NCERT Class 12 Chemistry Part 2 Chapter 8 Aldehydes Ketones and Carboxylic Acids

Question 8.1

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 (viii) Imine (ix) 2,4-DNP-derivative (x) Schiff’s base.

(i) Cyanohydrin: Produced by reaction of aldehydes/ketones with HCN. Example: \(RCHO + HCN \to RCH(OH)CN\).
(ii) Acetal: Gem-dialkoxy compound formed by reaction of aldehyde with 2 equivalents of alcohol in dry HCl gas. Example: \(RCH(OR’)_2\).
(iii) Semicarbazone: Derivative formed by condensation of aldehyde/ketone with semicarbazide (\(NH_2NHCONH_2\)). Structure: \(>C=N-NHCONH_2\).
(iv) Aldol: \(\beta\)-hydroxy aldehyde/ketone formed by condensation of two molecules of carbonyl compound containing \(\alpha\)-H.
(v) Hemiacetal: Formed by reaction of aldehyde with 1 equivalent of alcohol. Structure: \(RCH(OH)(OR’)\).
(vi) Oxime: Formed by reaction with Hydroxylamine (\(NH_2OH\)). Structure: \(>C=N-OH\).
(vii) Ketal: Gem-dialkoxy compound from ketone + ethylene glycol (cyclic ketal).
(viii) Imine: Compound containing \(>C=N-\) bond, formed by reaction with Ammonia derivatives.
(ix) 2,4-DNP-derivative: Orange ppt formed with 2,4-Dinitrophenylhydrazine. Used for identification.
(x) Schiff’s base: Substituted imine formed by aldehyde + primary amine. Structure: \(R-CH=N-R’\).

Question 8.2

Name the following compounds according to IUPAC system of nomenclature:
(i) \(CH_3CH(CH_3)CH_2CH_2CHO\)
(ii) \(CH_3CH_2COCH(C_2H_5)CH_2CH_2Cl\)
(iii) \(CH_3CH=CHCHO\)
(iv) \(CH_3COCH_2COCH_3\)
(v) \(CH_3CH(CH_3)CH_2C(CH_3)_2COCH_3\)
(vi) \((CH_3)_3CCH_2COOH\)
(vii) \(OHCC_6H_4CHO-p\)

(i): 4-Methylpentanal.
(ii): 6-Chloro-4-ethylhexan-3-one.
(iii): But-2-enal (Crotonaldehyde).
(iv): Pentane-2,4-dione.
(v): 3,3,5-Trimethylhexan-2-one.
(vi): 3,3-Dimethylbutanoic acid.
(vii): Benzene-1,4-dicarbaldehyde (Terephthalaldehyde).

Question 8.3

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.

(i) CH3-CH(CH3)-CH2-CHO (ii) O2N-C6H4-CO-CH2-CH3 (iii) H3C-C6H4-CHO (iv) (CH3)2C=CH-CO-CH3 (Mesityl oxide) (v) CH3-CH(Cl)-CH2-CO-CH3 (vi) CH3-CH(Ph)-CH(Br)-CH2-COOH (vii) HO-C6H4-CO-C6H4-OH (viii) CH3-C≡C-CH=CH-COOH

Question 8.4

Write the IUPAC names of the ketones and aldehydes possible for the formula \(C_2H_8O\). (Correction: Likely C4H8O or C5H10O based on context, NCERT usually asks isomerism). Let’s assume question is “Possible Aldehydes and Ketones for C5H10O”.

For \(C_5H_{10}O\):

Aldehydes: Pentanal, 3-Methylbutanal, 2-Methylbutanal, 2,2-Dimethylpropanal.
Ketones: Pentan-2-one, Pentan-3-one, 3-Methylbutan-2-one.

Question 8.5

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

(i): \(C_6H_5-CH=N-NH-C_6H_3(NO_2)_2\).
(ii): Cyclopropane ring attached to \(=N-OH\).
(iii): \(CH_3-CH(OCH_3)_2\).
(iv): Cyclobutane ring attached to \(=N-NH-CO-NH_2\).
(v): Hexane chain with cyclic dioxolane ring at C3.
(vi): \(H-CH(OH)(OCH_3)\).

Question 8.6

Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents:
(i) \(PhMgBr\) then \(H_3O^+\)
(ii) Tollens’ reagent
(iii) Semicarbazide and weak acid
(iv) Excess ethanol and acid
(v) Zinc amalgam and dilute hydrochloric acid.

(i): Cyclohexylphenylmethanol (Secondary alcohol via Grignard).
(ii): Cyclohexanecarboxylate ion (Oxidation + Silver mirror).
(iii): Cyclohexanecarbaldehyde semicarbazone.
(iv): Cyclohexanecarbaldehyde diethyl acetal.
(v): Methylcyclohexane (Clemmensen Reduction – \(C=O \to CH_2\)).

Question 8.7

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.

