NCERT Exercise Solutions – Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids
12.1 What is meant by the following terms ? Give an example of the reaction in each case.
(i) Cyanohydrin (ii) Acetal (iii) Semicarbazone
(iv) Aldol (v) Hemiacetal (vi) Oxime
(vii) Ketal (vii) Imine (ix) 2,4-DNP-derivative
(x) Schiff’s base
Ans- (i) Cyanohydrins are gem-Hydroxynitriles, which are substances with hydroxyl and cyano groups on the same carbon atom. In a weakly basic media, they are made by adding HCN to aldehydes or ketones.
(ii) Acetals are dialkoxy compounds that have two alkoxy groups attached to the terminal carbon atom. These are created when two equivalents of a monohydric alcohol react with an aldehyde in the presence of dry HCl gas.
(iii) Semicarbazides, which are derived from aldehydes and ketones, act on them in an acidic solution to form semicarbazones.
(iv) P-hydroxy aldehydes or ketones, known as aldols, are created when two molecules of the same compound or one molecule of each of two separate compounds are combined in the presence of a diluted aqueous base.
(v)Alkoxyalcohols are referred to as hemiacetals in gem. These are made by mixing one molecule of a monohydric alcohol with an aldehyde while dry HCl gas is present.
(vi)Oximes are created when hydroxyl amine and aldehydes or ketones combine in a mildly acidic media.
12.2 Name the following compounds according to IUPAC system of nomenclature:
Ans- (i) 4-Methyl pentanal
(vi) 3,3-Dimethyl butanoic acid
12.3 Draw the structures of the following compounds.
(vi) 3-Bromo-4-phenylpentanoic acid
(viii) Hex-2-en-4-ynoic acid
Ans– Structures are:
12.4 Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.
(i) CH3CO(CH2 )4CH3
(ii) CH3CH2CHBrCH2CH(CH3 )CHO
(v) CHO (vi) PhCOPh
Ans – (i) CH3CO(CH2)4CH3 – IUPAC name is: Heptan-2-one
(ii) CH3CH2CHBrCH2CH(CH3)CHO – IUPAC name is: 4-Bromo-2-methylhaxanal
(iii) CH3(CH2)5CHO – IUPAC name is: Heptanal
(iv) Ph-CH=CH-CHO – IUPAC name is: 3-phenylprop-2-enal
(v)IUPAC name is: Cyclopentanecarbaldehyde
(vi)PhCOPh – IUPAC name is: Diphenylmethanone
12.5 Draw structures of the following derivatives.
(i) The 2,4-dinitrophenylhydrazone of benzaldehyde
(ii) Cyclopropanone oxime
(iv) The semicarbazone of cyclobutanone
(v) The ethylene ketal of hexan-3-one
(vi) The methyl hemiacetal of formaldehyde
12.6 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
12.7 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 happens with aldehydes and ketones that include at least one -hydrogen. One or more -hydrogen atoms can be found in the compounds (ii), 2-methylpentanal, (v), cyclohexanone, (vi), 1-phenylpropanone, and (vii), phenylacetaldehyde. These hence go through aldol condensation.
Cannizzaro reactions occur with aldehydes that don’t include -hydrogen atoms. Methanal, benzaldehyde, and 2, 2-dimethylbutanal are the only three substances that lack a -hydrogen. These therefore experience cannizzaro reactions.
Compound (iv) (iv) Benzophenone is an alcohol and compound (viii) butan-1-ol is a ketone with no -hydrogen atom. As a result, neither aldol condensation nor cannizzaro reactions occur in these molecules.
12.8 How will you convert ethanal into the following compounds?
(iii) But-2-enoic acid
Ans- On treatment with dilute alkali
12.9 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.
Ans- Propanal and butane both contain a-hydrogen atoms. These can go through both self- and cross-aldol condensation to produce the following four compounds:
12.10 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- The provided substance, C9H10O, must be an aldehyde because it reduces Tollen’s reagent and creates a derivative of 2,4-DNP. Because of the Cannizzaro reaction, the CHO group is consequently immediately linked to the benzene ring.
Since it produces 1, 2-benzene dicarboxylic acid upon severe oxidation, it must be an ortho- substituted benzaldehyde. O-ethyl benzaldehyde is the only o-substituted aromatic aldehyde with the chemical formula C9H10O. On the basis of this framework, all of the reactions can now be explained.
12.11 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- Since the carboxylic acid B and alcohol C are produced by the hydrolysis of an ester A with the chemical formula C8H16O2, and the acid B is produced by the oxidation of C with chromic acid, both the carboxylic acid B and alcohol C must have the same number of carbon atoms.
Furthermore, because ester A has eight carbon atoms, the carboxylic acid B and the alcohol C must each have four carbon atoms.
Since dehydration of alcohol C results in but-l-ene, C must be a straight chain alcohol, such as butan-l-ol.
If C is butanol, then the ester A must be butyl butanoate and the acid B must be butanoic acid.
