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NCERT Solutions for Class 12 Chemistry In Hindi Chapter 12 Aldehydes, Ketones and Carboxylic Acids In Hindi Mediem

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NCERT Solutions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids in Hindi Mediem

Download the Class 12 Chemistry NCERT Solutions in Hindi medium and English medium as well offered by the leading e-learning platform Vedantu. If you are a student of Class 12, you have reached the right platform. The NCERT Solutions for Class 12 Chemistry in Hindi provided by us are designed in a simple, straightforward language, which are easy to memorise. You will also be able to download the PDF file for NCERT Solutions for Class 12 Chemistry  in Hindi from our website at absolutely free of cost.


NCERT, which stands for The National Council of Educational Research and Training, is responsible for designing and publishing textbooks for all the classes and subjects. NCERT textbooks covered all the topics and are applicable to the Central Board of Secondary Education (CBSE) and various state boards.


Class:

NCERT Solutions for Class 12

Subject:

Class 12 Chemistry

Chapter Name:

Chapter 12 - Aldehydes Ketones And Carboxylic Acids

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2024-25

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes



We, at Vedantu, offer free NCERT Solutions in English medium and Hindi medium for all the classes as well. Created by subject matter experts, these NCERT Solutions in Hindi are very helpful to the students of all classes.

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Access NCERT Solutions for Class-12 Chemistry Chapter 12 – एल्डिहाइड, कीटोन तथा कार्बोक्सिलिक अम्ल

अभ्यास के अन्तर्गत दिए गए प्रश्नोत्तर

1. निम्नलिखित यौगिकों की संरचना लिखिए –

(i) $\alpha$-मेथॉक्सीप्रोपिऑनऐल्डिहाइड

उत्तरः 

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(ii) 3-हाइड्रॉक्सीब्यूटेनल

उत्तरः 

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(iii) 2-हाइड्रॉक्सीसाइक्लोपेन्टेन कार्बोल्डिहाइड

उत्तरः 

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(iv) 4-ऑक्सोपेन्टेनल

उत्तरः 

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(v) डाइ-द्वितीयक-ब्यूटिल कीटोन

उत्तरः 

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(vi) 4-क्लोरोऐसीटोफीनोन।

उत्तरः 

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2. निम्नलिखित अभिक्रियाओं के उत्पादों की संरचना लिखिए –

(i) 

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उत्तरः 

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(ii) $\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}\right)_{2} \mathbf{C d}+2 \mathrm{CH}_{3} \mathrm{COCl} \longrightarrow$

उत्तरः 

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(iii) $\mathbf{H}_{3} \mathbf{C}-\mathbf{C} \equiv \mathbf{C}-\mathbf{H} \stackrel{\mathbf{H g}^{2+}, \mathbf{H}_{2} \mathbf{S O}_{4}}{\longrightarrow}$

उत्तरः 

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(iv) 

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उत्तरः 

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3. निम्नलिखित यौगिकों को उनके क्वथनांकों के बढ़ते क्रम में व्यवस्थित कीजिए –
$\mathrm{CH}_{3} \mathrm{CHO}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}, \mathrm{CH}_{3} \mathrm{OCH}_{3}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2}$

उत्तर: यौगिकों के मोलर द्रव्यमान तुलनात्मक हैं- $\mathrm{CH}_{3} \mathrm{CHO}(44), \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(46), \mathrm{CH}_{3} \mathrm{OCH}_{3}$ (46), यह संयुक्त अणुओं के रूप में पाया जाता है। अतः इसका क्वथनांक उच्चतम होता है (351 K)।

$\mathrm{CH}_{3} \mathrm{CHO}$ के द्विध्रुव आघूर्ण $(2.72 \mathrm{D})$ का मान $\mathrm{CH}_{3} \mathrm{OCH}_{3}(1.18 \mathrm{D})$ से उच्च होता है, अतिएव $\mathrm{CH}_{3} \mathrm{CHO}$ में द्विध्रुव-द्विधुव अन्योन्यक्रियाएँ $\mathrm{CH}_{3} \mathrm{OCH}_{3}$ से प्रबल होती हैं। अत्त: $\mathrm{CH}_{3} \mathrm{CHO}$ का क्रृथनांक $\mathrm{CH}_{3} \mathrm{OCH}_{3}$ से उच्च होता है। $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}$ केवल दुर्बल वाण्डर वाल बलों को प्रदर्शित करता है। $\mathrm{CH}_{3} \mathrm{OCH}_{3}$ में कुछ प्रबल द्विध्रुव-द्विध्रुव अन्योन्यक्रियाएँ होती हैं। अत: $\mathrm{CH}_{3} \mathrm{OCH}_{3}$ का क्रथनांक $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}$ से अधिक होता है, अतएव यौगिकों के क्रथनांकों का बढ़ता क्रम निम्रवत् है $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}<\mathrm{CH}_{3} \mathrm{OCH}_{3}<\mathrm{CH}_{3} \mathrm{CHO}<\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$

