GSEB Class 12 Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

GSEB Class 12 Chemistry Aldehydes, Ketones and Carboxylic Acids InText Questions and Answers

Question 1.Write the structures of the following compounds.
(i) α -Methoxypropionaldehyde
(ii) 3-Hydroxybutanal
(iii) 2-Hydroxycyclopentane carbaldehyde
(iv) 4-Oxopentanal
(v) Di-sec. butyl ketone
(vi) 4-Fluoroacetophenone

Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation):(i) यह संरचना 2-मेथॉक्सीप्रोपेनल को दर्शाती है, जहाँ प्रोपेनल श्रृंखला में दूसरे कार्बन पर मेथॉक्सी समूह जुड़ा है।
(ii) यह 3-हाइड्रॉक्सीब्यूटानल की संरचना है, जिसमें ब्यूटानल श्रृंखला के तीसरे कार्बन पर हाइड्रॉक्सी समूह मौजूद है।
(iii) यह 2-हाइड्रॉक्सीसाइक्लोपेन्टेनकार्बाल्डिहाइड की संरचना है, जहाँ साइक्लोपेन्टेन वलय से जुड़े एल्डिहाइड समूह के दूसरे कार्बन पर हाइड्रॉक्सी समूह है।
(iv) यह 4-ऑक्सोपेन्टेनल की संरचना है, जिसमें पेन्टेनल श्रृंखला के चौथे कार्बन पर ऑक्सो (कीटोन) समूह है।
(v) यह डाई-सेक-ब्यूटाइल कीटोन की संरचना है, जहाँ कार्बोनिल समूह से दो सेक-ब्यूटाइल समूह जुड़े हैं।
(vi) यह 4-फ्लोरोएसीटोफेनोन की संरचना है, जिसमें बेंजीन वलय पर एसीटोफेनोन समूह के पैरा स्थिति में फ्लोरीन परमाणु उपस्थित है।
In simple words: The answer provides the chemical structures for six different organic compounds, illustrating how various functional groups and substituents are arranged in each molecule.

🎯 Exam Tip: Correctly drawing organic structures requires a strong understanding of IUPAC nomenclature and functional group placement. Pay attention to numbering, parent chain identification, and substituent positions for accuracy.

Question 2.Write the structures of products of the following reactions.

Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation):(i) इस अभिक्रिया में बेंजीन C₂H₃Cl के साथ निर्जल AlCl₃ और CS₂ की उपस्थिति में अभिकृत होता है, जिससे एथिलफेनिल कीटोन (एसीटोफेनोन) बनता है।
(ii) यह अभिक्रिया डाईबेंजाइलकैडमियम और एसीटिल क्लोराइड के बीच है, जिससे कीटोनिक उत्पाद (डाइबेंजाइल एसीटोन) बनता है।
(iii) यह प्रोपाइन का हाइड्रेशन है जो \( \text{Hg}^{2+}, \text{H}_2\text{SO}_4 \) की उपस्थिति में एसीटोन बनाता है।
(iv) यह टोल्यूनि का क्रोमिल क्लोराइड (CrO₂Cl₂) और जल के साथ ऑक्सीकरण है, जिससे बेंजलडिहाइड बनता है।
In simple words: The products of various organic reactions are depicted, including Friedel-Crafts acylation, reaction with organocadmium reagents, alkyne hydration, and Etard reaction, showing the resulting compounds.

🎯 Exam Tip: Recognizing common reagents and their specific reaction types (e.g., Friedel-Crafts, hydration, oxidation) is crucial. Practice predicting products based on reactant functional groups and reaction conditions.

Question 3.Arrange the following compounds in increasing order of their boiling points. CH₃CHO, CH₃CH₂OH, CH₃OCH₃, CH₃CH₂CH₃

Answer:The increasing order of boiling points is: CH₃CH₂CH₃ < CH₃OCH₃ < CH₃CHO < CH₃CH₂OH. This sequence is determined by the intermolecular forces acting between the molecules. These compounds possess comparable molecular masses. Ethanol (CH₃CH₂OH) exhibits the strongest hydrogen bonding among them. In methoxymethane (CH₃OCH₃) and ethanal (CH₃CHO), ethanal has stronger dipole-dipole attraction because it is more polar than methoxymethane, resulting in a higher boiling point for ethanal than CH₃-O-CH₃. Propane (CH₃CH₂CH₃) is non-polar, thus, it only has weak van der Waals' forces.
In simple words: Boiling points increase with stronger intermolecular forces; hydrogen bonding in ethanol is highest, followed by dipole-dipole interactions in ethanal and methoxymethane, and finally weak van der Waals' forces in non-polar propane.

🎯 Exam Tip: When comparing boiling points, prioritize intermolecular forces in this order: hydrogen bonding > dipole-dipole interactions > London dispersion forces (van der Waals). Molecular size and branching also play a role.

Question 4.Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions.
(i) Ethanal, Propanal, Propanone, Butanone.
(ii) Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone.
Hint: Consider steric effect and electronic effect.

Answer:(i) The electron-withdrawing inductive effect and steric hindrance follow the order: Butanone > Propanone > Propanal > Ethanal. Consequently, the reactivity towards nucleophilic addition reactions decreases in the sequence: Butanone < Propanone < Propanal < Ethanal.
(ii) Acetophenone is a ketone, making it the least reactive among the given compounds, which are predominantly aldehydes. The methyl group (-CH₃) in p-tolualdehyde acts as an electron-donating group, while the nitro group (-NO₂) in p-nitrobenzaldehyde is electron-withdrawing. Therefore, p-tolualdehyde is less reactive, and p-nitrobenzaldehyde is more reactive than benzaldehyde. The overall reactivity order is: Acetophenone < p-tolualdehyde < benzaldehyde < p-nitrobenzaldehyde.
In simple words: Nucleophilic addition reactivity is influenced by steric hindrance and electron density at the carbonyl carbon; less hindered and more electron-deficient carbonyls are more reactive.

🎯 Exam Tip: For nucleophilic addition, aldehydes are generally more reactive than ketones due to less steric hindrance and weaker electron-donating inductive effects. Electron-withdrawing groups enhance reactivity by increasing the electrophilicity of the carbonyl carbon.

Question 5.Predict the products of the following reactions:

Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation):(i) साइक्लोपेन्टेनोन की हाइड्रॉक्सिलैमाइन से अभिक्रिया से साइक्लोपेन्टेनोन ऑक्सिम बनता है।
(ii) साइक्लोहेक्सानोन की 2,4-डाइनिट्रोफेनिलहाइड्राजीन से अभिक्रिया से 2,4-डाइनिट्रोफेनिलहाइड्राज़ोन व्युत्पन्न बनता है।
(iii) एल्डिहाइड (R-CH=CH-CHO) की सेमीकार्बाज़ाइड से अभिक्रिया से संगत सेमीकार्बाज़ोन बनता है।
(iv) एसीटोफेनोन की एथिलएमीन से अभिक्रिया से शिफ़ बेस (इमाइन) बनता है।
In simple words: The structures illustrate the products formed from various carbonyl compounds reacting with nitrogen-containing reagents like hydroxylamine, 2,4-dinitrophenylhydrazine, semicarbazide, and ethylamine, forming oximes, hydrazones, semicarbazones, and imines respectively.

🎯 Exam Tip: Carbonyl compounds (aldehydes and ketones) readily undergo addition-elimination reactions with ammonia derivatives (like hydroxylamine, hydrazines) to form C=N compounds, with water as a byproduct.

Question 6.Give the IUPAC names of the following compounds:

Answer:
(i) 3-Phenylpropanoic acid
(ii) 3-Methylbut-2-enoic acid
(iii) 2-Methylcyclopentanecarboxylic acid
(iv) 2,4,6-Trinitrobenzoic acid
In simple words: The given chemical structures are systematically named according to IUPAC rules, identifying the parent carboxylic acid and its substituents.

🎯 Exam Tip: Mastering IUPAC nomenclature for carboxylic acids involves correctly identifying the longest carbon chain containing the -COOH group, assigning 'oic acid' suffix, and numbering substituents from the carboxyl carbon.

Question 7.Show how each of the following compounds can be converted to benzoic acid,
(i) Ethylbenzene
(ii) Acetophenone
(iii) Bromobenzene
(iv) Phenylethene (Styrene)

Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation):(i) एथिलबेंजीन को KMnO₄/KOH के साथ अभिकृत करके बेंजोइक अम्ल में परिवर्तित किया जाता है, जिसके बाद अम्लीय हाइड्रोलिसिस (H₃O⁺) होता है।
(ii) एसीटोफेनोन को KMnO₄/KOH के साथ अभिकृत करके बेंजोइक अम्ल में परिवर्तित किया जाता है, जिसके बाद अम्लीय हाइड्रोलिसिस (H₃O⁺) होता है।
(iii) ब्रोमोबेंजीन को Mg के साथ अभिकृत करके फेनिलमैग्नीशियम ब्रोमाइड बनाया जाता है। इसे कार्बन डाइऑक्साइड (CO₂) से उपचारित करने के बाद, जल (HOH) से हाइड्रोलिसिस करके बेंजोइक अम्ल प्राप्त किया जाता है।
(iv) फेनिलेथेन (स्टाइरीन) को KMnO₄/KOH के साथ अभिकृत करके बेंजोइक अम्ल में परिवर्तित किया जाता है, जिसके बाद अम्लीय हाइड्रोलिसिस (H₃O⁺) होता है।
In simple words: This section illustrates the conversion of various aromatic compounds, including alkylbenzenes, ketones, aryl halides, and alkenes, into benzoic acid, primarily through strong oxidation or Grignard reactions followed by carboxylation and hydrolysis.

🎯 Exam Tip: Alkyl groups attached to a benzene ring can be oxidized to a carboxylic acid group by strong oxidizing agents like KMnO₄, regardless of the length of the alkyl chain, as long as it has at least one benzylic hydrogen.

Question 8.Which acid of each pair shown here would you expect to be stronger?
(i) CH₃CO₂H or CH₂FCO₂H
(ii) CH₂FCO₂H or CH₂CH₂ClCO₂H
(iii) CH₂FCH₂CH₂CO₂H or CH₃CHFCH₂CO₂H
(iv) F₃C-COOH or H₂C-COOH

Answer:(i) Fluorochloroacetic acid (FCH₂COOH) is a stronger acid due to the -I (electron-withdrawing inductive) effect of fluorine.
(ii) Fluorochloroacetic acid (FCH₂CO₂H) is significantly stronger than 3-chloropropanoic acid (CH₂CH₂ClCO₂H) because the -I effect of fluorine is much stronger than that of chlorine.
(iii) 3-Fluoro-2-methylpropanoic acid (CH₃CHFCH₂COOH) is stronger than 4-fluorobutanoic acid (CH₂FCH₂CH₂CO₂H). This is because the inductive effect diminishes with increasing distance; thus, the fluorine atom closer to the carboxyl group exerts a greater acid-strengthening effect.
(iv) Trifluoroacetic acid (F₃C-COOH) is stronger. The three highly electronegative fluorine atoms strongly withdraw electrons, significantly increasing the acidity, whereas the two hydrogens in CH₂COOH have no such effect.
In simple words: Acidity of carboxylic acids increases with the presence of electron-withdrawing groups (like F or Cl) and their proximity to the carboxyl group, as they stabilize the conjugate base.

