CBSE Class 12 Chemistry Aldehydes Ketones and Carboxylic Acids Important Questions Set 03

Very Short Answer Questions (1 Mark)

Teacher's Note:
1. Number the chain from the –CHO end (C-1) for aldehydes; for ketones give the C=O the lowest locant, then the double bond.

Question 2. Write the IUPAC name of the following:
CH₃–CH₂–CHO [CBSE Comptt. OD 2015]
Answer:
Propanal.

Teacher's Note:
1. Count the CHO carbon in the chain: three carbons in all, so the parent is propane → propanal.

Question 3. Write the IUPAC name of CH₃–CH(NH₂)–CHO [CBSE OD 2014]
Answer:
2-Aminopropanal.

Teacher's Note:
1. The –CHO carbon is always C-1, so the amino group on the next carbon gets locant 2.

Question 4. Write the IUPAC name of:
CH₃–CH₂–CH=CH–CHO [CBSE OD 2012]
Answer:
Pent-2-enal.

Teacher's Note:
1. For an unsaturated aldehyde the ending is ‘-enal’; the locant 1 for –al is understood and not written.

Question 5. Draw the structure of 3-methylpentanal. [CBSE Comptt. Delhi 2015]
Answer:
CH₃–CH₂–CH(CH₃)–CH₂–CHO

Teacher's Note:
1. Start numbering from the CHO carbon: C-1 is CHO, so the methyl sits on the third carbon from it.

Question 6. Write the structure of 3-methylbutanal. [CBSE Delhi 2013]
Answer:
CH₃–CH(CH₃)–CH₂–CHO

Teacher's Note:
1. Check: four carbons in the main chain including CHO, plus one CH₃ branch – total C₅H₁₀O.

Question 7. Write the structure of 4-chloropentan-2-one. [CBSE Delhi 2013]
Answer:
CH₃–CH(Cl)–CH₂–CO–CH₃

Teacher's Note:
1. The keto group takes the lowest number (C-2), so the Cl lands on C-4 of the five-carbon chain.

Question 8. Write the structure of p-methyl benzaldehyde molecule. [CBSE Delhi 2013]
Answer:
p-CH₃–C₆H₄–CHO (CHO and CH₃ at positions 1 and 4 of the benzene ring):

Teacher's Note:
1. para means positions 1 and 4 – draw the two groups directly opposite each other on the ring.

Question 9. Draw the structure of the compound named 4-methylpent-3-en-2-one. [CBSE Comptt. Delhi 2013]
Answer:
CH₃–C(CH₃)=CH–CO–CH₃

Teacher's Note:
1. This compound is mesityl oxide, the dehydration product of the aldol of acetone.

Question 10. What type of aldehydes undergo Cannizzaro reaction? [CBSE Comptt. Delhi 2017]
Answer:
Having no α-hydrogen.

Teacher's Note:
1. Examples: HCHO, C₆H₅CHO, (CH₃)₃C–CHO; aldehydes with α-H give aldol condensation instead.

Short Answer Questions (2 Marks)

Question 11. Name the reagents used in the following reactions: [CBSE Delhi 2015]
(i) CH₃–CO–CH₃ \( \xrightarrow{?} \) CH₃–CH(OH)–CH₃
(ii) C₆H₅–CH₂–CH₃ \( \xrightarrow{?} \) C₆H₅–COO⁻K⁺
Answer:
(i) Sodium borohydride (NaBH₄) or Lithium aluminium hydride (LiAlH₄)
(ii) Alkaline potassium permanganate (KMnO₄–KOH)

Teacher's Note:
1. Whatever the length of the side chain, alkaline KMnO₄ oxidises it to a single –COOH (here as the potassium salt).

Question 12. Name the electrophile produced in the reaction of benzene with benzoyl chloride in the presence of anhydrous AlCl₃. Name the reaction also. [NCERT Exemplar]
Answer:
C₆H₅C⁺O (Benzoylinium cation)
Friedel–Crafts acylation reaction

Teacher's Note:
1. AlCl₃ removes Cl⁻ from the acyl chloride to generate the acylium ion, which is the attacking electrophile.