Analysis (\(\alpha\)-Hydrogen)

Aldol Condensation (Has \(\alpha\)-H): 2-Methylpentanal, Cyclohexanone, 1-Phenylpropanone, Phenylacetaldehyde.
Cannizzaro Reaction (No \(\alpha\)-H): Methanal, Benzaldehyde, 2,2-Dimethylbutanal.
Neither: Benzophenone (Ketone with no \(\alpha\)-H, usually doesn’t give Cannizzaro easily, no aldol), Butan-1-ol (Alcohol).

Example Products

Methanal (Cannizzaro): Methanol + Potassium formate.
Cyclohexanone (Aldol): 2-(1-Hydroxycyclohexyl)cyclohexanone \(\to\) Dehydration product.

Question 8.8

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

(i) Butane-1,3-diol

Ethanal \(\xrightarrow{Dil. NaOH}\) 3-Hydroxybutanal (Aldol) \(\xrightarrow{NaBH_4}\) Butane-1,3-diol.

(ii) But-2-enal (Crotonaldehyde)

Ethanal \(\xrightarrow{Dil. NaOH}\) Aldol \(\xrightarrow{\Delta, -H_2O}\) But-2-enal.

(iii) But-2-enoic acid

But-2-enal (from ii) \(\xrightarrow{Tollens’ Reagent}\) But-2-enoic acid (Selective oxidation of CHO).

Question 8.9

Write structural formulas and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde serves as nucleophile and which as electrophile.

This is a Cross Aldol Condensation. Four products are formed (2 Self-Aldol, 2 Cross-Aldol).

Product 1 (Self Propanal): 2-Methylpent-2-enal. (Prop nuc + Prop elec).
Product 2 (Self Butanal): 2-Ethylhex-2-enal. (But nuc + But elec).
Product 3 (Cross): 2-Methylhex-2-enal. (Propanal nuc + Butanal elec).
Product 4 (Cross): 2-Ethylpent-2-enal. (Butanal nuc + Propanal elec).

Question 8.10

An organic compound with the molecular formula \(C_9H_{10}O\) forms 2,4-DNP derivative, reduces Tollens’ reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound.

Deductions

Forms 2,4-DNP \(\to\) Carbonyl Group.
Reduces Tollens’ \(\to\) Aldehyde.
Undergoes Cannizzaro \(\to\) No \(\alpha\)-Hydrogen (Aldehyde group directly on benzene ring).
Oxidation gives 1,2-benzenedicarboxylic acid (Phthalic acid) \(\to\) Ortho-substituted benzene ring.
Formula \(C_9H_{10}O\). Benzene ring (\(C_6H_4\)) + CHO + Ethyl group (\(C_2H_5\)) fits the formula.

Compound: 2-Ethylbenzaldehyde.

Question 8.11

An organic compound (A) (molecular formula \(C_8H_{16}O_2\)) 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.

Analysis

(A) is an Ester. Hydrolysis gives Acid (B) + Alcohol (C).
Oxidation of (C) gives (B). This means (B) and (C) have the same number of carbon atoms. Total carbons = 8. So each has 4 carbons.
Alcohol (C) dehydrates to but-1-ene. (C) is Butan-1-ol.
Acid (B) is Butanoic Acid.
Ester (A) is Butyl butanoate.

CH3CH2CH2COOCH2CH2CH2CH3 + H2O -> CH3CH2CH2COOH + CH3CH2CH2CH2OH

Question 8.12

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).

(i) Reactivity (Steric Hindrance): Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde.
(ii) Acid Strength (-I effect): (CH3)2CHCOOH < CH3CH2CH2COOH < CH3CH(Br)CH2COOH < CH3CH2CH(Br)COOH. (Br closer to COOH increases acidity).
(iii) Acid Strength (EWG/EDG): 4-Methoxybenzoic acid < Benzoic acid < 4-Nitrobenzoic acid < 3,4-Dinitrobenzoic acid.

Question 8.13

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.

(i) Propanal/Propanone: Tollens’ Test (Propanal gives silver mirror).
(ii) Acetophenone/Benzophenone: Iodoform Test (Acetophenone gives yellow ppt due to \(CH_3-CO-\)).
(iii) Phenol/Benzoic Acid: \(\text{NaHCO}_3\) Test (Benzoic acid gives effervescence of \(CO_2\)).
(iv) Benzoic Acid/Ethyl Benzoate: \(\text{NaHCO}_3\) Test.
(v) Pentan-2-one/Pentan-3-one: Iodoform Test (Pentan-2-one positive).
(vi) Benzaldehyde/Acetophenone: Tollens’ (Benzaldehyde +ve) or Iodoform (Acetophenone +ve).
(vii) Ethanal/Propanal: Iodoform Test (Ethanal positive).

Question 8.14

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 (v) p-Nitrobenzaldehyde.

(i) Benzene -> Bromobenzene -> Phenyl MgBr -> Benzoic Acid -> Methyl Benzoate. (ii) Benzene -> Benzoic Acid -> m-Nitrobenzoic acid (COOH is meta directing). (iii) Benzene -> Toluene -> p-Nitrotoluene -> p-Nitrobenzoic acid. (iv) Benzene -> Toluene -> Benzyl bromide -> Benzyl cyanide -> Phenylacetic acid. (v) Benzene -> Toluene -> p-Nitrotoluene -> p-Nitrobenzaldehyde (Etard reaction).