🎯 Exam Tip: The strength of carboxylic acids is primarily governed by the stability of their conjugate base. Electron-withdrawing groups stabilize the carboxylate ion, increasing acidity, while electron-donating groups destabilize it, decreasing acidity. The magnitude of the inductive effect depends on the electronegativity and number of substituents, and its effect diminishes with distance.

GSEB Class 12 Chemistry Aldehydes, Ketones and Carboxylic Acids Text Book Questions and Answers

Question 1.Name the following compounds according to the IUPAC system of nomenclature:
(i) CH₃CH(CH₃)CH₂CH₃CHO
(ii) CH₃CH₂COCH(C₂H₅)CH₂CH₂Cl
(iii) CH₃CH = CHCHO
(iv) CH₃COCH₂COCH₃
(v) CH₃CH(CH₃)CH C(CH₃)₂COCH₃
(vi) (CH₃)₃CCH₂COOH
(vii) OHCC₆H₄CHO-p

Answer:The IUPAC names for the compounds are:
(i) 4-Methylpentanal
(ii) 6-Chloro-4-ethylhexan-3-one
(iii) But-2-enal
(iv) Pentane-2,4-dione
(v) 3,3,5-Trimethylhexan-2-one
(vi) 3,3-Dimethylbutanoic acid
(vii) Benzene-1,4-dicarbaldehyde
In simple words: The compounds are systematically named following IUPAC rules, identifying the longest carbon chain, functional groups, and substituents.

🎯 Exam Tip: For complex molecules, prioritize the main functional group, identify the longest continuous carbon chain, and correctly number the carbons to give substituents the lowest possible numbers. Pay attention to stereochemistry if applicable.

Question 2.Draw the structures of the following compounds.
(i) 3-Methylbutanal
(ii) p-Nitropropiophenone
(iii) p-Methylbenzaldehyde
(iv) 4-Methylpent-3-en-2-one
(v) 4-Chloropentan-2-one
(vi) 3-Bromo-4-phenylpentanoic acid
(vii) p, p' -Dihydroxybenzophenone
(viii) Hex-2-en-4-ynoic acid

Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation):(i) 3-मेथिलब्यूटेनल: ब्यूटेनल श्रृंखला में तीसरे कार्बन पर एक मेथिल समूह जुड़ा है।
(ii) p-नाइट्रोप्रोपियोफेनोन: प्रोपियोफेनोन के फिनाइल वलय पर पैरा स्थिति में एक नाइट्रो समूह है।
(iii) p-मेथिलबेंजलडिहाइड: बेंजलडिहाइड के फिनाइल वलय पर पैरा स्थिति में एक मेथिल समूह है।
(iv) 4-मेथिलपेन्ट-3-एन-2-वन: एक पेन्टेन श्रृंखला जिसमें दूसरे कार्बन पर कीटोन, तीसरे कार्बन पर दोहरा बंध और चौथे कार्बन पर एक मेथिल समूह है।
(v) 4-क्लोरोपेन्टेन-2-वन: एक पेन्टेन श्रृंखला जिसमें दूसरे कार्बन पर कीटोन और चौथे कार्बन पर एक क्लोरो समूह है।
(vi) 3-ब्रोमो-4-फेनिलपेन्टेनोइक अम्ल: एक पेन्टेनोइक अम्ल श्रृंखला जिसमें तीसरे कार्बन पर ब्रोमो और चौथे कार्बन पर एक फेनिल समूह है।
(vii) p,p'-डाइहाइड्रॉक्सीबेंजोफेनोन: बेंजोफेनोन के दोनों फिनाइल वलयों पर पैरा स्थिति में हाइड्रॉक्सी समूह हैं।
(viii) हेक्स-2-एन-4-आइनोइक अम्ल: एक हेक्सानोइक अम्ल जिसमें दूसरे कार्बन पर दोहरा बंध और चौथे कार्बन पर तिहरा बंध है।
In simple words: The structures for eight different organic compounds are provided, demonstrating their specific arrangements of carbon atoms, functional groups, and substituents as indicated by their IUPAC names.

🎯 Exam Tip: Accurately drawing chemical structures from IUPAC names requires careful attention to the main functional group, the numbering of the parent chain, and the placement of all substituents (alkyl groups, halogens, other functional groups) at their specified positions, along with correct representation of double and triple bonds.

Question 3.Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.
(i) CH₃CO(CH₂)₄CH₃
(ii) CH₃CH₂CHBrCH₂CH(CH₃)CHO
(iii) CH₃(CH₂)₅CHO
(iv) Ph-CH=CH-CHO
(v) Cyclopentanecarbaldehyde
(vi) PhCOPh

Answer:The IUPAC names are:
(i) Heptan-2-one
(ii) 4-Bromo-2-methylhexanal
(iii) Heptanal
(iv) 3-Phenylpropenal
(v) Cyclopentanecarbaldehyde
(vi) Diphenylmethanone
In simple words: This list provides the systematic IUPAC names for various ketones and aldehydes, which indicate their structural composition and functional groups.

🎯 Exam Tip: When naming aldehydes and ketones, identify the longest carbon chain containing the carbonyl group. For aldehydes, use the suffix '-al'; for ketones, use '-one' and indicate the position of the carbonyl group if necessary. Prioritize the carbonyl carbon for numbering.

Question 4.Draw structures of the following derivatives.
(i) The 2,4-dinitrophenylhydrazone ofbenzaldehyde
(ii) Cyclopropanone oxime
(iii) Acetaldehydedimethylacetal
(iv) The semicarbazone of cyclobutanone
(v) The ethylene ketal of hexan-3-one
(vi) The methyl hemiacetal of formaldehyde

Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation):(i) बेंजलडिहाइड के 2,4-डाइनिट्रोफेनिलहाइड्राज़ोन की संरचना, जो बेंजलडिहाइड और 2,4-डाइनिट्रोफेनिलहाइड्राजीन की अभिक्रिया से बनती है।
(ii) साइक्लोप्रोपेनोन ऑक्सिम की संरचना, जो साइक्लोप्रोपेनोन और हाइड्रॉक्सिलैमाइन की अभिक्रिया से बनती है।
(iii) एसिटाल्डिहाइड डाइमेथिलऐसीटल की संरचना, जो एसिटाल्डिहाइड और मेथनॉल की अभिक्रिया से बनती है।
(iv) साइक्लोब्यूटेनोन सेमीकार्बाज़ोन की संरचना, जो साइक्लोब्यूटेनोन और सेमीकार्बाज़ाइड की अभिक्रिया से बनती है।
(v) हेक्सेन-3-ओन के एथिलीन कीटल की संरचना, जो हेक्सेन-3-ओन और एथिलीन ग्लाइकॉल की अभिक्रिया से बनती है।
(vi) फॉर्मेल्डिहाइड के मेथिल हेमीऐसीटल की संरचना, जो फॉर्मेल्डिहाइड और मेथनॉल की अभिक्रिया से बनती है।
In simple words: This section provides the structural diagrams for various derivatives of aldehydes and ketones, including hydrazones, oximes, acetals, semicarbazones, ketals, and hemiacetals, formed by reactions with specific reagents.

🎯 Exam Tip: Understanding the formation of derivatives like oximes, hydrazones, acetals, and ketals is essential. These reactions typically involve carbonyl compounds reacting with alcohols or nitrogen-containing compounds, often under acid catalysis.

Question 5.Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents.
(i) PhMgBr and then H₃O⁺
(ii) Tollens' reagent
(iii) Semicarbazide and weak acid
(iv) Excess ethanol and acid
(v) Zinc amalgam and dilute hydrochloric acid

Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation):(i) साइक्लोहेक्सेन्कार्बाल्डिहाइड की फेनिलमैग्नीशियम ब्रोमाइड और उसके बाद हाइड्रोनियम आयन (H₃O⁺) से अभिक्रिया से साइक्लोहेक्सिलफेनिल कार्बिनॉल बनता है।
(ii) साइक्लोहेक्सेन्कार्बाल्डिहाइड की टॉलेन्स अभिकर्मक से अभिक्रिया से साइक्लोहेक्सेन कार्बोक्सिलेट आयन और सिल्वर धातु बनती है।
(iii) साइक्लोहेक्सेन्कार्बाल्डिहाइड की सेमीकार्बाज़ाइड और दुर्बल अम्ल से अभिक्रिया से साइक्लोहेक्सेन्कार्बाल्डिहाइड सेमीकार्बाज़ोन बनता है।
(iv) साइक्लोहेक्सेन्कार्बाल्डिहाइड की अतिरिक्त इथेनॉल और अम्ल से अभिक्रिया से साइक्लोहेक्सेन्कार्बाल्डिहाइड डाइएथिल एसिटल बनता है।
(v) साइक्लोहेक्सेन्कार्बाल्डिहाइड की जिंक अमलगम और तनु हाइड्रोक्लोरिक अम्ल (क्लीमेंसन अपचयन) से अभिक्रिया से मेथिल साइक्लोहेक्सेन बनता है।
In simple words: This answer illustrates the products of cyclohexanecarbaldehyde undergoing various reactions: Grignard addition, Tollens' oxidation, semicarbazone formation, acetal formation, and Clemmensen reduction.

🎯 Exam Tip: Aldehydes are highly versatile functional groups. They can undergo nucleophilic addition (Grignard), oxidation (Tollens'), condensation (semicarbazide), acetal formation (alcohols), and reduction (Clemmensen). Understand the specific conditions and products for each type of reaction.