Question 13. An aromatic compound ‘A’ on treatment with CHCl₃ and KOH gives two compounds, both of which give same product ‘B’ when distilled with zinc dust. Oxidation of ‘B’ gives ‘C’ with molecular formula C₇H₆O₂. Sodium salt of ‘C’ on heating with soda lime gives ‘D’ which may also be obtained by distilling ‘A’ with zinc dust. Identify ‘A’, ‘B’, ‘C’ and ‘D’. [CBSE 2019]
Answer:
A = C₆H₅OH (Phenol), B = C₆H₅CHO (Benzaldehyde), C = C₆H₅COOH (Benzoic acid), D = C₆H₆ (Benzene)

Teacher's Note:
1. The two compounds from Reimer-Tiemann reaction are o- and p-hydroxybenzaldehyde; zinc dust removes the –OH group from both, giving the same product.

Question 14. Write the reagents used in the following reactions: [CBSE Ajmer 2015]
(i) C₆H₅–CO–CH₃ \( \xrightarrow{?} \) C₆H₅–CH₂–CH₃
(ii) CH₃COOH \( \xrightarrow{?} \) CH₃–COCl
Answer:
(i) Zn–Hg, conc. HCl or H₂NNH₂ and KOH/ethylene glycol, Heat
(ii) PCl₅ or SOCl₂

Teacher's Note:
1. SOCl₂ is preferred in practice because its by-products, SO₂ and HCl, are gases and escape.

Question 15. Identify the compounds A, B and C in the following reaction: [NCERT Exemplar]
CH₃–Br \( \xrightarrow{\text{Mg/Ether}} \) (A) \( \xrightarrow[(ii)\,\text{Water}]{(i)\,CO_2} \) (B) \( \xrightarrow[\Delta]{CH_3OH/H^+} \) (C)
Answer:
A = CH₃MgBr (Methyl magnesium bromide), B = CH₃–COOH (Ethanoic acid), C = CH₃–CO–O–CH₃ (Methyl ethanoate)

Teacher's Note:
1. The last step is Fischer esterification: acid + alcohol in the presence of H⁺ gives an ester.

Question 16. Arrange the following compounds in increasing order of their property as indicated:
(i) CH₃COCH₃, C₆H₅–CO–C₆H₅, CH₃CHO (reactivity towards nucleophilic addition reaction)
(ii) Cl–CH(Cl)–COOH, Cl–CH₂–COOH, CCl₃–COOH (acidic character) [CBSE Bhubaneshwar 2015]
Answer:
(i) C₆H₅COC₆H₅ < CH₃COCH₃ < CH₃CHO
(ii) Cl–CH₂–COOH < Cl–CH(Cl)–COOH < CCl₃–COOH

Teacher's Note:
1. Benzophenone is the least reactive: two bulky phenyl groups crowd the carbonyl carbon and resonance reduces its electrophilicity.

Question 17. Give reasons:
(i) Oxidation of aldehydes is easier than ketones. [CBSE 2019]
OR
Oxidation of propanal is easier than propanone. [CBSE 2020]
(ii) CH₂=CH–COOH is more acidic than CH₃CH₂–COOH. [CBSE East 2016]
Answer:
(i) As aldehydes contain H atom on the carbonyl group but ketones do not. Cleavage of C–H bond in aldehydes is easier than cleavage of C–C bond in ketones.
(ii) This is because in CH₂=CH–COOH, the carbonyl group attached to sp² hybridised carbon atom which is more electronegative and makes release of H⁺ ion easy.

Teacher's Note:
1. Greater s-character makes a carbon more electronegative: sp > sp² > sp³, so acids with an unsaturated α-carbon are stronger.

Question 18. Arrange the following in the decreasing order of their acidic character.
(i) C₆H₅COOH, FCH₂COOH, NO₂CH₂COOH
(ii) CH₃CH₂OH, CH₃COOH, ClCH₂COOH, FCH₂COOH, C₆H₅CH₂COOH
Answer:
(i) NO₂CH₂COOH > FCH₂COOH > C₆H₅COOH
(ii) FCH₂COOH > ClCH₂COOH > C₆H₅CH₂COOH > CH₃COOH > CH₃CH₂OH

Teacher's Note:
1. –NO₂ has a stronger −I effect than –F, so nitroacetic acid is the strongest acid in set (i).

Question 19. Arrange the following in order of property indicated for each set.
(i) CH₃CHO, CH₃CH₂OH, CH₃OCH₃, CH₃CH₂CH₃ (increasing order of boiling points)
(ii) (CH₃)₂CHCOOH, CH₃CH₂CH(Br)COOH, CH₃CH(Br)CH₂COOH (increasing order of their acid strengths)
Answer:
(i) CH₃CH₂CH₃ < CH₃OCH₃ < CH₃CHO < CH₃CH₂OH
(ii) (CH₃)₂CHCOOH < CH₃–CH(Br)–CH₂–COOH < CH₃–CH₂–CH(Br)–COOH

Teacher's Note:
1. The −I effect weakens rapidly with distance, so Br on the α-carbon raises acidity much more than Br on the β-carbon.

