CBSE Class 12 Chemistry Aldehydes Ketones and Carboxylic Acids Important Questions Set 02

Very Short Answer Questions (1 Mark)

Question 1. Give IUPAC name of the following compound:
(CH₃)₂C=CHCOCH₃ [CBSE (F) 2010]
Answer:
4-Methylpent-3-en-2-one

Teacher's Note:
1. Number the chain from the end that gives the C=O group the lowest locant; the double bond position comes next.

Question 2. Write the structure of the following compound: 3-oxopentanal. [CBSE (F) 2011]
Answer:
CH₃–CH₂–CO–CH₂–CHO

Teacher's Note:
1. When –CHO and a ketone C=O are both present, the aldehyde is the principal group (C-1) and the ketone is named as the prefix “oxo”.

Question 3. Write the IUPAC name of the following compound:
CH₂=CH–CO–CH₃ [CBSE 2019]
Answer:
But-3-en-2-one

Teacher's Note:
1. The carbonyl carbon gets the lower number (C-2), so the double bond ends up at C-3.

Question 4. Give the name of the reagent that bring the following transformation: But-2-ene to ethanal.
Answer:
O₃/H₂O–Zn dust

Teacher's Note:
1. Reductive ozonolysis (with Zn dust) stops at the aldehyde stage; without Zn, H₂O₂ formed would oxidise it to the acid.

Question 5. Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions: ethanal, propanal, propanone, butanone. [CBSE Delhi 2012]
Answer:
Butanone < Propanone < Propanal < Ethanal

Teacher's Note:
1. Larger and more numerous alkyl groups mean more +I effect and more steric hindrance, so reactivity falls.

Question 6. What is Tollens’ reagent? Write one usefulness of this reagent. [CBSE (AI) 2010]
Answer:
Ammonical silver nitrate (AgNO₃ + NH₄OH) solution is known as Tollens’ reagent. It is used to detect the presence of –CHO group in an organic compound.

Teacher's Note:
1. The active species is [Ag(NH₃)₂]⁺; it is reduced to metallic silver, which appears as a silver mirror.

Question 7. What do you mean by Schiff’s base? Give an example.
Answer:
Aldehydes and ketones react with primary aliphatic or aromatic amines to form azomethines or Schiff’s bases.
R–CH=O (Aldehyde) + H₂N–R′ (1° Amine) \( \xrightarrow[\text{Heat}]{\text{Trace of }H^+} \) R–CH=N–R′ (Schiff’s base) + H₂O

Teacher's Note:
1. Do not confuse Schiff’s base (an imine, C=N–R) with Schiff’s reagent (used to test for aldehydes).

Question 8. Name the aldehyde which does not give Fehling’s solution test.
Answer:
Benzaldehyde

Teacher's Note:
1. Aromatic aldehydes give Tollens’ test but not Fehling’s test.

Question 9. How will you prepare benzyl alcohol from benzaldehyde without using a reducing agent? [HOTS]
Answer:
2C₆H₅CHO (Benzaldehyde) + NaOH (50%) \( \xrightarrow{\text{Cannizzaro reaction}} \) C₆H₅CH₂OH (Benzyl alcohol) + C₆H₅COONa (Sodium benzoate)
Here, a concentrated solution of NaOH is used which is not a reducing agent.

Teacher's Note:
1. In Cannizzaro reaction one benzaldehyde molecule itself acts as the reducing agent by transferring a hydride ion.

Question 10. Write two important uses of formalin. [CBSE Sample Paper 2011]
Answer:
Formalin is used in the
(i) preservation of biological specimens.
(ii) manufacture of bakelite.

Teacher's Note:
1. Formalin is a 40% aqueous solution of formaldehyde.

Question 11. Write the structural formula and IUPAC name of terephthalic acid.
Answer:
  Benzene-1,4-dicarboxylic acid

Teacher's Note:
1. Phthalic acid is the 1,2-isomer and terephthalic acid the 1,4-isomer; terephthalic acid is a monomer of terylene.

Short Answer Questions (2 Marks)

Question 12. Write structures of main compounds A and B in each of the following reactions:
(a) CH₃CH₂OH \( \xrightarrow{\text{PCC}} \) A \( \xrightarrow{\text{CH}_{3}\text{OH/dry HCl(g)}} \) B
(b) C₆H₅COCH₃ \( \xrightarrow{\text{NaOI}} \) A + B [CBSE, Delhi Set 3, 2019]
Answer:
(a) A = CH₃CHO,   B = CH₃CH(OH)OCH₃
CH₃CH₂OH (Ethanol) \( \xrightarrow{\text{PCC}} \) CH₃CHO (Ethanal) (A) \( \xrightarrow{\text{CH}_{3}\text{OH/dry HCl(g)}} \) CH₃CH(OH)OCH₃ (1-methoxy ethanol) (B)
(b) A and B = CHI₃, C₆H₅COONa
C₆H₅COCH₃ (Acetophenone) \( \xrightarrow{\text{NaOI}} \) CHI₃ (Iodoform) (A) + C₆H₅COONa (Sodium benzoate) (B)

Teacher's Note:
1. With one mole of alcohol and dry HCl an aldehyde gives a hemiacetal; with excess alcohol the final product is the acetal CH₃CH(OCH₃)₂.