Question 6.Which of the following compounds would undergo aldol condensation, with the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction.
(i) Methanal
(ii) 2-Methylpentanal
(iii) Benzaldehyde
(iv) Benzophenone
(v) Cyclohexanone
(vi) 1-Phenylpropanone
(vii) Phenylacetaldehyde
(viii) Butan-1-ol
(ix) 2,2-Dimethylbutanal

Answer:Compounds undergoing aldol condensation: (ii) 2-Methylpentanal, (v) Cyclohexanone, (vi) 1-Phenylpropanone, (vii) Phenylacetaldehyde.
Compounds undergoing Cannizzaro reaction: (i) Methanal, (iii) Benzaldehyde, (ix) 2,2-Dimethylbutanal.
Compounds undergoing neither: (iv) Benzophenone, (viii) Butan-1-ol.
ℹ️ चित्र व्याख्या (Diagram Explanation):(i) मेथेनल (2 HCHO) की NaOH के साथ कैनिज़ारो अभिक्रिया से मेथनॉल (CH₃OH) और सोडियम फॉर्मेट (HCOONa) बनता है।
(ii) 2-मेथिलपेन्टेनल (2CH₃CH₂CH₂CH(CH₃)CHO) की NaOH के साथ एल्डोल संघनन अभिक्रिया से 2,2-मेथिल-3-हाइड्रॉक्सी-4-एथिलहेप्टेनाल बनता है।
(iii) बेंजलडिहाइड (2 C₆H₅CHO) की NaOH के साथ कैनिज़ारो अभिक्रिया से बेंजाइल अल्कोहल (C₆H₅CH₂OH) और सोडियम बेंजोएट (C₆H₅COONa) बनता है।
(iv) बेंजोफेनोन में α-हाइड्रोजन नहीं होता और यह एक कीटोन है, इसलिए यह न तो एल्डोल संघनन और न ही कैनिज़ारो अभिक्रिया करेगा।
(v) साइक्लोहेक्सानोन की NaOH के साथ एल्डोल संघनन अभिक्रिया से 2-(1-हाइड्रॉक्सीसाइक्लोहेक्सिल)साइक्लोहेक्सानोन बनता है।
(vi) 1-फेनिलप्रोपेनोन की NaOH के साथ एल्डोल संघनन अभिक्रिया से 3-हाइड्रॉक्सी-3-फेनिल-1-पेन्टेनोन बनता है।
(vii) फेनिलएसिटाल्डिहाइड की NaOH के साथ एल्डोल संघनन अभिक्रिया से 2,4-डाइफेनिल-3-हाइड्रॉक्सीब्यूटानल बनता है।
(viii) ब्यूटेन-1-ओएल एक अल्कोहल है, न कि एल्डिहाइड या कीटोन, इसलिए यह न तो एल्डोल संघनन और न ही कैनिज़ारो अभिक्रिया करेगा।
(ix) 2,2-डाइमेथिलब्यूटेनल (2CH₃CH₂C(CH₃)₂CHO) में α-हाइड्रोजन नहीं होते, इसलिए इसकी NaOH के साथ कैनिज़ारो अभिक्रिया से 2,2-डाइमेथिल-1-ब्यूटेनॉल और सोडियम 2,2-डाइमेथिलब्यूटेनोएट बनता है।
In simple words: Aldol condensation occurs for carbonyl compounds with α-hydrogens, forming β-hydroxy aldehydes/ketones, while Cannizzaro reaction occurs for aldehydes lacking α-hydrogens, undergoing disproportionation to an alcohol and a carboxylic acid salt.

🎯 Exam Tip: The presence or absence of α-hydrogens is the key to differentiating between aldol condensation (requires α-hydrogens) and Cannizzaro reaction (no α-hydrogens). Ketones typically undergo aldol condensation if they have α-hydrogens but do not undergo Cannizzaro reaction.

Question 7.How will you convert ethanal into the following compounds?
(i) Butane-1,3-diol
(ii) But-2-enal
(iii) But-2-enoic acid

Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation):(i) एथेनल (CH₃CHO) का एल्डोल संघनन कराने पर 3-हाइड्रॉक्सीब्यूटेनल बनता है। इस उत्पाद को LiAlH₄ ईथर की उपस्थिति में अपचयित करने पर ब्यूटेन-1,3-डायोल प्राप्त होता है।
(ii) एथेनल (2CH₃CHO) का एल्डोल संघनन और उसके बाद जल निष्कासन (डेहाइड्रेशन) करने पर ब्यूट-2-एनाल प्राप्त होता है।
(iii) ब्यूट-2-एनाल को क्लोरीन (Cl₂) के साथ CCl₄ की उपस्थिति में अंधेरे में उपचारित करने पर 2,3-डाइक्लोरोब्यूटेनाल प्राप्त होता है। फिर इसका ऑक्सीकरण करने पर ब्यूट-2-एनोइक अम्ल बनता है।
In simple words: This section demonstrates the conversion of ethanal into various compounds: butane-1,3-diol via aldol condensation and reduction; but-2-enal via aldol condensation and dehydration; and but-2-enoic acid via aldol condensation, dehydration, halogenation, and oxidation.

🎯 Exam Tip: Ethanals are versatile starting materials. Aldol condensation is a crucial step for chain lengthening, followed by reduction to diols or dehydration to α,β-unsaturated aldehydes. Oxidation of these aldehydes or their halogenated derivatives can lead to carboxylic acids.

Question 8.An organic compound (A) (molecular formula C₈H₁₆O₂) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene. Write equations for the reactions involved.

Answer:Compound 'A' can be either butyl butanoate or (2-methyl propyl)-2-methyl propanoate.
a. When butyl butanoate \( (\text{CH}_3\text{CH}_2\text{CH}_2\text{COOCH}_2\text{CH}_2\text{CH}_2\text{CH}_3) \) is hydrolyzed with dilute \( \text{H}_2\text{SO}_4 \), it yields butanoic acid (B) \( (\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}) \) and butan-1-ol (C) \( (\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}) \). Oxidation of butan-1-ol (C) produces butanoic acid (B). Dehydration of butan-1-ol (C) gives but-1-ene \( (\text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2) \).
b. When (2-methyl propyl)-2-methyl propanoate \( (\text{CH}_3\text{CH}(\text{CH}_3)\text{COOCH}_2\text{CH}(\text{CH}_3)_2) \) is hydrolyzed with dilute \( \text{H}_2\text{SO}_4 \), it yields 2-methylpropanoic acid (B) \( (\text{CH}_3\text{CH}(\text{CH}_3)\text{COOH}) \) and 2-methylpropan-1-ol (C) \( (\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{OH}) \). Oxidation of 2-methylpropan-1-ol (C) produces 2-methylpropanoic acid (B). Dehydration of 2-methylpropan-1-ol (C) gives but-1-ene \( (\text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2) \).
ℹ️ चित्र व्याख्या (Diagram Explanation):(a) ब्यूटेन-1-ओइक अम्ल (B) और ब्यूटेन-1-ऑल (C) बनाने के लिए ब्यूटाइल ब्यूटेनोएट (A) का जलीय अपघटन। ब्यूटेन-1-ऑल (C) के ऑक्सीकरण से ब्यूटेन-1-ओइक अम्ल (B) प्राप्त होता है, और ब्यूटेन-1-ऑल (C) के निर्जलीकरण से ब्यूट-1-इन प्राप्त होता है।
(b) 2-मेथिल प्रोपाइल)-2-मेथिल प्रोपायनेट (A) का जलीय अपघटन 2-मेथिल प्रोपायनिक अम्ल (B) और 2-मेथिल प्रोपेन-1-ऑल (C) बनाने के लिए। 2-मेथिल प्रोपेन-1-ऑल (C) के ऑक्सीकरण से 2-मेथिल प्रोपायनिक अम्ल (B) प्राप्त होता है, और 2-मेथिल प्रोपेन-1-ऑल (C) के निर्जलीकरण से ब्यूट-1-इन प्राप्त होता है।
In simple words: Compound A (an ester) is hydrolyzed into a carboxylic acid B and an alcohol C. Alcohol C can be oxidized back to acid B, and dehydrates to but-1-ene, allowing for the identification of the specific ester and its hydrolysis products.

🎯 Exam Tip: This type of problem often involves ester hydrolysis, where identifying the products (acid and alcohol) and then tracing their reactions (oxidation, dehydration) helps deduce the original ester's structure. Pay attention to molecular formulas and reaction conditions.

 

Question 11. How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom,
(i) Methyl benzoate
(ii) m-Nitrobenzoic acid
(iii) p-Nitrobenzoic acid
(iv) Phenylacetic acid
(v) p-Nitrobenzaldehyde.
Answer:
(i) Methyl benzoate can be synthesized from benzene. First, benzene reacts with \( \text{CH}_3\text{Cl} \) in the presence of anhydrous \( \text{AlCl}_3 \) to form toluene (\( \text{C}_6\text{H}_5\text{CH}_3 \)). This toluene is then oxidized using \( \text{KMnO}_4/\text{OH}^- \) to yield benzoic acid (\( \text{C}_6\text{H}_5\text{COOH} \)). Finally, benzoic acid reacts with methanol (\( \text{CH}_3\text{OH} \)) in the presence of \( \text{H}_2\text{SO}_4 \) to produce methyl benzoate (\( \text{C}_6\text{H}_5\text{COOCH}_3 \)).
(ii) To obtain m-nitrobenzoic acid, benzoic acid is treated with a mixture of concentrated \( \text{HNO}_3 \) and concentrated \( \text{H}_2\text{SO}_4 \). The nitro group directs to the meta position.
(iii) p-Nitrobenzoic acid is not directly prepared from benzoic acid by simple nitration as the carboxylic acid group is meta-directing. Instead, start with toluene, nitrate it to get p-nitrotoluene (a mixture with ortho-nitrotoluene is formed, which can be separated), then oxidize the methyl group to a carboxylic acid group using \( \text{KMnO}_4/\text{OH}^- \).
(iv) Phenylacetic acid synthesis begins with benzene reacting with \( \text{CH}_3\text{Cl} \) in the presence of anhydrous \( \text{AlCl}_3 \) to give toluene. Toluene then undergoes chlorination with \( \text{Cl}_2 \) under UV light to form benzyl chloride (\( \text{C}_6\text{H}_5\text{CH}_2\text{Cl} \)). Benzyl chloride is subsequently treated with \( \text{NaCN} \) to produce benzyl cyanide (\( \text{C}_6\text{H}_5\text{CH}_2\text{CN} \)). Finally, hydrolysis of benzyl cyanide with 70% \( \text{H}_2\text{SO}_4 \) reflux yields phenylacetic acid (\( \text{C}_6\text{H}_5\text{CH}_2\text{COOH} \)).
(v) For p-nitrobenzaldehyde, toluene is first nitrated with concentrated \( \text{HNO}_3 \) and \( \text{H}_2\text{SO}_4 \) to yield p-nitrotoluene (and o-nitrotoluene). The p-nitrotoluene is then subjected to oxidation using \( \text{CrO}_2\text{Cl}_2 \) (Etard reaction) or air diluted with \( \text{N}_2 \) over \( \text{V}_2\text{O}_5 \) at 773 K to convert the methyl group into an aldehyde group, forming p-nitrobenzaldehyde.
In simple words: This question outlines how to synthesize various substituted benzoic acids and aldehydes from a benzene starting material using a series of specific chemical reactions like Friedel-Crafts alkylation, oxidation, nitration, and hydrolysis.
🎯 Exam Tip: When dealing with multi-step organic syntheses, remember the directing nature of existing groups (ortho/para vs. meta) and choose reagents appropriate for selective oxidation or functional group interconversion.

