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

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. In (CH₃)₃C–COOH the α-carbon is attached to three CH₃ groups and carries no hydrogen.
2. HVZ reaction requires an α-hydrogen, so the absence of α-H explains why it does not occur.

Teacher's Note:
1. Pivalic acid, (CH₃)₃CCOOH, is the standard example of an acid that fails the HVZ reaction.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Phosphorus converts a little of the acid into acid chloride which is more reactive than the parent carboxylic acid. Thus, it is the acid chloride, not the acid itself, that undergoes chlorination at the α-carbon.
2. P + X₂ forming PX₃ is a correct statement, but it does not by itself explain the α-halogenation described in the assertion.

Teacher's Note:
1. Red P is needed only in small amount because the PX₃ it forms converts a little acid into acyl halide, which is the species halogenated at the α-carbon.

Answer: (d) A is false, but R is true.

Explanation:
1. Bromination occurs at α-positions.
2. CH₃–CH₂–COOH Br₂/P
⟶
–HBr CH₃CHBr–COOH Br₂/P
⟶
–HBr CH₃–CBr₂–COOH

Teacher's Note:
1. HVZ halogenates only the α-carbon; the β-CH₃ group is never touched.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. β-ketoacids are unstable acids. They readily undergo decarboxylation through a cyclic transition state.
2. C₆H₅–CO–CH₂–COOH → C₆H₅–C(OH)=CH₂ + CO₂ → C₆H₅–CO–CH₃

Teacher's Note:
1. Only acids with a C=O group β to the –COOH (β-keto acids, malonic acids) lose CO₂ easily on gentle heating.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. CH₃–CH(CH₃)–CH₂COOH NaOH/CaO
⟶
Δ CH₃–CH(CH₃)–CH₃ + Na₂CO₃

Teacher's Note:
1. Decarboxylation removes exactly one carbon, so a C₅ acid gives a C₄ alkane.

Answer: (a) Isopropyl alcohol

Explanation:
1. Iodoform test is given by the organic compounds having CH₃–CO– or CH₃–CH(OH)– group.
2. CH₃–CH(OH)–CH₃ : Isopropyl alcohol; CH₃CH₂CHO : Propionaldehyde; C₂H₅–CO–C₆H₅ : Ethylphenyl ketone; C₆H₅–CH₂–OH : Benzyl alcohol
3. Therefore, isopropyl alcohol will give positive iodoform test.

Teacher's Note:
1. NaOI is an oxidant: it first oxidises CH₃CH(OH)– to CH₃CO–, which is why such alcohols also give iodoform.

Answer: (b) ICH₂COCH₂I

Explanation:
1. Iodoform reaction of acetone occurs in following steps :
2. CH₃–CO–CH₃ + NaOI → CH₃–CO–CH₂I + NaOH
3. CH₃–CO–CH₂I + NaOI → CH₃–CO–CHI₂ + NaOH
4. CH₃–CO–CHI₂ + NaOI → CH₃–CO–CI₃ + NaOH
5. CH₃–CO–CI₃ + NaOH → CH₃COONa + CHI₃

Teacher's Note:
1. Once one iodine is on a carbon, that carbon’s remaining H atoms become more acidic, so the same carbon is iodinated again and again.

Answer: (c) 2-acetyl-2-methylcyclohexanone (cyclohexanone with both –CH₃ and –COCH₃ at C-2)

Explanation:
1. Given reagents indicate the presence of –COCH₃ group in the starting compound A. Further, since the –COOH group introduced in B due to iodoform reaction is absent in the final product, B should be a β-keto acid. Hence, A should have structure given in option (c).
2. 2-acetyl-2-methylcyclohexanone (A) (i) NaOI
⟶
(ii) H⁺ 1-methyl-2-oxocyclohexanecarboxylic acid (B) Heat
⟶
(–CO₂) 2-methylcyclohexanone

Teacher's Note:
1. Option (d) would also form a β-keto acid, but its decarboxylation gives plain cyclohexanone – the methyl group in the product must already be on C-2.

