CBSE Class 12 Chemistry Aldehydes Ketones and Carboxylic Acids Important Questions Set 01

Very Short Answer Questions (1 Mark)

Question 1. 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

Teacher's Note:
1. Carboxylic acids boil highest because they form hydrogen-bonded dimers; aldehydes cannot hydrogen bond among themselves.

Question 2. Write chemical equations for the following reactions :
Benzoyl chloride is hydrogenated in presence of Pd/BaSO₄.
Answer:
C₆H₅COCl H₂
⟶
Pd-BaSO₄ C₆H₅CHO (Benzaldehyde)

Teacher's Note:
1. This is Rosenmund reduction; the poisoned catalyst stops the reaction at the aldehyde stage.

Question 3. Write structures of compounds A and B in each of the following reactions.
C₆H₅–CH₂CH₃ KMnO₄ – KOH
⟶ A H₃O⁺
⟶ B
Answer:
C₆H₅–CH₂CH₃ KMnO₄–KOH
⟶ C₆H₅COOK (A) (Potassium benzoate) H₃O⁺
⟶ C₆H₅COOH (B) (Benzoic acid)

Teacher's Note:
1. In alkaline KMnO₄ the acid is obtained as its potassium salt; acidification is needed to get the free acid.

Question 4. Write structures of compounds A and B in each of the following reactions :
Cyclohexanol CrO₃
⟶ A H₂N–NH–CONH₂
⟶ B
Answer:
Cyclohexanol CrO₃
⟶ Cyclohexanone (A), C₆H₁₀=O H₂N–NH–CONH₂
⟶ C₆H₁₀=NNHCONH₂ (B) (Cyclohexanone semicarbazone)

Teacher's Note:
1. A secondary alcohol is oxidised to a ketone, and the ketone condenses with semicarbazide through its terminal –NH₂ group.

Question 5. Give reasons :
Chloroacetic acid is stronger than acetic acid.
Answer:
Chloroacetic acid has lower pK_a 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.

Teacher's Note:
1. Mention both points in the exam: the –I effect of Cl and the greater stability of the chloroacetate ion.

Question 6. Write the IUPAC name of the following compound :

Answer:
2-Hydroxybenzaldehyde

Teacher's Note:
1. The –CHO group gets priority over –OH, so the compound is named as a benzaldehyde with C-1 bearing –CHO.

Question 7. 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.

Teacher's Note:
1. Aldehydes and ketones have no H attached to oxygen, so they show only dipole–dipole attraction, which is weaker than hydrogen bonding.

Question 8. Complete the following reactions :
CH₃–CH(CH₃)–COOH (i) Br₂/Red P₄
⟶
(ii) H₂O
Answer:
CH₃–CH(CH₃)–COOH (i) Br₂/Red P₄
⟶
(ii) H₂O CH₃–C(CH₃)(Br)–COOH (2-Bromo-2-methylpropanoic acid)

Teacher's Note:
1. The only α-hydrogen here is on C-2, so a single bromine enters at C-2 (HVZ reaction).

Question 9. Write the IUPAC name of the following :
CH₃–C≡C–CH=CH–CO–OH
Answer:
Hex-2-en-4-ynoic acid

Teacher's Note:
1. Number from the –COOH carbon (C-1); the double bond gets locant 2 and the triple bond 4, and ‘en’ is written before ‘yn’.

Question 10. Write the equation involved in Etard reaction.
Answer:
Etard reaction :
C₆H₅–CH₃ (Toluene) + CrO₂Cl₂ CS₂
⟶ C₆H₅–CH(OCrOHCl₂)₂ (Chromium complex) H₃O⁺
⟶ C₆H₅–CHO (Benzaldehyde)

Teacher's Note:
1. Etard reaction stops at the aldehyde because the chromium complex protects it from further oxidation until hydrolysis.

Short Answer Questions (2 Marks)

Question 11. 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 :
(C₂H₃OCl) A H₂/Pd-BaSO₄
⟶ B dil. NaOH
⟶ C Heat
⟶ D
Answer:
CH₃–CO–Cl (A) H₂/Pd-BaSO₄
⟶ CH₃–CHO (B) dil. NaOH
⟶ CH₃–CH(OH)–CH₂–CHO (C) Δ
⟶ CH₃–CH=CH–CHO (D)
A = CH₃COCl, B = CH₃CHO, C = CH₃CH(OH)CH₂CHO, D = CH₃CH=CHCHO

Teacher's Note:
1. C₂H₃OCl with a Rosenmund step points straight to acetyl chloride; then follow aldol addition and dehydration.

