CBSE Class 12 Chemistry Aldehydes Ketones and Carboxylic Acids Important Questions Set 07

Very Short Answer Questions (1 Mark)

Question 1. What happens when calcium ethanoate is dry distilled ?
Answer:
(CH₃COO)₂Ca \( \xrightarrow{Dry\ distil} \) (CH₃)₂C=O + CaCO₃
Cal. ethanoate gives propanone (acetone).

Teacher's Note:
1. Dry distillation of a calcium salt of a monocarboxylic acid gives a ketone; with calcium formate added, an aldehyde is obtained instead.

Question 2. What is the product of the reaction of ethanoyl chloride with diethylcadmium followed by acid hydrolysis ?
Answer:
2 CH₃COCl + (CH₃CH₂)₂Cd \( \xrightarrow{Dry\ ether} \) 2 CH₃COCH₂CH₃ + CdCl₂
Ethanoyl chloride + Diethylcadmium → Butanone

Teacher's Note:
1. Dialkylcadmiums are less reactive than Grignard reagents, so the reaction stops at the ketone and does not go on to a tertiary alcohol.

Question 3. How is acetone obtained from 2-bromopropane ?
Answer:
(CH₃)₂CH–Br \( \xrightarrow[(Hydrolysis)]{Aq.\ KOH} \) (CH₃)₂CHOH \( \xrightarrow[(Dehydrogenation)]{Cu,\ 573\ K} \) (CH₃)₂C=O + H₂
2-Bromopropane → 2-Propanol → Acetone

Teacher's Note:
1. Dehydrogenation over hot copper converts a 2° alcohol into a ketone without the risk of over-oxidation.

Question 4. Write the product of the following reaction :
CH₃–CH=CH–CH₂CN \( \xrightarrow[(ii)\ H_2O]{(i)\ DiBAL\text{-}H} \) (CBSE 2016)
Answer:
DiBAL-H reduces –CN group to –CHO group but does not reduce ethylenic double bond. For example,
CH₃–CH=CH–CH₂CN \( \xrightarrow[(ii)\ H_2O]{(i)\ DiBAL\text{-}H} \) CH₃–CH=CH–CH₂CHO
Pent-3-en-1-nitrile → Pent-3-en-1-al

Teacher's Note:
1. DiBAL-H is chosen when a C=C bond must survive; LiAlH₄ would take a nitrile all the way to an amine.

Question 5. Write one chemical reaction to illustrate Wacker’s process for converting ethylene into ethanal.
Answer:
CH₂=CH₂ + H₂O + PdCl₂ \( \xrightarrow[Air\ or\ O_2]{CuCl_2} \) CH₃CH=O + Pd + 2 HCl
Ethylene → Ethanal

Teacher's Note:
1. CuCl₂ and air re-oxidise Pd back to PdCl₂, so palladium works as a catalyst in the overall process.

Question 6. How is acetone obtained from ethanol ? (CBSE 2007 ; Jharkhand Board 2011)
Or How is acetic acid converted into acetone ? (J & K Board 2011)
Answer:
CH₃CH₂OH \( \xrightarrow[(Oxidation)]{K_2Cr_2O_7/H_2SO_4} \) CH₃COOH \( \xrightarrow[-CO_2,\ -H_2O]{CaCO_3} \) (CH₃COO)₂Ca \( \xrightarrow[-CaCO_3]{Dry\ distil.} \) (CH₃)₂C=O
Ethanol → Ethanoic acid (Acetic acid) → Cal. ethanoate → Acetone

Teacher's Note:
1. The second half of this sequence (acid → calcium salt → dry distillation) also answers ‘acetic acid to acetone’.

Question 7. Name one reagent used to convert toluene into benzaldehyde.
Answer:
Chromyl chloride in CS₂ (CrO₂Cl₂/CS₂) or chromium trioxide in acetic anhydride [CrO₃/(CH₃CO)₂O] followed by acid hydrolysis.

Teacher's Note:
1. The CrO₂Cl₂ route is the Étard reaction; strong oxidants like KMnO₄ would carry toluene all the way to benzoic acid.

