NCERT Solutions for Class 12 Chemistry : The NCERT solutions provided here will enhance the concepts of the students, as well as suggest alternative methods to solve particular problems to the teachers. The target is to direct individuals towards problem solving strategies, rather than solving problems in one prescribed format. The links below provide the detailed solutions for all the Class 12 Chemistry problems.
Chemistry is much more than the language of Science. We have made sure that our solutions reflect the same. We aim to aid the students with answering the questions correctly, using logical approach and methodology. The NCERT Solutions provide ample material to enable students to form a good base with the fundamentals of the subject.
Table of Contents
ToggleNCERT Solutions for Class 12 Chemistry Chapter :12 Aldehydes, Ketones and Carboxylic Acids
INTEXT Questions
Question 1.
Write the structures of the following compounds.
- α-Methoxypropionaldehyde
- 3-Hydroxybutanal
- 2-Hydroxycyclopentanecarbaldehyde
- 4-Oxopentanal
- Di-sec.butyl ketone
- 4-Fluoroacetophenone
Solution:
Question 2.
Write the structures of the products of the following reactions :
Solution:
Question 3.
Arrange the following compounds in increasing order of their boiling points.
CH3CHO, CH3CH2OH, CH3OCH3, CH3CH2CH3
Solution:
CH3CH2CH3 < CH3OCH3 < CH3CHO < CH3CH2OH
Question 4.
Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions.
- Ethanal, Propanal, Propanone, Butanone.
- Benzaldehyde, p-Tolualdehyde,
p – Nitrobenzaldehyde, Acetophenone. Hint: Consider steric effect and electronic effect.
Solution:
- Butanone < Propanone < Propanal < Ethanal
- Acetophenone < p-Tolualdehyde, Benzaldehyde < p-Nitrobenzaldehyde
Question 5.
Predict the products of the following reactions :
Solution:
Question 6.
Give the IUPAC names of the following compounds :
Solution:
- 3-Phenylpropanoic acid
- 3-Methylbut-2-enoic acid
- 2-Methylcyclopentanecarboxylic acid
- 2,4,6-Trinitrobenzoic acid
Question 7.
Show how each of the following compounds can be converted to benzoic acid.
- Ethylbenzene
- Acetophenone
- Bromobenzene
- Phenylethene (Styrene)
Solution:
Question 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) H2CFCOOH will be stronger of the two. The presence of electronegative F atom at the α-C causes electron withdrawal from the COOH and facilitates the release of H+.
(ii) CH2FCO2H is a stronger acid for the same reason as stated above. F is more electronegative than Cl, so it withdraws electrons from the carboxyl group to a greater extent.
(iii) CH3CHFCH2COOH is stronger. Although both the givenacidshaveFatomin them, it is the proximity of F in CH3CHFCH2COOH to the COOH group which makes it more acidic.
NCERT Exercises
Question 1.
What is meant by the following terms? Give an example of the reaction in each case.
- Cyanohydrin
- Acetal
- Semicarbazone
- Aldol
- Hemiacetal
- Oxime
- Ketal
- Imine
- 2,4-DNP derivative
- Schiff’s base
Solution:
Question 2.
Name the following compounds according to IUPAC system of nomenclature :
- CH3CH(CH3)CH2CH2CHO
- CH3CH2COCH(C2H5)CH2CH2Cl
- CH3CH=CHCHO
- CH3COCH2COCH3
- CH3CH(CH3)CH2C(CH3)2COCH3
- (CH3)3CCH2COOH
- OHCC6H4CHO-p
Solution:
- 4-Methylpentanal
- 6-Chloro-4-ethylhexan-3-one
- But-2-en-1-al
- Pentane-2, 4-dione
- 3, 3, 5-Trimethylhexan-2-one
- 3, 3-Dimethylbutanoic acid
- Benzene-1, 4-dicarbaldehyde
Question 3.
Draw the structures of the following compounds.
- 3-Methylbutanal
- p-Nitropropiophenone
- p-Methylbenzaldehyde
- 4-Methylpent-3-en-2-one
- 4-Chloropentan-2-one
- 3-Bromo-4-phenylpentanoic acid
- p, p’-Dihydroxybenzophenone
- Hex-2-en-4-ynoic acid
Solution:
Question 4.
Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.
Solution:
Question 5.
