Please refer to Aldehydes, Ketones and Carboxylic Acids Class 12 Chemistry Important Questions with answers below. These solved questions for Chapter 12 Aldehydes, Ketones and Carboxylic Acids in NCERT Book for Class 12 Chemistry have been prepared based on the latest syllabus and examination guidelines issued by CBSE, NCERT, and KVS. Students should learn these solved problems properly as these will help them to get better marks in your class tests and examinations. You will also be able to understand how to write answers properly. Revise these questions and answers regularly. We have provided Notes for Class 12 Chemistry for all chapters in your textbooks.

Important Questions Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids

All Aldehydes, Ketones and Carboxylic Acids Class 12 Chemistry Important Questions provided below have been prepared by expert teachers of Standard 12 Chemistry. Please learn them and let us know if you have any questions.

Very Short Answer Questions :

Question. (a) Give a plausible explanation for each one of the following :
(i) There are two –NH₂ groups in semicarbazide. However, only one such group is involved in the formation of  semicarbazones
(ii) Cyclohexanone forms cyanohydrins in good yield but 2,4,6-trimethylcyclohexanone does not.
(b) An organic compound with molecular formula C₉H₁₀O forms 2,4-DNP derivative, reduces Tollen’s reagent and undergoes Cannizzaro’s reaction. On vigorous oxidation it gives 1,2-benzene-di-carboxylic acid. Identify the compound.  
Answer : (a) (i) Semicarbazide has the following resonance structures arising due to the electron withdrawing nature of the O atom.

Lone pairs of N-1 and N-2 are involved in conjugation with >C=O group while that of N-3 is not involved in resonance thus, it is involved in the formation of semicarbazone.
(ii) Formation of cyanohydrin involves the nucleophilic attack of cyanide ions (CN^–) at the carbonyl carbon. In cyclohexanone, reaction proceeds but in 2,4,6-trimethylcyclohexanone, the methyl groups cause steric hindrance and yields are poor.

(b) The compound forms 2,4-DNP derivative. It shows that it is a carbonyl compound. Further it reduces Tollens’ reagent which shows that it contains aldehydic group. It undergoes Cannizzaro reaction indicating that aldehyde group is without any a-hydrogen. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid which shows that there are two carbon residues on benzene ring. Since the molecular formula is C₉H₁₀O, it fits into the
structure, 2-ethylbenzaldehyde.

Question. Why carboxylic acid does not give reactions of carbonyl group?  
Answer : The carbonyl group in —COOH is inert and does not show nucleophilic addition reaction like carbonyl compound. It is due to resonance stabilisation of carboxylate ion :

Question. Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Give two reasons.  
Answer : Phenoxide ion has the following resonating structures :

(i) Phenoxide ion is a resonance hybrid of structures I to V, where each structure has a contribution of 20% in the resonance hybrid. On the other hand, each of the two contributing structures of carboxylate ion contribute 50% towards the resonance hybrid. Therefore, the carboxylate ion tends to be more stable than the phenoxide ion and
hence has higher acidity.
(ii) The negative charge rests on the electronegative O atom in carboxylate ion. 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 carbon has negative charge on it. For the above two reasons carboxylate ion is more stable and has higher acidity than phenol.

Question. Write the structures of A, B, C, D and E in the following reactions :

Answer : 

Question. Describe the following giving chemical equation :
De-carboxylation reaction 
Answer : Decarboxylation : Sodium or potassium salt of carboxylic acids on heating with soda lime (NaOH and CaO), loses a molecule of carbon dioxide and alkanes are obtained as products.

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

Answer : 

Question. (a) Identify A, B and C in the following sequence of reactions :

(b) Predict the structures of the products formed when benzaldehyde is treated with
(i) conc. NaOH
(ii) HNO₃/H₂SO₄ (at 273– 383 K) 
Answer :

Question. Write the IUPAC name of

Answer : 3-Bromo-5-chloro Benzoic acid

Question. Identify A to E in the following reactions :

Answer : 

Question. Write the main product in the following equation :
CH₃—COOH →^PCI₅ 
Answer : CH₃—COOH  →^PCI₅  CH₃—COCl + POCl₃ + HCl

Question. How will you carry out the following conversions?
(i) Acetylene to Acetic acid
(ii) Toluene to m-nitrobenzoic acid 
Answer : 