Question 8. CH₃CHO is more reactive than CH₃COCH₃ towards reaction with HCN ? Explain. (CBSE 2019)
Or Propanone is less reactive than ethanal towards nucleophilic addition reactions. Give reasons. (CBSE (Foreign) 2017)
Answer:
Due to electron-donating inductive effect (+I-effect) of the two CH₃ groups in CH₃COCH₃ as compared to smaller +I-effect of one CH₃ group in CH₃CHO, the magnitude of the +ve charge on the carbon atom of the carbonyl group in CH₃CHO is more than in CH₃COCH₃. As a result, nucleophilic addition reactions occurs more readily in CH₃CHO than in CH₃COCH₃.
CH₃CHO : smaller +I-effect, more +ve charge on the carbonyl carbon, more reactive.
CH₃COCH₃ : stronger +I-effect, less +ve charge on the carbonyl carbon, less reactive.

Teacher's Note:
1. In exams, mention the steric factor too: two CH₃ groups also crowd the carbonyl carbon and hinder the approach of the nucleophile.

Question 9. (a) C₆H₅COCH₃ is much less reactive than CH₃COCH₃ towards nucleophilic addition reactions. Explain.
(b) Benzaldehyde is less reactive than propanal in nucleophilic addition reactions. (CBSE 2022)
Answer:
(a) In C₆H₅COCH₃, the electron-donating resonance effect (+R-effect) of the benzene ring reduces the +ve charge on the carbon atom of the carbonyl group. Since +R-effect of benzene ring is more pronounced than +I-effect of the CH₃ groups, therefore, magnitude of the +ve charge on the carbon atom of the carbonyl group in C₆H₅COCH₃ is reduced to a much greater extent than in CH₃COCH₃. As a result, C₆H₅COCH₃ is much less reactive than CH₃COCH₃.
C₆H₅–C(CH₃)=O ↔ ⁺C₆H₅=C(CH₃)–O⁻ etc. (the ring pushes electrons into the C=O group)
(b) In C₆H₅CHO, the electron-donating resonance effect (+R-effect) of benzene ring reduces the +ve charge on the carbon atom of the carbonyl group. Since +R-effect of benzene ring is more pronounced than +I-effect of the CH₃CH₂ group, therefore, magnitude of the +ve charge on the carbon atom of the carbonyl group in C₆H₅CHO is reduced to a much greater extent than in CH₃CH₂CHO. As a result, C₆H₅CHO is much less reactive than CH₃CH₂CHO.
C₆H₅CHO : less +ve charge on the carbonyl carbon, less reactive. CH₃CH₂→CH=O : more +ve charge on the carbonyl carbon, more reactive in nucleophilic addition reactions.

Teacher's Note:
1. Aromatic aldehydes and ketones are always less reactive than comparable aliphatic ones because the ring donates electrons to the C=O by resonance.

Question 10. Name two reagents which can be used to distinguish acetaldehyde from acetone ? (Pb. Board 2012)
Answer:
Acetaldehyde reduces Tollens’ reagent to produce shining silver mirror and produce red ppt. of Cu₂O with Fehling’s solution.
CH₃CHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → CH₃COO⁻ + 2Ag↓ + 4NH₃ + 2H₂O
Acetaldehyde + Tollens’ reagent → Acetate ion + Silver mirror
CH₃CHO + 2Cu²⁺ + 5OH⁻ → CH₃COO⁻ + Cu₂O↓ + 3H₂O
Fehling’s reagent gives a red ppt. of Cu₂O.
Acetone, on the other hand, does not give these tests.

Teacher's Note:
1. Note that the iodoform test cannot be used here, because both acetaldehyde and acetone contain the CH₃CO– group.

Short Answer Questions (2 Marks)

Question 11. Complete the following equations and name the products formed : C₆H₅COCH₃ \( \xrightarrow{Zn/Hg\text{-}HCl} \) (HP Board 2011)
Answer:
C₆H₅COCH₃ \( \xrightarrow{Zn\text{-}Hg/HCl} \) C₆H₅CH₂CH₃ (Ethylbenzene) + H₂O (Clemmensen reduction)

Teacher's Note:
1. This two-step route (acylation, then Clemmensen) is the clean way to make ethylbenzene without polyalkylation.

