Chapter 8 Aldehydes, Ketones and Carboxylic Acids Intext Questions NCERT Solutions Class 12 Chemistry- PDF Download
Page No. 231
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
(i) α-Methoxypropionaldehyde
(ii) 3-Hydroxybutanal
(iii) 2-Hydroxycyclopentane carbaldehyde
(iv) 4-Oxopentanal
(v) Di-sec-butyl ketone
(vi) 4-Fluoroacetophenone
Page No. 234
8.2. Write the structures of products of the following reactions:
Solution
(i) Benzene reacts with C2H5COCl, to yield propiophenone (Friedel Craft’s acylation).
(ii) (C6H5CH2)2Cd + 2CH3COCl →
(iii) Propyne reacts to form Propanone
(iv) p-Nitrotoluene reacts to form p-Nitrobenzaldehyde
Page No. 236
8.3. Arrange the following compounds in increasing order of their boiling points.
CH3CHO, CH3CH2OH, CH3OCH3, CH3CH2CH3
Solution
The molecular masses of the given compounds are in the range 44 to 46. CH3CH2OH undergoes extensive intermolecular H-bonding, resulting in the association of molecules. Therefore, it has the highest boiling point. CH3CHO is more polar than CH3OCH3 and so CH3CHO has stronger intermolecular dipole−dipole attraction than CH3OCH3⋅CH3CH2CH3 has only weak van der Waals force.
Thus, the arrangement of the given compounds in the increasing order of their boiling points is given by:
CH3CH2CH3 < CH3OCH3 < CH3CHO < CH3CH2OH
Page No. 243
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
(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
8.5. Predict the products of the following reactions:
Solution
(i)
(ii)
(iii)
(iv)
Page No. 244
8.6. Give the IUPAC names of the following compounds:
(i) PhCH2CH2COOH
(ii) (CH3)2C=CHCOOH
Solution
(i) 3-Phenylpropanoic acid
(ii) 3-Methylbut-2-enoic acid
(iii) 2-Methylcyclopentanecarboxylic acid
(iv) 2,4,6-Trinitrobenzoic acid
Page No. 248
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
(i) Ethylbenzene to benzoic acid
(ii) Acetophenone to benzoic acid
(iii) Bromobenzene to benzoic acid
(iv) Phenylethene (Styrene) to benzoic acid
Page No. 254
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
Solution
(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).