Ncert Chemistry Class 12 intext questions solutions chapter 8 Aldehydes Ketones and Carboxylic Acids

Aldehydes Ketones and Carboxylic Acids intext questions solutions: Class 12 Chemistry Chapter 8 intext questions solutions

Textbook	Ncert
Class	Class 12
Subject	Chemistry
Chapter	Chapter 8
Chapter Name	Aldehydes Ketones and Carboxylic Acids Class 12 intext questions solutions
Category	Intext questions
Medium	English

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Question 8.1: Write the structures of the following compounds.

• (i) α-Methoxypropionaldehyde
• (ii) 3-Hydroxybutanal
• (iii) 2-Hydroxycyclopentane carbaldehyde
• (iv) 4-Oxopentanal
• (v) Di-sec-butyl ketone
• (vi) 4-Fluoroacetophenone

Solution 8.1:

(i) α-Methoxypropionaldehyde

(ii) 3-Hydroxybutanal

(iii) 2-Hydroxycyclopentane carbaldehyde

(iv) 4-Oxopentanal

(v) Di-sec-butyl ketone

(vi) 4-Fluoroacetophenone

Question 8.2: Write the structures of products of the following reactions;

(i)

(ii)

(iii)

(iv)

Solution 8.2:

(i)

(ii)

(iii)

(iv)

Question 8.3: Arrange the following compounds in increasing order of their boiling points.
CH₃CHO, CH₃CH₂OH, CH₃OCH₃, CH₃CH₂CH₃

Solution 8.3: To arrange the compounds in increasing order of their boiling points, let’s consider the types of intermolecular forces that affect boiling points.

• CH₃CHO (acetaldehyde): This molecule contains a polar carbonyl group (C=O), so it experiences dipole-dipole interactions. However, it does not form hydrogen bonds.
• CH₃CH₂OH (ethanol): This molecule has a hydroxyl group (-OH), which allows it to form hydrogen bonds, the strongest type of intermolecular force among these molecules.
• CH₃OCH₃ (dimethyl ether): This molecule is polar, but it does not form hydrogen bonds because there is no hydrogen attached to oxygen. It primarily experiences dipole-dipole interactions and weaker London dispersion forces.
• CH₃CH₂CH₃ (propane): This molecule is nonpolar and only experiences London dispersion forces, the weakest of the intermolecular forces.

Order of Intermolecular Forces (affecting boiling points):

1. London dispersion forces (weakest) – propane (CH₃CH₂CH₃)
2. Dipole-dipole interactions – acetaldehyde (CH₃CHO) and dimethyl ether (CH₃OCH₃)
3. Hydrogen bonding (strongest) – ethanol (CH₃CH₂OH)

Predicted boiling points (increasing order):

1. CH₃CH₂CH₃ (propane) – Weak London forces
2. CH₃OCH₃ (dimethyl ether) – Dipole-dipole forces
3. CH₃CHO (acetaldehyde) – Stronger dipole-dipole forces due to the carbonyl group
4. CH₃CH₂OH (ethanol) – Strong hydrogen bonding

Final Order (increasing boiling point):

CH₃CH₂CH₃ < CH₃OCH₃ < CH₃CHO < CH₃CH₂OH

Question 8.4: Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions

• (i) Ethanal, Propanal, Propanone, Butanone.
• (ii) Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone.

Solution 8.4: (i) The +I effect of the alkyl group increases in the order:

Ethanal < Propanal < Propanone < Butanone

The electron density at the carbonyl carbon increases with the increase in the +I effect. As a result, the chances of attack by a nucleophile decrease. Hence, the increasing order of the reactivities of the given carbonyl compounds in nucleophilic addition reactions is:

Butanone < Propanone < Propanal < Ethanal

(ii) The +I effect is more in ketone than in aldehyde. Hence, acetophenone is the least reactive in nucleophilic addition reactions. Among aldehydes, the +I effect is the highest in p-tolualdehyde because of the presence of the electron-donating −CH₃ group and the lowest in p-nitrobezaldehyde because of the presence of the electron-withdrawing −NO₂ group.

Hence, the increasing order of the reactivities of the given compounds is:

Acetophenone < p-tolualdehyde < Benzaldehyde < p-Nitrobenzaldehyde

Question 8.5: Predict the products of the following reactions:

(i) 

(ii) 

(iii) 

(iv) 

Solution 8.5:

(i)

(ii)

(iii)

(iv)

Question 8.6: Give the IUPAC names of the following compounds:

(i) Ph CH₂CH₂COOH

(ii) (CH₃)₂C=CHCOOH

Solution 8.6:

• (i) 3-Phenylpropanoic acid
• (ii) 3-Methylbut-2-enoic acid
• (iii) 2-Methylcyclopentanecarboxylic acid
• (iv) 2,4,6-Trinitrobenzoic acid

Question 8.7: Show how each of the following compounds can be converted to benzoic acid.

• (i) Ethylbenzene
• (ii) Acetophenone
• (iii) Bromobenzene
• (iv) Phenylethene (Styrene)

Solution 8.7:

(i) Ethylbenzene to benzoic acid

(ii) Acetophenone to benzoic acid

(iii) Bromobenzene to benzoic acid

(iv) Phenylethene (Styrene) to benzoic acid

Question 8.8: Which acid of each pair shown here would you expect to be stronger?

(i) CH₃CO₂H or CH₂FCO₂H

(ii) CH₂FCO₂H or CH₂ClCO₂H

(iii) CH₂FCH₂CH₂CO₂H or CH₃CHFCH₂CO₂H

(iv) 

Solution 8.8: (i)

The +I effect of −CH₃ group increases the electron density on the O−H bond. Therefore, release of proton becomes difficult. On the other hand, the −I effect of F decreases the electron density on the O−H bond. Therefore, proton can be released easily. Hence, CH₂FCO₂H is a stronger acid than CH₃CO₂H.

(ii)

F has stronger −I effect than Cl. Therefore, CH₂FCO₂H can release proton more easily than CH₂ClCO₂H. Hence, CH₂FCO₂H is stronger acid than CH₂ClCO₂H.

(iii)

Inductive effect decreases with increase in distance. Hence, the +I effect of F in CH₃CHFCH₂CO₂H is more than it is in CH₂FCH₂CH₂CO₂H. Hence, CH₃CHFCH₂CO₂H is stronger acid than CH₂FCH₂CH₂CO₂H.

(iv)

Due to the −I effect of F, it is easier to release proton in the case of compound (A). However, in the case of compound (B), release of proton is difficult due to the +I effect of −CH₃ group. Hence, (A) is a stronger acid than (B).