Question 12. 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
Answer:
(i) 2(CH₃)₂C=O Ba(OH)₂
⟶ CH₃–C(OH)(CH₃)–CH₂COCH₃ (4-Hydroxy-4-methylpentan-2-one)
(ii) C₆H₅COCH₃ (Acetophenone) + 4[H] Zn(Hg)/Conc.HCl
⟶ C₆H₅CH₂CH₃ (Ethylbenzene) + H₂O

Teacher's Note:
1. Ketones also give aldol addition with dilute base; the product of acetone is diacetone alcohol.

Question 13. Give reasons :
(i) Electrophilic substitution in benzoic acid takes place at meta-position.
(ii) Carboxylic acids do not give the characteristic reactions of carbonyl group.
Answer:
(i) Electrophilic substitution in benzoic acid takes place at meta-position. Due to resonance in benzoic acid, the ortho and para positions carry a positive charge, so the electron density is relatively higher at meta-position. Therefore, electrophilic substitution in benzoic acid takes place at meta-position.
e.g., C₆H₅COOH + NO₂⁺ → m-O₂N–C₆H₄–COOH + H⁺
(ii) 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 :
R–C(=O)–O⁻ <→ R–C(–O⁻)=O

Teacher's Note:
1. Because of resonance the carbonyl carbon of –COOH is much less electrophilic, so reagents like HCN, NaHSO₃ and NH₂OH do not add to it.

Question 14. Describe how the following conversions can be brought about :
(i) Ethylbenzene to benzoic acid
(ii) Bromobenzene to benzoic acid
Answer:
(i) C₆H₅CH₂CH₃ (Ethylbenzene) KMnO₄/OH⁻
⟶
Vigorous oxidation C₆H₅COOH (Benzoic acid)
(ii) C₆H₅Br (Bromobenzene) Mg/dry ether
⟶
Grignard reaction C₆H₅MgBr (Phenylmagnesium bromide) (i) Dry ice, (ii) H₃O⁺
⟶ C₆H₅COOH (Benzoic acid)

Teacher's Note:
1. Any alkyl side chain with a benzylic H is oxidised to –COOH; the Grignard + CO₂ route adds one carbon.

Question 15. Which acid of each pair shown here would you expect to be stronger?
(i) F–CH₂–COOH or Cl–CH₂–COOH
(ii) C₆H₅OH (phenol) or CH₃COOH
Answer:
(i) F–CH₂COOH > Cl–CH₂COOH
(ii) CH₃COOH is stronger than C₆H₅OH (phenol)

Teacher's Note:
1. F has the stronger –I effect; and the carboxylate ion (charge on two O atoms) is more stable than the phenoxide ion (charge partly on less electronegative C atoms).

Question 16. The reaction of carbonyl compound with pure HCN is very slow and becomes fast in presence of a base.
Answer:
With pure HCN reaction occurs very slowly because it is a weak nucleophile. With base it produces CN⁻ ion which is a strong nucleophile and readily adds to the carbonyl compound.
HCN + ⁻OH → H₂O + CN⁻

Teacher's Note:
1. Write the equation showing CN⁻ formation – examiners look for the stronger nucleophile being generated by the base.

Question 17. A compound having the molecular formula C₃H₆O forms a crystalline white ppt. with sodium bisulphite and reduces Fehling’s solution. Suggest the structural formula and IUPAC name of this compound. Name an isomer for it from a group other than its own.
Answer:
Since the compound forms crystalline white precipitate with sodium bisulphite, it contains a carbonyl group. The compound reduces Fehling’s solution so, the carbonyl group is an aldehyde.
Structure : CH₃–CH₂–CHO
IUPAC name : Propanal
Isomer : CH₃–CO–CH₃ (Propanone)

Teacher's Note:
1. Propanal and propanone are functional isomers – same formula C₃H₆O, different functional groups.