Question 13. Write the equations involved in the following reactions:
(i) Wolff-Kishner reduction
(ii) Etard reaction. [CBSE, Delhi Set 1, 2017]
Answer:
(i) >C=O \( \xrightarrow[\text{−H}_{2}\text{O}]{\text{NH}_{2}\text{NH}_{2}} \) >C=NNH₂ \( \xrightarrow[\text{heat}]{\text{KOH/ethylene glycol}} \) >CH₂ + N₂
Wolff-Kishner reduction method is used to reduce a carbonyl compound like aldehyde or ketone to a hydrocarbon. The reduction reaction takes place when the carbonyl compound is heated with a mixture of hydrazine and a strong base like potassium hydroxide at a temperature range of 453 K to 473 K in ethylene glycol solvent.
(ii) C₆H₅CH₃ (Toluene) + CrO₂Cl₂ \( \xrightarrow{\text{CS}_{2}} \) C₆H₅CH(OCrOHCl₂)₂ (Chromium complex) \( \xrightarrow{\text{H}_{3}\text{O}^{+}} \) C₆H₅CHO (Benzaldehyde)
Etard reaction is a reaction in which chromyl chloride oxidises methyl group to a chromium complex which gives benzaldehyde upon hydrolysis.

Teacher's Note:
1. Do not mix up the two reductions: Wolff-Kishner uses hydrazine and KOH (basic); Clemmensen uses Zn-Hg and conc. HCl (acidic).

Question 14. Write the reactions involved in the following reaction:
(i) Clemmensen reduction
(ii) Cannizzaro reaction [CBSE, Delhi Set 3, 2017]
Answer:
(i) >C=O \( \xrightarrow[\text{HCl}]{\text{Zn–Hg}} \) >CH₂ + H₂O
Clemmensen reduction is the process by which the carbonyl group of aldehydes and ketones is reduced to CH₂ group on treatment with zinc-amalgam and concentrated hydrochloric acid.
(ii) 2HCHO (Formaldehyde) + Conc. KOH \( \xrightarrow{\Delta} \) CH₃OH (methanol) + HCOOK (Potassium formate)
Cannizzaro reaction is one in which aldehydes which do not have an α-hydrogen atom, undergo self-oxidation and reduction (disproportionation) reaction on treatment with a concentrated alkali.

Teacher's Note:
1. In Cannizzaro reaction one molecule is oxidised (to the acid salt) and the other is reduced (to the alcohol); write both products.

Question 15. (i) What type of aldehydes undergo Cannizzaro reaction?
(ii) Arrange the following compounds in increasing order of their property as indicated:
(a) CH₃COCH₃, C₆H₅COCH₃, CH₃CHO (reactivity towards nucleophilic addition reaction)
(b) Cl–CH₂–COOH, F–CH₂–COOH, CH₃–COOH (acidic character)
Answer:
(i) Having no α-hydrogen
(ii) (a) C₆H₅COCH₃ < CH₃COCH₃ < CH₃CHO
(b) CH₃COOH < Cl–CH₂–COOH < F–CH₂–COOH

Teacher's Note:
1. For (b), the more electronegative the halogen, the stronger its −I effect and the stronger the acid; F beats Cl.

Question 16. Give simple chemical tests to distinguish between the following pairs of compounds:
(a) Benzaldehyde and Benzoic acid
(b) Propanal and Propanone
Answer:
(a) Benzoic acid reacts with NaHCO₃ to give brisk effervescence of CO₂ while benzaldehyde does not.
C₆H₅COOH (Benzoic acid) + NaHCO₃ → C₆H₅COONa (Sodium benzoate) + CO₂↑ + H₂O
(b) Propanal being aldehyde when heated with Tollens’ reagent to give silver mirror but propanone being a ketone does not.
CH₃CH₂CHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → CH₃CH₂COO⁻ + 2Ag↓ (Silver mirror) + 4NH₃ + 2H₂O

Teacher's Note:
1. For a full-mark test, name the reagent, state what each compound shows, and write the equation for the one that reacts.

Question 17. How will you convert the following:
(a) Propanone to propan-2-ol
(b) Ethanal to 2-hydroxy propanoic acid
Answer:
(a) CH₃–CO–CH₃ (Propanone) \( \xrightarrow[\text{Reduction}]{\text{LiAlH}_{4}} \) CH₃–CH(OH)–CH₃ (Propan-2-ol)
(b) CH₃CHO (Ethanal) + HCN → CH₃CH(OH)CN \( \xrightarrow[\text{HCl}]{\text{H}_{2}\text{O}} \) CH₃CH(OH)COOH (2-Hydroxy propanoic acid)

Teacher's Note:
1. Cyanohydrin formation adds one carbon; acid hydrolysis then turns –CN into –COOH, giving an α-hydroxy acid.

Question 18. Ketones are less reactive than aldehydes. Why?
Answer:
Ketones are less reactive than aldehydes due to following facts:
(i) Electron releasing effect: In ketones, the carbonyl carbon is attached to alkyl groups which are electron releasing in nature. These alkyl groups push electrons towards carbonyl carbon and therefore, decrease the magnitude of positive charge on it and make it less reactive toward nucleophilic attack.
(ii) Steric effect: In ketones, the presence of two bulky alkyl groups also hinders the approach of the nucleophile to the carbonyl carbon.

