 4 | H + s X ' ------------------------------------------------------------------------------------------------- I : 2 Q n the Claisen-Schmidt reaction to prepare, dibenzalacetone from 5.3 g of benzaldehyde, a total of 3.51 g of product was obtained. The percentage yield in this reaction was _______ %. JEE Main 2025 (Online) 29th January Evening Shift Ans: 60 Solution: Mass of benzaldehyde = 5.3 g Mass of product = 3.51 g Claisen - Schmidt reaction : Condensation of aromatic aldehydes / ketones (without alpha-hydrogen) with diphatic aldehydes /ketones (with alpha-hydrogen) in the presence of weak base to form -unsaturated aldehydes/ketones. Given benzaldehyde dibenzalacetone Formula of percent yield Percent yield Actual yield = 3.51 g From the reaction, the stoichiometric ratio between benzaldehyde and dibenzal acetone is 2 : 1 benzaldehyde : dibenzalacetone Moles of benzaldehyde: = 0.04994 mol Molarmass of benzaldehyde = 106.12 g/mol For 1 mole of benzaldehyde, moles product (dibenzalacetone) is formed. So, for 0.04994 mol benzaldehyde, mol dibenzalacetone is formed. So, moles of dibenzalacetone = 0.02497 mol Mass of product dibenzalacetone (Theoretical yield) Mass = moles molarmass mol g/mol = 5.850 g Percent yield ------------------------------------------------------------------------------------------------- I : 3 Q dentify the structure of the final product (D) in the following sequence of the reactions : a , ß ? = a c t u a l y i e l d T h e o r e t i c a l y i e l d × 1 0 0 2 : 1 1 : 1 2 M o l e s = M a s s M o l a r m a s s = 5 . 3 g 1 0 6 . 1 2 g / m o l 1 2 1 2 × 0 . 0 4 9 9 4 × = 0 . 0 2 4 9 7 × 2 3 4 . 2 9 = 3 . 5 1 g 5 . 8 5 g × 1 0 0 = 0 . 6 × 1 0 0 = 6 0 % Total number of hybridised carbon atoms in product D is ____________. s p 2 Page 3 A : 1 Q ' absorbs 2 moles of hydrogen and ' X ' upon oxidation wit h gives The total number of bonds present in the compound ' ' is __________. JEE Main 2025 (Online) 23rd January Evening Shift Ans 7 2 : Solution: K M n O 4 | H + s X ' ------------------------------------------------------------------------------------------------- I : 2 Q n the Claisen-Schmidt reaction to prepare, dibenzalacetone from 5.3 g of benzaldehyde, a total of 3.51 g of product was obtained. The percentage yield in this reaction was _______ %. JEE Main 2025 (Online) 29th January Evening Shift Ans: 60 Solution: Mass of benzaldehyde = 5.3 g Mass of product = 3.51 g Claisen - Schmidt reaction : Condensation of aromatic aldehydes / ketones (without alpha-hydrogen) with diphatic aldehydes /ketones (with alpha-hydrogen) in the presence of weak base to form -unsaturated aldehydes/ketones. Given benzaldehyde dibenzalacetone Formula of percent yield Percent yield Actual yield = 3.51 g From the reaction, the stoichiometric ratio between benzaldehyde and dibenzal acetone is 2 : 1 benzaldehyde : dibenzalacetone Moles of benzaldehyde: = 0.04994 mol Molarmass of benzaldehyde = 106.12 g/mol For 1 mole of benzaldehyde, moles product (dibenzalacetone) is formed. So, for 0.04994 mol benzaldehyde, mol dibenzalacetone is formed. So, moles of dibenzalacetone = 0.02497 mol Mass of product dibenzalacetone (Theoretical yield) Mass = moles molarmass mol g/mol = 5.850 g Percent yield ------------------------------------------------------------------------------------------------- I : 3 Q dentify the structure of the final product (D) in the following sequence of the reactions : a , ß ? = a c t u a l y i e l d T h e o r e t i c a l y i e l d × 1 0 0 2 : 1 1 : 1 2 M o l e s = M a s s M o l a r m a s s = 5 . 3 g 1 0 6 . 1 2 g / m o l 1 2 1 2 × 0 . 0 4 9 9 4 × = 0 . 0 2 4 9 7 × 2 3 4 . 2 9 = 3 . 5 1 g 5 . 8 5 g × 1 0 0 = 0 . 6 × 1 0 0 = 6 0 % Total number of hybridised carbon atoms in product D is ____________. s p 2 JEE Main 2025 (Online) 7th April Evening Shift Ans: 7 Solution: Number of -atoms in product NTA Ans. ALLEN Ans. ------------------------------------------------------------------------------------------------- T : 4 Q he compounds which give positive Fehling's test are : Choose the correct answer from the options given below : ? s p 2 C D = 7 = 7 = 7 Page 4 A : 1 Q ' absorbs 2 moles of hydrogen and ' X ' upon oxidation wit h gives The total number of bonds present in the compound ' ' is __________. JEE Main 2025 (Online) 23rd January Evening Shift Ans 7 2 : Solution: K M n O 4 | H + s X ' ------------------------------------------------------------------------------------------------- I : 2 Q n the Claisen-Schmidt reaction to prepare, dibenzalacetone from 5.3 g of