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

Single Correct Answer

Question 1. BrMgCH₂CH₂MgBr + CH₃–CO–CH₂–CO–CH₃ → A H₂O
⟶ B. The product B is (JEE Main 2021)
(a)
(b) HO–C₆H₄–OH (p-)
(c)
(d)
Answer:
Ans. (a)
The two C–MgBr ends of the di-Grignard reagent add to the two C=O groups of CH₃COCH₂COCH₃ (in dry ether), closing a five-membered ring; A is the bis(OMgBr) compound.
A H₂O
⟶
−2 Mg(OH)Br 1,3-dimethylcyclopentane-1,3-diol (B). Thus, option (a) is correct.

Teacher's Note:
1. Each ketone C=O that reacts with a Grignard reagent becomes a tertiary alcohol carbon carrying its original groups.

Question 2. What is the major product of the following reaction ?
(CH₃)₂CH–CHO OH⁻
⟶
H₂O (JEE Main 2022)
(a)
(b)
(c)
(d)
Answer:
Ans. (b)
OH⁻ removes the α-H of 2-methylpropanal (−H₂O) to give the carbanion (CH₃)₂C⁻–CHO.
This carbanion attacks the C=O group of a second molecule; the alkoxide formed takes H⁺ from H₂O to give the aldol, (CH₃)₂CH–CH(OH)–C(CH₃)₂–CHO. Thus, option (b) is correct.

Teacher's Note:
1. The aldol here cannot dehydrate to an α,β-unsaturated aldehyde because its α-carbon carries no hydrogen.

Question 3. Reaction between benzaldehyde and acetophenone in presence of dilute NaOH is known as (NEET 2020)
(a) Cannizzaro’s reaction
(b) Cross Cannizzaro’s reaction
(c) Cross Aldol condensation
(d) Aldol condensation
Answer:
Ans. (c)
C₆H₅–CH=O + H₂CH–CO–C₆H₅ −H₂O
⟶ C₆H₅CH=CH–CO–C₆H₅ (Cross-Aldol condensation)
Thus, option (c) is correct.

Teacher's Note:
1. Two different carbonyl compounds, only one of which has α-H, give a cross aldol (Claisen–Schmidt) condensation, not a Cannizzaro reaction.

Question 4. The compound A in the following reaction is
A (i) CH₃MgBr/H₂O
⟶
(ii) Conc. H₂SO₄/Δ B (i) O₃
⟶
(ii) Zn/H₂O C + D
C (i) Conc. KOH
⟶
(ii) Δ C₆H₅COO⁻K⁺ + C₆H₅CH₂OH
D Ba(OH)₂
⟶
Δ (CH₃)₂C=CH–CO–CH₃ (JEE Main 2020)
(a) C₆H₅–CO–CH(CH₃)₂
(b) C₆H₅–CO–CH₂CH₃
(c) C₆H₅–CH₂–CO–CH₃
(d) C₆H₅–CO–CH₃
Answer:
Ans. (c)
(i) Since compound (D) on treatment with Ba(OH)₂ followed by heating undergoes aldol condensation to give mesityl oxide, (CH₃)₂C=CH–COCH₃, therefore (D), must be acetone.
2CH₃COCH₃ Ba(OH)₂
⟶
Aldol condensation (CH₃)₂C(OH)–CH₂–CO–CH₃ Δ
⟶
−H₂O (CH₃)₂C=CH–CO–CH₃ (Mesityl oxide)
(ii) Since (C) on treatment with conc. KOH, followed by heating undergoes Cannizzaro reaction to give a mixture of pot. benzoate and benzyl alcohol, therefore, (C) must be benzaldehyde.
(iii) Since benzaldehyde (C) and acetone (D) are obtained by reductive ozonolysis of (B), therefore, (B) must be C₆H₅CH=C(CH₃)₂.
C₆H₅CH=C(CH₃)₂ (B) (i) O₃
⟶
(ii) Zn/H₂O C₆H₅CH=O (C) + O=C(CH₃)₂ (D)
(iv) Since B [C₆H₅CH=C(CH₃)₂] is obtained from compound (A) by addition of CH₃MgBr followed by heating with conc. H₂SO₄, therefore, (A) must be benzyl methyl ketone.
C₆H₅CH₂COCH₃ (A) (i) CH₃MgBr
⟶
(ii) H₂O C₆H₅CH₂C(CH₃)₂OH Conc. H₂SO₄
⟶
(−H₂O) C₆H₅CH=C(CH₃)₂ (B)
Thus, option (c) is correct.

