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NCERT Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids
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Aldehydes, Ketones and Carboxylic Acids
Chapter: 12
| Part – II |
1. What is meant by the following terms? Give an example of the reaction in each case.
(i) Cyanohydrin.
Ans: Cyanohydrins are also known as hydroxynitriles. Cyanohydrins are organic compounds containing a cyano group (-CN) and a hydroxyl group (-OH) attached to the same carbon atom.
Example: The reaction of acetone with hydrogen cyanide (HCN) forms acetone cyanohydrin.
(ii) Acetal.
Ans: Acetals also known as Polyoxymethylene (POM). Acetals are common carbonyl compound derivatives that are often used in Organic Synthesis as protecting groups for aldehydes and ketones, as well as in many other reactions.
Example: The reaction of acetaldehyde (CH₃CHO) with ethanol forms acetal.
(iii) Semicarbazone.
Ans: A semicarbazone is a derivative of imines formed by a condensation reaction between a ketone or aldehyde and semicarbazide.
Example: Acetone reacts with semicarbazide to form acetone semicarbazone.
(iv) Aldol.
Ans: The aldol reaction (aldol addition) is a reaction in organic chemistry that combines two carbonyl compounds to form a new β-hydroxy carbonyl compound.
Example: Acetaldehyde undergoes aldol condensation to form 3-hydroxybutanal.
(v) Hemiacetal.
Ans: When an alcohol adds to an aldehyde, the result is called a hemiacetal; when an alcohol adds to a ketone the resulting product is a hemiketal.
Example: Acetaldehyde reacts with one molecule of ethanol to form a hemiacetal.
(vi) Oxime.
Ans: An oxime is a chemical compound belonging to the imines, with the general formula RR1C=NOH, where R is a organic side-chain.
Example: Acetone reacts with hydroxylamine to form acetone oxime.
(vii) Ketal.
Ans: Ketal is a functional group that is derived from a ketone. Ketal is also referred to as an acetal derivative of ketone. Ketals are used in industry as solvents, plasticizers, and intermediates.
Example: Acetone reacts with ethanol to form an ethyl ketal.
(vii) Imine.
Ans: A compound containing the NH group or its substituted form NR that is derived from ammonia by replacement of two hydrogen atoms by a hydrocarbon group or other nonacid organic group. They are used to synthesise dyes and coordination polymers.
Example: Acetone reacts with methylamine to form an imine.
(ix) 2,4-DNP-derivative.
Ans: 2,4-Dinitrophenylhydrazine can be used for the qualitative identification of ketone or aldehyde functional group carbonyl functionality. A positive result is shown by the formation of a precipitate called dinitrophenylhydrazone, which can be yellow, orange, or red.
Example: Acetone reacts with 2,4-DNP to form acetone 2,4-DNP derivative.
(x) Schiff’s base.
Ans: Schiff’s bases are imines where the nitrogen atom is connected to an aryl or alkyl group rather than a hydrogen.
Example: Benzaldehyde reacts with aniline to form a Schiff’s base.
2. Name the following compounds according to IUPAC system of nomenclature:
(i) CH3CH(CH3)CH2CH2CHO
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(ii) CH3CH2COCH(C2H5)CH2CH2Cl
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(iii) CH3CH = CHCHO
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(iv) CH3COCH2COCH3
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(v) CH3CH(CH3)CH2C(CH3)2CH3
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(vi) (CH3)3CCH2COOH
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(vii) OH C6H4 CHO-p
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3. Draw the structures of the following compounds.
(i) 3-Methylbutanal.
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(ii) p-Nitropropiophenone.
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(iii) p-Methylbenzaldehyde.
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(iv) 4-Methylpent-3-en-2-one.
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(v) 4-Chloropentan-2-one.
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(vi) 3-Bromo-4-phenylpentanoic acid.
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(vii) p,p’-Dihydroxybenzophenone.
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(viii) Hex-2-en-4-ynoic acid.
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4. Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.
(i) CH3CO(CH2)4CH3
Ans: IUPAC Name: 2-Heptanone Common Name: Methyl n-pentyl ketone
(ii) CH3CH2CHBrCH2CH(CH3)CHO
Ans: IUPAC Name: 4-Bromo-3-methylhexane Common Name: 4-Bromo-3-methyl-n-hexaldehyde
(iii) CH3(CH2)5CHO
Ans: IUPAC Name: Heptanal Common Name: n-Heptaldehyde.