benzaldehyde, a total of 3.51 g of product was obtained. The percentage yield in this reaction was _______ %. JEE Main 2025 (Online) 29th January Evening Shift Ans: 60 Solution: Mass of benzaldehyde = 5.3 g Mass of product = 3.51 g Claisen - Schmidt reaction : Condensation of aromatic aldehydes / ketones (without alpha-hydrogen) with diphatic aldehydes /ketones (with alpha-hydrogen) in the presence of weak base to form -unsaturated aldehydes/ketones. Given benzaldehyde dibenzalacetone Formula of percent yield Percent yield Actual yield = 3.51 g From the reaction, the stoichiometric ratio between benzaldehyde and dibenzal acetone is 2 : 1 benzaldehyde : dibenzalacetone Moles of benzaldehyde: = 0.04994 mol Molarmass of benzaldehyde = 106.12 g/mol For 1 mole of benzaldehyde, moles product (dibenzalacetone) is formed. So, for 0.04994 mol benzaldehyde, mol dibenzalacetone is formed. So, moles of dibenzalacetone = 0.02497 mol Mass of product dibenzalacetone (Theoretical yield) Mass = moles molarmass mol g/mol = 5.850 g Percent yield ------------------------------------------------------------------------------------------------- I : 3 Q dentify the structure of the final product (D) in the following sequence of the reactions : a , ß ? = a c t u a l y i e l d T h e o r e t i c a l y i e l d × 1 0 0 2 : 1 1 : 1 2 M o l e s = M a s s M o l a r m a s s = 5 . 3 g 1 0 6 . 1 2 g / m o l 1 2 1 2 × 0 . 0 4 9 9 4 × = 0 . 0 2 4 9 7 × 2 3 4 . 2 9 = 3 . 5 1 g 5 . 8 5 g × 1 0 0 = 0 . 6 × 1 0 0 = 6 0 % Total number of hybridised carbon atoms in product D is ____________. s p 2 JEE Main 2025 (Online) 7th April Evening Shift Ans: 7 Solution: Number of -atoms in product NTA Ans. ALLEN Ans. ------------------------------------------------------------------------------------------------- T : 4 Q he compounds which give positive Fehling's test are : Choose the correct answer from the options given below : ? s p 2 C D = 7 = 7 = 7 JEE Main 2025 (Online) 22nd January Morning Shift Options: A. (A), (C) and (D) Only B. (A), (B) and (C) Only C. (A), (D) and (E) Only D. (C), (D) and (E) Only Answer: D Solution: Compound (C), (D) and (E) will give positive Fehling’s test. ------------------------------------------------------------------------------------------------- : 5 Q Residue (A) + HCl (dil) Compound (B) Structure of residue (A) and compound (B) formed respectively is : JEE Main 2025 (Online) 22nd January Evening Shift Options: A. B. C. ? Page 5 A : 1 Q ' absorbs 2 moles of hydrogen and ' X ' upon oxidation wit h gives The total number of bonds present in the compound ' ' is __________. JEE Main 2025 (Online) 23rd January Evening Shift Ans 7 2 : Solution: K M n O 4 | H + s X ' ------------------------------------------------------------------------------------------------- I : 2 Q n the Claisen-Schmidt reaction to prepare, dibenzalacetone from 5.3 g of benzaldehyde, a total of 3.51 g of product was obtained. The percentage yield in this reaction was _______ %. JEE Main 2025 (Online) 29th January Evening Shift Ans: 60 Solution: Mass of benzaldehyde = 5.3 g Mass of product = 3.51 g Claisen - Schmidt reaction : Condensation of aromatic aldehydes / ketones (without alpha-hydrogen) with diphatic aldehydes /ketones (with alpha-hydrogen) in the presence of weak base to form -unsaturated aldehydes/ketones. Given benzaldehyde dibenzalacetone Formula of percent yield Percent yield Actual yield = 3.51 g From the reaction, the stoichiometric ratio between benzaldehyde and dibenzal acetone is 2 : 1 benzaldehyde : dibenzalacetone Moles of benzaldehyde: = 0.04994 mol Molarmass of benzaldehyde = 106.12 g/mol For 1 mole of benzaldehyde, moles product (dibenzalacetone) is formed. So, for 0.04994 mol benzaldehyde, mol dibenzalacetone is formed. So, moles of dibenzalacetone = 0.02497 mol Mass of product dibenzalacetone (Theoretical yield) Mass = moles molarmass mol g/mol = 5.850 g Percent yield ------------------------------------------------------------------------------------------------- I : 3 Q dentify the structure of the final product (D) in the following sequence of the reactions : a , ß ? = a c t u a l y i e l d T h e o r e t i c a l y i e l d × 1 0 0 2 : 1 1 : 1 2 M o l e s = M a s s M o l a r m a s s = 5 . 3 g 1 0 6 . 1 2 g / m o l 1 2 1 2 × 0 . 0 4 9 9 4 × = 0 . 0 2 4 9 7 × 2 3 4 . 2 9 = 3 . 5 1 g 5 . 8 5 g × 1 0 0 = 0 . 