Answer: (d) E = Ph–CO–^*CH₃ ; F = Ph–COO⁻Na⁺ ; G = ^*CHI₃

Explanation:
1. Ph–CO–^*CH₂–COOH (β-keto acid) Heat
⟶ Ph–CO–^*CH₃ (E) + CO₂
2. Ph–CO–^*CH₃ I₂
⟶
NaOH Ph–COO⁻Na⁺ (F) + ^*CHI₃ (G)

Teacher's Note:
1. In the haloform reaction the methyl carbon of CH₃CO– becomes CHX₃, while the carbonyl carbon stays in the carboxylate.

Answer: (c) propanone and 2,6-dimethyl-2,5-hepta-dien-4-one

Explanation:
1. Since compound A (C₃H₆O) undergoes iodoform test, it must be CH₃COCH₃ (propanone). Further, the compound ‘B’ obtained from ‘A’ has three times more the number of carbon atoms as in ‘A’ (propanone), ‘B’ must be phorone, i.e., 2,6-dimethyl-2,5-heptadien-4-one.
2. (CH₃)₂C=O + H₃CCOCH₃ + O=C(CH₃)₂ (A, propanone, 3 molecules) HCl
⟶ (CH₃)₂C=CHCOCH=C(CH₃)₂ (2,6-dimethyl-2,5-heptadien-4-one)

Teacher's Note:
1. Mesityl oxide (C₆H₁₀O) comes from two acetone molecules; phorone (C₉H₁₄O) from three – use the carbon count to decide.

Answer: (c) Hydrolysis

Explanation:
1. Condensation reaction is the reverse of hydrolysis, which splits a chemical entity into two parts through the action of the polar water molecule.

Teacher's Note:
1. A condensation joins two molecules with loss of a small molecule such as water; hydrolysis adds water back to split them.

Answer: (b) CH₃CH=CHCOCH₃

Explanation:
1. CH₃CHO + CH₃COCH₃ → CH₃CH(OH)CH₂COCH₃ Δ
⟶
–H₂O CH₃CH=CHCOCH₃

Teacher's Note:
1. In a cross aldol, the more reactive aldehyde is usually the one attacked, while the ketone supplies the carbanion.

Answer: (a) Acetophenone and Formaldehyde

Explanation:
1. C₆H₅COCH₃ + HCHO → C₆H₅COCH₂CH₂OH Δ
⟶
–H₂O C₆H₅COCH=CH₂
2. The carbanion of acetophenone adds to formaldehyde (no α-H), and dehydration gives phenyl vinyl ketone.

Teacher's Note:
1. Split the C=C of the enone: the carbon next to C=O came from the ketone’s α-carbon, the other from the aldehyde.

Answer: (d) CH₃CH₂CHO

Explanation:
1. Propanal has α-hydrogens, so it undergoes aldol condensation.
2. HCHO and C₆H₅CHO have no α-hydrogen, and CH₃CH₂OH is not a carbonyl compound.

Teacher's Note:
1. Two conditions are needed for aldol condensation: a carbonyl group and at least one α-hydrogen.

Answer: (d) C₆H₅CHO

Explanation:
1. Benzaldehyde (C₆H₅CHO) with no α-hydrogen cannot undergo aldol condensation.

Teacher's Note:
1. Aldehydes with no α-H, such as benzaldehyde, undergo Cannizzaro reaction with conc. alkali instead.

Answer: (c) A is true, but R is false.

Explanation:
1. In benzaldehyde the +R effect of the phenyl group outweighs its –I effect, so the electron density on the carbonyl carbon increases (it becomes less electrophilic).
2. Hence benzaldehyde is less reactive than ethanal towards nucleophilic attack; the reason as stated is wrong.

Teacher's Note:
1. Resonance of the benzene ring with C=O reduces the partial positive charge on the carbonyl carbon – this is why all aromatic aldehydes are less reactive than aliphatic ones.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. HSO₃⁻ is a bulky nucleophile, hence, cannot attack on sterically hindered ketones. Acetone forms the bisulphite addition product, while di-tert-butyl ketone does not.