Question 18. Account for the following :
(a) Aromatic carboxylic acids do not undergo Friedel–Crafts reaction.
(b) pK_a value of 4-nitrobenzoic acid is lower than that of benzoic acid.
Answer:
(a) Due to presence of electron withdrawing group (—COOH) in aromatic carboxylic acids, they do not undergo Friedel-Crafts reaction.
(b) Due to presence of strong electron withdrawing group (—NO₂), 4-nitrobenzoic acid is more acidic than benzoic acid and therefore, pK_a value is lower.

Teacher's Note:
1. For (a) you can also add that the –COOH group gets bonded to the AlCl₃ catalyst; remember lower pK_a means stronger acid.

Question 19. (i) What is the advantage of using DIBAL-H as reducing agent?
(ii) Which of the following can be nitrated more easily and why? Benzoic acid or phenol.
Answer:
(i) DIBAL–H reduces alkynes to alkenes but does not reduce ethylenic double bonds and hence this reagent can be used to reduce unsaturated nitriles to the corresponding unsaturated aldehydes.
(ii) Phenol gets easily nitrated than benzoic acid. Because carboxyl group is ring deactivating group whereas hydroxyl group is ring activating group.

Teacher's Note:
1. DIBAL-H also reduces esters to aldehydes at low temperature – a selective route to –CHO without touching C=C.

Question 20. Account for the following :
(i) CH₃CHO is more reactive than CH₃COCH₃ towards reaction with HCN.
(ii) There are two –NH₂ groups in semicarbazide (H₂NNHCONH₂). However, only one is involved in the formation of semicarbazone.
Answer:
(i) 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.
(ii) Semicarbazide has the following resonance structures arising due to the electron withdrawing nature of the O atom.
H₂N⁽¹⁾–CO–N⁽²⁾H–N⁽³⁾H₂ <→ H₂N⁺=C(–O⁻)–NH–NH₂ <→ H₂N–C(–O⁻)=N⁺H–NH₂
Lone pairs of N-1 and N-2 are involved in conjugation with the >C=O group while that of N-3 is not involved in resonance; thus, only the –NH₂ group (N-3) attached to NH is free to act as a nucleophile and is involved in the formation of semicarbazone.

Teacher's Note:
1. Only the –NH₂ not attached to C=O is nucleophilic; the amide –NH₂ lone pair is delocalised into the carbonyl group.

Question 21. (i) Write the IUPAC name of CH₃–CH=CH–CHO
(ii) Write the IUPAC name of the following: CH₃–CH₂–CHO
Answer:
(i) ⁴CH₃–³CH=²CH–¹CHO
IUPAC name = But-2-enal
(ii) Propanal.

Teacher's Note:
1. Number from the –CHO carbon; the double bond locant is written before “en” and the suffix “-al” needs no number.

Question 22. (i) Write the IUPAC name of CH₃–CH(NH₂)–CHO
(ii) Draw the structure of 3-methylpentanal.
Answer:
(i) 2-Aminopropanal.
(ii) CH₃–CH₂–CH(CH₃)–CH₂–CHO

Teacher's Note:
1. When drawing from a name, count the parent chain carbons including the CHO carbon (pentanal = 5 C) and put the methyl on C-3 counted from CHO.

Question 23. (i) What type of aldehydes undergo Cannizzaro reaction?
(ii) An aromatic organic compound ‘A’ with molecular formula C₈H₈O gives positive DNP and iodoform tests. It neither reduces Tollens’ reagent nor does it decolourise bromine water. Write the structure of ‘A’.
Answer:
(i) Having no α-hydrogen.
(ii) C₆H₅COCH₃
‘A’ gives positive DNP test. Therefore, it is an aldehyde or a ketone. Since it does not reduce Tollens’ reagent, ‘A’ must be a ketone. ‘A’ responds to iodoform test. Therefore, it should be a methyl ketone. The molecular formula of ‘A’ indicates high degree of unsaturation, yet it does not decolourise bromine water. This indicates the presence of unsaturation due to an aromatic ring. The molecular formula of ‘A’ indicates that it should be phenyl methyl ketone (acetophenone).

Teacher's Note:
1. In such identification questions, write one deduction per test (DNP → carbonyl, no Tollens’ → ketone, iodoform → CH₃CO–); each deduction earns marks.