 

Question 12. How will you bring about the following conversions in not more than two steps?
(i) Propanone to Propene
(ii) Benzoic acid to Benzaldehyde
(iii) Ethanol to 3-Hydroxybutanal
(iv) Benzene to m-Nitroacetophenone
(v) Benzaldehyde to Benzophenone
(vi) Bromobenzene to 1-Phenylethanol
(vii) Benzaldehyde to 3-Phenylpropan-1-ol
(viii) Benzaldehyde to \(\alpha\)-Hydroxyphenylacetic acid
(ix) Benzoic acid to m-Nitrobenzyl alcohol
Answer:
(i) Propanone to Propene:
\( \text{CH}_3\text{COCH}_3 \xrightarrow[\text{ether}]{\text{LiAlH}_4} \text{CH}_3-\text{CH}(\text{OH})-\text{CH}_3 \xrightarrow[\Delta]{\text{H}_2\text{SO}_4} \text{CH}_3-\text{CH}=\text{CH}_2 \)
(ii) Benzoic acid to Benzaldehyde:
\( \text{C}_6\text{H}_5\text{COOH} \xrightarrow[\text{PCl}_3/\text{PCl}_5]{\text{-HCl}} \text{C}_6\text{H}_5\text{COCl} \xrightarrow[\text{Rosenmund's reduction}]{\text{H}_2/\text{Pd-BaSO}_4} \text{C}_6\text{H}_5\text{CHO} \)
(iii) Ethanol to 3-Hydroxybutanal:
\( 2\text{CH}_3\text{CH}_2\text{OH} \xrightarrow[\text{CH}_2\text{Cl}_2]{\text{PCC}} 2\text{CH}_3\text{CHO} \xrightarrow[\text{Aldol condensation}]{\text{dil. NaOH}} \text{CH}_3-\text{CH}(\text{OH})-\text{CH}_2\text{CHO} \)
(iv) Benzene to m-Nitroacetophenone:
\( \text{C}_6\text{H}_6 \xrightarrow[\text{anhyd. AlCl}_3]{\text{CH}_3\text{COCl}} \text{C}_6\text{H}_5\text{COCH}_3 \xrightarrow[\text{conc. H}_2\text{SO}_4]{\text{conc. HNO}_3} \text{m-NO}_2\text{C}_6\text{H}_4\text{COCH}_3 \)
(v) Benzaldehyde to Benzophenone:
\( \text{C}_6\text{H}_5\text{CHO} \xrightarrow[\text{catalyst absent}]{+\text{Cl}_2} \text{C}_6\text{H}_5\text{COCl} \xrightarrow[\text{anhyd. AlCl}_3]{\text{Benzene}} \text{C}_6\text{H}_5\text{COC}_6\text{H}_5 \)
(vi) Bromobenzene to 1-Phenylethanol:
\( \text{C}_6\text{H}_5\text{Br} \xrightarrow[\text{THF}]{\text{Mg}} \text{C}_6\text{H}_5\text{MgBr} \xrightarrow[\text{(ii) H}_2\text{O}/\text{H}^+]{\text{(i) CH}_3\text{CHO}} \text{C}_6\text{H}_5\text{CH}(\text{OH})\text{CH}_3 \)
(vii) Benzaldehyde to 3-Phenylpropan-1-ol:
\( \text{C}_6\text{H}_5\text{CHO} + \text{CH}_3\text{CHO} \xrightarrow[\text{heat}]{\text{C}_2\text{H}_5\text{ONa}} \text{C}_6\text{H}_5-\text{CH}=\text{CH}-\text{CHO} \xrightarrow[\text{KOH/glycol}]{\text{NH}_2\text{NH}_2} \text{C}_6\text{H}_5-\text{CH}_2-\text{CH}_2-\text{CH}_2\text{OH} \)
(viii) Benzaldehyde to \(\alpha\)-Hydroxyphenylacetic acid:
\( \text{C}_6\text{H}_5\text{CHO} \xrightarrow{\text{HCN}} \text{C}_6\text{H}_5\text{CH}(\text{OH})\text{CN} \xrightarrow{\text{H}_2\text{O}/\text{H}^+} \text{C}_6\text{H}_5\text{CH}(\text{OH})\text{COOH} \)
(ix) Benzoic acid to m-Nitrobenzyl alcohol:
\( \text{C}_6\text{H}_5\text{COOH} \xrightarrow[\text{HNO}_3]{\text{conc. HNO}_3/\text{H}_2\text{SO}_4} \text{m-NO}_2\text{C}_6\text{H}_4\text{COOH} \xrightarrow[\text{LiAlH}_4]{\text{ether}} \text{m-NO}_2\text{C}_6\text{H}_4\text{CH}_2\text{OH} \)
In simple words: This section provides step-by-step reaction sequences to convert one organic compound into another, typically involving two or fewer steps to achieve the target molecule.
🎯 Exam Tip: For conversions, always identify the functional groups involved and the required changes. Consider common reactions like reduction, oxidation, nucleophilic addition, and electrophilic substitution, keeping in mind their regioselectivity.

 

Question 13. Describe the following:
(i) Acetylation
(ii) Cannizzaro reaction
(iii) Cross aldol condensation
(iv) Decarboxylation
Answer:
(i) Acetylation – Friedel-Crafts acylation reaction: Aromatic ketones are synthesized by reacting aromatic hydrocarbons with acid chlorides (e.g., acetyl chloride) in the presence of an anhydrous \( \text{AlCl}_3 \) catalyst.
\( \text{C}_6\text{H}_6 + \text{CH}_3\text{COCl} \xrightarrow{\text{anhyd. AlCl}_3} \text{C}_6\text{H}_5\text{COCH}_3 + \text{HCl} \)
(ii) Cannizzaro reaction – Aldehydes lacking an \(\alpha\)-hydrogen atom (e.g., formaldehyde, benzaldehyde) undergo a disproportionation reaction (self-oxidation and reduction) when treated with a concentrated alkali solution. This results in the formation of the corresponding alcohol and a salt of the carboxylic acid.
\( 2\text{HCHO} + \text{NaOH} \rightarrow \text{CH}_3\text{OH} + \text{HCOONa} \)
\( 2\text{C}_6\text{H}_5\text{CHO} + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{CH}_2\text{OH} + \text{C}_6\text{H}_5\text{COONa} \)
(iii) Cross aldol condensation – This is an aldol condensation reaction that occurs between two different aldehydes, two different ketones, or an aldehyde and a ketone, provided they all contain \(\alpha\)-hydrogens. When a mixture of two distinct aldehydes is treated with a base, four different aldol products can be formed due to self-condensation and cross-condensation.
(iv) Decarboxylation – Decarboxylation is the process of removing a carbon dioxide molecule (\( \text{CO}_2 \)) from a carboxylic acid or its derivative.
(a) When sodium salts of carboxylic acids are heated with soda lime (\( \text{NaOH} + \text{CaO} \)), alkanes are formed.
\( \text{CH}_3\text{COONa} + \text{NaOH} \xrightarrow[\Delta]{\text{CaO}} \text{CH}_4 + \text{Na}_2\text{CO}_3 \)
(b) Heating calcium salts of carboxylic acids yields aldehydes or ketones.
(c) Kolbe's electrolysis: Electrolysis of aqueous solutions of sodium or potassium salts of carboxylic acids produces higher alkanes.
\( 2\text{CH}_3\text{COO}^- \xrightarrow{\text{anode}} \text{CH}_3\text{CH}_3 + 2\text{CO}_2 + 2\text{e}^- \)
In simple words: This question asks for definitions and examples of important organic reactions: acetylation (adding an acetyl group), Cannizzaro reaction (disproportionation of aldehydes without alpha-H), cross aldol condensation (aldol reaction with two different reactants), and decarboxylation (removal of CO2 from a carboxylic acid).
🎯 Exam Tip: Understand the conditions, reactants, and products for each named reaction. Pay special attention to mechanistic details, like the requirement of \(\alpha\)-hydrogens for aldol condensation or their absence for the Cannizzaro reaction.

 

Question 14. Complete each synthesis by giving missing starting material, reagent or products.
Answer:
i. Initial reactant: Benzoic acid. Reagents: \( \text{KMnO}_4/\text{KOH} \), then heat. Product: Potassium benzoate (\( \text{C}_6\text{H}_5\text{COOK} \)).
ii. Initial reactant: Benzoic acid. Reagent: \( \text{SOCl}_2 \), then heat. Product: Benzoyl chloride (\( \text{C}_6\text{H}_5\text{COCl} \)).
iii. Initial reactant: \( \text{CH}_3\text{CHO} \). Reagent: \( \text{H}_2\text{NCONHNH}_2 \). Product: Ethanal semicarbazone (\( \text{CH}_3\text{CH}=\text{NCONH}_2 \)).
iv. Initial reactant: Benzaldehyde (\( \text{C}_6\text{H}_5\text{CHO} \)). Reagent: \( \text{NaCN/HCl} \). Product: Benzaldehyde cyanohydrin (\( \text{C}_6\text{H}_5\text{CH}(\text{OH})\text{CN} \)).
v. Initial reactant: Cyclohexanone. Reagent: Tollen's reagent (\( \text{[Ag(NH}_3\text{)}_2\text{]}^+ \)). Product: Cyclohexanecarboxylate ion (\( \text{C}_6\text{H}_{11}\text{COO}^- \)).
vi. Initial reactant: Benzoic acid. Reagent: \( \text{BH}_3 \), then \( \text{H}_2\text{O}_2/\text{OH}^- \), then \( \text{PCC} \). This is a series of reactions not directly leading to a single product here, but if the goal is reduction to alcohol and then oxidation to aldehyde, it would be a multi-step synthesis. Assuming a direct conversion based on usual questions, if starting from benzoic acid, a reduction with \( \text{LiAlH}_4 \) followed by mild oxidation could yield an aldehyde. The image context points to a reduction. Given the context, the initial reactant here is likely a ketone, which is reduced to an alcohol, then oxidized to a different product. Let's assume the question expects the product of the given reagents on an unspecified starting material. If the input is something like a ketone, then \( \text{BH}_3 \) would reduce it to an alcohol.
vii. Initial reactant: Propanal (\( \text{CH}_3\text{CHO} \)) reacting with propanal (\( \text{CH}_3\text{CH}_2\text{CHO} \)). Reagent: Dilute \( \text{NaOH} \). Product: Cross aldol products like 2-ethyl-3-hydroxybutanal.
viii. Initial reactant: \( \text{CH}_3\text{COCH}_2\text{COOC}_2\text{H}_5 \). Reagents: (i) \( \text{NaBH}_4 \), (ii) \( \text{H}^+ \). Product: Ethyl 3-hydroxybutanoate.
ix. Initial reactant: Cyclohexanol. Reagent: \( \text{CrO}_3 \). Product: Cyclohexanone.
x. Initial reactant: Cyclohexanone. Reagents: (i) \( \text{O}_3 \), (ii) \( \text{Zn-H}_2\text{O} \). Product: A dicarboxylic acid resulting from ring cleavage.
xi. Initial reactant: 2-Methylpropene. Reagents: (i) \( \text{BH}_3 \), (ii) \( \text{H}_2\text{O}_2/\text{OH}^- \). Product: Isobutanol (2-methylpropan-1-ol).
In simple words: This task involves identifying the missing components (reactants, reagents, or products) in a series of chemical reactions, demonstrating knowledge of various organic transformations.
🎯 Exam Tip: For synthesis problems, always analyze the change in carbon skeleton and functional groups. Knowing common reagents and their specific roles (oxidation, reduction, dehydration, etc.) is crucial.

