Aldehydes Ketones and Carboxylic Acids Class 12 Chemistry Important Questions

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Important Questions of Aldehydes Ketones and Carboxylic Acids Class 12

Very Short Answer Questions

Question. Draw the structure of the following derivatives :
(i) Propanone oxime
(ii) Semicarbazone of the CH₃CHO
Answer.

Question. Give simple chemical tests to distinguish between the following pairs of compounds :
(i) Benzaldehyde and benzoic acid
(ii) Propanal and propanone.
Answer. (i) Benzaldehyde and benzoic acid can be distinguished by sodium bicarbonate test.
Benzoic acid will give effervescence with NaHCO₃ but benzaldehyde will not react.
(ii) Propanal and propanone can be distinguished by their reactions with Tollens’ reagent.
Propanal will form the silver mirror, but propanone does not react.

Question. Account for the following :
CH₃CHO is more reactive than CH₃COCH₃
towards reaction with HCN.
Answer. It is a nucleophilic addition reaction, in which CN – acts as a nucleophile. CH₃CHO undergoes nucleophilic addition reactions faster than CH₃COCH₃ as in CH₃COCH₃ there are two electron releasing methyl groups attached to the carbonyl carbon that hinders the approach of nucleophile to carbonyl carbon and reduce the electrophilicity of the carbonyl group while in CH₃CHO, there is only one methyl group attached to carbonyl carbon.

Question. Write the IUPAC name of the following :

Answer.

Question. Write the product in the following reaction :

Answer.

Question. How do you convert the following :
Ethyne to Ethanal.
Answer.

Question. Write the IUPAC name of the following compound :

Answer.

Question. Draw the structure of 3-methylpentanal.
Answer.

Question. Write the structure of 3-methyl butanal.
Answer.

Question. Write the structure of p-Methylbenzaldehyde molecule.
Answer.

Question. Draw the structure of the compound named 4-methylpent-3-en-2-one.
Answer.

Question. Write the IUPAC name of the following

Answer.

Question. Distinguish between the following :
(i) C₆H₅—COCH₃ and C₆H₅—CHO
Answer. Benzaldehyde and acetophenone can be distinguished by Tollens’ test.
Benzaldehyde will form silver mirror, on treatment with Tollens’ reagent whereas acetophenone will not show Tollens’ Test.

Question. Name the reagents used in the following reactions :

Answer. (i) Lithium aluminium hydride (LiAlH₄).

Question. Write the structure of 3-oxopentanal.
Answer.

Question. How is following obtained?
Benzaldehyde from toluene.
Answer.

Question. How would you convert :
Ethanol to acetone.
Answer.

Question. Aldehydes and Ketones have lower boiling points than corresponding alcohols. Why?
Answer. The boiling points of aldehydes and ketones are lower than that of corresponding alcohols and acids due to absence of intermolecular H–bonding in aldehydes and ketones.

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

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. Write the product in the following reaction :

Answer.

Question. How will you obtain the following :
Benzaldehyde from Phenol.
Answer.

Question. Give simple chemical tests to distinguish between the following pair of compounds.
Propanal and Butan-2-one.
Answer. Propanal and Butan-2-one can be distinguised by their reactions with tollen’s reagent. Propanal will form the silver mirror, but Butan-2-one does not react.

Question. Write the IUPAC name of the following :
CH₃ — CH₂ — CHO
Answer.

Question. Write the IUPAC name of the compound :

Answer.

Question. Write the structural formula of 1-phenylpentan-1-one.
Answer.

Question. Write the IUPAC name of the following :

Answer.

Question. Write the IUPAC name of the following compound

Answer.

Question. Write the IUPAC name of the following :

Answer.

Question. Write the IUPAC name of
Ph — CH = CH — CHO
Answer.

Question. Write the IUPAC name of the following :

Answer.

Question. Draw the structural formula of 1-phenyl Propan-1-one molecule.
Answer.

Question. Write the structure of 4-chloropentan-2-one.
Answer.

Question. Write the IUPAC name of the following compound :

Answer.

Question. Give simple chemical tests to distinguish between the following pair of compounds :
Ethanal and Propanal
Answer. Ethanal and propanal can be distinguished by iodoform test.
Yellow precipitate of iodoform will be formed from ethanal on heating with iodine and sodium hydroxide solution.