Question 12. Complete the following equations and name the products formed : CH₃COCH₃ \( \xrightarrow[(ii)\ Dil.\ HCl]{(i)\ Mg\text{-}Hg,\ dry\ ether} \) (West Bengal Board 2013)
Answer:
Magnesium amalgam brings about bimolecular reduction of acetone to give pinacol.
2CH₃COCH₃ \( \xrightarrow[(ii)\ dil.\ HCl]{(i)\ Mg\text{-}Hg,\ dry\ ether} \) (CH₃)₂C(OH)–C(OH)(CH₃)₂ (Pinacol or 2, 3-Dimethylbutane-2, 3-diol)

Teacher's Note:
1. Two acetone molecules are joined through their carbonyl carbons, so the product has six carbons.

Question 13. Complete the following equations and name the products formed : 2CH₃CHO \( \xrightarrow{Dil.\ NaOH} \) ? \( \xrightarrow[-H_2O]{\Delta} \) ? (Hr. Board 2013)
Answer:
2CH₃CHO \( \xrightarrow{Dil.\ NaOH} \) CH₃CH(OH)CH₂CHO (3-Hydroxybutanal) \( \xrightarrow[-H_2O]{\Delta} \) CH₃CH=CHCHO (But-2-enal)

Teacher's Note:
1. The OH is on the β-carbon of the aldol, which is why it is easily lost as water with the α-H.

Question 14. Complete the following equations and name the products formed : HCHO \( \xrightarrow{Conc.\ NaOH\ solution} \) (HP Board 2011, 2013)
Answer:
2HCHO + NaOH (conc.) → CH₃OH (Methanol) + HCOONa (Sodium formate) (Cannizzaro reaction)

Teacher's Note:
1. Two molecules of HCHO are needed: one is oxidised and one is reduced.

Question 15. Complete the following equations and name the products formed : 2C₆H₅CHO + Conc. NaOH (conc.) → (HP Board 2011 ; CBSE 2014; CBSE Sample Paper 2017)
Answer:
2C₆H₅CHO + NaOH (conc.) → C₆H₅CH₂OH (Benzyl alcohol) + C₆H₅COONa (Sodium benzoate) (Cannizzaro reaction)

Teacher's Note:
1. Benzaldehyde has no α-H, so concentrated alkali gives Cannizzaro, not aldol, products.

Question 16. How will you convert : acetophenone to ethylbenzene
Answer:
C₆H₅COCH₃ \( \xrightarrow[(Clemmensen\ reduction)]{Zn\text{-}Hg/conc.\ HCl} \) C₆H₅CH₂CH₃

Teacher's Note:
1. Wolff-Kishner reduction (NH₂NH₂, then KOH/glycol) gives the same product.

Question 17. How will you convert : propanone to propane (CBSE 2013)
Answer:
CH₃COCH₃ \( \xrightarrow{Zn\text{-}Hg/conc.\ HCl\ or\ NH_2NH_2,\ KOH/glycol} \) CH₃CH₂CH₃

Teacher's Note:
1. Name the reaction you use (Clemmensen or Wolff-Kishner); it earns the mark for the condition.

Question 18. How will you convert : acetophenone to benzoic acid
Answer:
C₆H₅COCH₃ \( \xrightarrow[(Iodoform\ reaction)]{I_2/NaOH} \) C₆H₅COONa + CHI₃ ; C₆H₅COONa \( \xrightarrow{H_3O^+} \) C₆H₅COOH
Alternatively, C₆H₅COCH₃ \( \xrightarrow{KMnO_4/KOH,\ \Delta;\ then\ H_3O^+} \) C₆H₅COOH

Teacher's Note:
1. The CH₃ of the methyl ketone is lost as CHI₃, leaving the benzoate.

Question 19. How will you convert : acetaldehyde to isopropyl alcohol
Answer:
CH₃CHO + CH₃MgBr \( \xrightarrow{Dry\ ether} \) CH₃CH(OMgBr)CH₃ \( \xrightarrow{H^+/H_2O} \) CH₃CH(OH)CH₃

Teacher's Note:
1. Any aldehyde other than HCHO gives a 2° alcohol with a Grignard reagent.