Question 24. (i) (CH₃)₃C–CHO does not undergo aldol condensation. Comment.
(ii) Out of CH₃CH₂COCH₂CH₃ and CH₃CH₂CH₂COCH₃, which gives iodoform test.
Answer:
(i) No α-H is present.
(ii) CH₃CH₂CH₂COCH₃ will give iodoform test as it has a terminal ketomethyl group.

Teacher's Note:
1. (CH₃)₃CCHO has no α-hydrogen, so with conc. alkali it undergoes Cannizzaro reaction instead of aldol condensation.

Question 25. (i) Draw the resonating structures of carboxylic acid.
(ii) Carboxylic acids behave as fairly strong acids. Comment.
Answer:
(i) –C(=O)–Ö–H ↔ –C⁺(–O⁻)–Ö–H ↔ –C(–O⁻)=O⁺–H
(ii) Carboxylic acids are quite strong acids due to the presence of polar O–H group. They ionize to give H⁺ ions and therefore, behave as acids.
R–C(=O)–OH ↔ R–C(–O⁻)=O⁺–H
Carboxylic acids as well as carboxylate ion both are stabilised by resonance.

Teacher's Note:
1. The carboxylate ion has two equivalent resonance structures, which is why it is more stable than the acid and why the acid ionises readily.

Question 26. Which bond C–OH or CO–H of carboxylic acid is broken when
(i) Acid reacts with alcohol
(ii) Acid reacts with Sodium
Answer:
(i) C–OH
(ii) CO–H

Teacher's Note:
1. In esterification the acid loses its OH (shown by ¹⁸O labelling of the alcohol); with sodium only the acidic H is replaced, giving H₂.

Question 27. Write structures of main compounds A and B in each of the following reactions:
(i) CH₃CH₂CN \( \xrightarrow{\text{CH}_{3}\text{MgBr/H}_{3}\text{O}^{+}} \) A \( \xrightarrow{\text{LiAlH}_{4}} \) B
(ii) Toluene (C₆H₅CH₃) \( \xrightarrow[\text{(ii) H}_{3}\text{O}^{+}\text{/}\Delta]{\text{(i) CrO}_{3}\text{/(CH}_{3}\text{CO)}_{2}\text{O}} \) A \( \xrightarrow{\text{H}_{2}\text{N–NH}_{2}} \) B [CBSE, Delhi Set 3, 2019]
Answer:
(i) A ⇒ CH₃CH₂CO–CH₃,   B ⇒ CH₃CH₂–CH(OH)–CH₃
CH₃CH₂CN \( \xrightarrow{\text{CH}_{3}\text{MgBr/H}_{3}\text{O}^{+}} \) CH₃CH₂COCH₃ (A) \( \xrightarrow{\text{LiAlH}_{4}} \) CH₃CH₂CH(OH)CH₃ (B)
(ii) A ⇒ C₆H₅CHO,   B ⇒ C₆H₅–CH=N–NH₂
C₆H₅CH₃ \( \xrightarrow[\text{(ii) H}_{3}\text{O}^{+}]{\text{(i) CrO}_{3}\text{/(CH}_{3}\text{CO)}_{2}\text{O}} \) C₆H₅CHO (A) \( \xrightarrow{\text{H}_{2}\text{N–NH}_{2}} \) C₆H₅CH=N–NH₂ (B) + H₂O

Teacher's Note:
1. Hydrazine alone gives the hydrazone (C=N–NH₂); only with KOH/ethylene glycol and heat does it go on to the hydrocarbon (Wolff-Kishner).

Short Answer Questions (3 Marks)

Question 28. Write the equation of the reactions of ethanal with
(i) Fehling’s solution (ii) Phenylhydrazine (iii) Hydroxylamine.
Answer:
(i) CH₃CHO + 2Cu²⁺ + 5OH⁻ → CH₃COO⁻ + Cu₂O (Red ppt.) + 3H₂O
(ii) CH₃CH=O + H₂NNH–C₆H₅ (Phenyl hydrazine) → CH₃CH=N–NH–C₆H₅ + H₂O
(iii) CH₃CH=O + NH₂OH (Hydroxylamine) → CH₃CH=N–OH (Ethanal oxime) + H₂O

Teacher's Note:
1. Reactions with ammonia derivatives are addition–elimination reactions: always show the H₂O eliminated.