Question 20. Name the reagents used in the following reactions:
(i) CH₃–CO–CH₃ → ? → CH₃–CH(OH)–CH₃
(ii) C₆H₅–CH₂–CH₃ → ? → C₆H₅–COO⁻K⁺ [CBSE Delhi 2015]
Answer:
(i) LiAlH₄ / NaBH₄ / H₂, Pt
(ii) KMnO₄, KOH

Teacher's Note:
1. Alkaline KMnO₄ oxidises the whole side chain of an alkylbenzene to –COO⁻, however long it is, as long as there is a benzylic H.

Question 21. Write the reagents required in the following reactions:
(i) CH₂=CH–CH₂OH → ? → CH₂=CH–CHO
(ii) CH₃–COOH → ? → CH₃–CONH₂ [CBSE OD 2015]
Answer:
(i) PCC / Cu at 573 K
(ii) NH₃, Δ(heat)

Teacher's Note:
1. PCC stops the oxidation at the aldehyde and does not touch the C=C double bond.

Question 22. Complete the following reactions:
(i) 2HCHO → [Conc. KOH] → ?
(ii) C₆H₅CHO → [HNO₃/H₂SO₄, 273–283 K] → ? [CBSE Delhi 2013]
Answer:
(i) 2HCHO (Methanal) → [Conc. KOH] → CH₃OH (Methanol) + HCOOK (Cannizzaro reaction)
(ii) C₆H₅CHO (Benzaldehyde) → [HNO₃/H₂SO₄, 273–283 K] → m-NO₂C₆H₄CHO (m-nitrobenzaldehyde)

Teacher's Note:
1. –CHO is a deactivating, meta-directing group, so nitration of benzaldehyde gives the meta product.

Question 23. Give simple chemical test to distinguish between the following pairs of compounds.
(i) Ethanal and propanal
(ii) Benzoic acid and phenol. [CBSE Delhi 2013]
Answer:
(i) Iodoform test: Ethanal gives this test due to the presence of CH₃CO group
CH₃CHO (Ethanal) + 4NaOH + 3I₂ → CHI₃ (Iodoform, yellow ppt) + HCOONa + 3NaI + 3H₂O
CH₃CH₂CHO + 4NaOH + 3I₂ → No reaction.
(ii) Sodium bicarbonate test: Benzoic acid gives brisk effervescence of CO₂ when react with NaHCO₃, while phenol does not give it, as it is a weak acid.
C₆H₅COOH + NaHCO₃ → C₆H₅COONa + H₂O + CO₂↑
C₆H₅OH + NaHCO₃ → No reaction.

Teacher's Note:
1. Ethanal is the only aldehyde that gives the iodoform test, because it alone has a CH₃ attached to the –CHO carbon.

Question 24. Write the structures and IUPAC names of the cross aldol condensation products only of ethanal and propanal. [CBSE SQP 2018-2019]
Answer:
(i) 2-Methylbut-2-enal: CH₃–CH=C(CH₃)–CHO
(ii) Pent-2-enal: CH₃–CH₂–CH=CH–CHO

Teacher's Note:
1. In a cross aldol, decide which aldehyde gives the carbanion (from its α-carbon) and which gives the carbonyl; swapping roles gives the second cross product.

Question 25. How will you carry out the following conversions?
(i) Acetylene to acetic acid
(ii) Toluene to m-nitro benzoic acid. [CBSE Delhi 2013]
Answer:
(i) CH≡CH → [dil H₂SO₄, Hg²⁺] → CH₂=CHOH → [Tautomerisation] → CH₃CHO → [K₂Cr₂O₇, H₂SO₄] → CH₃COOH
(ii) C₆H₅CH₃ → [(i) KMnO₄/KOH (ii) H₃O⁺] → C₆H₅COOH → [HNO₃/H₂SO₄, Nitration] → m-O₂N–C₆H₄–COOH

Teacher's Note:
1. Order matters in (ii): oxidise first so that –COOH (a meta director) is present before nitration; nitrating toluene first would give o/p products.

Question 26. Identify the reaction and write the IUPAC name of the product formed:
(a) CH₃–CH₂–COOH → [(i) Br₂/Red phosphorus] →
(b) C₆H₅COCl → [H₂, Pd–BaSO₄] → [CBSE SQP 2018-2019]
Answer:
(a) Reaction: Hell-Volhard-Zelinsky reaction. IUPAC: 2-Bromopropanoic acid.
(b) Reaction: Rosenmund reduction reaction. IUPAC: Benzaldehyde.