Question 20. How will you convert : acetone to t-butyl alcohol or 2-methyl-2- propanol
Answer:
CH₃COCH₃ + CH₃MgBr \( \xrightarrow{Dry\ ether} \) (CH₃)₃COMgBr \( \xrightarrow{H^+/H_2O} \) (CH₃)₃COH

Teacher's Note:
1. Hydrolysis is done with dilute acid so that the basic Mg(OH)Br dissolves.

Question 21. How will you convert : formaldehyde to methyl alcohol ?
Answer:
HCHO + H₂ \( \xrightarrow{Ni\ (or\ LiAlH_4)} \) CH₃OH ; or 2HCHO + NaOH (conc.) → CH₃OH + HCOONa (Cannizzaro reaction)

Teacher's Note:
1. Cannizzaro gives only half of the formaldehyde as methanol; direct reduction converts all of it.

Question 22. How will you convert : HCHO into urotropine
Answer:
6HCHO + 4NH₃ → (CH₂)₆N₄ (Urotropine) + 6H₂O

Teacher's Note:
1. Urotropine (hexamine) is used as a urinary antiseptic.

Question 23. How will you convert : benzene to benzophenone.
Answer:
C₆H₆ + C₆H₅COCl \( \xrightarrow{Anhyd.\ AlCl_3} \) C₆H₅COC₆H₅ + HCl

Teacher's Note:
1. Benzoyl chloride itself can be made from benzoic acid with SOCl₂.

Question 24. How will you convert : benzene to acetophenone. (CBSE 2008 ; Assam Board 2012)
Answer:
C₆H₆ + CH₃COCl \( \xrightarrow{Anhyd.\ AlCl_3} \) C₆H₅COCH₃ + HCl

Teacher's Note:
1. Acetic anhydride, (CH₃CO)₂O, can be used in place of acetyl chloride.

Question 25. How will you convert : ethanal to propanone. (CBSE Sample Paper 2018-19)
Answer:
CH₃CHO \( \xrightarrow[(ii)\ H_3O^+]{(i)\ CH_3MgBr} \) CH₃CH(OH)CH₃ \( \xrightarrow{K_2Cr_2O_7/H_2SO_4\ or\ PCC} \) CH₃COCH₃

Teacher's Note:
1. Adding one carbon to an aldehyde and then oxidising is the standard aldehyde-to-ketone route.

Question 26. How will you convert : ethanal to 2-hydroxy-3-butenoic acid.
Answer:
CH₃CHO + HCHO \( \xrightarrow[(Cross\ aldol\ condensation)]{Dil.\ NaOH} \) HOCH₂–CH₂CHO (3-Hydroxypropanal) \( \xrightarrow[(-H_2O)]{H^+,\ \Delta} \) CH₂=CH–CHO (2-Propenal or Acrolein)
CH₂=CH–CHO \( \xrightarrow[(1,2\text{-}Addition)]{HCN} \) CH₂=CH–CH(OH)–CN (2-Hydroxy-3-butenenitrile) \( \xrightarrow[(Hydrolysis)]{H^+/H_2O} \) CH₂=CH–CH(OH)–COOH (2-Hydroxy-3-butenoic acid)

Teacher's Note:
1. Work backwards: an α-hydroxy acid comes from a cyanohydrin, so first make the aldehyde with one carbon less (acrolein).

Question 27. How will you convert : methanal into ethanol.
Answer:
HCHO + CH₃MgBr \( \xrightarrow{Dry\ ether} \) CH₃CH₂OMgBr \( \xrightarrow{H^+/H_2O} \) CH₃CH₂OH

Teacher's Note:
1. Formaldehyde is the only aldehyde that gives a 1° alcohol with a Grignard reagent.