Teacher's Note:
1. Bisulphite addition works for aldehydes and most methyl ketones only; bulky groups around C=O block the large nucleophile.

Answer: (d) A is false, but R is true.

Explanation:
1. Aromatic aldehydes and ketones are less reactive than the corresponding aliphatic analogues towards nucleophilic addition reactions due to the +R effect of benzene ring. Further, aldehydes are more reactive than ketones due to +I effect and steric effect of alkyl group.
2. Therefore, the ease of nucleophilic addition will follow the order : CH₃–CO–H > CH₃–CO–CH₃ > C₆H₅COCH₃

Teacher's Note:
1. Aryl ketones are the least reactive carbonyl compounds – they rank below simple aliphatic ketones.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Formation of cyanohydrin from an aldehyde or ketone occurs very slowly with pure HCN because it is feebly ionised. This reaction is catalysed by a base. Base generates CN⁻ ion which is a stronger nucleophile and readily adds to carbonyl compound.
2. OH⁻ + HCN ⇌ CN⁻ + H₂O; >C=O + CN⁻ ⇌ >C(O⁻)CN H⁺
⟶ >C(OH)CN

Teacher's Note:
1. HCN is a weak acid, so very little CN⁻ is present without a base; the base raises the concentration of the real nucleophile.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Electron withdrawing group (–NO₂) increases the reactivity towards nucleophilic addition reactions, whereas electron donating group (–CH₃) decreases the reactivity towards nucleophilic addition reactions.

Teacher's Note:
1. Ring substituents change the electrophilicity of the C=O carbon: –NO₂ increases it, –CH₃ and –OCH₃ decrease it.

Answer: (a) benzyl alcohol and sodium formate

Explanation:
1. It is an example of cross Cannizzaro reaction where aromatic aldehyde gets reduced to alcohol and aliphatic aldehyde gets oxidised to its sodium salt (both aldehydes must not contain any α-hydrogen).
2. C₆H₅CHO + NaOH + HCHO → C₆H₅CH₂OH + HCOONa

Teacher's Note:
1. In a cross Cannizzaro reaction HCHO is always the one oxidised, because it is the most reactive towards OH⁻.

Answer: (c) C₆H₅CHO

Explanation:
1. C₆H₅CHO has no α-hydrogen, so it undergoes Cannizzaro reaction with concentrated NaOH.
2. CH₃CHO and C₆H₅CH₂CHO have α-hydrogens (they give aldol condensation) and CH₃COCH₃ is a ketone.

Teacher's Note:
1. Only aldehydes without α-H give Cannizzaro reaction; ketones never do.

Answer: (a) 2, 2, 2-trichloroethanol

Explanation:
1. The Cannizzaro product of given reaction yields 2, 2, 2-trichloroethanol.
2. 2Cl₃C–CHO NaOH
⟶ Cl₃C–COO⁻ (trichloroacetate ion) + Cl₃C–CH₂OH (2, 2, 2-trichloroethanol)

Teacher's Note:
1. In Cannizzaro reaction one molecule is oxidised to the acid salt and the other is reduced to the alcohol with the same carbon skeleton.

Answer: (b) the transfer of hydride to the carbonyl group

Explanation:
1. Hydride transfer is the slowest step.
2. Ph–CH(OH)–O⁻ + Ph–CO–H slow step
⟶ Ph–CO–OH + Ph–CH₂O⁻

Teacher's Note:
1. Proton transfers are fast; transferring a hydride ion from carbon to carbon is the rate-determining step.

Answer: (a) Cannizzaro reaction

Explanation:
1. C–C bond is not formed in Cannizzaro reaction while other reactions result in the formation of C–C bond.

Teacher's Note:
1. Cannizzaro reaction is a redox (disproportionation) reaction – it only moves a hydride, it does not join carbon skeletons.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. HVZ reaction needs an α-hydrogen; the α-carbon of (CH₃)₃CCOOH carries three CH₃ groups and no hydrogen, so R explains A.