(iv) Ph – CH = CH – CHO
Ans: IUPAC Name: 3-Phenylprop-2-enal. Common Name: Cinnamaldehyde
(v)
Ans: IUPAC Name: Cyclopentene carbaldehyde. Common Name: cyclopentanecarboxaldehyde.
(vi) PhCOPh.
Ans: IUPAC Name: Diphenylmethanone. Common Name: Benzophenone
5. Draw structures of the following derivatives.
(i) The 2,4-dinitrophenylhydrazone of benzaldehyde.
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(ii) Cyclopropanone oxime.
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(iii) Acetaldehyde Dimethyl Acetal.
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(iv) The semicarbazone of cyclobutanone.
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(v) The ethylene ketal of hexan-3-one.
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(vi) The methyl hemiacetal of formaldehyde.
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6. Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents.
(i) PhMgBr and then H3O+
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(ii) Tollens’ reagent.
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(iii) Semicarbazide and weak acid.
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(iv) Excess ethanol and acid.
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(v) Zinc amalgam and dilute hydrochloric acid.
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7. Which of the following compounds would undergo aldol condensation, which 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.
Ans: Aldol Condensation: (ii), (v), (vi), (vii). The products are:
(ii)
(v)
(vi)
(vii)
Cannizzaro Reaction: (i), (iii), (ix)
(i) CH3OH + HCOONa
(iii) C6H4CH2OH + C6H5COONa
(ix)
Neither: (iv), (viii).
8. How will you convert ethanal into the following compounds?
(i) Butane-1,3-diol.
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(ii) But-2-enal.
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(iii) But-2-enoic acid.
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9. Write structural formulas and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde acts as nucleophile and which as electrophile.
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10. An organic compound with the molecular formula C9H10O forms 2,4-DNP derivative, reduces Tollens’ reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound.
Ans: An organic molecule with the chemical formula C9H10O produces 2,4-DNP derivative while also reducing Tollens reagent. As a result, So it is an aldehyde.
(ii) It undergoes Cannizzaro reaction, so the aldehyde group should be directly attached to the benzene ring.
(iii) The chemical goes through the Cannizzaro reaction. As a result, it lacks an alpha H atom. It produces 1,2-benzene dicarboxylic acid after severe oxidation. As a result, the-CHO group is directly linked to the benzene ring, resulting in ortho di-substituted benzene. 2-ethyl benzaldehyde is the compound. For molecular formula C₂H₂O, the possibility is only O-ethyl benzaldehyde.
11. 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.
Ans: (i) An organic compound A with molecular formula C8H16O2 gives a carboxylic acid (B) and an alcohol (C) on hydrolysis with dilute sulphuric acid. Thus, compound A must be an ester.
(ii) Alcohol (C) on oxidation produces acid [B]. It means both [B] and [C] have same number of carbon atoms, i.e., four each.
(iii) The equations for all the above reactions are:
12. Arrange the following compounds in increasing order of their property as indicated:
(i) Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN)
Ans: (i) The reactivity of a compound depends on the steric hindrance due to the groups present around the carboxyl group. Increased steric hindrance decreases compound reactivity.
Reactivity towards HCN follows this order:
Di-tertiary butyl ketone < Methyl tertiary butyl ketone < Acetone < Acetaldehyde.
(ii) Alkyl group with + I effect decreases the acidic
(ii) CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH, (CH3 )2 CHCOOH, CH3CH2CH2COOH (acid strength)
Ans: Alkyl group with + I effect decreases the acidic strength whereas with – I effect increases the acidic strength -I effect decreases with distance.