Teacher's Note:
1. Always give both factors, electronic and steric; one factor alone usually earns only half the marks.

Question 19. Write structures of compound A and B in each of the following reactions:
(i) Ethylbenzene (C₆H₅CH₂CH₃) \( \xrightarrow{\text{KMnO}_{4}\text{-KOH}} \) A \( \xrightarrow{\text{H}_{3}\text{O}^{+}} \) B
(ii) Cyclohexanol \( \xrightarrow{\text{CrO}_{3}} \) A \( \xrightarrow{\text{H}_{2}\text{N-NH-CONH}_{2}} \) B [CBSE Delhi Set-1 2019]
Answer:
(i) A = C₆H₅COOK (potassium benzoate),   B = C₆H₅COOH (benzoic acid)
Ethyl benzene when treated with KMnO₄ and KOH, undergoes oxidation to produce potassium benzoate. This potassium benzoate when treated with an acid forms benzoic acid.
(ii) A = cyclohexanone, (CH₂)₅C=O;   B = cyclohexanone semicarbazone, (CH₂)₅C=N–NH–CO–NH₂
Cyclohexanol is oxidized by CrO₃ to cyclohexanone. Cyclohexanone when treated with semicarbazide produces cyclohexanone semicarbazone.

Teacher's Note:
1. Any alkyl side chain on benzene, whatever its length, is oxidised by alkaline KMnO₄ to –COOH (as the salt), provided it has a benzylic hydrogen.

Question 20. Write structures of main compounds A and B in each of the following reactions:
(a) C₆H₅COOH \( \xrightarrow{\text{PCl}_{5}} \) A \( \xrightarrow{\text{H}_{2}\text{/Pd–BaSO}_{4}} \) B
(b) CH₃CN \( \xrightarrow[\text{(ii) H}_{3}\text{O}^{+}]{\text{(i) CH}_{3}\text{MgBr}} \) A \( \xrightarrow{\text{Zn(Hg)/conc. HCl}} \) B [CBSE OD Set-2 2019]
Answer:
(a) A = C₆H₅COCl, B = C₆H₅CHO
C₆H₅COOH \( \xrightarrow{\text{PCl}_{5}} \) C₆H₅COCl (Benzoyl chloride) [A] \( \xrightarrow{\text{H}_{2}\text{/Pd–BaSO}_{4}} \) C₆H₅CHO (Benzaldehyde) [B]
(b) A = CH₃COCH₃, B = CH₃CH₂CH₃
CH₃CN \( \xrightarrow[\text{(ii) H}_{3}\text{O}^{+}]{\text{(i) CH}_{3}\text{MgBr}} \) CH₃–CO–CH₃ (Acetone) [A] \( \xrightarrow{\text{Zn(Hg)/conc. HCl}} \) CH₃CH₂CH₃ (Propane) [B] + H₂O

Teacher's Note:
1. H₂/Pd–BaSO₄ is Rosenmund reduction; the poisoned catalyst stops the reduction at the aldehyde stage.

Question 21. How do you convert the following?
(a) Ethanal to Propanone
(b) Toluene to Benzoic acid
Answer:
(a) CH₃CHO \( \xrightarrow{\text{(i) CH}_{3}\text{MgBr, Dry ether (ii) H}_{2}\text{O/H}^{+}} \) CH₃CH(OH)CH₃ \( \xrightarrow{\text{CrO}_{3}} \) CH₃COCH₃
(b) C₆H₅CH₃ \( \xrightarrow[\text{H}_{3}\text{O}^{+}]{\text{KMnO}_{4}\text{–KOH}} \) C₆H₅COOH

Teacher's Note:
1. To add one carbon to an aldehyde and get a ketone, use a Grignard reagent to form the 2° alcohol, then oxidise it.

Question 22. Write the reactions involved in the following:
(i) Hell-Volhard-Zelinsky reaction
(ii) Decarboxylation reaction [CBSE, Delhi Set 2, 2020]
Answer:
(i) R–CH₂–COOH \( \xrightarrow[\text{(ii) H}_{2}\text{O}]{\text{(i) X}_{2}\text{/Red P}} \) R–CH(X)–COOH   (X = Cl, Br)
In Hell-Volhard-Zelinsky (HVZ) reaction, carboxylic acid having an α-hydrogen is halogenated at α-position on treatment with chlorine or bromine in the presence of red phosphorus to give α-halogenated carboxylic acid.
(ii) R–COONa \( \xrightarrow[\text{Heat}]{\text{NaOH and CaO}} \) R–H + Na₂CO₃
In Decarboxylation reaction, carboxylic acid loses CO₂ to form hydrocarbons when their sodium salts are heated with sodalime (NaOH and CaO) in the ratio 3 : 1.

Teacher's Note:
1. Write all reagents and conditions on the arrow (X₂/red P; NaOH + CaO, heat) and show the side product Na₂CO₃.