Teacher's Note:
1. Both are name reactions that appear every year – learn the reagent that identifies each: Br₂/red P for HVZ, H₂/Pd–BaSO₄ for Rosenmund.

Teacher's Note:
1. Link each change to its reagent type: C=O to CH₂ is Clemmensen, adding a carbon is Grignard, ketone to alcohol is LiAlH₄, ester to aldehyde is DIBAL-H.

Question 28. How will you convert:
(i) C₂H₂ → CH₃COOH
(ii) C₆H₅CONH₂ → C₆H₅COOH
Answer:
(i) HC≡CH –[H₂SO₄, HgSO₄]→ CH₃CHO –[KMnO₄, dil. H₂SO₄]→ CH₃COOH
(ii) C₆H₅CONH₂ + HNO₂ → C₆H₅COOH + N₂ + H₂O

Teacher's Note:
1. Ethyne is the only alkyne that gives an aldehyde on hydration; all other alkynes give ketones.

Short Answer Questions (3 Marks)

Question 29. Write chemical equations for the following reactions:
(i) Propanone is treated with dilute Ba(OH)₂.
(ii) Acetophenone is treated with Zn(Hg)/Conc. HCl
(iii) Benzoyl chloride is hydrogenated in presence of Pd/BaSO₄. [CBSE Delhi Set-1, 2019]
Answer:
(i) 2CH₃–CO–CH₃ (Propanone) \( \xrightarrow{\text{Ba(OH)}_{2}} \) (CH₃)₂C(OH)–CH₂–CO–CH₃ (Ketol, 4-hydroxy-4-methylpentan-2-one) \( \xrightarrow[\text{−H}_{2}\text{O}]{\text{Heat}} \) (CH₃)₂C=CH–CO–CH₃ (4-methylpent-3-en-2-one)
(ii) C₆H₅COCH₃ (Acetophenone) \( \xrightarrow{\text{Zn(Hg)/Conc HCl}} \) C₆H₅CH₂CH₃ (Ethylbenzene)
(iii) C₆H₅COCl (Benzoyl chloride) \( \xrightarrow[\text{Pd/BaSO}_{4}]{\text{H}_{2}} \) C₆H₅CHO (Benzaldehyde)

Teacher's Note:
1. Ba(OH)₂ is the dilute base that brings about aldol condensation of acetone; the C–C bond forms between the α-carbon of one molecule and the C=O carbon of the other.

Question 30. Write the structures of the main products in the following reactions:
(i) Methyl (2-oxocyclohexyl)acetate (a cyclohexanone ring carrying –CH₂–CO–OCH₃ on the carbon next to C=O) \( \xrightarrow{\text{NaBH}_{4}} \)
(ii) C₆H₅CH=CH₂ + H₂O \( \xrightarrow{\text{H}^{+}} \)
(iii) C₆H₅OC₂H₅ + HI → [CBSE Delhi/Outside Delhi, 2018]
Answer:
(i) Methyl (2-hydroxycyclohexyl)acetate: the ring C=O is reduced to CH–OH while the ester group –CH₂–CO–OCH₃ remains unchanged.
(ii) C₆H₅CH(OH)CH₃
(iii) C₂H₅I + C₆H₅OH

Teacher's Note:
1. NaBH₄ is a mild reducing agent: it reduces aldehydes and ketones but not esters, so only the ring ketone is reduced.

Question 31. Predict the products of the following reactions:
(i) CH₃–CO–CH₃ \( \xrightarrow[\text{(ii) KOH/Glycol, }\Delta]{\text{(i) H}_{2}\text{N–NH}_{2}} \) ?
(ii) C₆H₅–CO–CH₃ \( \xrightarrow{\text{NaOH/I}_{2}} \) ? + ?
(iii) CH₃COONa \( \xrightarrow[\Delta]{\text{NaOH/CaO}} \) ? [CBSE Delhi 2015]
Answer:
(i) CH₃CH₂CH₃
(ii) C₆H₅COONa + CHI₃
(iii) CH₄

Teacher's Note:
1. Each part tests one named reaction: Wolff-Kishner reduction, iodoform reaction and soda-lime decarboxylation; name them to show you know why.