Increasing order of acidic strength is
(CH₃)2 CHCOOH < CH3CH2CH2COOH < CH3CH (Br) CH2COOH < CH3CH2 CH(Br) COOH
Increasing order of acidic strength is
(CH₃)₂ CH COOH < CH3 CH2 CH2 COOH < CH3CH (Br) CH2 COOH < CH3 CH2 CH(Br)COOH
(iii) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)
Ans: The strength of acids is diminished by electron donating groups, whereas the strength of acids is increased by electron-withdrawing groups, as we saw in the prior situation. Benzoic acid is stronger than 4-methoxybenzoic acid because the methoxy group donates electrons, reducing acidity. The nitro group, being electron-withdrawing, increases acidity. Since 3,4-dinitrobenzoic acid has two nitro groups, it is even stronger than 4-nitrobenzoic acid. Therefore, acidity increases accordingly.
Increasing order of acidic strength is: 4-Methoxy benzoic acid < benzoic acid < 4-nitro benzoic acid < 3,4-dinitrobenzoic acid.
13. Give simple chemical tests to distinguish between the following pairs of compounds.
(i) Propanal and Propanone.
Ans: Boil both compounds separately with iodine and aqueous sodium carbonate solution. Propanone gives a yellow precipitate in the iodoform test, indicating the presence of a methyl ketone group. On the other hand, propanal, being an aldehyde without a methyl group attached to the carbonyl carbon, does not react in the iodoform test.
(ii) Acetophenone and Benzophenone.
Ans: Acetophenone and Benzophenone: Aceto-phenone and benzophenone can be distinguished by iodoform test. Acetophenone forms a yellow iodoform precipitate with I₂ and NaOH, while benzophenone does not react.
(iii) Phenol and Benzoic acid.
Ans: Phenol and benzoic acid can be distinguished using the sodium bicarbonate test, as only benzoic acid produces effervescence. To the solutions of both benzoic acid and phenol, add a small amount of solid bicarbonate. Whereas benzoic acid gives brisk effervescence of carbon dioxide which turns lime water milky, phenol does not react. Phenol reacts with neutral FeCl3 to form an iron-phenol complex giving violet colouration.
(iv) Benzoic acid and Ethyl benzoate.
Ans: (a)Tollens’ Test: Benzaldehyde (an aldehyde) gives a positive Tollens’ test (silver mirror), while acetophenone (a ketone) does not.
(b) Benzoic acid produces vapours of benzene when heated with soda lime (CaO + NaOH) while ethyl benzoate does not.
(v) Pentan-2-one and Pentan-3-one.
Ans: Both are ketones; however, pentan-2-one is a methyl ketone, while pentan-3-one does not have a methyl group. Therefore, they can be distinguished with the help of Iodoform test which is given by pentan-2-one and not by pentan-3-one. But pentan-3-one not being a methyl ketone does not respond to this test.
(vi) Benzaldehyde and Acetophenone.
Ans: Benzaldehyde and Acetophenone: Acetophenone, on treatment with I, and NaOH gives yellow ppt. of iodoform while benzaldehyde does not give this test.
(vii) Ethanal and Propanal.
Ans: Both aldehydes give a positive Tollens’ test, but they can be distinguished by their boiling points or using 2,4-DNP derivatives, as Ethanal will form acetaldehyde derivative, and Propanal will form propanal derivative, which can be separated based on melting points.
14. 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.
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(ii) m-Nitrobenzoic acid.
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(iii) p-Nitrobenzoic acid.
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(iv) Phenylacetic acid.
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(v) p-Nitrobenzaldehyde.
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15. How will you bring about the following conversions in not more than two steps?
(i) Propanone to Propene.
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(ii) Benzoic acid to Benzaldehyde.
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(iii) Ethanol to 3-Hydroxybutanal.
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(v) Benzaldehyde to Benzophenone.
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(iv) Benzene to m-Nitroacetophenone.
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(vi) Bromobenzene to 1-Phenylethanol.
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(vii) Benzaldehyde to 3-Phenylpropan-1-ol.
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(viii) Benzaldehyde to-Hydroxyphenylacetic acid.
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(ix) Benzoic acid to m-Nitrobenzyl alcohol.
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16. Describe the following:
(i) Acetylation.
Ans: Acetylation is a chemical reaction that adds an acetyl group to a chemical compound, replacing a hydrogen atom. The acetyl group is a chemical functional group that consists of a carbonyl group and a methyl group. Deacetylation is the reverse process where an acetyl group is removed from a molecule, often catalysed by enzymes.