Question 23. Do the following conversions in not more than two steps:
(i) Propene to Acetone
(ii) Propanoic acid to 2-hydroxypropanoic acid [CBSE Foreign Set-1, 2, 3 2017]
Answer:
(i) CH₃CH=CH₂ \( \xrightarrow[\text{H}^{+}]{\text{H}_{2}\text{O}} \) CH₃CH(OH)CH₃ \( \xrightarrow[\text{CrO}_{3}]{\text{[O]}} \) CH₃COCH₃
(ii) CH₃CH₂COOH \( \xrightarrow{\text{Br}_{2}\text{/Red P}} \) CH₃CH(Br)COOH \( \xrightarrow[\text{(ii) H}^{+}]{\text{(i) aq KOH or NaOH}} \) CH₃CH(OH)COOH
(or any other suitable method)

Teacher's Note:
1. Acid-catalysed hydration of propene follows Markovnikov’s rule, giving the 2° alcohol needed for a ketone.

Question 24. Would you expect benzaldehyde to be more reactive or less reactive in nucleophilic addition reactions than propanal? Explain your answer.
Answer:
The carbon atom of the carbonyl group of benzaldehyde is less electrophilic than carbon atom of the carbonyl group present in propanal. The polarity of the carbonyl group is reduced in benzaldehyde due to resonance as shown below and hence it is less reactive than propanal.

Teacher's Note:
1. Resonance of the C=O group with the benzene ring feeds electron density into the carbonyl carbon – this is the key point examiners look for.

Question 25. Complete the following reaction sequence: [NCERT Exemplar]
CH₃–CO–CH₃ \( \xrightarrow[(ii)\,H_2O]{(i)\,CH_3MgBr} \) (A) \( \xrightarrow[\text{Ether}]{\text{Na metal}} \) (B) \( \xrightarrow{+CH_3{-}Br} \) (C)
Answer:
A = (CH₃)₃C–OH (2-Methylpropan-2-ol)
B = (CH₃)₃C–O⁻Na⁺ (Sodium tert-butoxide)
C = (CH₃)₃C–OCH₃ (2-Methoxy-2-methylpropane)

Teacher's Note:
1. The last step is Williamson synthesis: the bulky alkoxide must be combined with a methyl (primary) halide, not the other way round.

Question 26. Oxidation of ketones involves carbon–carbon bond cleavage. Name the products formed on oxidation of 2,5-dimethylhexan-3-one. [NCERT Exemplar]
Answer:
(CH₃)₂CH–CO–CH₂–CH(CH₃)₂ \( \xrightarrow{[O]} \) (CH₃)₂CH–COOH (2-Methylpropanoic acid) + (CH₃)₂CH–CH₂–COOH (3-Methylbutanoic acid) + CH₃–CO–CH₃ (Propan-2-one)
CH₃–CO–CH₃ (Acetone) \( \xrightarrow{[O]} \) CH₃COOH (Ethanoic acid) + HCOOH (Methanoic acid)

Teacher's Note:
1. Cleavage can occur on either side of the C=O group, so an unsymmetrical ketone gives a mixture of acids.

Question 27. An organic compound ‘A’ with molecular formula C₅H₈O₂ is reduced to n-pentane on treatment with Zn–Hg/HCl. ‘A’ forms a dioxime with hydroxylamine and gives a positive iodoform test and Tollens’ test. Identify the compound A and deduce its structure. [HOTS]
Answer:
As ‘A’ gives positive iodoform test, so it has CH₃–CO– group.
As ‘A’ gives positive Tollens’ test, so it must have –CHO group.
So ‘A’ is CH₃–CO–CH₂CH₂CHO (4-Oxopentanal).

Teacher's Note:
1. A dioxime means two carbonyl groups; reduction to n-pentane confirms a straight five-carbon chain.

Question 28. What product will be formed on reaction of propanal with 2-methylpropanal in the presence of NaOH? Write the name of the reaction also. [NCERT Exemplar] [HOTS]
Answer:
The reaction taking place is cross aldol condensation. Since both aldehydes have α-hydrogens, a mixture of four aldols is formed:
(i) CH₃CH₂CHO + CH₃CH₂CHO \( \xrightarrow{\text{Dil. NaOH}} \) CH₃CH₂CH(OH)–CH(CH₃)–CHO (3-Hydroxy-2-methylpentanal) \( \xrightarrow[-H_2O]{\Delta/H^+} \) CH₃CH₂CH=C(CH₃)–CHO (2-Methylpent-2-enal)
(ii) (CH₃)₂CHCHO + CH₃CH₂CHO \( \xrightarrow{\text{Dil. NaOH}} \) (CH₃)₂CH–CH(OH)–CH(CH₃)–CHO (3-Hydroxy-2,4-dimethylpentanal) \( \xrightarrow[-H_2O]{\Delta} \) (CH₃)₂CH–CH=C(CH₃)–CHO (2,4-Dimethylpent-2-enal)
(iii) CH₃CH₂CHO + (CH₃)₂CHCHO \( \xrightarrow{\text{Dil. NaOH}} \) CH₃CH₂CH(OH)–C(CH₃)₂–CHO (3-Hydroxy-2,2-dimethylpentanal)
(iv) (CH₃)₂CHCHO + (CH₃)₂CHCHO \( \xrightarrow{\text{Dil. NaOH}} \) (CH₃)₂CH–CH(OH)–C(CH₃)₂–CHO (3-Hydroxy-2,2,4-trimethylpentanal)
Aldols (iii) and (iv) have no α-hydrogen left and cannot be dehydrated.