 

Question 15. Give possible explanation for each of the following:
(i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not.
(ii) There are two -\(\text{NH}_2\) groups in semicarbazide. However, only one is involved in the formation of semicarbazones.
(iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.
Answer:
(i) The presence of three electron-donating methyl groups in 2,2,6-trimethylcyclohexanone significantly decreases the electrophilicity of the carbonyl carbon, making it less susceptible to nucleophilic addition. Additionally, these bulky methyl groups create substantial steric hindrance, physically impeding the attack of nucleophiles like cyanide, thus preventing cyanohydrin formation in good yield.
(ii) Semicarbazide contains two primary amine (-\(\text{NH}_2\)) groups and one secondary amide group. Among the two \(\text{NH}_2\) groups, the nitrogen atom of the terminal amino group (which is more distant from the carbonyl group) has a higher tendency to donate its lone pair electrons because its electron density is not delocalized. In contrast, the nitrogen atom of the \(\text{NH}_2\) group directly adjacent to the carbonyl group (\(\text{C}=\text{O}\)) experiences a partial positive charge due to resonance with the carbonyl oxygen, which reduces its electron-donating ability and therefore its reactivity in nucleophilic attack.
(iii) Esterification, the reaction between a carboxylic acid and an alcohol to form an ester and water, is a reversible reaction. According to Le Chatelier's principle, to drive a reversible reaction towards completion and maximize product yield, it is necessary to remove one of the products from the reaction mixture as it forms. Therefore, continuously removing either water (e.g., by distillation) or the ester shifts the equilibrium to the right, favoring ester formation.
In simple words: This question explains why certain reactions occur or are hindered: steric hindrance and electron donation affect reactivity of carbonyls, resonance dictates which amino group in semicarbazide is reactive, and Le Chatelier's principle guides product removal in reversible reactions like esterification.
🎯 Exam Tip: Always consider steric hindrance and electronic effects (inductive, resonance) when explaining reactivity. For equilibrium reactions, remember Le Chatelier's principle and its application in practical synthesis to maximize yield.

 

Question 16. An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollens' reagent but forms an addition compound with sodium hydrogen sulphite and gives positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.
Answer:
To determine the empirical formula, first find the percentage of oxygen:
Percentage of Oxygen = \( 100\% - (69.77\% + 11.63\%) = 18.60\% \)
Now, calculate the simplest ratio of atoms:

Element	%	Atomic mass	% / Atomic mass	Simplest ratio
C	69.77%	12	\( \frac{69.77}{12} = 5.81 \)	\( \frac{5.81}{1.16} = 5 \)
H	11.63%	1	\( \frac{11.63}{1} = 11.63 \)	\( \frac{11.63}{1.16} = 10 \)
O	18.60%	16	\( \frac{18.60}{16} = 1.16 \)	\( \frac{1.16}{1.16} = 1 \)

The empirical formula is \( \text{C}_5\text{H}_{10}\text{O} \).
Empirical formula mass = \( 5 \times 12 + 10 \times 1 + 1 \times 16 = 60 + 10 + 16 = 86 \).
Given molecular mass = 86.
\( \text{n} = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{86}{86} = 1 \).
Therefore, the molecular formula is also \( \text{C}_5\text{H}_{10}\text{O} \).
(a) Since the compound forms a sodium hydrogen sulphite addition product, it indicates the presence of a carbonyl group (\( \text{C}=\text{O} \)), meaning it is either an aldehyde or a ketone.
(b) The compound does not reduce Tollens' reagent, which rules out an aldehyde functional group. Thus, the compound must be a ketone.
(c) A positive iodoform test confirms that the compound is a methyl ketone (i.e., contains the \( \text{CH}_3\text{CO}- \) group).
(d) Vigorous oxidation yields a mixture of ethanoic acid (\( \text{CH}_3\text{COOH} \)) and propanoic acid (\( \text{CH}_3\text{CH}_2\text{COOH} \)). This specific outcome from oxidation of a ketone suggests that the original ketone must have been pentan-2-one.
When pentan-2-one (\( \text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3 \)) undergoes vigorous oxidation, the carbon-carbon bond on either side of the carbonyl group can break. Breakage between C2 and C3 would yield ethanoic acid and propanoic acid.
Possible structure: Pentan-2-one (\( \text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3 \)).
\( \text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3 \xrightarrow[\text{H}_2\text{SO}_4]{\text{K}_2\text{Cr}_2\text{O}_7} \text{CH}_3\text{COOH} + \text{CH}_3\text{CH}_2\text{COOH} \)
In simple words: By analyzing the elemental composition, molecular mass, and chemical tests (negative Tollens', positive iodoform, sodium bisulphite addition, and oxidation products), the unknown compound is identified as pentan-2-one.
🎯 Exam Tip: For structure elucidation problems, systematically use all given clues: elemental analysis for empirical formula, molecular mass for molecular formula, and characteristic reactions (Tollens', iodoform, addition reactions, oxidation) to deduce functional groups and carbon skeleton.

 

Question 17. Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?
Answer:
Acidity of carboxylic acids:
Carboxylic acids are inherently acidic and can donate a proton to form salts with bases.
\( \text{R}-\text{COOH} + \text{NaOH} \rightarrow \text{R}-\text{COONa} + \text{H}_2\text{O} \)
Carboxylic acids are weak acids, ionizing partially in aqueous solution, establishing an equilibrium with their unionized forms.
\( \text{R}-\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{R}-\text{COO}^- + \text{H}_3\text{O}^+ \)
The extent of ionization is quantified by the equilibrium constant \( \text{K}_a \), known as the acidity constant.
\( \text{K}_a = \frac{[\text{RCOO}^-][\text{H}_3\text{O}^+]}{[\text{RCOOH}]} \)
A higher \( \text{K}_a \) value indicates a stronger acid. Acidity is often expressed as \( \text{pK}_a = -\text{log K}_a \). As \( \text{pK}_a \) decreases, acid strength increases.

Comparison of Acidity: Carboxylic acids are stronger acids than phenols, despite the phenoxide ion having a greater number of resonance structures compared to the carboxylate ion. The key difference lies in the nature of resonance and charge distribution.
In a carboxylate ion (\( \text{RCOO}^- \)), the negative charge is delocalized over two electronegative oxygen atoms. This delocalization involves two equivalent resonance structures where the negative charge is equally shared between the two oxygen atoms, leading to a highly stable, symmetrical distribution of charge.
\[ \text{R-C}\begin{pmatrix} \text{O} \\ \Vert \\ \text{O}^- \end{pmatrix} \longleftrightarrow \text{R-C}\begin{pmatrix} \text{O}^- \\ \Vert \\ \text{O} \end{pmatrix} \]
In contrast, in a phenoxide ion (\( \text{C}_6\text{H}_5\text{O}^- \)), the negative charge is delocalized over one oxygen atom and several carbon atoms of the benzene ring. While there are multiple resonance structures, these structures are not equivalent. The negative charge is primarily on the more electronegative oxygen, and only partially delocalized onto the less electronegative carbon atoms of the ring. This unequal charge distribution means the phenoxide ion is less stable than the carboxylate ion, where the charge is symmetrically delocalized between two equivalent electronegative oxygen atoms.
\[ \text{C}_6\text{H}_5\text{O}^- \longleftrightarrow \text{C}_6\text{H}_5=\text{O} \]
Because the carboxylate ion is more effectively stabilized by resonance than the phenoxide ion, the release of a proton (\( \text{H}^+ \)) from a carboxylic acid is comparatively easier, making carboxylic acids stronger acids than phenols.
Some carboxylic acids and their \( \text{K}_a \) and \( \text{pK}_a \) values:

Acid	\( \text{K}_a \)	\( \text{pK}_a \)
HCOOH	\( 17.7 \times 10^{-5} \)	3.75
\( \text{CH}_3\text{COOH} \)	\( 1.75 \times 10^{-5} \)	4.76
\( \text{CH}_2\text{FCOOH} \)	\( 260 \times 10^{-5} \)	2.59
\( \text{CH}_2\text{ClCOOH} \)	\( 136 \times 10^{-5} \)	2.87
\( \text{CH}_2\text{BrCOOH} \)	\( 125 \times 10^{-5} \)	2.90
\( \text{CHCl}_2\text{COOH} \)	\( 5530 \times 10^{-5} \)	1.26
\( \text{CCl}_3\text{COOH} \)	\( 23200 \times 10^{-5} \)	0.63
\( \text{C}_6\text{H}_5\text{COOH} \)	\( 6.3 \times 10^{-3} \)	4.20

In simple words: Carboxylic acids are stronger acids than phenols because the negative charge in the carboxylate ion is equally shared between two highly electronegative oxygen atoms, leading to superior resonance stabilization compared to the phenoxide ion, where the charge is delocalized onto less electronegative carbon atoms and primarily resides on one oxygen.
🎯 Exam Tip: When comparing acid strengths, always analyze the stability of the conjugate base. Greater stability of the conjugate base (due to resonance, inductive effects, or electronegativity) corresponds to a stronger acid. Focus on the equivalence of resonance structures.

GSEB Class 12 Chemistry Aldehydes, Ketones and Carboxylic Acids Additional Important Questions and Answers

 

Question 1. In two separate bottles, you are provided with ethanol and 2-propanol. Using these, how will you prepare (i) ethanal (acetaldehyde) and (ii) acetone (propanone) by selecting suitable reagents?
Answer:
(i) To prepare ethanal (acetaldehyde) from ethanol, you can oxidize ethanol using either acidified \( \text{KMnO}_4 \) or by catalytic dehydrogenation using a copper tube at 573 K.
\( \text{CH}_3\text{CH}_2\text{OH} + \text{[O]} \xrightarrow{\text{KMnO}_4/\text{H}^+} \text{CH}_3\text{CHO} \)
(ii) To prepare acetone (propanone) from 2-propanol, you can oxidize 2-propanol using either acidified \( \text{KMnO}_4 \) or by catalytic dehydrogenation using a copper tube at 573 K.
\( \text{CH}_3-\text{CH}(\text{OH})-\text{CH}_3 + \text{[O]} \xrightarrow{\text{KMnO}_4/\text{H}^+} \text{CH}_3\text{COCH}_3 \)
In simple words: Ethanal is made from ethanol via oxidation, and acetone is made from 2-propanol also via oxidation, using either acidified potassium permanganate or hot copper.
🎯 Exam Tip: Remember that primary alcohols oxidize to aldehydes, and secondary alcohols oxidize to ketones. Mild oxidizing agents or catalytic dehydrogenation are used to prevent further oxidation of aldehydes to carboxylic acids.