Acetylation: Acetylation is the replacement of H atom of alcohols, amines and other such compounds with acetyl group, e.g.,
(ii) Cannizzaro reaction.
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Cannizzaro’s Reaction: In this reaction two molecules of aldehydes (without the a-hydrogen atom) react in presence of concentrated alkali by mutual oxidation-reduction forming one molecule of each of the corresponding alcohol and acid. When methanol is heated with alkali, it produces methanol and methanoic acid through a chemical reaction.
(iii) Cross aldol condensation Chemistry 256
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Cross aldol condensation is a reaction that occurs when two different carbonyl compounds, such as aldehydes or ketones, undergo aldol condensation. The reaction usually produces a mixture of four products, which can be synthetically useless.
(iv) Decarboxylation.
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Decarboxylation: Decarboxylation is a chemical reaction that removes a carboxyl group from a compound and releases carbon dioxide (CO2). It’s most often used to describe the reaction of carboxylic acids, where a carbon atom is removed from a carbon chain. Carboxylic acid lose carbon dioxide, when their sodium salts are heated with soda lime (NaOH + CaO). The reaction is known as Decarboxylation.
17. Complete each synthesis by giving missing starting material, reagent or products:
(i)
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(ii)
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(iii)
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(iv)
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(v)
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(vi)
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(vii)
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(viii)
Ans: CH3CH(OH)CH2COOC2H5
(ix)
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(x)
Ans: 1. R2BH;
2. H2O2/OH;
3. PCC
(xi)
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(xii)
Ans: C6H5COCL-AICI3
18. Give plausible explanation for each of the following:
(i) Cyclohexanone forms cyanohydrin in good yield but 2, 2, 6-trimethylcyclo hexanone does not.
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In 2,2,6 trimethyl cyclohexanone, three methyl groups are present at a-position with respect to the ketonic (> C = O) group. Therefore, these groups cause steric hindrance during the nucleophilic attack of CN-ion, so cyanohydrin is not formed. Cyclohexanone forms cyanohydrin without steric hindrance due to the absence of methyl groups, allowing smooth reaction.
(ii) There are two–NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones.
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Semicarbazide has two amino groups, but one participates in resonance, reducing its electron density. Consequently, this NH group has lower nucleophilicity and does not act as a nucleophile in reactions.
But the other NH₂ group (attached to NH) has a lone pair of electrons which are not involved in resonance. So, this pair is available for the nucleophilic attack on the carbonyl group (> C = O) of aldehydes and/or ketones.
(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.
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An ester and water form reversibly from a carboxylic acid and alcohol in the presence of an acid catalyst. If either water or ester is not removed as soon as it is formed, then it reacts to give back the reactants as the reaction is reversible. Therefore, to shift the equilibrium in the forward direction i.e., to produce more ester, either of the two should be removed.
19. 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 additional compound with sodium hydrogen sulphite and give positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.
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| Element | Percentage | At.Mass | Reactive No. of Atoms | Simplest Ratio |
| Carbon (C) | 69.77 | 12 | 69.77 / 12 = 5.814 | 5.8/1.16 = 5 |
| Hydrogen (H) | 11.63 | 1 | 11.63 / 1 = 11.63 | 11.63/1.63 = 10 |
| Oxygen (O) | 18.60 | 16 | 18.60 / 16 = 1.162 | 1.16/1.16 = 1 |
Empirical formula = C5H10O
Empirical formula mass = 5 х 12 + 10 х 1 + 16 = 60 + 10 + 16 = 86
Molecular mass = 86
N = (Molecules formula mass / Empirical formula mass)
= 86/86 = 1
Molecular formula = C5H10O
Structure of the compound CH3CH2COCH2CH3.
20. Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?
Ans: The negative charge in the carboxylate ion is delocalized over two oxygen atoms, while the negative charge in the phenoxide ion is delocalized over one oxygen atom. This makes the carboxylate ion more stable than the phenoxide ion. Carboxylic acid is stronger than phenol because the carboxylate ion has more effective resonance stabilisation, with the negative charge equally distributed between two oxygen atoms. In contrast, phenoxide ion’s resonance is less stabilising, as the negative charge is less effectively delocalized across the aromatic ring.