Short Answer Questions (3 Marks)

Question 28. Complete the following by giving missing starting material, reagent or products:
(i) ............ –[O₃, Zn–H₂O]→ 2 cyclohexanone
(ii) Methylenecyclohexane (C₆H₁₀=CH₂) –[............]→ cyclohexanecarbaldehyde (C₆H₁₁–CHO)
(iii) C₆H₅CH₂OH –[KMnO₄, KOH, Δ]→ ............
Answer:
(i) C₆H₁₀=C₆H₁₀, cyclohexylidenecyclohexane (ozonolysis)
(ii) B₂H₆/THF, H₂O₂/OH⁻, then PCC (hydroboration-oxidation followed by oxidation)
(iii) C₆H₅COOK, potassium benzoate (oxidation)

Teacher's Note:
1. To work backwards from ozonolysis products, join the two C=O carbons by a C=C double bond after removing the oxygens.

Question 29. How can the following be converted:
(i) Ethanol → Acetone
(ii) Benzene → Acetophenone
(iii) Benzoic acid → Benzaldehyde
Answer:
(i) CH₃CH₂OH –[(O), CrO₃ or PCC]→ CH₃CHO –[CH₃MgBr, dry ether]→ CH₃–CH(CH₃)–OMgBr –[H₃O⁺]→ CH₃–CH(OH)–CH₃ –[CrO₃]→ CH₃–CO–CH₃
(ii) C₆H₆ + CH₃COCl –[anhydrous AlCl₃]→ C₆H₅COCH₃ + HCl
(iii) C₆H₅COOH –[PCl₅]→ C₆H₅COCl; C₆H₅COCl + H₂ –[Pd/BaSO₄]→ C₆H₅CHO + HCl

Teacher's Note:
1. For (i) use PCC (or Cu/573 K) in the first step so that ethanol stops at ethanal instead of going on to ethanoic acid.

Question 30. Give reasons for the following:
(i) Carboxylic acids do not give characteristic reactions of carbonyl group.
(ii) Treatment of C₆H₅CHO with HCN gives a mixture of two isomers which cannot be separated even by fractional distillation.
(iii) Sodium bisulphite is used for purification of ketones and aldehydes.
Answer:
(i) R–C(=O)–OH ↔ R–C(–O⁻)=O⁺H. Because of resonance, the position of >C=O group is changing.
(ii) C₆H₅CHO + HCN → C₆H₅CH(OH)CN. The product has a chiral carbon and is formed as a mixture of two optical isomers, d(+) and l(−). Since the two optical isomers have the same boiling point, fractional distillation is not possible.
(iii) Due to formation of addition compound with NaHSO₃ whereas impurities do not. The crystalline addition compound is separated and decomposed by dilute acid (or alkali) to regenerate the pure carbonyl compound:
CH₃CHO + NaHSO₃ → CH₃CH(OH)SO₃Na –[H₂O/H⁺]→ CH₃CHO + NaHSO₃

Teacher's Note:
1. The cyanohydrin is a racemic mixture because CN⁻ can attack the planar C=O equally from either face.

Question 31. Write tests to distinguish between:
(i) CH₃CHO and C₆H₅CHO
(ii) C₆H₅–OH and CH₃COOH
(iii) Pentanal and Pentan-2-one
Answer:
(i) CH₃CHO gives brick red ppt. with Fehling's solution, C₆H₅CHO does not.
(ii) Phenol does not give brisk effervescence with NaHCO₃ but CH₃COOH gives this test (CO₂ evolved).
(iii) Pentanal forms silver mirror with Tollens' reagent but pentan-2-one does not.

Teacher's Note:
1. NaHCO₃ separates carboxylic acids from phenols because only acids stronger than carbonic acid release CO₂.

Question 32. Convert:
(i) Benzaldehyde to acetophenone
(ii) Malonic acid to acetic acid
(iii) Acetaldehyde to 2-butanol
Answer:
(i) C₆H₅CHO –[(O), KMnO₄/H₂SO₄]→ C₆H₅COOH –[NaOH]→ C₆H₅COONa –[NaOH/CaO, Δ]→ C₆H₆ –[CH₃COCl, AlCl₃]→ C₆H₅COCH₃ + HCl
(ii) CH₂(COOH)₂ (malonic acid) –[Δ]→ CH₃COOH + CO₂
(iii) CH₃CHO –[C₂H₅MgBr, dry ether]→ CH₃CH(OMgBr)C₂H₅ –[H₃O⁺]→ CH₃CH(OH)CH₂CH₃ (2-butanol)

Teacher's Note:
1. Aldehyde + Grignard reagent gives a 2° alcohol; choose the alkyl group of RMgX to supply the missing carbons.