Question. Write the IUPAC name of the compound :

Answer.

Question. Write the IUPAC name of the compound :

Answer.

Question. Given reason :
pH of reaction should be carefully controlled while preparing ammonia derivatives of carbonyl compounds.
Answer. In strongly acidic medium ammonia derivatives being basic will react with acids and will not react with carbonyl compound. In basic medium, OH– will attack carbonyl group. Therefore, pH of a reaction should be carefully controlled.

Short Answer Questions

Question. Predict the products of the following reactions :

Answer.

Question. How will you bring about the following conversions?
(i) Ethanal to but-2-enal
(ii) Propanone to propene
Answer.

Question. How will you bring about the following conversions :
(i) Ethanol to 3-hydroxybutanal
(ii) Benzaldehyde to Benzophenone
Answer.

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. How will you convert the following :
(i) Propanone to Propan-2-ol
(ii) Ethanal to 2-hydroxy propanoic acid.
Answer.

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. Which acid of each pair shown here would you expect to be stronger?
(i) F—CH₂—COOH or Cl—CH₂—COOH

Answer.

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

Question. Write a suitable chemical equation to complete each of the following transformations :
(i) Butan-1-ol to butanoic acid
Answer.

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. Give simple chemical test to distinguish between :
(a) Pentan-2-one and Pentan-3-one.
(b) Ethanal and Propanal.
Answer.

Question. Write the products of the following reactions :

Answer.

Question. A compound ‘A’ of molecular formula C₂H₃OCl undergoes a series of reactions as shown below. Write the structure of A, B, C and D in the following reactions

Answer.

Question. Describe the following reactions :
(i) Acetylation
(ii) Aldol condensation
Answer. (i) Acetylation : Introduction of acetyl group

(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 β-hydroxyaldehyde or β-hydroxyketone which upon heating give a,β-unsaturated aldehyde or ketone.

Long Answer Questions

Question. Write the structures of the main products when acetone (CH3 — CO — CH₃) reacts with the following reagents :
(i) Zn Hg/conc. HCl
(ii) H₂N NHCONH₂/H⁺
(iii) CH₃MgBr and then H₃O⁺
Answer.

Question. An organic compound (A) which has characteristic odour, on treatment with NaOH forms two compounds (B) and (C). Compound
(B) has the molecular formula C₇H₈O which on oxidation with CrO₃ gives back compound (A). Compound (C) is the sodium salt of the acid. Compound (C) when heated with soda lime yields an aromatic hydrocarbon (D). Deduce
Answer. Benzaldehyde has a characteristic odour. Reaction of ‘A’ with NaOH appears to be Cannizzaro reaction which gives ‘B’ (benzyl alcohol) and ‘C’ (Sodium salt of benzoic acid). Oxidation of alcohols gives aldehydes. Sodium salt of benzoic acid on heating with soda lime given benzene (D).

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.

Therefore, (A) : Propanone
(B) : Propan-2-ol
(C) : 2-Bromopropane
(D) : sec-Propyl magnesium bromide
(E) : 2,3-Dimethylbutan-2-ol

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

Answer.

Question. Identify A to E in the following reactions :

Answer.

Question. An organic compound A has the molecular formula C₈H₁₆O₂. It gets hydrolysed with dilute sulphuric acid and gives a carboxylic acid B and an alcohol C. Oxidation of C with chromic acid also produced B. C on dehydration reaction gives but-1-ene. Write equations for the reactions involved.
Answer. Compound ‘A’ (C₈H₁₆O₂) on hydrolysis gives an acid ‘B’ and an alcohol ‘C’. It shows that ‘A’ is an ester. Since the oxidation of alcohol ‘C’ also gives the acid ‘B’ indicates that ‘B’ and ‘C’ both contain same number of carbon atoms, i.e, four carbon atoms each and same arrangement of atoms. Formation of but-1-ene on dehydration of ‘C’ indicates it to be butan-1-ol, so the possible structure for ‘A’ could be

Question. Identify A, B, C, D and E in following sequence of reactions :

Answer.

Question. 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-benzenedicarboxylic acid. Identify the compound.
Answer. 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.