Question. How will you bring about the following conversion?
Benzoic acid to Benzaldehyde 
Answer : 

Question. Write the IUPAC name of the following compound:

Answer : Ethyl-4-chlorobenzoate

Question. Write the product in the following reaction :
CH₃—CH—CH—CH₂CN →^{(i) DIBAL-H}_{(ii) H2O} 
Answer : CH₃—CH=CH—CH₂CN →^{(i) DIBAL-H}_{(ii) H2O} CH₃—CH=CH—CH₂CHO

Question. Account for the following :
Carboxyl ic acids do not give reactions of carbonyl group.  
Answer : In carboxylic acid C O is in resonance and not available for reaction.

Question. Write the products formed when ethanal reacts with the following reagents :
(i) CH₃MgBr and then H₃O⁺
(ii) Zn-Hg/conc. HCl
(iii) C₆H₅CHO in the presence of dilute NaOH 
Answer : 

Short Answer Questions :

Question. How will you obtain the following :
Benzoic acid from Aniline 
Answer : 

Question. Write the chemical equation to illustrate the following name reaction :
Hell-Volhard-Zelinsky reaction 
Answer : Hell-Volhard-Zelinsky reaction : Carboxylic acids react with chlorine or bromine in the presence of phosphorous to give compounds in which a-hydrogen atom is replaced by halogen atom.

Question. Account for the following :
Cl—CH₂COOH is a stronger acid than CH₃COOH. 
Answer : Chloroacetic acid has lower pKa value than acetic acid; ‘Cl’ in chloroacetic acid shows –I effect, it creates less electron density on oxygen of carboxylic acid. Thus, release of proton becomes easier. In case of acetic acid, the state of affair is just opposite. Hence, chloroacetic acid is stronger than acetic acid.

Question. An organic compound (A) on treatment with acetic acid in the presence of sulphuric acid roduces an ester (B). (A) on mild oxidation gives (C). (C) with 50% KOH followed by  acidification with dilute HCl generates (A) and
(D). (D) with PCl₅ followed by reaction with ammonia gives (E). (E) on dehydration produces hydrocyanic acid. Identify the compounds A, B, C, D and E. 
Answer : 

Question. Write the products of the following reaction :

Answer : 

Question. Arrange the following compounds in an increasing order of their indicated property :
(i) Benzoic acid, 4-Nitrobenzoic acid, 3,
4-Dinitrobenzoic acid.
4-Methoxybenzoic acid (acid strength)
(ii) CH₃CH₂CH(Br)COOH, CH₃CH(Br)CH₂COOH
(CH₃)₂CHCOOH, CH₃CH₂CH₂COOH
(acid strength)  
Answer : (i) 4-methoxybenzoic acid < benzoic acid < 4-nitrobenzoic acid < 3,4-dinitrobenzoic acid.
(ii) The overall acidic strength increases in the order :
(CH₃)₂CHCOOH < CH₃CH₂CH₂COOH < CH₃CH(Br)CH₂COOH < CH₃CH₂CH(Br)COOH

Question. Name the reagent used in the following reaction :
CH₃—COOH →^? CH₃—COCl 
Answer : CH₃—COOH  →^PCI₅  CH₃—COCl + POCl₃ + HCl

Question. Write one chemical equation for each to illustrate the following reaction :
Fischer esterification. 
Answer : Fischer esterification :
RCOOH + R′OH = RCOOR′ + H₂O
 Acid         Alcohol

Question. State reasons for the following :
(i) Monochloroethanoic acid has a higher pK_a value than dichloroethanoic acid.
(ii) Ethanoic acid is a weaker acid than benzoic acid.  
Answer : (i) The strength of an acid is indicated by pKa value, where, pKa = – log K_a
Since monochloroethanoic acid is weaker than dichloroethanoic acid so it has lower value of dissociation constant K_a.
Therefore, it has higher value of pK_a.
(ii) The —COOH group in benzoic acid is attached to sp² – carbon of the phenyl ring and is more acidic than acetic acid in which —COOH group is attached to sp³ – carbon atom of CH₃ group. So, benzoic acid is stronger than acetic or acetic acid is weaker acid than benzoic acid.