The following are the chemical equations:
12.12 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)
(ii) CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH, (CH3)2CHCOOH, CH3CH2CH2COOH (acid strength)
(iii) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)
Ans- (i)A nucleophile, or CN, attacks a chemical when HCN reacts with it. As a result, the compound’s reactivity with HCN reduces as its negative charge rises. The +I effect grows in the compounds that are presented, as is demonstrated below. Steric impediment can be seen to rise in the similar way.
(ii) We are aware that an alkyl group with the +I action weakens the acidity. Isopropyl groups have a greater +I impact than n-propyl groups. Similar to this, bromine (Br) with the -I action strengthens the acid. The greater the proximity of the carbon atom to the carboxyl (COOH) group, the greater the -I-effect and the more potent the acid. As a result, the ascending order of acidic strength is as follows:
(CH3)2CHCOOH< CH3CH2CH2COOH < CH3CH(Br)CH2COOH < CH3CH2CH(Br) COOH
(iii) Since the acidic strength is decreased by electron-donating groups, 4-methoxy benzoic acid is weaker than benzoic acid. Additionally, 3,4-dinitrobenzoic acid and 4-nitrobenzoic acid are stronger acids than benzoic acid because electron withdrawing groups increase the acidic strength. Additionally, 3,4-dinitrobenzoic acid is a stronger acid than 4-nitrobenzoic acid due to the existence of an additional -NO2 group at the /w-position with respect to a -COOH group. This results in an increase in total acidity in the following order: 4-methoxybenzoic acid, benzoic acid, 4-nitrobenzoic acid, and 3,4-dinitrobenzoic acid.
12.13 Give simple chemical tests to distinguish between the following pairs of compounds.
(i) Propanal and Propanone
(ii) Acetophenone and Benzophenone
(iii) Phenol and Benzoic acid
(iv) Benzoic acid and Ethyl benzoate
(v) Pentan-2-one and Pentan-3-one
(vi) Benzaldehyde and Acetophenone
(vii) Ethanal and Propanal
12.14 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
(ii) m-Nitrobenzoic acid
(iii) p-Nitrobenzoic acid
(iv) Phenylacetic acid
12.15 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) Benazaldehyde to α-Hydroxyphenylacetic acid
(ix) Benzoic acid to m- Nitrobenzyl alcohol
Ans- (i) Propanone to propene
12.16 Describe the following:
(ii) Cannizzaro reaction
(iii) Cross aldol condensation
Ans- (i) Acetylation is the process of adding an acetyl group to a molecule, namely by replacing an active hydrogen atom with an acetyl group. The presence of a base, such as pyridine, dimethylanitine, etc., is typically required for acetylation to occur.
12.17 Complete each synthesis by giving missing starting material, reagent or products
12.18 Give plausible explanation for each of the following:
(i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not.
(ii) There are two –NH2 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.
Ans- (i) The assault of a CN” ion (nucleophile) at the carbonyl carbon atom in cyclohexanone is simple to carry out. However, the three CH3 groups in 2, 4, and 6-trimethylcyclohexanone, which release electrons by nature (+ I effect), will significantly increase the electron density on the carbonyl carbon atom, making a nucleophile attack unlikely. Additionally, the two —CH3 substituents in the ortho locations will prevent the CN- ion from attacking the carbonyl group.
Despite the fact that semicarbazide has two – NH2 groups, only one of them—the one that is directly connected to C = O—is engaged in resonance, as seen above. As a result, this -NH2 group’s N has less electron density, which prevents it from acting as a nucleophile. The lone pair of electrons on the N atom of the other -NH2 group, on the other hand, is not involved in resonance and is therefore available for nucleophilic attack on the C = O group of aldehydes and ketones.
(iii) Ester along with water is formed reversibly from a carboxylic acid and an alcohol in presence of an acid.
12.19 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 addition compound with sodium hydrogensulphite and give positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.
Since the substance produces sodium hydrogen sulphite, it must either be an aldehyde or a methyl/cyclic ketone. It is impossible for the substance to be an aldehyde because it does not reduce Tollens’ reagent. The supplied substance is a methyl ketone because the iodoform test results for the molecule are positive. The methyl ketone is pentan-2-one because the provided molecule undergoes strong oxidation to produce a combination of ethanoic acid and propanoic acid.
12.20 Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?
In the instance of the phenoxide ion, structures (V–VI) have a negative charge on the carbon atom that is less electronegative. They therefore make only a very minor contribution to the resonance stability of the phenoxide ion.
The negative charge on the oxygen atom is delocalized over two oxygen atoms in structures I and II (carboxylate ions), but it is still localised in structures III and IV just the electrons of the benzene ring are delocalized. The carboxylate ion is much more resonance stabilised than the phenoxide ion because delocalization of benzene electrons has a far smaller impact on the stability of the phenoxide ion. Therefore, it is significantly simpler to release a proton from carboxylic acids than from phenols. In other words, phenols are weaker acids than carboxylic acids.