Chapter 8: Aldehydes, Ketones, and Carboxylic Acids

A is an organic compound with the molecular formula C₈H₁₆O₂. This gives carboxylic acid (B) and alcohol (C) on hydrolysis with dilute sulphuric acid. Thus, compound A must be an ester.

Further, the oxidation of alcohol (C) with chromic acid gives acid B. Thus, B and C must contain an equal number of carbon atoms.

A total of 8 carbon atoms are present in compound A, and each of B and C contain 4 carbon atoms.

Again, alcohol C gives but-1-ene on dehydration. Therefore, C is of straight-chain, and hence, it is butan-1-ol.

On oxidation, Butan-1-ol gives butanoic acid. Hence, acid B is butanoic acid.

Hence, the ester with the molecular formula C₈H₁₆O₂ is butylbutanoate.

All the given reactions can be explained by the following equations.

(i) When HCN reacts with a compound, the attacking species is a nucleophile, CN^–. Therefore, the reactivity with HCN decreases when the negative charge on the compound increases. The +I effect increases in the given compound. Steric hindrance also increases in the same.

Hence, the given compounds can be arranged according to their increasing reactivities toward HCN as follows:

Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde

(ii) After losing a proton, carboxylic acids gain a negative charge, as shown below:

Now, the stability of the carboxyl ion increases by any group that helps to stabilise the negative charge, which will increase the strength of the acid. Thus, groups having the −I effect will increase the strength of the acids and groups having the +I effect will decrease the strength of the acids. In the given compounds, Br-group has a −I effect, and the −CH₃ group has a +I effect. Thus, acids containing Br− are stronger.

Now, the +I effect of the isopropyl group is more than that of the n-propyl group. Hence, CH₃CH₂CH₂COOH is a stronger acid than (CH₃)₂CHCOOH.

Also, as the distance increases, the +I effect grows weaker. Hence, CH₃CH₂CH(Br)COOH is a stronger acid than CH₃CH(Br)CH₂COOH.

Hence, the strengths of the given acids increase as:

(CH₃)₂CHCOOH < CH₃CH₂CH₂COOH < CH₃CH(Br)CH₂COOH < CH₃CH₂CH(Br)COOH

(iii) As we have seen in the previous case, the strength of acids is decreased by the electron-donating group, while the strength of acids increases by electron-withdrawing groups. As the methoxy group is an electron-donating group, benzoic acid is a stronger acid than 4-methoxybenzoic acid. The Nitro group is an electron-withdrawing group and will increase the strength of acids. As 3,4-dinitrobenzoic acid contains two nitro groups, it is a slightly stronger acid than 4-nitrobenzoic acid. Hence, the strengths of the given acids increase as:

4-Methoxybenzoic acid < Benzoic acid < 4-Nitrobenzoic acid < 3, 4-Dinitrobenzoic acid

(i) Propanal and propanone can be distinguished by the following tests.

(a) Tollen’s test

Tollen’s reagent is reduced by propanal as it is an aldehyde. But, Tollen’s reagent is not reduced as propanone is a ketone.

(b) Fehling’s test

Aldehydes respond to Fehling’s test, but ketones do not.

Propanone being a ketone, does not reduce Fehling’s solution to a red-brown precipitate of Cu₂O, but Propanal being an aldehyde, does the reaction.

(c) Iodoform test:

At least one methyl group should be present in aldehydes and ketones linked to the carbonyl carbon atom to respond to the iodoform test. They are oxidised by sodium hypoiodite (NaOI) to give iodoforms. Propanone being a methyl ketone, responds to this test, but propanal does not.

(ii) Acetophenone and Benzophenone can be distinguished using the iodoform test.

Iodoform test:

yellow ppt. of iodoform is produced when methyl ketones are oxidised by sodium hypoiodite.

Acetophenone being a methyl ketone, responds to this test, but benzophenone does not.

(iii) Phenol and benzoic acid can be distinguished by the ferric chloride test.

Ferric chloride test:

Phenol reacts with neutral FeCl₃ to form an iron-phenol complex giving violet colouration.

A buff-coloured ppt of ferric benzoate is produced when benzoic acid reacts with neutral FeCl₃.

(iv) Benzoic acid and ethyl benzoate can be distinguished by the sodium bicarbonate test.

Sodium bicarbonate test:

Brisk effervescence is produced when acids react with NaHCO₃ due to the evolution of CO₂ gas.

Benzoic acid, being an acid, responds to this test, but ethyl benzoate does not.

(v) Pentan-2-one and pentan-3-one can be distinguished by the iodoform test.

Iodoform test:

Pentan-2-one responds to this test as it is a methyl ketone. But pentan-3-one, not being a methyl ketone, does not respond to this test.

(vi) Benzaldehyde and acetophenone can be distinguished by the following tests.

(a) Tollen’s Test

Aldehydes respond to Tollen’s test. The red-brown precipitate of Cu₂O is produced by Benzaldehyde reduces Tollen’s reagent, but acetophenone being a ketone, does not.