Question. Identify A and E in the following series of reactions :

Answer : 

Question. An organic compound with molecular formula C₅H₁₀O does not reduce Tollens’ reagent but forms an addition compound with sodium hydrogen sulphite and gives a positive  iodoform test. On vigorous oxidation, it gives ethanoic acid and propanoic acid. Identify the compound and write all chemical equations for the reactions.  
Answer : The given compound does not reduce Tollens’ reagent, so it is not an aldehyde but the formation of addition compound with sodium hydrogen sulphite indicates it to be a carbonyl compound. Since this compound gives positive iodoform test, so it should

On oxidation, this compound gives ethanoic and propanoic acids which confirm its structure to be I.

Question. An organic compound A(C₃H₆O) is resistant to oxidation but forms compound B(C₃H₈O) on reduction. B reacts with HBr to form the compound C. C with Mg forms Grignard  reagent D which reacts with A to form a product which on hydrolysis gives E. Identify A and E. 
Answer : Ketones are oxidised under vigrous conditions.

Question. Predict the products of the following reaction :
CH₃COONa →_Δ^NaOH/CaO ? 
Answer : CH₃COONa →_Δ^NaOH/CaO CH₄ + Na₂CO₃
                                                  Methane

Question. Write the mechanism of esterification of carboxylic acids. 
Answer : Esterification : Carboxylic acids react with alcohols or phenols in the presence of a mineral acid like concentrated H₂SO₄ or HCl gas as catalyst and give ester.

Question. How are the following conversions carried out?
(i) Ethyl cyanide to ethanoic acid.
(ii) Butan-1-ol to butanoic acid.
(iii) Benzoic acid to m-bromobenzoic acid. 
Answer : 

Question. An organic compound (A) with molecular formula C₈H₈O forms an orange red precipitate with 2, 4-DNP reagent and gives yellow precipitate on heating with I₂ and NaOH. It neither reduces Tollens’ reagent nor Fehling’s reagent, nor does it decolourise bromine water or Baeyer’s reagent. On drastic oxidation with chromic acid, it gives a carboxylic acid (B) having molecular formula C₇H₆O₂. Identify the compounds (A) and (B) and explain the reactions involved. 
Answer : (A) forms, 2, 4-DNP derivative. Therefore, it is an aldehyde or a ketone. Since it does not reduce Tollen’s or Fehling reagent, (A) must be a ketone. (A) responds to iodoform test.
Compound (B), being an oxidation product of a ketone should be a carboxylic acid. The molecular formula of (B) indicates that it should be benzoic acid and compound (A) should, therefore, be a mono-substituted aromatic methyl ketone.

Question. How would you convert :
Benzoic acid to benzamide.  
Answer : Benzoic acid to benzamide :

Question. Name the reagents used in the following reaction :
C₆H₅—CH₂—CH₃ →? C₆H₅—COO^–K⁺ 
Answer : Alkaline potassium permanganate (KMnO₄,KOH)

Question. Distinguish between
CH₃COOH and HCOOH  
Answer : Add Tollens’ reagent to formic acid and warm. Silver mirror is formed.

Acetic acid does not give this test.

Question. Two moles of organic compound ‘A’ on treatment with a strong base gives two compound ‘B’ and ‘C’. Compound ‘B’ on dehydrogenation with Cu gives ‘A’ while acidication of ‘C’ yields carboxylic acid ‘D’ with molecular formula of CH₂O₂. Identify the compounds A, B, C and D and write all chemical reactions involved. 
Answer : Since the molecular formula of D is CH₂O₂, thus, D is HCOOH (formic acid). D is obtained by the acidification of C, so, C is sodium formate (HCOONa).
Thus, A must be formaldehyde (as it undergoes Cannizzaro reaction with a strong base).

Question. (a) Write the product of the following reaction :
CH₃COOH →^CI₂/P
(b) Give simple chemical tests to distinguish between the following pairs of compounds :
Benzaldehyde and benzoic acid 
Answer : 

(b) Benzaldehyde when treated with ammoniacal silver nitrate gives silver mirror.