4. निम्नलिखित यौगिकों को नाभिकरागी योगज अभिक्रियाओं में उनकी बढ़ती हुई अभिक्रियाशीलता के क्रम में व्यवस्थित कीजिए –

  1. एथेनल, प्रोपेनल, प्रोपेनोन, ब्यूटेनोन

  2. बेन्जेल्डिहाइड, p-टॉलूऐल्डिहाइड, p-नाइट्रोबेन्जेल्डिहाइड, ऐसीटोफीनोन।

[संकेत– त्रिविम प्रभाव व इलेक्ट्रॉनिक प्रभाव को ध्यान में रखें।]

उत्तर

1. कार्बोनिल यौगिकों की नाभिकरागी योगज अभिक्रियाओं के प्रति क्रियाशीलता का बढ़ता क्रम है –

ब्यूटेनोन < प्रोपेनोन < प्रोपनल < एथेनल

2. क्रियाशीलता का बढ़ता क्रम है –

ऐसीटोफीनोन < p-टॉलूऐल्डिहाइड < बेन्जेल्डिहाइड < p-नाइट्रोबेन्जेल्डिहाइड

ऐसीटोफीनोन कीटोन है, जबकि अन्य सदस्य ऐल्डिहाइड हैं। अत: यह सबसे कम क्रियाशील होता है। p-टॉलूऐल्डिहाइड में CH3 समूह कार्बोनिल समूह के सापेक्ष p:स्थान पर है जो कार्बोनिल समूह के कार्बन पर अतिसंयुग्मन (hyperconjugation) प्रभाव के कारण इलेक्ट्रॉन घनत्व बढ़ाता है और इसे बेन्जेल्डिहाइड से कम क्रियाशील बनाता है।

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दूसरी ओर p-नाइट्रोबेन्जेल्डिहाइड में -NO2 समूह शक्तिशाली इलेक्ट्रॉन निष्कासक समूह है। यह अनुनाद के कारण इलेक्ट्रॉन निष्कासित करता है। अत: कार्बोनिल समूह के कार्बन परमाणु पर इलेक्ट्रॉन घनत्व घटाता है। यह नाभिकस्नेही के आक्रमण की सुविधा प्रदान करता है तथा इसे बेन्जेल्डिहाइड की तुलना में अधिक क्रियाशील बनाता है।


5. निम्नलिखित अभिक्रियाओं के उत्पादों को पहचानिए –

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उत्तरः

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6. निम्नलिखित यौगिकों के आई०यू०पी०ए०सी० नाम दीजिए –

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उत्तरः (i) 3-फेनिलप्रोपेनोइक अम्ल

(ii) 3-मेथिलब्यूट-2-इनोइक अम्ल

(iii) 2-मेथिलसाइक्लोपेन्टेनकार्बोक्सिलिक अम्ल ।

(iv) 2, 4, 6-ट्राइनाइट्रोबेन्जोइक अम्ल


7. निम्नलिखित यौगिकों को बेन्जोइक अम्ल में कैसे परिवर्तित किया जा सकता है?
(i) एथिल बेन्जीन

उत्तरः

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(ii) ऐसीटोफीनोन

उत्तरः

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(iii) ब्रोमोबेन्जीन

उत्तरः

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(iv) फेनिलएथीन (स्टाइरीन)।

उत्तरः 

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8. नीचे प्रदर्शित अम्लों के प्रत्येक युग्म में कौन-सा अम्ल अधिक प्रबल है?

(i) $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$ अथवा $\mathrm{CH}_{2} \mathrm{FCO}_{2} \mathrm{H}$

उत्तरः

$\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$

$\mathrm{CH}_{2} \mathrm{FCO}_{2} \mathrm{H}$

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प्रोटॉन का निकलना कठिन होता है; क्योंकि $-\mathrm{CH}_{3}$ समूह का $+\mathrm{I}$ प्रभाव $\mathrm{O}-\mathrm{H}$ आबन्ध में इलेक्ट्रॉन-घनत्व बढ़ा देता है।

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प्रोटॉन का निकलना सरल होता है; क्योंकि $-\mathrm{F}$ का -I प्रभाव $\mathrm{O}-\mathrm{H}$ आबन्ध में इलेक्ट्रॉन-घनत्व घटा देता है।

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ऋणावेश की सघनता से +I प्रभाव  कार्बोक्सिलेट आयन को स्थायी कर देता है। 

ऋणावेश के फैल जाने से -I प्रभाव कार्बोक्सिलेट आयन को अस्थायी कर देता है। 


अत: $\mathrm{O}-\mathrm{H}$ आबन्ध में कम इलेक्ट्रॉन-घनत्व तथा $\mathrm{FCH}_{2} \mathrm{COO}^{-}$आयन के उच्च स्थायित्व के कारण $\mathrm{FCH}_{2} \mathrm{COOH}, \mathrm{CH}_{3} \mathrm{COOH}$ की अर्पेक्षा एक प्रबल अम्ल है।