Teacher's Note:
1. Always locate the α-carbon (the one attached to –COOH) and count its hydrogens before predicting HVZ.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Phosphorus converts a little of the acid into acid chloride which is more reactive than the parent carboxylic acid. Thus, it is the acid chloride, not the acid itself, that undergoes chlorination at the α-carbon.
2. Phosphorus does this by first reacting with the halogen to form PX₃, which converts the acid into the acyl halide; so R explains the use of red P in A.

Teacher's Note:
1. Red P is needed only in small amount: the PX₃ formed makes some acyl halide, which enolises and is halogenated at the α-position.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. β-Ketoacids are unstable acids. They readily undergo decarboxylation through a cyclic transition state:
2. C₆H₅–CO–CH₂–COOH → C₆H₅–C(OH)=CH₂ (enol) + CO₂ → C₆H₅–CO–CH₃

Teacher's Note:
1. C₆H₅COCOOH is an α-keto acid; only β-keto acids can form the six-membered cyclic transition state that loses CO₂ easily.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. CH₃–CH(CH₃)–CH₂COOH \( \xrightarrow[\Delta]{\text{NaOH/CaO}} \) CH₃–CH(CH₃)–CH₃ (isobutane) + Na₂CO₃
2. The composition of soda lime is correct but it does not explain why isobutane is formed.

Teacher's Note:
1. Decarboxylation removes one carbon: a C₅ acid gives a C₄ alkane with the same branching.

Answer: (d) All of the above

Explanation:
1. Both aldehydes and ketones can be prepared by all these methods: acyl chlorides give aldehydes with H₂/Pd-BaSO₄ and ketones with dialkylcadmium, alkenes give either on ozonolysis, and 1°/2° alcohols give aldehydes/ketones on oxidation.

Teacher's Note:
1. Ozonolysis gives an aldehyde or a ketone depending on how many alkyl groups sit on each alkene carbon.

Answer: (a) Fehling’s test

Explanation:
1. On heating an aldehyde with Fehling’s reagent, a reddish brown precipitate is obtained. Aldehydes are oxidised to corresponding carboxylate anion. Aromatic aldehydes do not respond to this test.
2. R–CHO + 2Cu²⁺ + 5OH⁻ → R–COO⁻ + Cu₂O (Reddish brown ppt) + 3H₂O

Teacher's Note:
1. Tollens’ test cannot be used here, since both aliphatic and aromatic aldehydes give the silver mirror.

Answer: (c) CH₃COCH₃ [A], CH₃CH₂CH₃ [B]

Explanation:
1. CH₃CN \( \xrightarrow[\text{(ii) H}_{3}\text{O}^{+}]{\text{(i) CH}_{3}\text{MgBr}} \) CH₃–CO–CH₃ (Acetone) [A]
2. CH₃COCH₃ \( \xrightarrow{\text{Zn(Hg)/conc. HCl}} \) CH₃CH₂CH₃ (Propane) [B] + H₂O

Teacher's Note:
1. Nitriles + Grignard reagent followed by hydrolysis give ketones; Clemmensen reduction then keeps all three carbons.

Answer: (c) Fehlings’ solution

Explanation:
1. Fehling’s solution does not react with acetone and benzaldehyde as aromatic aldehydes and ketones do not react with Fehling’s solution.

Teacher's Note:
1. Fehling’s solution reacts only with aliphatic aldehydes (and reducing sugars); ketones and aromatic aldehydes give no red precipitate.

Answer: (a) Grignard reaction

Explanation:
1. The number of C-atoms can be increased in the chain by Grignard reaction.
2. HCHO + CH₃MgBr → H–CH(CH₃)–OMgBr \( \xrightarrow{\text{H}_{2}\text{O/H}^{+}} \) CH₃CH₂OH + Mg(Br)OH

Teacher's Note:
1. Grignard reagents form a new C–C bond; Cannizzaro, Clemmensen and HVZ reactions keep the carbon count unchanged.