Question. Arrange the following compounds in an increasing order of their property as indicated :
(i) Benzoic acid, 3,4-dinitrobenzoic acid, 4-methoxybenzoic acid (acid strength)
(ii) CH₃CH₂CH(Br)COOH, CH₃CH(Br)CH₂COOH, (CH₃)₂CHCOOH
                                                                           (acid strength) 
Answer : (i) The overall acid strength increases in the order.
4-methoxybenzoic acid < benzoic acid < 3,4-dinitrobenzoic acid.
(ii) We know that + I-effect decreases while –I-effect increases the acid strength of carboxylic acids. The overall acid strength increases in the order.
(CH₃)₂CHCOOH < CH₃CH(Br)CH₂COOH < CH₃CH₂CH(Br)COOH

Question. How will you convert the following :
Ethanal to 2-hydroxy propanoic acid 
Answer :

Question. Write the chemical equations to illustrate the following name reactions :
(i) Wolff–Kishner reduction
(ii) Aldol condensation
(iii) Cannizzaro reaction 
Answer : (i) Wolff-Kishner reduction reaction : The carbonyl group of aldehydes and ketones is reduced to CH₂ group on treatment with hydrazine followed by heating with potassium hydroxide in a high boiling solvent such as ethylene glycol.

(ii) Aldol condensation : Two molecules of an aldehyde or ketones having at least one a-hydrogen atom condense in the presence of a dilute alkali to give b-hydroxyaldehyde or b-hydroxyketone which upon heating give a,b-unsaturated aldehyde or ketone.

(iii) Cannizzaro’s reaction : Aldehydes which do not contain a-H atom undergo disproportionation when heated with concentrated (50 %) NaOH.

Question. How are the following conversions carried out :
(i) Ethylcyanide to ethanoic acid
(ii) Butan-1-o1 to butanoic acid
(iii) Methylbenzene to benzoic acid Write Chemical equations for the involved reactions.  
Answer : 

Question. An organic compound (A) on treatment with ethyl alcohol gives a carboxylic acid (B) and compound (C). Hydrolysis of (C) under acidified conditions gives (B) and (D). Oxidation of (D) with KMnO₄ also gives (B). (B) on heating with Ca(OH)₂ gives (E) having molecular formula C₃H₆O. (E) does not give Tollen’s test and does not reduce Fehling’s solution but forms a 2, 4-dinitrophenylhydrazone. Identify (A), (B), (C), (D) and (E).  
Answer : 

Question. How is the following obtained?
Benzoic acid from ethyl benzene. 
Answer : 

Question. How are the following conversions carried out Acetic acid to methylamine. 
Answer : Acetic acid to methyl amine :

Question. Arrange the following in the increasing order of their boiling points.
CH₃CHO, CH₃COOH, CH₃CH₂OH 
Answer : Increasing order of boiling point :
CH₃—CHO < C₂H₅OH < CH₃—COOH

Question. (a) Giving a chemical equation for the following process :
Decarboxylation
(b) State chemical tests to distinguish between the following pairs of compounds :
Phenol and Benzoic acid  
Answer : (i) Decarboxylation : Sodium or potassium salt of carboxylic acids on heating with soda lime (NaOH and CaO), loses a molecule of carbon dioxide and alkanes are obtained as products.

(ii) Phenol and benzoic acid can be distinguished by their reactions with sodium bicarbonate solution. Benzoic acid will give effervescence with NaHCO₃ but phenol will not react.

Question. Write the IUPAC name of the following :

Answer : Hex-2-en-4yn-oic acid

Question. Give reasons :
Chloroacetic acid is stronger than acetic acid. 
Answer : Chloroacetic acid has lower pKa value than acetic acid; ‘Cl’ in chloroacetic acid shows –I effect, it creates less electron density on oxygen of carboxylic acid. Thus, release of proton becomes easier. In case of acetic acid, the state of affair is just opposite. Hence, chloroacetic acid is stronger than acetic acid.

Question. Give chemical tests to distinguish between the following pairs of compounds :
(i) Methyl acetate and ethyl acetate.
(ii) Benzaldehyde and benzoic acid. 
Answer : (i) Ethylacetate is hydrolysed slowly by water to form ethyl alcohol while methyl acetate gives methyl alcohol.

(ii) Benzaldehyde when treated with ammoniacal silver nitrate gives silver mirror.

Question. Predict the organic products of the following reaction :

Answer : 

Question. Give simple chemical tests to distinguish between the following pairs of compounds:
Benzoic acid and Ethyl benzoate 
Answer : Benzoic acid and ethyl benzoate can be distinguished by their reactions with sodium bicarbonate solution. Benzoic acid will give effervescence with NaHCO₃.