Question 29. Illustrate the following name reactions giving a chemical equations in each case :
(i) Clemmensen reaction
(ii) Cannizzaro reaction
Answer:
(i) Clemmensen reduction : The carbonyl group of aldehydes and ketones is reduced to CH₂ group on treatment with zinc amalgam and concentrated hydrochloric acid.
(CH₃)₂C=O (Propanone) + 4[H] Zn – Hg
⟶
HCl (CH₃)₂CH₂ (Propane) + H₂O
(ii) Cannizzaro reaction : Aldehydes which do not contain α-H atom undergo disproportionation when heated with concentrated (50%) NaOH.
2HCHO (Methanal) + NaOH (50%) → HCOONa (Sodium formate) + CH₃OH (Methanol)

Teacher's Note:
1. Write a definition line and one balanced example for every name reaction – both carry marks.

Question 30. (i) Write the structures of compounds A, B and C in each of the following reactions :
(a) C₆H₅Br Mg/dry ether
⟶ A (a) CO₂(g)
⟶
(b) H₃O⁺ B PCl₅
⟶ C
(b) CH₃CN (a) SnCl₂/HCl
⟶
(b) H₃O⁺ A dil. NaOH
⟶ B Δ
⟶ C
(ii) Do the following conversion in not more than two steps :
Benzoic acid to benzaldehyde
Answer:
(i) (a) C₆H₅Br Mg/dry ether
⟶ C₆H₅MgBr (A) (a) CO₂(g)
⟶
(b) H₃O⁺ C₆H₅COOH (B) PCl₅
⟶ C₆H₅COCl (C)
(b) CH₃–C≡N (a) SnCl₂/HCl
⟶
(b) H₃O⁺ CH₃–CHO (A) dil. NaOH
⟶ CH₃–CH(OH)–CH₂–CHO (B) Δ
⟶ CH₃CH=CH–CHO (C)
(ii) C₆H₅COOH (Benzoic acid) SOCl₂
⟶ C₆H₅COCl (Benzoyl chloride) H₂/Pd-S, BaSO₄
⟶
Rosenmund’s reduction C₆H₅CHO (Benzaldehyde)

Teacher's Note:
1. CH₃CN with SnCl₂/HCl is Stephen reduction (nitrile to aldehyde); the aldehyde then undergoes aldol condensation.

Question 31. During practical exams, lab assistant provided two test tubes containing 5 mL benzoic acid and 5 mL acetaldehyde to every student. A student found that test tubes given to him were unlabelled. He informed the teacher before performing any experiment with the given chemicals.
How can the chemicals be distinguished for correct labelling?
Answer:
Chemicals can be distinguished by sodium bicarbonate test and iodoform test.
Benzoic acid will give brisk effervescence due to evolution of carbon dioxide gas with sodium bicarbonate solution while acetaldehyde does not.
Acetaldehyde will give yellow precipitate of iodoform with iodine and sodium hydroxide solution while benzoic acid does not.

Teacher's Note:
1. NaHCO₃ test is the quickest test for a –COOH group; ethanal is the only aldehyde that gives the iodoform test.

Question 32. (i) Write the equations involved in the following reactions :
(a) Stephen reaction
(b) Etard reaction
(ii) Distinguish between CH₃COOH and HCOOH.
Answer:
(i) (a) Stephen reduction :
R–CN + SnCl₂ + HCl → R–CH=NH H₃O⁺
⟶ R–CHO
(b) Etard reaction :
C₆H₅–CH₃ (Toluene) + CrO₂Cl₂ CS₂
⟶ C₆H₅–CH(OCrOHCl₂)₂ (Chromium complex) H₃O⁺
⟶ C₆H₅–CHO (Benzaldehyde)
(ii) Add Tollens’ reagent to formic acid and warm. Silver mirror is formed.
HCOOH (Formic acid) + 2[Ag(NH₃)₂]⁺ + 2OH⁻ Warm
⟶ 2Ag (Silver mirror) + CO₂ + 2NH₃ + 2NH₄OH
Acetic acid does not give this test.

Teacher's Note:
1. Formic acid behaves like an aldehyde because of its H–C=O unit, which is the basis of this distinction.