Question 32. Predict the products of the following reactions:
(i) CH₃–CO–CH₃ \( \xrightarrow{\text{H}_{2}\text{N–NH}_{2}} \) ?
(ii) C₆H₅–CH₃ \( \xrightarrow[\text{(b) H}^{+}]{\text{(a) KMnO}_{4}\text{/KOH}} \) ?
(iii) C₆H₅COOH \( \xrightarrow{\text{Br}_{2}\text{/FeBr}_{3}} \) ? [CBSE OD 2015]
Answer:
(i) (CH₃)₂C=N–NH₂
(ii) C₆H₅COOH (Benzoic acid)
(iii) m-Bromobenzoic acid (3-BrC₆H₄COOH)

Teacher's Note:
1. –COOH is a meta-directing group, so bromination of benzoic acid gives the meta product.

Question 33. How will you bring about the following conversions:
(i) Propanone to propane
(ii) Benzoyl chloride to benzaldehyde.
(iii) Ethanal to but-2-enal.
Answer:
(i) CH₃–CO–CH₃ (Propanone) \( \xrightarrow[\text{Zn(Hg) + HCl (Conc.)}]{\text{Clemmensen Reduction}} \) CH₃–CH₂–CH₃ (Propane)
(ii) C₆H₅COCl (Benzoyl chloride) \( \xrightarrow[\text{+ H}_{2}]{\text{BaSO}_{4}\text{/Pd}} \) C₆H₅CHO (Benzaldehyde)
(iii) 2CH₃–CHO (Ethanal) \( \xrightarrow[\text{Aldol Condensation}]{\text{dil. NaOH}} \) CH₃–CH(OH)–CH₂–CHO (3-hydroxybutanal) \( \xrightarrow[\text{−H}_{2}\text{O}]{\Delta} \) CH₃–CH=CH–CHO (But-2-enal)

Teacher's Note:
1. The aldol (3-hydroxybutanal) dehydrates easily on warming because the C=C formed is conjugated with C=O.

Question 34. (a) Give reasons:
(i) Benzoic acid is a stronger acid than acetic acid.
(ii) Methanal is more reactive towards nucleophilic addition reaction than ethanal.
(b) Give a simple chemical test to distinguish between propanal and propanone. [CBSE, Outside Delhi set 2, 2019]
Answer:
(a) (i) Due to greater electronegativity of sp² hybridised carbon to which carboxyl carbon is attached / Due to greater resonance stabilization of carboxylate ion with the benzene ring.
(ii) Because carbonyl carbon of methanal is more electrophilic than that of ethanal / due to +I effect of methyl group in ethanal, reactivity decreases.
CH₃ group present in ethanal decreases the positive charge on carbonyl carbon by +I effect which is not possible in methanal. As Nu⁻ attack is favourable with more positive charge and less hindrance at carbonyl carbon, therefore methanal is more reactive than ethanal.
(b) On heating with Tollens’ reagent / [Ag(NH₃)₂]⁺, propanal forms silver mirror whereas propanone does not. (or any other suitable chemical test)
CH₃CH₂CHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → CH₃CH₂COO⁻ + 2Ag↓ (Silver mirror) + 4NH₃ + 2H₂O

Teacher's Note:
1. The iodoform test also works for (b): propanone (a methyl ketone) gives yellow CHI₃, propanal does not.

Question 35. Write structures of compounds A, B and C in each of the following reaction:
(i) C₆H₅Br \( \xrightarrow{\text{Mg/dry ether}} \) A \( \xrightarrow[\text{(b) H}_{3}\text{O}^{+}]{\text{(a) CO}_{2}\text{(g)}} \) B \( \xrightarrow{\text{PCl}_{5}} \) C
(ii) CH₃CN \( \xrightarrow[\text{(b) H}_{3}\text{O}^{+}]{\text{(a) SnCl}_{2}\text{/HCl}} \) A \( \xrightarrow{\text{dil. NaOH}} \) B \( \xrightarrow{\Delta} \) C [CBSE Marking Scheme, 2017]
Answer:
(i) A : C₆H₅MgBr,   B : C₆H₅COOH,   C : C₆H₅COCl
C₆H₅Br \( \xrightarrow{\text{Mg/dry ether}} \) C₆H₅MgBr (A) \( \xrightarrow{\text{(a) CO}_{2}\text{ (g)}} \) C₆H₅–CO–OMgBr \( \xrightarrow{\text{(b) H}_{3}\text{O}^{+}} \) C₆H₅COOH (B) \( \xrightarrow{\text{PCl}_{5}} \) C₆H₅–CO–Cl (C)
(ii) A : CH₃CHO,   B : CH₃CH(OH)CH₂CHO,   C : CH₃CH=CHCHO
CH₃CN \( \xrightarrow[\text{(b) H}_{3}\text{O}^{+}]{\text{(a) SnCl}_{2}\text{/HCl}} \) CH₃CHO (A) \( \xrightarrow{\text{dil. NaOH}} \) CH₃–CH(OH)–CH₂–CH=O (B) \( \xrightarrow{\Delta} \) CH₃–CH=CH–CH=O (C)

