NCERT Solutions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids in Hindi Mediem
FAQs on NCERT Solutions for Class 12 Chemistry In Hindi Chapter 12 Aldehydes, Ketones and Carboxylic Acids In Hindi Mediem
1. Where can I find reliable, step-by-step NCERT Solutions for Class 12 Chemistry, Chapter 12 (Aldehydes, Ketones and Carboxylic Acids)?
Vedantu provides comprehensive and step-by-step NCERT Solutions for Class 12 Chemistry Chapter 12, meticulously prepared by subject matter experts. These solutions are fully aligned with the latest 2025-26 CBSE syllabus and are designed to help you understand the correct method for solving every question in the textbook.
2. Do the NCERT Solutions for Chapter 12 cover both the in-text questions and the end-of-chapter exercises?
Yes, the NCERT Solutions for Aldehydes, Ketones and Carboxylic Acids provide detailed answers for all questions. This includes the in-text questions that appear throughout the chapter, as well as every problem from the main end-of-chapter exercise, ensuring complete coverage of the NCERT textbook.
3. How do the NCERT Solutions explain the method for comparing the boiling points of aldehydes, alcohols, and ethers with similar molecular masses?
The solutions explain this by focusing on intermolecular forces. The correct method involves these steps:
- Alcohols: Identify that they have the highest boiling points due to strong intermolecular hydrogen bonding.
- Aldehydes/Ketones: Note their boiling points are intermediate due to dipole-dipole interactions from the polar carbonyl group.
- Ethers & Hydrocarbons: These have the lowest boiling points as they only exhibit weak van der Waals forces.
4. What is the correct approach, as per NCERT Solutions, to arrange aldehydes and ketones in increasing order of their reactivity towards nucleophilic addition?
The correct approach involves analysing two main factors:
- Steric Hindrance: Aldehydes are generally more reactive than ketones because the hydrogen atom attached to the carbonyl carbon in aldehydes causes less steric hindrance compared to the bulkier alkyl/aryl groups in ketones.
- Electronic Effects: Electron-donating groups (like alkyl groups) decrease the positive charge on the carbonyl carbon, making it less electrophilic and thus less reactive. Electron-withdrawing groups increase reactivity.
5. Why is p-nitrobenzaldehyde more reactive than benzaldehyde in nucleophilic addition reactions, a concept explained in Chapter 12 solutions?
This is due to electronic effects. The nitro group (-NO₂) at the para position is a powerful electron-withdrawing group. It pulls electron density away from the benzene ring and the attached carbonyl group through the -R (resonance) effect. This intensifies the positive charge on the carbonyl carbon, making it a much better target for nucleophiles and thus increasing its reactivity compared to unsubstituted benzaldehyde.
6. How should one solve NCERT questions that ask to compare the acidic strength of different substituted carboxylic acids, like fluoroacetic acid and chloroacetic acid?
To solve these problems, you must analyse the stability of the carboxylate anion formed after donating a proton. The key principle is the inductive effect (-I effect) of the substituent.
- Identify the electron-withdrawing group (e.g., F, Cl).
- A stronger electron-withdrawing group will pull electron density more effectively, dispersing the negative charge and stabilising the carboxylate anion.
- Since fluorine is more electronegative than chlorine, its -I effect is stronger. Therefore, fluoroacetic acid is a stronger acid than chloroacetic acid because its conjugate base is more stable.
7. The NCERT solution for Chapter 12 highlights the inductive effect. How does the position of a substituent, like fluorine, impact the acidic strength of a carboxylic acid?
The inductive effect is distance-dependent; its influence decreases significantly as the distance from the carboxyl group increases. For instance, in comparing 3-fluorobutanoic acid and 4-fluorobutanoic acid:
- In 3-fluorobutanoic acid, the fluorine atom is closer to the -COOH group and exerts a stronger electron-withdrawing effect.
- In 4-fluorobutanoic acid, the fluorine is farther away, and its stabilising -I effect on the carboxylate anion is much weaker.
8. What is the step-by-step mechanism for converting a compound like bromobenzene into benzoic acid using a Grignard reagent, as detailed in the NCERT solutions?
The conversion follows a two-step process as per the correct NCERT methodology:
- Step 1: Formation of Grignard Reagent: Bromobenzene is treated with magnesium metal in the presence of dry ether. This forms phenylmagnesium bromide (C₆H₅MgBr), the Grignard reagent.
- Step 2: Carboxylation and Acidification: The Grignard reagent, a strong nucleophile, attacks the electrophilic carbon of solid carbon dioxide (dry ice). This forms a magnesium carboxylate salt. Subsequent acidic hydrolysis (using H₃O⁺) protonates the salt to yield the final product, benzoic acid (C₆H₅COOH).

