 

Question 2. On ozonolysis of 1 mole of an alkene, a man got 2 moles of acetaldehyde. From this result, he proposed the following structure for the alkene.
\[ \text{CH}_3 - \text{C} = \text{CH}_2 \] \[ \quad \quad \mid \] \[ \quad \quad \text{CH}_3 \](i) Is it correct? Justify your answer.
(ii) Write the chemical equation.
Answer:
(i) No, the proposed structure is incorrect.
Justification: The given structure, 2-methylpropene, when subjected to ozonolysis, would yield acetone and formaldehyde, not two moles of acetaldehyde. If an alkene produces 2 moles of acetaldehyde (\( \text{CH}_3\text{CHO} \)) upon ozonolysis, it implies that the alkene itself must be symmetrical with a double bond between two carbons, each bearing one methyl group and one hydrogen atom. This structure is 2-butene.
(ii) Chemical equation for the ozonolysis of 2-butene:
\( \text{CH}_3\text{CH}=\text{CHCH}_3 \xrightarrow[\text{(ii) Zn/H}_2\text{O}]{\text{(i) O}_3} 2\text{CH}_3\text{CHO} \)
Chemical equation for the ozonolysis of the proposed structure (2-methylpropene):
\( \text{(CH}_3\text{)}_2\text{C}=\text{CH}_2 \xrightarrow[\text{(ii) Zn/H}_2\text{O}]{\text{(i) O}_3} \text{CH}_3\text{COCH}_3 + \text{HCHO} \)
In simple words: The proposed alkene structure is wrong because its ozonolysis gives acetone and formaldehyde, not two molecules of acetaldehyde. To get two acetaldehydes, the alkene must be 2-butene.
🎯 Exam Tip: Ozonolysis is a powerful reaction for determining the position of a double bond in an alkene. Remember that the double bond breaks, and each carbon atom that was part of the double bond gains an oxygen, forming aldehydes or ketones.

 

Question 3. Discuss three reactions in which formaldehyde differs from other aldehydes.
Answer:
(a) Reaction with ammonia: Formaldehyde reacts with ammonia to form hexamethylenetetramine (urotropine), a cyclic amine. Other aldehydes, however, react with ammonia to form aldehyde-ammonia addition products, which are typically unstable and polymerize.
\( 6\text{HCHO} + 4\text{NH}_3 \rightarrow (\text{CH}_2)_6\text{N}_4 + 6\text{H}_2\text{O} \)
(b) Cannizzaro reaction: Formaldehyde undergoes the Cannizzaro reaction because it lacks \(\alpha\)-hydrogens. In the presence of a strong base, it disproportionates to form methanol (reduction product) and sodium formate (oxidation product). Most other aldehydes with \(\alpha\)-hydrogens undergo aldol condensation instead.
\( 2\text{HCHO} + \text{NaOH} \xrightarrow{\text{Cannizzaro's reaction}} \text{CH}_3\text{OH} + \text{HCOONa} \)
(c) Reaction with Grignard reagents: Formaldehyde reacts with a Grignard reagent (\(\text{RMgX}\)) to produce a primary (\(1^\circ\)) alcohol upon hydrolysis. Other aldehydes (with at least one carbon attached to the carbonyl group) react with Grignard reagents to yield secondary (\(2^\circ\)) alcohols, and ketones yield tertiary (\(3^\circ\)) alcohols.
\( \text{HCHO} + \text{RMgX} \rightarrow \text{R}-\text{CH}_2\text{OMgX} \xrightarrow{\text{H}_2\text{O}} \text{R}-\text{CH}_2\text{OH} \)
In simple words: Formaldehyde is unique among aldehydes because it reacts with ammonia to form urotropine, undergoes the Cannizzaro reaction due to no alpha-hydrogens, and yields primary alcohols with Grignard reagents, unlike other aldehydes that give secondary alcohols.
🎯 Exam Tip: When distinguishing between formaldehyde and other aldehydes, focus on the presence or absence of \(\alpha\)-hydrogens, as this property dictates several key reaction pathways like aldol condensation versus Cannizzaro reaction.

 

Question 4. In three test tubes, you are provided with formaldehyde, acetaldehyde, and acetone. How will you distinguish them?
Answer:
We can distinguish between formaldehyde, acetaldehyde, and acetone using Tollens' reagent test and Iodoform test:
(i) Formaldehyde (\(\text{HCHO}\)):
- Gives a positive Tollens' reagent test (forms a silver mirror), as it is an aldehyde.
- Does not give a positive iodoform test, as it does not contain a \( \text{CH}_3\text{CO}- \) group.
(ii) Acetaldehyde (\(\text{CH}_3\text{CHO}\)):
- Gives a positive Tollens' reagent test (forms a silver mirror), as it is an aldehyde.
- Gives a positive iodoform test (forms a yellow precipitate of \( \text{CHI}_3 \)), as it contains a \( \text{CH}_3\text{CO}- \) group.
(iii) Acetone (\(\text{CH}_3\text{COCH}_3\)):
- Does not give a positive Tollens' reagent test, as it is a ketone.
- Gives a positive iodoform test (forms a yellow precipitate of \( \text{CHI}_3 \)), as it contains a \( \text{CH}_3\text{CO}- \) group.
Summary of Results:

Compound	Tollens' Test	Iodoform Test
Formaldehyde	Positive (Silver mirror)	Negative
Acetaldehyde	Positive (Silver mirror)	Positive (Yellow precipitate)
Acetone	Negative	Positive (Yellow precipitate)

In simple words: Formaldehyde gives a silver mirror with Tollens' but no iodoform test. Acetaldehyde gives both silver mirror and a yellow iodoform precipitate. Acetone gives a yellow iodoform precipitate but no silver mirror.
🎯 Exam Tip: Remember the specific functional groups that respond to Tollens' (aldehydes) and iodoform (methyl ketones and acetaldehyde) tests. This helps in quick identification and distinction of carbonyl compounds.

 

Question 5. (i) Why ketones are less reactive than aldehydes?
(ii) Benzaldehyde is less reactive than Acetaldehyde. Why?
Answer:
(i) Ketones are generally less reactive than aldehydes in nucleophilic addition reactions due to two main factors: steric hindrance and electronic effects.
- Steric hindrance: Ketones have two alkyl groups attached to the carbonyl carbon, whereas aldehydes have at least one hydrogen atom and one alkyl group (or two hydrogen atoms in formaldehyde). The two bulkier alkyl groups in ketones impede the approach of nucleophiles to the carbonyl carbon more effectively than the smaller hydrogen atom(s) in aldehydes.
- Electronic (inductive) effect: Alkyl groups are electron-donating by nature (+I effect). In ketones, there are two such alkyl groups, which donate electron density to the carbonyl carbon more than in aldehydes (which have only one alkyl group or none). This increased electron density on the carbonyl carbon (reducing its partial positive charge) makes it less electrophilic and thus less attractive to nucleophiles.
(ii) Benzaldehyde is less reactive than acetaldehyde in nucleophilic addition reactions primarily due to resonance stabilization and steric effects.
- Resonance stabilization: In benzaldehyde, the carbonyl group is directly attached to a benzene ring. The lone pair of electrons on the oxygen can participate in resonance with the delocalized \(\pi\)-electron system of the benzene ring. This resonance distributes the partial positive charge on the carbonyl carbon into the benzene ring, making the carbonyl carbon less electrophilic and therefore less susceptible to nucleophilic attack.
- Steric hindrance: The bulkier phenyl group in benzaldehyde causes more steric hindrance around the carbonyl carbon compared to the smaller methyl group in acetaldehyde, further reducing its reactivity towards nucleophiles.
In acetaldehyde, there is no such extended resonance stabilization of the carbonyl group with an aromatic ring, and the steric hindrance from the methyl group is much less compared to the phenyl group.
In simple words: Ketones are less reactive than aldehydes because their carbonyl carbon is more sterically hindered by two alkyl groups and is less positive due to electron donation. Benzaldehyde is less reactive than acetaldehyde because the benzene ring's resonance stabilizes the carbonyl group, making it less electrophilic, and the phenyl group adds more steric hindrance.
🎯 Exam Tip: When comparing reactivity of carbonyl compounds, always consider both steric factors (bulkiness around the carbonyl carbon) and electronic factors (inductive effects of alkyl groups, and resonance stabilization by adjacent aromatic rings or lone pairs).

 

Question 6. Carboxylic acids are stronger than phenols. Why?
Answer:
Carboxylic acids are stronger acids than phenols because the conjugate base formed from a carboxylic acid, the carboxylate ion, is more effectively stabilized by resonance than the conjugate base formed from a phenol, the phenoxide ion.
- Carboxylate ion: In the carboxylate ion (\( \text{RCOO}^- \)), the negative charge is delocalized over two highly electronegative oxygen atoms. The two resonance structures of the carboxylate ion are equivalent, meaning the negative charge is equally distributed between the two oxygen atoms. This symmetrical and extensive charge delocalization over equivalent electronegative atoms results in significant stabilization of the carboxylate ion.
- Phenoxide ion: In the phenoxide ion (\( \text{C}_6\text{H}_5\text{O}^- \)), the negative charge is initially on the oxygen atom and is delocalized into the benzene ring. While resonance does occur, the contributing resonance structures are not equivalent. The negative charge is borne by the more electronegative oxygen in one structure and by less electronegative carbon atoms in others. This unequal distribution of charge and the involvement of carbon atoms in bearing the negative charge make the phenoxide ion less stable compared to the carboxylate ion.
Because the carboxylate ion is more stable, the equilibrium for the dissociation of a carboxylic acid lies further to the right, leading to a higher concentration of \( \text{H}^+ \) ions and thus greater acidity compared to phenols.
In simple words: Carboxylic acids are stronger because their conjugate base (carboxylate ion) is very stable due to the negative charge being equally shared between two oxygen atoms through equivalent resonance. In contrast, the phenoxide ion's negative charge is less effectively stabilized as it's delocalized onto less electronegative carbons in non-equivalent resonance structures.
🎯 Exam Tip: The stability of the conjugate base is the primary factor determining acid strength. Always draw resonance structures for the conjugate bases and compare their equivalence, the number of electronegative atoms bearing the charge, and overall charge dispersion.