Teacher's Note:
1. SnCl₂/HCl followed by hydrolysis is Stephen reduction: a nitrile gives an aldehyde with the same number of carbons.

Question 36. Do the following conversions in not more than two steps:
(i) Benzoic acid to benzaldehyde
(ii) Ethyl benzene to benzoic acid
(iii) Propanone to propene [CBSE, Outside Delhi set 1, 2017]
Answer:
(i) C₆H₅COOH \( \xrightarrow{\text{SOCl}_{2}} \) C₆H₅COCl \( \xrightarrow{\text{H}_{2}\text{, Pd-BaSO}_{4}} \) C₆H₅CHO
(ii) C₆H₅C₂H₅ \( \xrightarrow{\text{KMnO}_{4}\text{ + KOH}} \) C₆H₅COOK \( \xrightarrow{\text{H}_{3}\text{O}^{+}} \) C₆H₅COOH
(iii) CH₃COCH₃ \( \xrightarrow{\text{NaBH}_{4}} \) CH₃CH(OH)CH₃ \( \xrightarrow{\text{conc. H}_{2}\text{SO}_{4}} \) CH₃CH=CH₂
(or any other correct method)

Teacher's Note:
1. Benzoic acid cannot be reduced directly to benzaldehyde; it is first converted into the acid chloride, which Rosenmund reduction stops at the aldehyde.

Long Answer Questions (5 Marks)

Question 37. (a) Write the structures of A and B in the following reactions:
(i) CH₃COCl \( \xrightarrow{\text{H}_{2}\text{, Pd–BaSO}_{4}} \) A \( \xrightarrow{\text{H}_{2}\text{N–OH}} \) B
(ii) CH₃MgBr \( \xrightarrow[\text{2. H}_{3}\text{O}^{+}]{\text{1. CO}_{2}} \) A \( \xrightarrow{\text{PCl}_{5}} \) B
(b) Distinguish between:
(i) C₆H₅–COCH₃ and C₆H₅–CHO
(ii) CH₃COOH and HCOOH
(c) Arrange the following in the increasing order of their boiling points:
CH₃CHO, CH₃COOH, CH₃CH₂OH
Answer:
(a) (i) A : CH₃CHO, B : CH₃CH=N–OH
CH₃COCl \( \xrightarrow{\text{H}_{2}\text{, Pd–BaSO}_{4}} \) CH₃CHO (Acetaldehyde) [A] \( \xrightarrow[\text{−H}_{2}\text{O}]{\text{H}_{2}\text{NOH}} \) CH₃–CH=N–OH (Acetaldoxime) [B]
(ii) A : CH₃COOH, B : CH₃COCl
CH₃MgBr \( \xrightarrow[\text{2. H}_{3}\text{O}^{+}]{\text{1. CO}_{2}} \) CH₃COOH \( \xrightarrow{\text{PCl}_{5}} \) CH₃COCl + HCl + POCl₃
(b) (i) Heat both compounds with NaOH and I₂, C₆H₅COCH₃ forms yellow ppt. of CHI₃ whereas C₆H₅CHO does not.
(ii) Add ammonical solution of silver nitrate (Tollens’ reagent) to both the compounds, HCOOH gives silver mirror but CH₃COOH does not. (or any other suitable test)
HCOOH + 2[Ag(NH₃)₂]⁺ + 2OH⁻ → 2Ag↓ + 2H₂O + CO₂ + 4NH₃
(c) CH₃CHO < CH₃CH₂OH < CH₃COOH

Teacher's Note:
1. Formic acid contains an aldehyde-like H–C=O unit, which is why it alone among carboxylic acids reduces Tollens’ reagent.