Teacher's Note:
1. In each aldol the α-carbon of the enolate bonds to the carbonyl carbon of the other molecule; write the enolate partner first to avoid wrong connections.

Short Answer Questions (3 Marks)

Question 29. 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:
(i) CH₃–CHO + CH₃MgBr H₃O⁺
⟶ CH₃–CH(OH)–CH₃
(ii) CH₃CHO Zn-Hg
⟶
Conc. HCl CH₃–CH₃
(iii) CH₃–CHO + C₆H₅CHO dil. NaOH
⟶ C₆H₅–CH(OH)–CH₂CHO

Teacher's Note:
1. In (iii) only ethanal has α-hydrogens, so it forms the carbanion and benzaldehyde is the acceptor; on heating the product dehydrates to cinnamaldehyde.

Question 30. (a) Draw the structures of the following :
(i) p-Methylbenzaldehyde
(ii) 4-Methylpent-3-en-2-one
(b) Describe how the following conversions can be brought about :
Cyclohexanol to cyclohexan 1-one
Answer:
(a) (i) p-Methylbenzaldehyde :
(ii) 4-Methylpent-3-en-2-one : CH₃–CO–CH=C(CH₃)–CH₃
(b) Cyclohexanol CrO₃, H₂SO₄
⟶
Oxidation Cyclohexan-1-one

Teacher's Note:
1. Number the chain from the end nearer the C=O group: the ketone gets C-2, the double bond C-3 and the methyl C-4.

Question 31. (A), (B) and (C) are three non-cyclic functional isomers of a carbonyl compound with molecular formula C₄H₈O. Isomers (A) and (C) give positive Tollens’ test whereas isomer (B) does not give Tollens’ test but gives positive iodoform test. Isomers (A) and (B) on reduction with Zn(Hg)/conc. HCl give the same product (D).
(a) Write the structures of (A), (B), (C) and (D).
(b) Out of (A), (B) and (C) isomers, which one is least reactive towards addition of HCN?
Answer:
(a) As (A) and (C) give positive Tollens’ test thus these two should be aldehyde while (B) should be a ketone (does not give Tollens’ test) with —CO—CH₃ group (as it gives positive iodoform test).
Three isomers are CH₃CH₂CH₂CHO (A), CH₃–CO–CH₂–CH₃ (B), CH₃–CH(CH₃)–CHO (C)
CH₃CH₂CH₂CHO (A) Zn(Hg)/conc. HCl
⟶ CH₃CH₂CH₂CH₃ (D)
CH₃–CO–CH₂–CH₃ (B) Zn(Hg)/conc. HCl
⟶ CH₃CH₂CH₂CH₃ (D)
(b) Out of (A), (B) and (C) isomers, (B) is least reactive towards addition of HCN.

Teacher's Note:
1. A and B must have the same unbranched skeleton because Clemmensen reduction gives the same alkane, so the branched aldehyde is C.

Question 32. In an industry aldehydes are being prepared by controlled oxidation of primary alcohol using acidified K₂Cr₂O₇ or aqueous or alkaline KMnO₄ as oxidant. It was suggested to the owner of the factory to use Collin’s reagent instead of acidic potassium dichromate. The yield of factory increased sharply.
Now answer the following questions :
(i) What is Collin’s reagent?
(ii) What are the advantages of using Collin’s reagent over conventional oxidising agent?
Answer:
(i) Collin’s reagent is a mixture of pyridine (C₅H₅N) and CrO₃ in dichloromethane (CH₂Cl₂).
(ii) Collin’s reagent is a mild oxidant. It oxidises 1°-alcohols to aldehydes and 2°-alcohols are oxidised to ketones.
In case of using acidic K₂Cr₂O₇ as oxidant, the aldehydes and ketones formed by the oxidation of alcohols undergo oxidation to give carboxylic acids.

Teacher's Note:
1. Strong oxidants take a 1° alcohol past the aldehyde to the acid; mild reagents like Collins’ reagent or PCC stop at the aldehyde.

Question 33. Identify A and E in the following series of reactions :
C₆H₅CH₃ + CrO₃ + (CH₃CO)₂O 273–283 K
⟶ A H₃O⁺
⟶ B Conc. NaOH
⟶ NaOOC–C₆H₅ + C
C₆H₅CH₃ KMnO₄, KOH
⟶
heat D Δ, H₃O⁺
⟶ E
Answer:
C₆H₅CH₃ CrO₃ + (CH₃CO)₂O
⟶
273 – 283 K C₆H₅CH(OCOCH₃)₂ (A) H₃O⁺
⟶ C₆H₅CHO (B)
2C₆H₅CHO (B) Δ, conc. NaOH
⟶ C₆H₅CH₂OH (C) + C₆H₅COONa
C₆H₅CH₃ KMnO₄, KOH
⟶
Δ C₆H₅COOK (D) H₃O⁺
⟶ C₆H₅COOH (E)
A = C₆H₅CH(OCOCH₃)₂ (benzylidene diacetate); E = C₆H₅COOH (benzoic acid)

Teacher's Note:
1. The gem-diacetate A protects the aldehyde from further oxidation; it is hydrolysed to benzaldehyde only in the next step.

