Page No 378: - Chapter 12 Aldehydes Ketones & Carboxylic Acids Exercise Solutions class 12 ncert solutions Chemistry - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Question 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

Answer:

(i)

(ii)

(iii)

(iv)

(v)

Question 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

Answer:

Aldehydes and ketones having at least one α-hydrogen undergo aldol condensation. The compounds (ii) 2−methylpentanal, (v) cyclohexanone, (vi) 1-phenylpropanone, and (vii) phenylacetaldehyde contain one or more α-hydrogen atoms. Therefore, these undergo aldol condensation.

Aldehydes having no α-hydrogen atoms undergo Cannizzaro reactions. The compounds (i) Methanal, (iii) Benzaldehyde, and (ix) 2, 2-dimethylbutanal do not have any α-hydrogen. Therefore, these undergo cannizzaro reactions.

Compound (iv) Benzophenone is a ketone having no α-hydrogen atom and compound (viii) Butan-1-ol is an alcohol. Hence, these compounds do not undergo either aldol condensation or cannizzaro reactions.

Aldol condensation

(ii)

(v)

(vi)

(vii)

Cannizzaro reaction

(i)

(iii)

(ix)

Question 12.8:

How will you convert ethanal into the following compounds?

(i) Butane-1, 3-diol (ii) But-2-enal (iii) But-2-enoic acid

Answer:

(i) On treatment with dilute alkali, ethanal produces 3-hydroxybutanal gives butane-1, 3-diol on reduction.

(ii) On treatment with dilute alkali, ethanal gives 3-hydroxybutanal which on heating produces but-2-enal.

(iii) When treated with Tollen’s reagent, But-2-enal produced in the above reaction produces but-2-enoic acid .

Question 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.

Answer:

(i) Taking two molecules of propanal, one which acts as a nucleophile and the other as an electrophile.

(ii) Taking two molecules of butanal, one which acts as a nucleophile and the other as an electrophile.

(iii) Taking one molecule each of propanal and butanal in which propanal acts as a nucleophile and butanal acts as an electrophile.

(iv) Taking one molecule each of propanal and butanal in which propanal acts as an electrophile and butanal acts as a nucleophile.

Question 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.

Answer:

It is given that the compound (with molecular formula C₉H₁₀O) forms 2, 4-DNP derivative and reduces Tollen’s reagent. Therefore, the given compound must be an aldehyde.

Again, the compound undergoes cannizzaro reaction and on oxidation gives 1, 2-benzenedicarboxylic acid. Therefore, the −CHO group is directly attached to a benzene ring and this benzaldehyde is ortho-substituted. Hence, the compound is 2-ethylbenzaldehyde.

The given reactions can be explained by the following equations.

Question 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.

Answer:

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

Since compound A contains a total of 8 carbon atoms, each of B and C contain 4 carbon atoms.

Again, on dehydration, alcohol C gives but-1-ene. 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 molecular formula C₈H₁₆O₂ is butylbutanoate.

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

Question 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)

Answer:

(i) When HCN reacts with a compound, the attacking species is a nucleophile, CN⁻. Therefore, as the negative charge on the compound increases, its reactivity with HCN decreases. In the given compounds, the +I effect increases as shown below. It can be observed that steric hindrance also increases in the same

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

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

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

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

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

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

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, electron-donating groups decrease the strengths of acids, while electron-withdrawing groups increase the strengths of acids. As methoxy group is an electron-donating group, 4-methoxybenzoic acid is a weaker acid than benzoic acid. Nitro group is an electron-withdrawing group and will increase the strengths 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

Question 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

Answer:

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

(a) Tollen’s test

Propanal is an aldehyde. Thus, it reduces Tollen’s reagent. But, propanone being a ketone does not reduce Tollen’s reagent.

(b) Fehling’s test

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

Propanal being an aldehyde reduces Fehling’s solution to a red-brown precipitate of Cu₂O, but propanone being a ketone does not.

(c) Iodoform test:

Aldehydes and ketones having at least one methyl group linked to the carbonyl carbon atom respond to iodoform test. They are oxidized 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:

Methyl ketones are oxidized by sodium hypoiodite to give yellow ppt. of iodoform. Acetophenone being a methyl ketone responds to this test, but benzophenone does not.

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

Ferric chloride test:

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

But benzoic acid reacts with neutral FeCl₃ to give a buff coloured ppt. of ferric benzoate.

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

Sodium bicarbonate test:

Acids react with NaHCO₃ to produce brisk effervescence due to the evolution of CO₂ gas.

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

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

Iodoform test:

Pentan-2-one is a methyl ketone. Thus, it responds to this test. 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. Benzaldehyde being an aldehyde reduces Tollen’s reagent to give a red-brown precipitate of Cu₂O, but acetophenone being a ketone does not.

(b) Iodoform test

Acetophenone being a methyl ketone undergoes oxidation by sodium hypoiodite (NaOI) to give a yellow ppt. of iodoform. But benzaldehyde does not respond to this test.

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

Iodoform test

Aldehydes and ketones having at least one methyl group linked to the carbonyl carbon atom responds to the iodoform test. Ethanal having one methyl group linked to the carbonyl carbon atom responds to this test. But propanal does not have a methyl group linked to the carbonyl carbon atom and thus, it does not respond to this state.

Question 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

(v) p-Nitrobenzaldehyde.

Answer:

(i)

(ii)

(iii)

(iv)

(v)

Question 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

Answer:

(i)

(ii)

(iii)

(iv)

(v) 

(vi)

(vii)

(viii)

(ix)

Question 12.16:

Describe the following:

(i) Acetylation

(ii) Cannizzaro reaction

(iii) Cross aldol condensation

(iv) Decarboxylation

Answer:

(i) Acetylation

The introduction of an acetyl functional group into an organic compound is known as acetylation. It is usually carried out in the presence of a base such as pyridine, dirnethylaniline, etc. This process involves the substitution of an acetyl group for an active hydrogen atom. Acetyl chloride and acetic anhydride are commonly used as acetylating agents.

For example, acetylation of ethanol produces ethyl acetate.

(ii) Cannizzaro reaction:

The self oxidation-reduction (disproportionation) reaction of aldehydes having no α-hydrogens on treatment with concentrated alkalis is known as the Cannizzaro reaction. In this reaction, two molecules of aldehydes participate where one is reduced to alcohol and the other is oxidized to carboxylic acid.

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

(iii) Cross-aldol condensation:

When aldol condensation is carried out between two different aldehydes, or two different ketones, or an aldehyde and a ketone, then the reaction is called a cross-aldol condensation. If both the reactants contain α-hydrogens, four compounds are obtained as products.

For example, ethanal and propanal react to give four products.

(iv) Decarboxylation:

Decarboxylation refers to the reaction in which carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with soda-lime.

Decarboxylation also takes place when aqueous solutions of alkali metal salts of carboxylic acids are electrolyzed. This electrolytic process is known as Kolbe’s electrolysis.