Question 38. (a) Write the chemical reaction involved in Wolff-Kishner reduction.
(b) Arrange the following in the increasing order of their reactivity towards nucleophilic addition reaction:
C₆H₅COCH₃, CH₃–CHO, CH₃COCH₃
(c) Why carboxylic acid does not give reactions of carbonyl group?
(d) Write the product in the following reaction.
CH₃CH₂CH=CH–CH₂CN \( \xrightarrow[\text{2. H}_{2}\text{O}]{\text{1. (i–Bu)}_{2}\text{AlH}} \) ?
(e) A and B are two functional isomers of compound C₃H₆O. On heating with NaOH and I₂, isomer B forms yellow precipitate of iodoform whereas isomer A does not form any precipitate. Write the formulae of A and B. [CBSE Delhi 2016]
Answer:
(a) >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
(b) C₆H₅COCH₃ < CH₃COCH₃ < CH₃CHO
(c) Because of resonance in carboxylic group, the carbonyl group, loses a double bond character.
Carboxylic acids do not give reactions of carbonyl groups as it enters into resonance with lone pair of –OH of –COOH groups thereby making the carbon atoms less electrophilic: R–C(=O)–ÖH ↔ R–C(–O⁻)=O⁺H
(d) CH₃CH₂CH=CH–CH₂CHO (Hex-3-enal)
(e) A : CH₃CH₂CHO,   B : CH₃COCH₃
CH₃CH₂CHO (Propanal) [A] + NaOH + I₂ → No yellow precipitate
CH₃–CO–CH₃ (Acetone) [B] + 4NaOH + 3I₂ \( \xrightarrow{\Delta} \) CHI₃ (Iodoform, yellow precipitate) + 3NaI + CH₃COONa + 3H₂O

Teacher's Note:
1. (i-Bu)₂AlH is DIBAL-H; it converts nitriles (and esters) into aldehydes without touching an isolated C=C.

Question 39. (a) Write the product(s) in the following reactions:
(i) Cyclohexanone + HCN → ?
(ii) C₆H₅COONa + NaOH \( \xrightarrow[\Delta]{\text{CaO}} \) ?
(iii) CH₃–CH=CH–CN \( \xrightarrow[\text{(b) H}_{2}\text{O}]{\text{(a) DIBAL-H}} \) ?
(b) Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Butanal and Butan-2-one
(ii) Benzoic acid and Phenol [CBSE Outside Delhi Set-1, 2017]
Answer:
(a) (i) Cyclohexanone cyanohydrin (1-hydroxycyclohexane-1-carbonitrile): the ring carbon of C=O now carries –OH and –CN.
(ii) C₆H₆ (Benzene)
(iii) CH₃–CH=CH–CHO
(b) (i) Tollen’s reagent test: Add ammoniacal solution of silver nitrate (Tollen’s reagent) in both the solutions. Butanal gives silver mirror whereas Butan-2-one does not.
(ii) Add neutral FeCl₃ in both the solutions, phenol forms violet colour but benzoic acid does not. (or any other correct test)

Teacher's Note:
1. Benzoic acid and phenol can also be told apart with NaHCO₃: only the carboxylic acid is strong enough to release CO₂.

Question 40. (a) Write the reactions involved in the following:
(i) Etard reaction   (ii) Stephen reduction
(b) How will you convert the following in not more than two steps:
(i) Benzoic acid to benzaldehyde
(ii) Acetophenone to benzoic acid
(iii) Ethanoic acid to 2-hydroxyethanoic acid [CBSE Outside Delhi Set-1, 2017]
Answer:
(a) (i) Etard reaction
C₆H₅CH₃ (Toluene) + CrO₂Cl₂ \( \xrightarrow{\text{CS}_{2}} \) C₆H₅CH(OCrOHCl₂)₂ (Chromium complex) \( \xrightarrow{\text{H}_{3}\text{O}^{+}} \) C₆H₅CHO (Benzaldehyde)
(ii) Stephen reaction
RCN + SnCl₂ + HCl → RCH=NH \( \xrightarrow{\text{H}_{3}\text{O}^{+}} \) RCHO
(b) (i) C₆H₅COOH (Benzoic acid) \( \xrightarrow{\text{SOCl}_{2}} \) C₆H₅COCl (Benzoyl chloride) \( \xrightarrow[\text{Pd/BaSO}_{4}]{\text{Rosenmund’s reduction}} \) C₆H₅CHO (Benzaldehyde)
(ii) C₆H₅COCH₃ (Acetophenone) \( \xrightarrow{\text{I}_{2}\text{/NaOH}} \) C₆H₅COONa (Sodium benzoate) \( \xrightarrow{\text{H}_{3}\text{O}^{+}} \) C₆H₅COOH (Benzoic acid)
(iii) CH₃COOH \( \xrightarrow{\text{Cl}_{2}\text{/P}} \) CH₂(Cl)COOH \( \xrightarrow{\text{KOH(aq)}} \) CH₂(OH)COOH
(or any other correct method)

Teacher's Note:
1. In Stephen reduction the nitrile is first reduced to an imine (as its hydrochloride), and it is the hydrolysis of this imine that gives the aldehyde.