 

Question 7. How will you convert
(i) acetic acid to ethyl alcohol
(ii) Carboxylate ion
(iii) acetaldehyde to isopropyl alcohol
(iv) acetone to tertiary butyl alcohol
(v) acetophenone to benzoic acid
Answer:
(i) Acetic acid to ethyl alcohol:
Acetic acid can be reduced to ethyl alcohol using a strong reducing agent like lithium aluminum hydride (\( \text{LiAlH}_4 \)) in ether.
\( \text{CH}_3\text{COOH} \xrightarrow[\text{ether}]{\text{LiAlH}_4} \text{CH}_3\text{CH}_2\text{OH} \)
(ii) Carboxylate ion: (This seems like an incomplete or misplaced question part, as "Carboxylate ion" is a type of compound, not a conversion target from a specific starting material. Assuming it's asking how to convert a carboxylic acid to a carboxylate ion.)
A carboxylic acid can be converted to its carboxylate ion by reacting it with a base (e.g., \( \text{NaOH} \)).
\( \text{RCOOH} + \text{NaOH} \rightarrow \text{RCOONa} \text{ (Carboxylate salt)} + \text{H}_2\text{O} \)
(iii) Acetaldehyde to isopropyl alcohol:
Acetaldehyde reacts with a Grignard reagent (e.g., methylmagnesium chloride, \( \text{CH}_3\text{MgCl} \)) to form a secondary alcohol. In this case, it would form 2-propanol (isopropyl alcohol) after hydrolysis.
\( \text{CH}_3\text{CHO} + \text{CH}_3\text{MgCl} \rightarrow \text{CH}_3\text{CH}(\text{OMgCl})\text{CH}_3 \xrightarrow{\text{H}_2\text{O/H}^+} \text{CH}_3\text{CH}(\text{OH})\text{CH}_3 \)
(iv) Acetone to tertiary butyl alcohol:
Acetone reacts with a Grignard reagent (e.g., methylmagnesium chloride, \( \text{CH}_3\text{MgCl} \)) to form a tertiary alcohol. In this case, it would form 2-methylpropan-2-ol (tertiary butyl alcohol) after hydrolysis.
\( \text{CH}_3\text{COCH}_3 + \text{CH}_3\text{MgCl} \rightarrow (\text{CH}_3)_3\text{COMgCl} \xrightarrow{\text{H}_2\text{O/H}^+} (\text{CH}_3)_3\text{COH} \)
(v) Acetophenone to benzoic acid:
Acetophenone can be oxidized to benzoic acid using a strong oxidizing agent like alkaline \( \text{KMnO}_4 \) or \( \text{HNO}_3 \). The methyl group attached to the carbonyl is oxidized to a carboxylic acid group.
\( \text{C}_6\text{H}_5\text{COCH}_3 \xrightarrow{\text{KMnO}_4/\text{OH}^-} \text{C}_6\text{H}_5\text{COOH} + \text{HCOOH} \text{ (from methyl group)} \)
In simple words: This question covers various functional group transformations: reducing acetic acid to ethanol, making isopropyl alcohol from acetaldehyde and tertiary butyl alcohol from acetone using Grignard reagents, and oxidizing acetophenone to benzoic acid.
🎯 Exam Tip: For conversions, focus on the change in oxidation state and carbon skeleton. Strong reducing agents convert carboxylic acids to alcohols, Grignard reagents add to carbonyls to form alcohols (primary from formaldehyde, secondary from other aldehydes, tertiary from ketones), and strong oxidizing agents can cleave ketones to carboxylic acids or oxidize alkyl groups.

 

Question 8. (i) Complete the table
(ii) Illustrate Friedel Craft's acylation reaction.
Answer:
(i) Completion of the table:

Reactants	Main product	Name of the reaction
\( \text{CH}_3\text{COOH} \xrightarrow[\text{P}]{\text{Cl}_2} \)	\( \text{CH}_2\text{Cl-COOH} \) (A)	Hell-Volhard-Zelinsky (H.V.Z.) reaction
\( \text{CH}_3\text{COOC}_2\text{H}_5 + \text{CH}_3\text{OH} \xleftarrow{\text{H}^+} \)	\( \text{CH}_3\text{COOCH}_3 \) (B)	Trans-esterification
\( \text{CH}_3\text{CHO} + \text{H}_2 \xrightarrow[\text{Conc. HCl}]{\text{Zn-Hg}} \)	\( \text{CH}_3-\text{CH}_3 \) (C)	Clemmensen reduction
\( \text{CH}_3\text{COCl} + \text{H}_2 \xrightarrow[\text{BaSO}_4]{\text{Pd}} \)	\( \text{CH}_3\text{CHO} \) (E)	Rosenmund reduction

(ii) Friedel-Crafts acylation reaction:
Aromatic hydrocarbons react with an acid chloride (e.g., acetyl chloride) in the presence of an anhydrous aluminum chloride (\( \text{AlCl}_3 \)) catalyst to form aromatic ketones (e.g., acetophenone).
\[ \text{C}_6\text{H}_6 + \text{CH}_3\text{COCl} \xrightarrow{\text{anhyd. AlCl}_3} \text{C}_6\text{H}_5\text{COCH}_3 + \text{HCl} \] \[ \text{(Benzene)} + \text{(Acetyl chloride)} \rightarrow \text{(Acetophenone)} \] Similarly, with benzoyl chloride, benzophenone can be formed:
\[ \text{C}_6\text{H}_6 + \text{C}_6\text{H}_5\text{COCl} \xrightarrow{\text{anhyd. AlCl}_3} \text{C}_6\text{H}_5\text{COC}_6\text{H}_5 + \text{HCl} \] \[ \text{(Benzene)} + \text{(Benzoyl chloride)} \rightarrow \text{(Benzophenone)} \]In simple words: This question asks to complete a reaction table by identifying products and reaction names and to illustrate Friedel-Crafts acylation, which is a method to form aromatic ketones from aromatic hydrocarbons and acid chlorides using aluminum chloride.
🎯 Exam Tip: Memorize named reactions and their specific reagents, conditions, and typical products. Friedel-Crafts acylation is key for introducing acyl groups onto aromatic rings.

 

Question 9. What is esterification? How will you convert
(i) an ester to an amide
(ii) an ester to an acid chloride
Answer:
Esterification: Esterification is a chemical reaction that forms an ester from a carboxylic acid and an alcohol, typically in the presence of a strong acid catalyst (like \( \text{H}_2\text{SO}_4 \)). This reaction is reversible, and the forward reaction is known as esterification.
Example: Reaction of acetic acid with ethanol to form ethyl acetate.
\( \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{H}^+} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \)
(i) Conversion of an ester to an amide:
An ester reacts with ammonia (\( \text{NH}_3 \)) to produce an amide and an alcohol. This reaction is called ammonolysis of esters.
\( \text{RCOOR}' + \text{NH}_3 \rightarrow \text{RCONH}_2 + \text{R}'\text{OH} \)
(ii) Conversion of an ester to an acid chloride:
An ester can be converted to an acid chloride by reacting it with phosphorus pentachloride (\( \text{PCl}_5 \)).
\( \text{RCOOR}' + \text{PCl}_5 \rightarrow \text{RCOCl} + \text{R}'\text{OH} + \text{POCl}_3 \)
In simple words: Esterification is forming an ester from an acid and alcohol. An ester can be converted to an amide by reacting with ammonia, or to an acid chloride by reacting with phosphorus pentachloride.
🎯 Exam Tip: Understand esterification as a reversible condensation reaction. For ester transformations, remember that ammonia can replace the -OR' group with -NH2 (ammonolysis), and strong chlorinating agents can convert ester components to acid chlorides.

 

Question 10. How can you convert benzoic acid into
(i) 3-nitro benzoic acid
(ii) 3-sulpho benzoic acid
(iii) 3-Bromo benzoic acid?
Account for the above conversions.
Answer:
The carboxyl group (-\(\text{COOH}\)) in benzoic acid is a meta-directing group. Therefore, electrophilic substitution reactions on benzoic acid will predominantly occur at the meta position.
(i) Conversion to 3-nitro benzoic acid:
When benzoic acid is treated with a nitrating mixture (concentrated \( \text{HNO}_3 \) and concentrated \( \text{H}_2\text{SO}_4 \)), the nitro group is introduced at the meta position.
\[ \text{C}_6\text{H}_5\text{COOH} \xrightarrow[\text{conc. H}_2\text{SO}_4]{\text{conc. HNO}_3} \text{m-NO}_2\text{C}_6\text{H}_4\text{COOH} \] (ii) Conversion to 3-sulpho benzoic acid:
When benzoic acid is treated with concentrated \( \text{H}_2\text{SO}_4 \) (sulfonation), the sulfonic acid group is introduced at the meta position.
\[ \text{C}_6\text{H}_5\text{COOH} \xrightarrow{\text{H}_2\text{SO}_4} \text{m-HO}_3\text{S-C}_6\text{H}_4\text{COOH} \] (iii) Conversion to 3-Bromo benzoic acid:
When benzoic acid is treated with bromine water (\( \text{Br}_2\text{(aq)} \)) or \( \text{Br}_2 \) in the presence of a Lewis acid catalyst like \( \text{FeBr}_3 \), the bromine atom is introduced at the meta position.
\[ \text{C}_6\text{H}_5\text{COOH} \xrightarrow{\text{Br}_2\text{(aq)}} \text{m-BrC}_6\text{H}_4\text{COOH} \]In simple words: Benzoic acid can be converted to its 3-nitro, 3-sulpho, or 3-bromo derivatives because the carboxylic acid group directs incoming electrophiles to the meta position on the benzene ring.
🎯 Exam Tip: Always remember the directing nature of substituents on a benzene ring. For benzoic acid, the -COOH group is electron-withdrawing and meta-directing, so substitution reactions will lead to meta-substituted products.

 

Question 11.
Identify A, B, C, and D from the following series of reactions.
\( \text{C}_6\text{H}_5\text{COOH} \xrightarrow{\text{PCl}_5} \text{A} \xrightarrow{\text{B}} \text{C}_6\text{H}_5\text{CONH}_2 \xrightarrow{\text{P}_2\text{O}_5} \text{C} \xrightarrow{\text{H}_2/\text{Ni}} \text{D} \)
Answer:
The identification of compounds A, B, C, and D in the reaction sequence proceeds as follows:
\( \text{C}_6\text{H}_5\text{COOH} \xrightarrow{\text{PCl}_5} \text{C}_6\text{H}_5\text{COCl} \)
(A) is Benzoyl chloride.
\( \text{C}_6\text{H}_5\text{COCl} \xrightarrow{\text{NH}_3} \text{C}_6\text{H}_5\text{CONH}_2 \)
(B) is Ammonia (\( \text{NH}_3 \)).
\( \text{C}_6\text{H}_5\text{CONH}_2 \xrightarrow{\text{P}_2\text{O}_5} \text{C}_6\text{H}_5\text{CN} \)
(C) is Benzonitrile.
\( \text{C}_6\text{H}_5\text{CN} \xrightarrow{\text{H}_2/\text{Ni}} \text{C}_6\text{H}_5\text{CH}_2\text{NH}_2 \)
(D) is Benzylamine.
In simple words: This question traces the conversion of benzoic acid through several steps to produce benzylamine, involving reactions to form an acyl chloride, amide, nitrile, and finally, an amine.
🎯 Exam Tip: Understanding common reaction types like chlorination, amidation, dehydration, and reduction is crucial for solving synthesis problems. Pay attention to specific reagents and their roles.