6 × 1 0 0 = 6 0 % Total number of hybridised carbon atoms in product D is ____________. s p 2 JEE Main 2025 (Online) 7th April Evening Shift Ans: 7 Solution: Number of -atoms in product NTA Ans. ALLEN Ans. ------------------------------------------------------------------------------------------------- T : 4 Q he compounds which give positive Fehling's test are : Choose the correct answer from the options given below : ? s p 2 C D = 7 = 7 = 7 JEE Main 2025 (Online) 22nd January Morning Shift Options: A. (A), (C) and (D) Only B. (A), (B) and (C) Only C. (A), (D) and (E) Only D. (C), (D) and (E) Only Answer: D Solution: Compound (C), (D) and (E) will give positive Fehling’s test. ------------------------------------------------------------------------------------------------- : 5 Q Residue (A) + HCl (dil) Compound (B) Structure of residue (A) and compound (B) formed respectively is : JEE Main 2025 (Online) 22nd January Evening Shift Options: A. B. C. ? D. Ans: C Solution: ------------------------------------------------------------------------------------------------- T : 6 Q he major product of the following reaction is : JEE Main 2025 (Online) 23rd January Morning Shift Options: A. C H 3 - C H 2 - C H 2 - O H B. C. D.Read More
FAQs on JEE Main Previous Year Questions (2025): Aldehydes Ketones and Carboxylic Acids
| 1. What are the main differences between aldehydes and ketones? |  |
Ans.Aldehydes and ketones are both carbonyl compounds characterized by the presence of a carbonyl group (C=O). The primary difference lies in the position of the carbonyl group: in aldehydes, the carbonyl group is located at the end of the carbon chain, while in ketones, it is situated within the chain. Aldehydes have the general formula RCHO, where R is a hydrocarbon group, whereas ketones have the formula RC(=O)R', where R and R' can be the same or different hydrocarbon groups. Additionally, aldehydes are typically more reactive than ketones due to the availability of hydrogen on the carbonyl carbon.
| 2. How are carboxylic acids formed from aldehydes and ketones? | |
Ans.Carboxylic acids can be formed from aldehydes and ketones through oxidation reactions. Aldehydes can be oxidized to form carboxylic acids by using oxidizing agents such as potassium permanganate (KMnO₄) or chromium trioxide (CrO₃). For ketones, the oxidation typically does not lead to carboxylic acids under mild conditions, but under strong oxidative conditions, they can be cleaved to produce carboxylic acids. The overall reaction involves the addition of oxygen to the carbonyl carbon, resulting in the formation of the carboxylic acid functional group (–COOH).
| 3. What are the common methods for the synthesis of aldehydes and ketones? | |
Ans.Aldehydes and ketones can be synthesized through several methods. One common method is the oxidation of primary and secondary alcohols, respectively. For aldehydes, primary alcohols are oxidized using oxidizing agents, while secondary alcohols yield ketones upon oxidation. Another method includes the reaction of Grignard reagents with carbon dioxide, which produces carboxylic acids that can be further reduced to aldehydes. Additionally, the Friedel-Crafts acylation reaction can be employed to synthesize ketones from aromatic compounds. Lastly, the direct hydration of alkynes can also yield aldehydes and ketones.
| 4. What are the physical properties and applications of carboxylic acids? | |
Ans.Carboxylic acids are characterized by their acidic nature due to the presence of the carboxyl group (–COOH). They have higher boiling points than aldehydes and ketones of comparable molecular weight because they can form hydrogen bonds. Carboxylic acids are typically soluble in water, especially those with shorter carbon chains. They are widely used in the production of polymers, pharmaceuticals, food preservatives, and as intermediates in organic synthesis. Common examples include acetic acid (used in vinegar), citric acid (found in citrus fruits), and benzoic acid (used as a food preservative).
| 5. What role do aldehydes and ketones play in biological systems? | |
Ans.Aldehydes and ketones play significant roles in various biological processes. For instance, glucose, a simple sugar, is an aldehyde (specifically an aldohexose) that serves as a primary energy source for cells. Ketones, such as acetoacetate and beta-hydroxybutyrate, are produced during the metabolism of fatty acids and serve as alternative energy sources during prolonged fasting or carbohydrate restriction. Both classes of compounds are vital in metabolic pathways and are involved in the synthesis of hormones, neurotransmitters, and other essential biomolecules.
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