Question 33. Write the structures of organic compounds A to F in the following sequence of reactions:
C₆H₅OH –[Zn, Δ]→ A –[conc. HNO₃, H₂SO₄]→ B –[Br₂]→ C
C –[H₂/Pt]→ D –[HNO₂, HCl]→ E –[H₃O⁺]→ F
Answer:
A = C₆H₆ (benzene); B = C₆H₅NO₂ (nitrobenzene)
C = m-Br–C₆H₄–NO₂ (1-bromo-3-nitrobenzene)
D = m-Br–C₆H₄–NH₂ (3-bromoaniline)
E = m-Br–C₆H₄–N₂⁺Cl⁻ (3-bromobenzenediazonium chloride)
F = m-Br–C₆H₄–OH (3-bromophenol)

Teacher's Note:
1. The –NO₂ group directs Br to the meta position; that meta relationship is carried through every later step.

Question 34. Complete the following:
(i) CH₃CONH₂ + HNO₂ →
(ii) CH₃CONH₂ + NaOH + Br₂ →
(iii) C₆H₅COOH –[HNO₃/H₂SO₄]→
Answer:
(i) CH₃CONH₂ + HNO₂ → CH₃COOH + N₂ + H₂O
(ii) CH₃CONH₂ + 4NaOH + Br₂ → CH₃NH₂ + Na₂CO₃ + 2NaBr + 2H₂O
(iii) C₆H₅COOH + HNO₃ → m-O₂N–C₆H₄–COOH (3-nitrobenzoic acid) + H₂O

Teacher's Note:
1. Hofmann bromamide degradation gives an amine with one carbon less than the amide – the lost carbon leaves as carbonate.

Question 35. Explain why dialkylcadmium is considered superior to Grignard reagent for the preparation of a ketone from an acid chloride?
Answer:
Since Cd (E.N. = 1.7) is less electropositive than Mg (E.N. = 1.2), therefore, dialkylcadmiums are less reactive than Grignard reagents towards nucleophilic addition reactions. As such, dialkylcadmiums react with the more reactive acid chlorides to give ketones but do not react further with the less reactive ketones thus formed to give tert-alcohols. In contrast, Grignard reagents being more reactive not only react with the acid chlorides but also with the ketones so formed to give tert-alcohols.
R′–CO–Cl (acid chloride) →^{R₂Cd, −RCdCl} R′–CO–R (ketone)
R′–CO–Cl →^{RMgX, −Mg(Cl)X} R′–CO–R (ketone) →^{(i) RMgX, (ii) H+/H₂O} R′–C(OH)R₂ (3° alcohol)

Teacher's Note:
1. Acid chloride + R₂Cd is the standard way to make a ketone; with RMgX the reaction runs on to a tertiary alcohol.

Long Answer Questions (5 Marks)

Question 36. (i) Carry out the following conversions: [CBSE 2019]
(a) p-nitrotoluene to 2-bromobenzoic acid
(b) Propanoic acid to acetic acid
(ii) An alkene with molecular formula C₅H₁₀ on ozonolysis gives a mixture of two compounds, B and C. Compound B gives positive Fehling test and also reacts with iodine and NaOH solution. Compound C does not give Fehling solution test but forms iodoform. Identify the compounds A, B and C.
Answer:
(i) (a) p-Nitrotoluene \( \xrightarrow{Br_2} \) 2-bromo-4-nitrotoluene \( \xrightarrow{Sn/HCl} \) 3-bromo-4-methylaniline \( \xrightarrow[273{-}278\,K]{NaNO_2/HCl} \) 3-bromo-4-methylbenzenediazonium chloride \( \xrightarrow[H_3PO_2]{H_2O} \) 2-bromotoluene \( \xrightarrow[OH^-]{KMnO_4} \) 2-bromobenzoic acid
(b) CH₃CH₂COOH \( \xrightarrow[\text{Heat}]{NH_3} \) CH₃CH₂CONH₂ \( \xrightarrow{Br_2/KOH} \) CH₃CH₂NH₂ \( \xrightarrow[273\,K]{HNO_2} \) CH₃CH₂OH \( \xrightarrow{KMnO_4/H^+\text{ or }OH^-} \) CH₃COOH
(ii) A: 2-Methylbut-2-ene / CH₃CH=C(CH₃)₂
B: Ethanal / Acetaldehyde / CH₃CHO
C: Propanone / Acetone / CH₃COCH₃

Teacher's Note:
1. In (a) the CH₃ group directs Br to its ortho position; the NO₂ group is then removed via the amine and diazonium salt (H₃PO₂ replaces N₂⁺ by H).