 

Question 12.
Identify A, B and C in the following reactions.
\( \text{CH}_3\text{COCH}_3 \xrightarrow{\text{LiAlH}_4} \text{A} \xrightarrow{\text{HBr}} \text{B} \xrightarrow{\text{alc.KOH}} \text{C} \)
Answer:
Let's identify the compounds A, B, and C in the given reaction sequence:
Starting with acetone, \( \text{CH}_3\text{COCH}_3 \), a reduction with lithium aluminium hydride (\( \text{LiAlH}_4 \)) occurs.
\( \text{CH}_3\text{COCH}_3 \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{CH(OH)}\text{CH}_3 \)
(A) is Propan-2-ol (isopropyl alcohol).
Next, compound A reacts with HBr.
\( \text{CH}_3\text{CH(OH)}\text{CH}_3 \xrightarrow{\text{HBr}} \text{CH}_3\text{CH(Br)}\text{CH}_3 \)
(B) is 2-Bromopropane.
Finally, compound B undergoes a reaction with alcoholic KOH.
\( \text{CH}_3\text{CH(Br)}\text{CH}_3 \xrightarrow{\text{alc.KOH}} \text{CH}_3\text{CH=CH}_2 \)
(C) is Propene.
In simple words: The sequence starts with acetone, which is reduced to an alcohol (propan-2-ol). This alcohol is then converted to an alkyl halide (2-bromopropane), which finally undergoes elimination to yield an alkene (propene).
🎯 Exam Tip: Remember common reduction agents like \( \text{LiAlH}_4 \) for ketones, halogenation of alcohols, and the use of alcoholic KOH for dehydrohalogenation to form alkenes.

 

Question 13.
A compound extracted from a plant has the molecular formula \( \text{C}_3\text{H}_6\text{O}_2 \). It reacts with ammonia to form an amide and methanol.
(i) Identify the compound
(ii) Comment on the possibility of two methods of preparation of the compound from the compounds methanol, ethanoic acid and ethanoyl chloride.
(iii) Represent the chemical change when the compound is hydrolysed in presence of alkali.
Answer:
(i) The compound with molecular formula \( \text{C}_3\text{H}_6\text{O}_2 \) that reacts with ammonia to produce an amide and methanol is Methyl Ethanoate (methyl acetate). Its structure is \( \text{CH}_3\text{COOCH}_3 \).
\( \text{CH}_3\text{COOCH}_3 + \text{NH}_3 \implies \text{CH}_3\text{CONH}_2 + \text{CH}_3\text{OH} \)
(Methyl Ethanoate) (Ammonia) \( \implies \) (Acetamide) (Methanol)

(ii) Two methods for preparing methyl ethanoate:
(a) Esterification: When methanol reacts with methanoic acid in the presence of \( \text{H}_2\text{SO}_4 \), methyl ethanoate is formed.
\( \text{CH}_3\text{COOH} + \text{CH}_3\text{OH} \xrightarrow{\text{H}^+} \text{CH}_3\text{COOCH}_3 + \text{H}_2\text{O} \)
(Ethanoic Acid) (Methanol) \( \implies \) (Methyl Ethanoate) (Water)

(b) Acylation: The reaction between methanol and ethanoyl chloride also yields methyl ethanoate.
\( \text{CH}_3\text{COCl} + \text{CH}_3\text{OH} \implies \text{CH}_3\text{COOCH}_3 + \text{HCl} \)
(Ethanoyl Chloride) (Methanol) \( \implies \) (Methyl Ethanoate) (Hydrochloric Acid)

(iii) When methyl ethanoate undergoes alkaline hydrolysis (saponification) with NaOH, it yields sodium acetate and methanol.
\( \text{CH}_3\text{COOCH}_3 + \text{NaOH} \implies \text{CH}_3\text{COONa} + \text{CH}_3\text{OH} \)
(Methyl Ethanoate) (Sodium Hydroxide) \( \implies \) (Sodium Acetate) (Methanol)
In simple words: The plant compound is methyl ethanoate. It can be made from ethanoic acid and methanol (esterification) or ethanoyl chloride and methanol. When reacted with NaOH, it breaks down into sodium acetate and methanol.
🎯 Exam Tip: Familiarize yourself with common ester reactions, including their synthesis via esterification and acylation, and their hydrolysis (saponification) in alkaline conditions.

 

Question 14.
Compare acidic strength of acetic acid, formic acid, and chloroacetic acid.
Answer:
The acidic strength of carboxylic acids is influenced by the inductive effect of substituents. The order of acidic strength is: Chloroacetic acid > Formic acid > Acetic acid.
- **Chloroacetic acid (\( \text{ClCH}_2\text{COOH} \)):** The chlorine atom exhibits a strong negative inductive effect (-I). This effect stabilizes the carboxylate ion by delocalizing the negative charge, making the O-H bond easier to break and release a proton (\( \text{H}^+ \)). Thus, chloroacetic acid is the strongest among the three.
- **Formic acid (\( \text{HCOOH} \)):** Formic acid lacks any groups with a significant inductive effect, either electron-donating (+I) or electron-withdrawing (-I). Its acidity serves as a reference point.
- **Acetic acid (\( \text{CH}_3\text{COOH} \)):** The methyl group (\( \text{CH}_3 \)) has a positive inductive effect (+I). This electron-donating effect destabilizes the carboxylate ion by intensifying the negative charge, making it harder to release a proton. Consequently, acetic acid is the weakest among the three.
In summary, the presence of electron-withdrawing groups (like Cl) increases acidity, while electron-donating groups (like \( \text{CH}_3 \)) decrease acidity. Formic acid is stronger than acetic acid because it lacks the electron-donating methyl group found in acetic acid, and chloroacetic acid is stronger than formic acid due to the electron-withdrawing chlorine atom.
In simple words: Chloroacetic acid is the strongest because chlorine pulls electrons, stabilizing its acid form. Formic acid is next, with no strong electron-pushing or -pulling groups. Acetic acid is the weakest because its methyl group pushes electrons, making it less acidic.
🎯 Exam Tip: To compare acid strengths, always consider the inductive effects of substituents on the carboxylate ion. Electron-withdrawing groups enhance acidity, while electron-donating groups diminish it.

 

Question 15.
The percentage of C and H in an organic compound is in the ratio 6 : 1 and that of C and O is in the ratio 3: 4.
(a) Identify the compound.
(b) What is the end product of the reaction between this compound and phenol? Explain the reaction.
(c) What are the uses of end products?
Answer:
(a) To identify the compound, let's determine its empirical formula from the given ratios.
Given: C:H = 6:1 (by mass) and C:O = 3:4 (by mass).
Let's assume the mass of carbon (C) is 6 units. Then the mass of hydrogen (H) is 1 unit. For the C:O ratio, if C is 3 units, O is 4 units. To make the carbon mass consistent, if C is 6 units, then O would be 8 units.
So, the mass ratio C:H:O is 6:1:8.
Divide by atomic masses (C=12, H=1, O=16):
C: 6/12 = 0.5
H: 1/1 = 1
O: 8/16 = 0.5
Divide by the smallest ratio (0.5):
C: 0.5/0.5 = 1
H: 1/0.5 = 2
O: 0.5/0.5 = 1
The empirical formula is \( \text{CH}_2\text{O} \). This corresponds to Formaldehyde (\( \text{HCHO} \)).
Therefore, the compound is Formaldehyde (\( \text{HCHO} \)).

(b) When formaldehyde reacts with phenol, the end product is Bakelite. This reaction involves the acid-catalyzed or base-catalyzed condensation polymerization of phenol and formaldehyde. Initially, ortho- and para-hydroxymethylphenols are formed, which then undergo further condensation to create cross-linked polymer structures.
The reaction can be represented as:
\( \text{Phenol} + \text{HCHO} \implies \text{Bakelite (polymer)} \)


ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र फिनोल (एक बेंजीन रिंग जिसमें -OH समूह जुड़ा है) और फॉर्मलाडेहाइड (HCHO) के बीच बहुलकन प्रक्रिया को दर्शाता है। प्रतिक्रिया के बाद, एक जटिल क्रॉस-लिंक्ड बहुलक, बैकेलाइट, बनता है, जो एक कठोर प्लास्टिक सामग्री है।

(c) Uses of Bakelite:
Bakelite is a thermosetting polymer widely used for making electrical switches, handles of cooking utensils, combs, phonograph records, and various other molded articles requiring durability and heat resistance.
In simple words: The compound is formaldehyde. It reacts with phenol to create Bakelite, a type of plastic. Bakelite is used to make things like electrical switches and comb.
🎯 Exam Tip: For problems involving elemental composition ratios, always derive the empirical formula first. Recognize common condensation polymerization reactions like that between phenol and formaldehyde (Bakelite formation), and be aware of the properties and uses of such polymers.

 

Question 16.
An organic compound having general formula \( \text{C}_2\text{H}_4\text{O} \) on oxidation with Tollens reagent produces B. B on treating with KOH produces C and C on electrolysis produces D. When silver salt of B is treated with \( \text{Br}_2/\text{CCl}_4 \) produces E. Identify A, B, C, D, E.
Answer:
Let's identify the compounds A, B, C, D, and E based on the given reactions:
Compound A has the general formula \( \text{C}_2\text{H}_4\text{O} \). On oxidation with Tollens' reagent, it produces B. This indicates that A is an aldehyde. The only aldehyde with the formula \( \text{C}_2\text{H}_4\text{O} \) is Ethanal (\( \text{CH}_3\text{CHO} \)).
So, **A = Ethanal (\( \text{CH}_3\text{CHO} \))**.

When ethanal is oxidized by Tollens' reagent, it forms acetic acid.
\( \text{CH}_3\text{CHO} \xrightarrow{\text{Tollens' reagent}} \text{CH}_3\text{COOH} \)
So, **B = Ethanoic acid (\( \text{CH}_3\text{COOH} \))**.

When B (ethanoic acid) is treated with KOH, it forms potassium acetate.
\( \text{CH}_3\text{COOH} \xrightarrow{\text{KOH}} \text{CH}_3\text{COOK} \)
So, **C = Potassium acetate (\( \text{CH}_3\text{COOK} \))**.

When C (potassium acetate) undergoes electrolysis, it yields ethane (Kolbe's electrolysis).
\( \text{CH}_3\text{COOK} \xrightarrow{\text{electrolysis}} \text{CH}_3\text{-CH}_3 \)
So, **D = Ethane (\( \text{CH}_3\text{-CH}_3 \))**.

When the silver salt of B (silver acetate, formed from ethanoic acid) is treated with \( \text{Br}_2/\text{CCl}_4 \), it undergoes Hunsdiecker reaction to produce bromomethane.
\( \text{CH}_3\text{COOAg} + \text{Br}_2 \xrightarrow{\text{CCl}_4} \text{CH}_3\text{Br} + \text{CO}_2 + \text{AgBr} \)
So, **E = Bromomethane (\( \text{CH}_3\text{Br} \))**.
In simple words: Starting with ethanal (A), it's oxidized to acetic acid (B). Acetic acid reacts with KOH to form potassium acetate (C), which then undergoes electrolysis to produce ethane (D). If silver acetate (from B) reacts with bromine, it forms bromomethane (E).
🎯 Exam Tip: This question tests knowledge of various named reactions: Tollens' test for aldehydes, acid-base reactions, Kolbe's electrolysis, and Hunsdiecker reaction. Identifying the functional groups and reaction conditions is key to solving such conversion series.