Question 33. 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 acidification 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).
2HCHO (A, Formaldehyde) NaOH
⟶ CH₃OH (B, Methanol) + HCOONa (C, Sodium formate)
CH₃OH (B) Cu (dehydrogenation)
⟶ HCHO (A, formaldehyde)
HCOONa (C) Acidification
⟶ HCOOH (D, formic acid)
Thus, A = Formaldehyde (HCHO), B = Methanol (CH₃OH), C = Sodium formate (HCOONa), D = Formic acid (HCOOH)

Teacher's Note:
1. Start from the smallest clue – CH₂O₂ can only be formic acid – and work backwards.

Question 34. (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) 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:
(a) Wolff-Kishner reduction : 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.
(CH₃)₂C=O (Acetone) NH₂NH₂
⟶
–H₂O (CH₃)₂C=NNH₂ KOH/ethylene glycol
⟶
heat (CH₃)₂CH₂ (Propane) + N₂
(b) Increasing order of reactivity towards nucleophilic addition reaction :
C₆H₅COCH₃ < CH₃COCH₃ < CH₃CHO
(c) Formula of compounds A and B is C₃H₆O. B forms yellow precipitate of iodoform. Hence, B must contain —COCH₃ group. Therefore, compound ‘B’ must be CH₃–CO–CH₃.
A does not give iodoform test and it is functional isomer of B thus, it may be CH₃CH₂CHO.

Teacher's Note:
1. Wolff–Kishner is the basic alternative to Clemmensen reduction; use it when the molecule has acid-sensitive groups.

Question 35. (a) Write the main product in the following equations :
(i) CH₃–CO–CH₃ LiAlH₄
⟶ ?
(ii) C₆H₅CHO HNO₃/H₂SO₄
⟶
273–283 K ?
(b) Write the product in the following reaction :
CH₃–CH=CH–CH₂CN (i) DIBAL-H
⟶
(ii) H₂O
Answer:
(a) (i) CH₃–CO–CH₃ LiAlH₄
⟶ CH₃–CH(OH)–CH₃
(ii) C₆H₅CHO (Benzaldehyde) HNO₃/H₂SO₄
⟶
273 – 283 K m-O₂N–C₆H₄–CHO (m-Nitrobenzaldehyde)
(b) CH₃–CH=CH–CH₂CN (i) DIBAL-H
⟶
(ii) H₂O CH₃–CH=CH–CH₂CHO

Teacher's Note:
1. –CHO is a meta-directing, deactivating group; DIBAL-H converts –CN to –CHO and leaves the C=C untouched.

Long Answer Questions (5 Marks)

Question 36. Write the structures of A, B, C, D and E in the following reactions :
C₆H₆ CH₃COCl
⟶
Anhyd. AlCl₃ A Zn–Hg/conc.HCl
⟶ B (i) KMnO₄ – KOH, Δ
⟶
(ii) H₃O⁺ C
A NaOI
⟶ D + E
Answer:
C₆H₆ CH₃COCl
⟶
Anhyd. AlCl₃ C₆H₅COCH₃ (A) Zn-Hg
⟶
Conc.HCl C₆H₅CH₂CH₃ (B) (i) KMnO₄ – KOH, Δ
⟶
(ii) H₃O⁺ C₆H₅COOH (C)
C₆H₅COCH₃ (A) NaOI
⟶ C₆H₅COONa (E) + CHI₃ (D)
A = Acetophenone, B = Ethylbenzene, C = Benzoic acid, D = Iodoform, E = Sodium benzoate

Teacher's Note:
1. Acetophenone is a methyl ketone, so with NaOI it gives iodoform and the sodium salt of the acid with one carbon less.

Question 37. Identify A to E in the following reactions :
C₆H₅COOH Conc. HNO₃
⟶
+Conc. H₂SO₄, Δ A SOCl₂
⟶ B (i) NaBH₄
⟶
(ii) H₃O⁺ C
C₆H₅COOH SOCl₂
⟶ D H₂, Pd, BaSO₄
⟶
+S or quinoline E
Answer:
C₆H₅COOH conc. HNO₃
⟶
conc. H₂SO₄ m-O₂N–C₆H₄–COOH (A) SOCl₂
⟶ m-O₂N–C₆H₄–COCl (B) (i) NaBH₄
⟶
(ii) H₃O⁺ m-O₂N–C₆H₄–CH₂OH (C)
C₆H₅COOH SOCl₂
⟶ C₆H₅COCl (D) H₂/Pd-BaSO₄
⟶
+S or quinoline C₆H₅CHO (E) (Benzaldehyde)
A = m-Nitrobenzoic acid, B = m-Nitrobenzoyl chloride, C = m-Nitrobenzyl alcohol, D = Benzoyl chloride, E = Benzaldehyde

Teacher's Note:
1. NaBH₄ reduces the acid chloride to the 1° alcohol but leaves the –NO₂ group; Rosenmund conditions stop at the aldehyde.