Draw structures of the following derivatives.
- The 2,4-dinitrophenylhydrazone of benz- aldehyde
- Cyclopropanone oxime
- Acetaldehydedimethylacetal
- The semicarbazone of cyclobutanone
- The ethylene ketal of hexan-3-one
- The methyl hemiacetal of formaldehyde
Solution:
Question 6.
Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents.
- PhMgBr and then H3O+
- Tollens’reagent
- Semicarbazide and weak acid
- Excess ethanol and acid
- Zinc amalgam and dilute hydrochloric acid
Solution:
Question 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.
- Methanal
- 2-Methylpentanal
- Benzaldehyde
- Benzophenone
- Cyclohexanone
- 1-Phenylpropanone
- Phenylacetaldehyde
- Butan-1-ol
- 2,2-Dimethylbutanal
Solution:
Aldol condensation is shown by those aldehydes or ketones which have at least one α-H atom while Cannizzaro reaction is undergone by aldehydes that have no α-H atom.
Question 8.
How will you convert ethanal into the following compounds?
- Butane-1,3-diol
- But-2-enal
- But-2-enoic acid
Solution:
Question 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 an electrophile
Solution:
The possible products of aldol condensation from propanal and butanal are
Question 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.
Solution:
(i) It is given that the compound reduces Tollens’ reagent. This proves that the compound is an aldehyde. Further, the fact that it undergoes Cannizzaro reaction shows that it lacks an α-H atom.
(ii) On oxidation it yields 1,2-benzenedicarboxylic acid. This shows that it is an o-substituted benzaldehyde. The only possible structure for the compound is :
Question 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 (8). (C) on dehydration gives but-1-ene. Write equations for the reactions involved.
Solution:
Question 12.
Arrange thefollowing compounds in increasing order of their property as indicated :
- Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl fert-butyl ketone (reactivity towards HCN)
- CH3CH2CH(Br)COOH,CH3CH(Br)CH2COOH, (CH3)2CHCOOH, CH3CH2CH2COOH (acid strength)
- Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)
Solution:
Question 13.
Give simple chemical tests to distinguish between the following pairs of compounds.
- Propanal and Propanone
- Acetophenone and Benzophenone
- Phenol and Benzoic acid
- Benzoic acid and Ethyl benzoate
- Pentan-2-one and Pentan-3-one
- Benzaldehyde and Acetophenone
- Ethanal and Propanal
Solution:
The given set of compounds may be distinguish by the following reaction.
Question 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
- Methyl benzoate
- m-Nitrobenzoic acid
- p-Nitrobenzoic acid
- Phenylacetic acid
- p-Nitrobenzaldehyde
Solution:
Question 15.
How will you bring about the following conversions in not more than two steps?
- Propanone to Propene
- Benzoic acid to Benzaldehyde
- Ethanol to 3-Hydroxybutanal
- Benzene to m-Nitroacetophenone
- Benzaldehyde to Benzophenone
- Bromobenzene to 1 -Phenylethanol
- Benzaldehyde to 3-Phenyipropan-1 -ol
- Benzaldehyde to a-Hydroxyphenylacetic acid
- Benzoic acid to m-Nitrobenzyl alcohol
Solution:
Question 16.
Describe the following :
- Acetylation
- Cannizzaro reaction
- Cross-aldol condensation
- Decarboxylation
Solution:
Question 17.
Complete each synthesis by giving missing starting material, reagents or products
Solution:
Question 18.
Give plausible explanation for each of the following :
- Cyclohexanone forms cyanohydrin in good yield but 2, 2,6 trimethylcyclohexa- none does not.
- There are two – NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones.
- During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.
Solution:
Question 19.
An organic compound contains 69.77% carbon, 11.63% hydrogen and rest, oxygen. The molecular mass of the compound 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.
Solution:
(a) The given compound does not reduce Tollens’ reagent which implies that it is not an aldehyde.
Question 20.
Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?
Solution:
(i) Phenoxide ion has the following resonating structures :
(iii) The negative charge that rests on the electronegative O atom in carboxylate ion. We know that the presence of negative charge on an electronegative atom makes the ion more stable. For the same reason RCOO– is more stable than the phenoxide ion where the oxygen has no negative charge on it. For the above two reasons carboxylate ion is more stable and has higher acidity than phenol.