Question 34. (i) How will you bring about the following conversions?
(a) Ethanal to but-2-enal
(b) Propanone to propene
(ii) Write the IUPAC name of the compound :

Answer:
(i) (a) 2CH₃–CHO (Ethanal) OH⁻
⟶
Aldol condensation CH₃–CH(OH)–CH₂–CHO Δ
⟶
H⁺ CH₃–CH=CH–CHO (But-2-enal)
(b) CH₃–CO–CH₃ (Propanone, Acetone) LiAlH₄
⟶ CH₃–CH(OH)–CH₃ (Propan-2-ol) Conc. H₂SO₄
⟶
443 K CH₃–CH=CH₂ (Propene)
(ii) 1-(4-Nitrophenyl)propan-1-one

Teacher's Note:
1. When the C=O is attached to a ring, name the chain as the parent ketone and the ring as a substituent: 1-(4-nitrophenyl)propan-1-one.

Question 35. Write the structures of the main products of the following reactions :
(i) C₆H₆ + C₆H₅COCl anhydrous AlCl₃
⟶
CS₂
(ii) H₃C–C≡C–H Hg²⁺, H₂SO₄
⟶
(iii) p-O₂N–C₆H₄–CH₃ 1. CrO₂Cl₂
⟶
2. H₂O
Answer:
(i) C₆H₆ + C₆H₅COCl Anhyd. AlCl₃
⟶
CS₂ C₆H₅–CO–C₆H₅ (Benzophenone)
(ii) CH₃–C≡CH Hg²⁺, H₂SO₄
⟶ CH₃–CO–CH₃ (Propanone)
(iii) p-O₂N–C₆H₄–CH₃ 1. CrO₂Cl₂
⟶
2. H₃O⁺ p-O₂N–C₆H₄–CHO (p-Nitrobenzaldehyde)

Teacher's Note:
1. Etard reaction (CrO₂Cl₂) converts a ring –CH₃ into –CHO without touching the –NO₂ group.

Question 36. Complete the following reactions:
(i) C₆H₅CHO \( \xrightarrow{\text{NaCN/HCl}} \)
(ii) (C₆H₅CH₂)₂Cd + 2CH₃COCl →
(iii) CH₃–CH(CH₃)–COOH \( \xrightarrow[\text{(ii) H}_{2}\text{O}]{\text{(i) Br}_{2}\text{/Red P}} \) [CBSE Delhi Set-1, 2019]
Answer:
(i) C₆H₅CHO (Benzaldehyde) \( \xrightarrow{\text{NaCN/HCl}} \) C₆H₅–CH(OH)–CN (Benzaldehyde cyanohydrin)
(ii) (C₆H₅CH₂)₂Cd + 2CH₃COCl → 2CH₃COCH₂C₆H₅ (1-phenylpropan-2-one) + CdCl₂
(iii) CH₃–CH(CH₃)–COOH (2-methylpropanoic acid) \( \xrightarrow[\text{(ii) H}_{2}\text{O}]{\text{(i) Br}_{2}\text{/Red P}} \) (CH₃)₂C(Br)–COOH (2-bromo-2-methylpropanoic acid)

Teacher's Note:
1. Dialkylcadmium converts an acyl chloride into a ketone: the acyl group CH₃CO– joins the alkyl group of the cadmium reagent.

Long Answer Questions (5 Marks)

Question 37. (a) Write the products formed when benzaldehyde reacts with the following reagents:
(i) CH₃CHO in presence of dilute NaOH
(ii) H₂N–NH–C₆H₅
(iii) Conc. NaOH
(b) Distinguish between following:
(i) CH₃–CH=CH–CO–CH₃ and CH₃–CH₂–CO–CH=CH₂
(ii) Benzaldehyde and Benzoic acid
Answer:
(a) (i) C₆H₅CHO (Benzaldehyde) + CH₃CHO (Acetaldehyde) \( \xrightarrow{\text{NaOH}} \) C₆H₅CH=CHCHO (Cinnamaldehyde) + H₂O
(ii) C₆H₅CHO + H₂N–NH–C₆H₅ (Phenyl hydrazine) → C₆H₅CH=N–NH–C₆H₅ + H₂O
(iii) 2C₆H₅CHO + NaOH(conc.) \( \xrightarrow{\text{Heat}} \) C₆H₅CH₂OH (Benzyl alcohol) + C₆H₅COONa (Sodium benzoate)
(b) (i) CH₃–CH=CH–CO–CH₃ gives iodoform test while CH₃–CH₂–CO–CH=CH₂ does not give.
CH₃–CH=CH–CO–CH₃ + 3NaOI → CHI₃ (Yellow ppt.) + 2NaOH + CH₃–CH=CHCOONa
CH₃CH₂COCH=CH₂ + 3NaOI → No ppt
(ii) (1) Benzaldehyde reacts with tollen’s reagent to form silver mirror. Benzoic acid does not give this reaction.
(2) With NaHCO₃ benzaldehyde does not react while benzoic acid produces brisk effervescence.