Question 38. (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 cyanohydrin in good yield but 2,4,6-trimethylcyclo-hexanone does not.
(b) An organic compound with molecular formula C₉H₁₀O forms 2,4-DNP derivative, reduces Tollens’ reagent and undergoes Cannizzaro reaction. On vigorous oxidation it gives 1,2-benzene-dicarboxylic acid. Identify the compound.
Answer:
(a) (i) Semicarbazide has the following resonance structures arising due to the electron withdrawing nature of the O atom.
H₂N⁽¹⁾–CO–N⁽²⁾H–N⁽³⁾H₂ <→ H₂N⁺=C(–O⁻)–NH–NH₂
H₂N⁽¹⁾–CO–N⁽²⁾H–N⁽³⁾H₂ <→ H₂N–C(–O⁻)=N⁺H–NH₂
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 α-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.
o-CH₃CH₂–C₆H₄–CHO (2-Ethylbenzaldehyde) Oxidation
⟶ o-HOOC–C₆H₄–COOH (1,2-Benzenedicarboxylic acid)

Teacher's Note:
1. In (a)(ii) the two methyl groups at C-2 and C-6 flank the C=O and block the approach of CN⁻; the methyl at C-4 has little effect.

Question 39. (a) Identify A, B and C in the following sequence of reactions :
CH₃CHO (i) C₂H₅MgCl
⟶
(ii) H₂O A conc. H₂SO₄
⟶ B HBr + Peroxide
⟶ C
(b) Predict the structures of the products formed when benzaldehyde is treated with
(i) conc. NaOH
(ii) HNO₃/H₂SO₄ (at 273–283 K)
Answer:
(a) CH₃CHO (i) C₂H₅MgCl
⟶ [CH₃–CH(OMgCl)–C₂H₅] (ii) H₂O
⟶ CH₃–CH(OH)–C₂H₅ (A) conc. H₂SO₄
⟶ CH₃CH=CHCH₃ (But-2-ene, B) HBr/peroxide
⟶ H₃C–CH₂–CH(Br)–CH₃ (2-bromobutane, C)
(b) (i) 2C₆H₅–CHO conc. NaOH
⟶
(Cannizzaro reaction) C₆H₅–CH₂OH (Benzyl alcohol) + C₆H₅–COO⁻Na⁺ (Sodium benzoate)
(ii) C₆H₅CHO (Benzaldehyde) HNO₃/H₂SO₄
⟶
273 – 283 K m-O₂N–C₆H₄–CHO (m-Nitrobenzaldehyde)

Teacher's Note:
1. But-2-ene is symmetrical, so with HBr the product is 2-bromobutane whether or not peroxide is present.

Question 40. 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:
(CH₃CO)₂O (A) + C₂H₅OH (Ethanol) → CH₃COOC₂H₅ (C, Ethyl ethanoate) + CH₃COOH (B, Ethanoic acid)
CH₃COOC₂H₅ + H₂O H⁺
⟶ CH₃COOH (B) + CH₃CH₂OH (D)
CH₃CH₂OH (D) KMnO₄
⟶ CH₃COOH (B) Ca(OH)₂
⟶
Dry distillation CH₃COCH₃ (E) + CaCO₃
E does not give Tollens’ test and does not reduce Fehling’s solution as it is ketone.
(CH₃)₂C=O + H₂N–NH–C₆H₃(NO₂)₂ → (CH₃)₂C=N–NH–C₆H₃(NO₂)₂ + H₂O (2,4-dinitrophenyl hydrazone)
A = (CH₃CO)₂O, B = CH₃COOH, C = CH₃COOC₂H₅, D = CH₃CH₂OH, E = CH₃COCH₃

Teacher's Note:
1. An anhydride reacting with an alcohol gives one molecule of ester and one of acid – that is the key to A here.