Question 37. (i) Write the structures of A, B, C and D in the following reactions:
C₆H₅COCl \( \xrightarrow{H_2/Pd{-}BaSO_4} \) A \( \xrightarrow{\text{conc. NaOH}} \) B + C;   A \( \xrightarrow{CH_3MgBr/H_3O^+} \) D
(ii) Distinguish between the following:
(a) C₆H₅–COCH₃ and C₆H₅–COCH₂CH₃
(b) Propanal and butan-2-one [CBSE (F) 2014]
(iii) Write the structure of 2-hydroxybenzaldehyde. [CBSE Chennai 2015]
Answer:
(i) A = C₆H₅CHO (Benzaldehyde), B or C = C₆H₅COONa (Sodium benzoate), C or B = C₆H₅CH₂–OH (Benzyl alcohol), D = C₆H₅–CH(OH)–CH₃ (1-Phenylethanol)
(ii) (a) C₆H₅COCH₃ being a methyl ketone gives iodoform test while C₆H₅COCH₂CH₃ does not.
C₆H₅COCH₃ (Acetophenone) + 3NaOI → C₆H₅COONa + CHI₃↓ (Iodoform, Yellow ppt.) + 2NaOH
(b) Propanal being an aldehyde reduces Tollens’ reagent to silver mirror but butan-2-one being a ketone does not.
CH₃CH₂CHO (Propanal) + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → CH₃CH₂COO⁻ + 2Ag↓ (Silver mirror) + 4NH₃ + 2H₂O
CH₃COCH₂CH₃ (Butan-2-one) \( \xrightarrow{\text{Tollens' reagent}} \) No silver mirror
(iii)

Teacher's Note:
1. Either test works for pair (b): only propanal gives the silver mirror, and only butan-2-one (a methyl ketone) gives iodoform.

Question 38. (i) Carry out the following conversions:
(a) Benzoic acid to aniline
(b) Bromomethane to ethanol
(ii) Write the structure of major product(s) in the following:
(a) CH₃–CH₂–CHO \( \xrightarrow[(b)\,KOH,\,\text{Glycol/heat}]{(a)\,H_2N{-}NH_2} \)
(b) (CH₃)₃C–CHO \( \xrightarrow{\text{conc. NaOH}} \)
(c) C₆H₅COOH \( \xrightarrow{NaOH} \) [CBSE 2019]
Answer:
(i) (a) C₆H₅COOH \( \xrightarrow[\text{Heat}]{NH_3} \) C₆H₅CONH₂ \( \xrightarrow{Br_2,\,KOH} \) C₆H₅NH₂
(b) CH₃Br \( \xrightarrow{KCN\,(alc.)} \) CH₃CN \( \xrightarrow[H^+/OH^-]{H_2O} \) CH₃COOH \( \xrightarrow[H_3O^+]{LiAlH_4/\text{Ether or }B_2H_6} \) CH₃CH₂OH
(ii) (a) CH₃–CH₂–CHO \( \xrightarrow[(b)\,KOH,\,\text{Glycol/heat}]{(a)\,H_2N{-}NH_2} \) CH₃CH₂CH₃
(b) (CH₃)₃C–CHO \( \xrightarrow{\text{conc. NaOH}} \) (CH₃)₃CCH₂OH + (CH₃)₃CCOONa
(c) C₆H₅COOH \( \xrightarrow{NaOH} \) C₆H₅COONa

Teacher's Note:
1. Hoffmann bromamide degradation removes one carbon (benzamide to aniline), while the cyanide route in (b) adds one carbon.