Teacher's Note:
1. For the iodoform test look for a CH₃CO– group: CH₃CH=CHCOCH₃ has one, but in CH₃CH₂COCH=CH₂ neither group on the C=O is methyl.

Question 38. (a) Write the final products in the following:
(i) (CH₃)₂C=O \( \xrightarrow[\text{Conc. HCl}]{\text{Zn/Hg}} \)
(ii) C₆H₅COONa \( \xrightarrow[\Delta]{\text{NaOH/CaO}} \)
(iii) CH₂=CH–CH₂–CN \( \xrightarrow[\text{(b) H}_{3}\text{O}^{+}]{\text{(a) DIBAL–H}} \)
(b) Arrange the following in the increasing order of their reactivity towards nucleophilic addition reaction:
CH₃COCH₃, HCHO, CH₃CHO, C₆H₅COCH₃
(c) Draw the structure of 2,4-DNP derivative of acetaldehyde. [CBSE Delhi Set-1, 2020]
Answer:
(a) (i) (CH₃)₂C=O + 4[H] \( \xrightarrow[\text{conc. HCl}]{\text{Zn/Hg}} \) CH₃CH₂CH₃ (Propane) + H₂O
(ii) C₆H₅COONa \( \xrightarrow[\Delta]{\text{NaOH/CaO}} \) C₆H₆ + Na₂CO₃
(iii) CH₂=CH–CH₂–CN \( \xrightarrow[\text{(b) H}_{3}\text{O}^{+}]{\text{(a) DIBAL–H}} \) CH₂=CH–CH₂–CHO
(b) C₆H₅COCH₃ < CH₃COCH₃ < CH₃CHO < HCHO
(c) CH₃CHO + H₂N–HN–C₆H₃(NO₂)₂ (2,4-DNP) → CH₃–CH=N–NH–C₆H₃(NO₂)₂ (acetaldehyde 2,4-dinitrophenylhydrazone; NO₂ groups at positions 2 and 4 of the ring) + H₂O

Teacher's Note:
1. DIBAL-H reduces –CN to –CHO but leaves the C=C untouched, so the double bond stays in the product.

Question 39. (a) An organic compound (A) having molecular formula C₄H₈O gives orange red precipitate with 2,4-DNP reagent. It does not reduce Tollens’ reagent but gives yellow precipitate of iodoform on heating with NaOH and I₂. Compound (A) on reduction with NaBH₄ gives compound (B) which undergoes dehydration reaction on heating with conc. H₂SO₄ to form compound (C). Compound (C) on ozonolysis gives two molecules of ethanal. Identify (A), (B) and (C) and write their structures. Write the reactions of compound (A) with (i) NaOH/I₂ and (ii) NaBH₄.
(b) Give reasons:
(i) Oxidation of propanal is easier than propanone.
(ii) α-hydrogen of aldehydes and ketones is acidic in nature.
Answer:
(a) Compound A (C₄H₈O) gives positive 2,4-DNP test, it must be carbonyl compound. It gives iodoform test.
CH₃–CO–CH₂–CH₃ (A, Butanone) \( \xrightarrow{\text{NaBH}_{4}} \) CH₃–CH(OH)–CH₂–CH₃ (B, Butan-2-ol) \( \xrightarrow{\text{conc. H}_{2}\text{SO}_{4}} \) CH₃CH=CHCH₃ (C, But-2-ene) \( \xrightarrow{\text{Ozonolysis}} \) 2CH₃CHO (Ethanal)
(i) CH₃–CO–CH₂–CH₃ (A) \( \xrightarrow{\text{NaOH/I}_{2}} \) C₂H₅COONa + CHI₃ (Iodoform)
(ii) CH₃–CO–CH₂–CH₃ (A) \( \xrightarrow{\text{NaBH}_{4}} \) CH₃–CH(OH)–CH₂–CH₃ (Butan-2-ol)
(b) (i) Oxidation of propanal is easier than propanone because aldehydes have one hydrogen atom attached to the carbonyl group while ketones have two alkyl or aryl groups attached to the carbonyl group. Propanal easily oxidised to form acid with same number of carbon atoms whereas propanone is difficult to be oxidise and form acids with less number of carbon atoms.
CH₃–CO–CH₃ (Propanone) \( \xrightarrow{\text{H}^{+}} \) CH₃–C(OH)=CH₂ \( \xrightarrow{\text{Oxidation}} \) CH₃COOH (Ethanoic acid) + CO₂ + H₂O
CH₃–CH₂–CHO \( \xrightarrow{\text{[O]}} \) CH₃CH₂COOH
(ii) α-hydrogen of aldehydes and ketones are acidic in nature because carbonyl group has the great electron withdrawing effect and the molecule thus formed after the removal of α-Hydrogen is resonance stabilised.
H–CH₂–CO–R \( \xrightarrow{\text{Base (OH}^{-}\text{)}} \) [⁻CH₂–CO–R ↔ CH₂=C(O⁻)–R]   (R = H, alkyl group)

Teacher's Note:
1. Work backwards from the ozonolysis clue: two CH₃CHO molecules come only from CH₃CH=CHCH₃, which fixes B as butan-2-ol and A as butanone.