Question 41. (i) Complete the following equations:
(a) C₆H₅CHO \( \xrightarrow{\text{conc. NaOH}} \)
(b) CH₃COCH₃ \( \xrightarrow{\text{LiAlH}_{4}} \)
(c) CH₃–COOH \( \xrightarrow[\text{(b) H}_{2}\text{O}]{\text{(a) Cl}_{2}\text{/P}} \)
(ii) Distinguish between:
(a) CH₃COOH and Phenol (C₆H₅OH)
(b) CH₃–CHO and CH₃–CO–CH₃ [CBSE Comptt. Delhi 2016]
Answer:
(i) (a) 2C₆H₅CHO \( \xrightarrow{\text{Conc. NaOH}} \) C₆H₅CH₂OH + C₆H₅COONa
(b) CH₃COCH₃ + 2[H] \( \xrightarrow{\text{LiAlH}_{4}} \) CH₃CH(OH)CH₃
(c) CH₃–COOH \( \xrightarrow[\text{(ii) H}_{2}\text{O}]{\text{(i) Cl}_{2}\text{/Red phosphorus}} \) CH₂(Cl)–COOH
(ii) (a) When CH₃COOH is added to an aqueous solution of sodium bicarbonate, brisk effervescence of CO₂ is evolved.
RCOOH (Carboxylic acid) + NaHCO₃ → RCOONa (Sodium carboxylate) + H₂O + CO₂
Phenol does not give this test.
(b) When CH₃CHO is heated with Tollen’s reagent, they form silver mirror while CH₃COCH₃ does not.
CH₃CHO + 2[Ag(NH₃)₂]⁺ (Tollens’ reagent) + 3OH⁻ → CH₃–COO⁻ + 2Ag↓ (Silver mirror) + 2H₂O + 4NH₃
(or any other suitable test)

Teacher's Note:
1. Both CH₃CHO and CH₃COCH₃ give the iodoform test, so it cannot separate this pair; use Tollens’ or Fehling’s test.

Question 42. (i) Describe the following giving chemical equations:
(a) Decarboxylation reaction
(b) Friedel-Crafts reaction
(ii) How will you bring about the following conversions?
(a) Benzoic acid to Benzaldehyde
(b) Benzene to m-Nitroacetophenone
(c) Ethanol to 3-Hydroxybutanal
Answer:
(i) (a) Carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with sodalime (NaOH and CaO).
R–COONa \( \xrightarrow[\text{Heat}]{\text{NaOH and CaO}} \) R–H + Na₂CO₃
(b) The alkyl / acyl group is introduced at ortho- and para- positions by reaction of anisole with alkyl halide / acyl halide in the presence of anhydrous aluminium chloride (a Lewis acid) as catalyst.
C₆H₅OCH₃ (Anisole) + CH₃Cl \( \xrightarrow[\text{CS}_{2}]{\text{Anhyd. AlCl}_{3}} \) 2-methylanisole (o-CH₃C₆H₄OCH₃) + 4-methylanisole (p-CH₃C₆H₄OCH₃)
(Award full marks if correct equation is given)
(ii) (a) C₆H₅COOH \( \xrightarrow{\text{PCl}_{5}} \) C₆H₅COCl \( \xrightarrow{\text{H}_{2}\text{/Pd-BaSO}_{4}} \) C₆H₅CHO
(b) C₆H₆ (Benzene) \( \xrightarrow{\text{CH}_{3}\text{COCl/AlCl}_{3}} \) C₆H₅COCH₃ (Acetophenone) \( \xrightarrow{\text{HNO}_{3}\text{/H}_{2}\text{SO}_{4}} \) m-NO₂C₆H₄COCH₃ (m-Nitroacetophenone)
(c) CH₃CH₂OH \( \xrightarrow{\text{CrO}_{3}} \) CH₃–CHO \( \xrightarrow{\text{dil. NaOH}} \) CH₃–CH(OH)–CH₂–CHO
(or any other correct method)

Teacher's Note:
1. Order of steps matters in (ii)(b): acylate first, because nitrobenzene is too deactivated for Friedel-Crafts reaction, and –COCH₃ then directs the nitro group to the meta position.