(b) Iodoform test

A yellow precipitate of iodoform is given by acetophenone (a methyl ketone) when it undergoes oxidation by sodium hypoiodite (NaOI). But benzaldehyde does not respond to this test.

(vii) Ethanal and propanal can be distinguished by the iodoform test.

Iodoform test

Carbonyl carbon atom having at least one methyl group in aldehydes and ketones responds to the iodoform test. Also, Ethanal having one methyl group linked to the carbonyl carbon atom, responds to this test. But, there are no methyl groups linked to the carbonyl carbon atom in propanal, and thus, it does not respond to this state.

Solution coming soon.

Solution coming soon.

(i) Acetylation

When an organic compound is introduced with an acetyl function group, it is known as acetylation. Bases such as pyridine, dimethylaniline, etc., are present when this process is carried out. An acetyl group is substituted with an active hydrogen atom in this process. Acetylating agents such as acetyl chloride and acetic anhydride are commonly used in the process.

For example, the acetylation of ethanol produces ethyl acetate.

(ii) Cannizzaro reaction:

Cannizzaro reaction is the self-disproportionation (oxidation-reduction) reaction of aldehydes with no α-hydrogens on treatment with concentrated alkalis.

One molecule of aldehyde is reduced to an alcohol, and the other is oxidised to carboxylic acid in this reaction.

For example, ethanol and potassium ethanoate are produced when ethanol is treated with concentrated potassium hydroxide.

(iii) Cross-aldol condensation:

Cross-aldol condensation is carried out between two different aldehydes or two different ketones, or an aldehyde and a ketone. Four compounds are obtained as products if both the reactants contain α-hydrogens.

For example, the reaction of ethanal and propanal gives four products.

(iv) Decarboxylation:

The reaction in which carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with soda-lime is called decarboxylation.

When aqueous solutions of alkali metal salts of carboxylic acids are electrolyzed also results in decarboxylation. This electrolytic process is known as Kolbe’s electrolysis.

Solution coming soon.

(i) Cyanohydrins are formed by cyclohexanones according to the following equation.

In this case, there will not be any steric hindrance; hence the nucleophile CN^– can easily attack. However, in the case of 2,2,6-trimethylcyclohexanone, methyl groups at α-positions offer steric hindrances, and as a result, CN− cannot attack effectively.

For this reason, it does not form a cyanohydrin.

(ii) Semicarbazide undergoes resonance involving only one of the two −NH₂ groups, which is attached directly to the carbonyl-carbon atom.

Therefore, the electron density on the −NH₂ group involved in the resonance also decreases.

As a result, it cannot act as a nucleophile. Since the other −NH₂ group is not involved in resonance, it can act as a nucleophile and attack carbonyl-carbon atoms of aldehydes and ketones to produce semicarbazones.

(iii) Ester, along with water, is formed reversibly from a carboxylic acid and an alcohol in the presence of an acid.

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.

% of carbon = 69.77 %

% of hydrogen = 11.63 %

% of oxygen = {100 − (69.77 + 11.63)}%

= 18.6 %

Thus, the ratio of the number of carbon, hydrogen, and oxygen atoms in the organic compound can be given as:

= 5.81:11.63:1.16

= 5:10:1

Therefore, the empirical formula of the compound is C₅H₁₀O. Now, the empirical formula mass of the compound can be given as:

= 5 x 12 + 10 x 1 + 1 x 16

= 86

The molecular mass of the compound = 86

Therefore, the molecular formula of the compound is given by C₅H₁₀O.

Tollen’s reagent is not reduced by the given compound; hence it is not an aldehyde. A positive iodoform test is given by the compound and also forms sodium hydrogen sulphate addition products. Since the compound is not an aldehyde, it must be a methyl ketone.

The given compound also gives a mixture of ethanoic acid and propanoic acid.

Hence, the given compound is pentan-2-ol.

The given reactions can be explained by the following equations:

Resonance structures of phenoxide ion are:

In II, III and IV, less electronegative carbon atoms carry a negative charge which is observed from the resonance structures of phenoxide ion. Therefore, the resonance stability of the phenoxide ion gets a negligible contribution from these three structures.

Hence, these structures can be eliminated. The more electronegative oxygen atom contains a negative charge by the structures I and V.

Resonance structures of carboxylate ion are as shown below:

In the case of carboxylate ion, resonating structures I′ and II′ contain a charge carried by a more electronegative oxygen atom.

Also, the negative charge is delocalised over two oxygen atoms in resonating structures I′ and II′. But in resonating structures I and V of the phenoxide ion, the negative charge is localised on the same oxygen atom. Therefore, its stability is determined more by the resonating structures of carboxylate ion than those of phenoxide ion. As a result, phenoxide ion is less resonance-stabilised than carboxylate ion. Hence, the carboxylic acid is a stronger acid than phenol.

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