Question 40. (a) Draw structures of the following derivatives:
(i) Cyanohydrin of cyclobutanone
(ii) Hemiacetal of ethanal
(b) Write the major product(s) in the following:
(i) CH₃–CH=CH–CH₂–CN \( \xrightarrow[\text{(ii) H}_{3}\text{O}^{+}]{\text{(i) DIBAL–H}} \)
(ii) CH₃–CH₂–OH \( \xrightarrow[\text{H}^{+}]{\text{CrO}_{3}} \)
(c) How can you distinguish between propanal and propanone? [CBSE Delhi Set-1, 2020]
Answer:
(a) (i) Cyanohydrin of cyclobutanone: a four-membered ring in which one ring carbon carries both –OH and –CN (1-hydroxycyclobutane-1-carbonitrile).
(ii) Hemiacetal of ethanal: CH₃–CH(OR)(OH)
(b) (i) CH₃–CH=CH–CH₂CN \( \xrightarrow[\text{(ii) H}_{3}\text{O}^{+}]{\text{(i) DIBAL–H}} \) CH₃–CH=CH–CH₂–CH=N–Al(iBu)₂ \( \xrightarrow{\text{H}_{3}\text{O}^{+}} \) CH₃–CH=CH–CH₂–CHO
(ii) CH₃–CH₂–OH \( \xrightarrow{\text{CrO}_{3}} \) [CH₃–CHO] → [CH₃–COOH]
(c) By iodoform test: Propanone on treatment with I₂/NaOH undergoes iodoform test to give a yellow ppt. of iodoform.
CH₃COCH₃ + 3NaOI → CHI₃ (Yellow ppt.) + CH₃COONa + 2NaOH
Propanal does not give this test.

Teacher's Note:
1. A hemiacetal carries both –OH and –OR on the same carbon; an acetal carries two –OR groups.

Question 41. Write the structures of A, B, C, D and E in the following reactions:
C₆H₆ \( \xrightarrow[\text{Anhyd. AlCl}_{3}]{\text{CH}_{3}\text{COCl}} \) A \( \xrightarrow{\text{Zn–Hg/Conc. HCl}} \) B \( \xrightarrow[\text{(ii) H}_{3}\text{O}^{+}]{\text{(i) KMnO}_{4}\text{–KOH, }\Delta} \) C
A \( \xrightarrow{\text{NaOH/I}_{2}} \) D + E
Answer:
A – C₆H₅COCH₃;   B – C₆H₅CH₂CH₃;   C – C₆H₅COOH;   D, E – C₆H₅COONa, CHI₃
C₆H₆ \( \xrightarrow[\text{Anhyd. AlCl}_{3}]{\text{CH}_{3}\text{COCl}} \) C₆H₅COCH₃ (Acetophenone, A) \( \xrightarrow{\text{Zn–Hg/Conc. HCl}} \) C₆H₅C₂H₅ (Ethylbenzene, B) \( \xrightarrow[\text{H}^{+}]{\text{KMnO}_{4}\text{–KOH}} \) C₆H₅COOH (Benzoic acid, C)
C₆H₅COCH₃ (A) \( \xrightarrow{\text{NaOH/I}_{2}} \) C₆H₅COO⁻Na⁺ (Sodium benzoate, D) + CHI₃↓ (Iodoform, yellow ppt., E)

Teacher's Note:
1. Read a reaction scheme branch by branch; here A is used twice, so identify A first and check it fits both arrows.

Question 42. (a) Write the chemical equation for the reaction involved in Cannizzaro reaction.
(b) Draw the structure of the semicarbazone of ethanal.
(c) Why pK_a of F–CH₂–COOH is lower than that of Cl–CH₂–COOH?
(d) Write the product in the following reaction
CH₃–CH=CH–CH₂CN \( \xrightarrow[\text{(ii) H}_{2}\text{O}]{\text{(i) DIBAL-H}} \) ?
(e) How can you distinguish between propanal and propanone? [CBSE Delhi 2016]
Answer:
(a) HCHO + HCHO \( \xrightarrow{\text{conc. NaOH}} \) HCOONa + CH₃OH (or any other example)
(b) CH₃CH=N–NHCONH₂
(c) Stronger −I effect of fluorine, stronger acid less pK_a / strong electron withdrawing power of fluorine.
In F–CH₂–COOH, fluorine is more electron withdrawing than chlorine in Cl–CH₂–COOH, so F–CH₂–COOH is more acidic than ClCH₂COOH. Hence, its pK_a value is lesser than ClCH₂COOH.
(d) CH₃CH=CHCH₂CHO (Pent-3-enal)
(e) Silver mirror formed on adding ammonical silver nitrate to propanal and not with propanone. (or any other correct test)
CH₃CH₂CHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → CH₃CH₂COO⁻ + 2Ag↓ (Silver mirror) + 4NH₃ + 2H₂O

Teacher's Note:
1. In Cannizzaro reaction one aldehyde molecule is oxidised to the acid salt and the other reduced to the alcohol; show both products.