(ii) $\mathrm{CH}_{2} \mathrm{FCO}_{2} \mathrm{H}$ अथवा $\mathrm{CH}_{2} \mathrm{CICO}_{2} \mathrm{H}$

उत्तर: $\mathrm{FCH}_{2} \mathrm{COO}^{-}$आयेन, Cl की तुलना में $\mathrm{F}$ के अधिक प्रबल -। प्रभाव के कारण $\mathrm{ClCH}_{2} \mathrm{COO}^{-}$आयन से अधिक स्थायी होता है। अत: $\mathrm{ClCH}_{2} \mathrm{COOH}$ की तुलना में $\mathrm{FCH}_{2} \mathrm{COOH}$ अधिक प्रबल अम्ल है।

(iii) $\mathrm{CH}_{2} \mathrm{FCH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}$ अथवा $\mathrm{CH}_{3} \mathrm{CHFCH}_{2} \mathrm{CO}_{2} \mathrm{H}$

उत्तरः

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प्रेरक प्रभाव दूरी के साथ घटता जाता है, इसालिए $\mathrm{F}^{-}$का -। प्रभाव, 4-फ्लुओरोब्यूटेनोइक अम्ल की तुलना में 3-फ्लुओरोब्यटेनोइक अम्ल में अधिक प्रबल होता है। इसलिए $\mathrm{FCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}$ की तुलना में $\mathrm{CH}_{3} \mathrm{CHFCH}_{2} \mathrm{COOH}$ प्रबल अम्ल है।

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उत्तरः

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$-\mathrm{CF}_{3}$ का -I प्रभाव प्रबल होता है, यह ऋणावेश को फैलाकर कार्बोंक्सिलेट आयन को स्थायित्व प्रदान करता है।

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$-\mathrm{CH}_{3}$ का $+\mathrm{I}$ प्रभाव दुर्बल होता है, यह ऋणावेश को सघन करके कार्बोक्सिलेट आयन को अस्थायी कर देता है।
इसीलिए, $\mathrm{CH}_{3}-\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{COO}^{-}(\mathrm{p})$ आयन से $\mathrm{F}_{3} \mathrm{C}-\mathrm{C}_{6} \mathrm{H}_{4}-\mathrm{COO}^{-}(\mathrm{p})$ आयन के अधिक स्थायी होने के कारण $\mathrm{F}_{3} \mathrm{C}-\mathrm{C}_{6} \mathrm{H}_{4}-\mathrm{COOH}(\mathrm{p})$ प्रबल अम्लीय है।


NCERT Solutions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids in Hindi

Chapter-wise NCERT Solutions are provided everywhere on the internet with an aim to help the students to gain a comprehensive understanding. Class 12 Chemistry Chapter 12 solution Hindi medium is created by our in-house experts keeping the understanding ability of all types of candidates in mind. NCERT textbooks and solutions are built to give a strong foundation to every concept. These NCERT Solutions for Class 12 Chemistry Chapter 12 in Hindi ensure a smooth understanding of all the concepts including the advanced concepts covered in the textbook.

NCERT Solutions for Class 12 Chemistry Chapter 12 in Hindi medium PDF download are easily available on our official website (vedantu.com). Upon visiting the website, you have to register on the website with your phone number and email address. Then you will be able to download all the study materials of your preference in a click. You can also download the Class 12 Chemistry Aldehydes, Ketones and Carboxylic Acids solution Hindi medium from Vedantu app as well by following the similar procedures, but you have to download the app from Google play store before doing that.

NCERT Solutions in Hindi medium have been created keeping those students in mind who are studying in a Hindi medium school. These NCERT Solutions for Class 12 Chemistry Aldehydes, Ketones and Carboxylic Acids in Hindi medium pdf download have innumerable benefits as these are created in simple and easy-to-understand language. The best feature of these solutions is a free download option. Students of Class 12 can download these solutions at any time as per their convenience for self-study purpose.

These solutions are nothing but a compilation of all the answers to the questions of the textbook exercises. The answers/solutions are given in a stepwise format and very well researched by the subject matter experts who have relevant experience in this field. Relevant diagrams, graphs, illustrations are provided along with the answers wherever required. In nutshell, NCERT Solutions for Class 12 Chemistry in Hindi come really handy in exam preparation and quick revision as well prior to the final examinations. 

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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.
Therefore, the order is generally Hydrocarbon < Ether < Aldehyde/Ketone < Alcohol.

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.
The general order of reactivity is: Ketones < Aldehydes.

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.
As a result, 3-fluorobutanoic acid is the stronger acid of the two.

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).