Question 39. (i) Give a simple chemical test to distinguish between benzaldehyde and ethanal. [CBSE (F) 2013]
(ii) Bring out the following conversions:
(a) 4-Nitrotoluene to 2-bromobenzoic acid
(b) Ethylcyanide to 1-phenyl propanone
(iii) 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.
Answer:
(i) Benzaldehyde and Ethanal
Ethanal reacts with NaOI (I₂/NaOH) to form yellow precipitate of iodoform while benzaldehyde does not give this test.
CH₃CHO + 3I₂ + 4NaOH → HCOO⁻Na⁺ + CHI₃↓ (Iodoform, Yellow ppt.) + 3NaI + 3H₂O
(ii) (a) 4-Nitrotoluene to 2-bromobenzoic acid
4-Nitrotoluene \( \xrightarrow{Br_2} \) 2-bromo-4-nitrotoluene \( \xrightarrow{Sn/HCl} \) 3-bromo-4-methylaniline \( \xrightarrow[273{-}278\,K]{NaNO_2/HCl} \) 3-bromo-4-methylbenzenediazonium chloride \( \xrightarrow[H_3PO_2]{H_2O} \) 2-bromotoluene \( \xrightarrow[OH^-]{KMnO_4} \) 2-bromobenzoic acid
(b) Ethylcyanide to 1-phenyl propanone
CH₃–CH₂–CN \( \xrightarrow[\text{Ether}]{+C_6H_5MgBr} \) CH₃CH₂–C(=NMgBr)–C₆H₅ \( \xrightarrow{H_3O^+} \) CH₃CH₂–CO–C₆H₅
(iii) A = CH₃–CH₂–CHO (Propanal); B = CH₃–CO–CH₃ (Propanone)
Reaction involved:
CH₃–CO–CH₃ (Propanone) + 3I₂ + 4NaOH \( \xrightarrow{\Delta} \) CHI₃ (Yellow ppt.) + CH₃–COO⁻Na⁺ + 3NaI + 3H₂O

Teacher's Note:
1. Ethanal is the only aldehyde that gives the iodoform test, which makes it the quickest way to tell it from benzaldehyde.

Question 40. Complete each synthesis by giving missing starting material, reagent or products:
(i) Cyclopentanone + HO–NH₂ \( \xrightarrow{H^+} \)
(ii) C₆H₅CH₂CH₃ \( \xrightarrow[\text{KOH, heat}]{KMnO_4} \)
(iii) Phthalic acid (benzene-1,2-dicarboxylic acid) \( \xrightarrow[\Delta]{SOCl_2} \)
(iv) Methylenecyclohexane → Cyclohexanecarbaldehyde
(v) Benzene → Benzophenone [CBSE Sample Paper 2017]
Answer:
(i) Cyclopentanone + HO–NH₂ \( \xrightarrow{H^+} \) C₅H₈=N–OH (Cyclopentanone oxime) + H₂O
(ii) C₆H₅CH₂–CH₃ (Ethylbenzene) \( \xrightarrow[\Delta]{KMnO_4{-}KOH} \) C₆H₅COO⁻K⁺ \( \xrightarrow{H_3O^+} \) C₆H₅COOH (Benzoic acid)
(iii) C₆H₄(COOH)₂ (Phthalic acid) \( \xrightarrow[\Delta]{SOCl_2} \) C₆H₄(COCl)₂ (Phthaloyl chloride)
(iv) C₆H₁₀=CH₂ (Methylenecyclohexane) \( \xrightarrow{B_2H_6} \) (C₆H₁₁–CH₂)₃B \( \xrightarrow[OH^-]{H_2O_2} \) C₆H₁₁–CH₂–OH \( \xrightarrow{PCC} \) C₆H₁₁–CHO (Cyclohexanecarbaldehyde)
(v) C₆H₆ (Benzene) + C₆H₅COCl (Benzoyl chloride) \( \xrightarrow{\text{Anhyd. }AlCl_3} \) C₆H₅COC₆H₅ (Benzophenone) + HCl

Teacher's Note:
1. Learn the reagent-product pairs as a set: NH₂OH gives oximes, NH₂NH₂ hydrazones, NH₂NHCONH₂ semicarbazones.