Question 8.15

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 \(\alpha\)-Hydroxyphenylacetic acid (ix) Benzoic acid to m-Nitrobenzyl alcohol.

(i) Propanone –(NaBH4)–> Propan-2-ol –(Conc. H2SO4)–> Propene. (ii) Benzoic acid –(SOCl2)–> Benzoyl chloride –(H2/Pd-BaSO4)–> Benzaldehyde. (iii) Ethanol –(Cu/573K)–> Ethanal –(Dil. NaOH)–> 3-Hydroxybutanal. (iv) Benzene –(CH3COCl/AlCl3)–> Acetophenone –(HNO3/H2SO4)–> m-Nitroacetophenone. (v) Benzaldehyde –(PhMgBr)–> Diphenylmethanol –(CrO3)–> Benzophenone.

Question 8.16

Describe the following: (i) Acetylation (ii) Cannizzaro reaction (iii) Cross aldol condensation (iv) Decarboxylation.

(i) Acetylation: Introduction of acetyl group (\(CH_3CO-\)) using acetyl chloride or acetic anhydride.
(ii) Cannizzaro: Disproportionation of aldehydes without \(\alpha\)-H in conc. alkali. Example: \(2HCHO \to CH_3OH + HCOO^-\).
(iii) Cross Aldol: Aldol condensation between two different aldehydes/ketones. Gives mixture of products if both have \(\alpha\)-H.
(iv) Decarboxylation: Removal of \(CO_2\) from carboxylic acid salt using Soda Lime (\(NaOH+CaO\)) to form alkane.

Question 8.17

Complete each synthesis by giving missing starting material, reagent or products.

(i) Ethylbenzene + KMnO4/KOH -> Benzoic Acid. (ii) Phthalic acid + SOCl2 -> Phthaloyl chloride. (iii) C6H5CHO + H2N-NH-CO-NH2 -> C6H5CH=N-NH-CO-NH2. (iv) Benzene + Acetyl Chloride/AlCl3 -> Acetophenone. (v) 4-Nitropropiophenone + Zn-Hg/HCl -> 1-Propyl-4-nitrobenzene.

Question 8.18

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, the ester is distilled off as soon as it is formed.

(i): Steric hindrance. In 2,2,6-trimethylcyclohexanone, methyl groups hinder the nucleophilic attack of \(CN^-\).
(ii): Resonance. One \(NH_2\) group (next to CO) is involved in resonance with the carbonyl group (\(NH_2-CO-NH-NH_2\)), making it less nucleophilic. The other \(NH_2\) is free to attack.
(iii): Esterification is reversible. Distilling off the ester shifts the equilibrium to the right (Le Chatelier’s principle) to increase yield.

Question 8.19

An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass 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.

Analysis

Empirical Formula: \(C: 69.77/12 = 5.81\), \(H: 11.63/1 = 11.63\), \(O: 18.6/16 = 1.16\). Ratio: \(C_5H_{10}O\). Molecular mass 86 matches formula mass.
Reactions: Forms \(NaHSO_3\) adduct \(\to\) Carbonyl. No Tollens’ \(\to\) Ketone. Positive Iodoform \(\to\) Methyl Ketone (\(CH_3-CO-\)).
Oxidation: Gives Ethanoic acid (\(2C\)) + Propanoic acid (\(3C\)). Total 5 carbons.
Structure: The ketone must break to give these acids. Pentan-2-one (\(CH_3-CO-CH_2-CH_2-CH_3\)). Oxidation cleaves \(C_2-C_3\) bond mainly.

Structure: Pentan-2-one.

Question 8.20

Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?

Reason: In the carboxylate ion (\(RCOO^-\)), the negative charge is delocalized over two equivalent electronegative oxygen atoms. This provides high stability.

In the phenoxide ion, the negative charge is delocalized over one oxygen and less electronegative carbon atoms of the benzene ring. Furthermore, the resonance structures of phenoxide are non-equivalent. Resonance stabilization in carboxylate is more effective, making carboxylic acid stronger.

Intext 8.1

Write the structures of the following compounds…

(Refer to textbook specific names for drawing structures. Examples: \(\alpha\)-Methoxypropionaldehyde, etc.)

Intext 8.2

Write the structures of products of the following reactions…

(i) Benzene + C2H5COCl/AlCl3 -> Propiophenone. (ii) (C6H5CH2)2Cd + 2CH3COCl -> 2 C6H5CH2COCH3 + CdCl2. (iii) CH3-C≡C-H + Hg2+,H2SO4 -> Acetone (CH3COCH3). (iv) Nitro-toluene + CrO2Cl2 -> p-Nitrobenzaldehyde.

Intext 8.4

Arrange in increasing order of boiling points: \(CH_3CHO, CH_3CH_2OH, CH_3OCH_3, CH_3CH_2CH_3\).

Propane (VdW) < Methoxyethane (Dipole) < Ethanal (Dipole) < Ethanol (H-bond).

NCERT Solutions