Teacher's Note:
1. Work backwards: identify the ozonolysis fragments first, join them through a C=C, then remove the CH₃ added by the Grignard reagent.

Question 5. C₆H₅CH₃ (i) CrO₂Cl₂
⟶
(ii) H⁺/H₂O X (C₂H₅O)₃Al
⟶ Y (Major Product)
Here, Y is
(a)
(b) C₆H₅–CO–OCH₂C₆H₅
(c)
(d)
Answer:
Ans. (b)
Toluene (i) CrO₂Cl₂, (ii) H⁺/H₂O
⟶
Etard reaction C₆H₅CHO (X)
With aluminium ethoxide two molecules of benzaldehyde undergo Tischenko reaction: [C₆H₅–CO–OH + H–O–CH₂–C₆H₅] −H₂O
⟶ C₆H₅–CO–OCH₂C₆H₅
Thus, option (b) is correct.

Teacher's Note:
1. In the Tischenko reaction one aldehyde molecule is oxidised and one reduced, as in Cannizzaro, but they end up joined as an ester.

Question 6. An unsaturated hydrocarbon X absorbs two hydrogen molecules on catalytic hydrogenation, and also gives following reactions :
X O₃
⟶
Zn/H₂O A [Ag(NH₃)₂]⁺
⟶ B (3-oxohexanedicarboxylic acid)
What is ‘X’ ? (JEE Main 2020)
(a)
(b)
(c)
(d)
Answer:
Ans. (a)
Since product (B) is a dicarboxylic acid with an oxo group at position-3, therefore, one of the double bonds is exocyclic which is at position-3 w.r.t. the endocyclic double bond.
X (i) O₃
⟶
(ii) Zn/H₂O OHC–CH₂–CO–CH₂–CH₂–CHO (A) + HCHO
A [Ag(NH₃)₂]⁺
⟶
oxidation HOOC–CH₂–CO–CH₂–CH₂–COOH (B, 3-oxohexanedioic acid)
Thus, option (a) is correct.

Teacher's Note:
1. The carbon that ends up as the ketone C=O is the one that carried the exocyclic =CH₂; the CH₂ group itself is lost as HCHO.

Question 7. Benzaldehyde and acetone can be best distinguished using
(a) Fehling’s solution
(b) sodium hydroxide solution
(c) 2, 4-DNP
(d) Tollens’ reagent
Answer:
Ans. (d)
Benzaldehyde reduces Tollens’ reagent but not Fehling’s solution.
Acetone, being a ketone, does not reduce Tollens’ reagent, while both give a precipitate with 2,4-DNP.

Teacher's Note:
1. Aromatic aldehydes give the Tollens’ test but not the Fehling’s test, so Fehling’s cannot tell benzaldehyde from a ketone.

Question 8. The product of acid hydrolysis of P and Q can be distinguished by
P = CH₂=C(CH₃)–OCOCH₃    Q = CH₃–CH=CH–OCOCH₃
(a) Lucas reagent
(b) 2, 4-DNP
(c) Fehling’s solution
(d) NaHSO₃
Answer:
Ans. (c)
P H⁺/H₂O
⟶ H₂C=C(OH)CH₃ Tautomerises
⟶ CH₃–CO–CH₃ (Acetone)
Q H⁺/H₂O
⟶ CH₃CH=CH–OH Tautomerises
⟶ H₃C–CH₂–CHO (Propanal)
Acetone (ketones are non-reducing) and propanal (aldehydes are reducing) can be distinguished by Fehling solution. Both of these, however, form NaHSO₃ adducts.

Teacher's Note:
1. Hydrolysis of an enol ester gives an enol, which immediately tautomerises to the carbonyl compound.

Question 9. Compound ‘A’ (molecular formula C₃H₈O) is treated with acidified potassium dichromate to form a product ‘B’ (molecular formula C₃H₆O). ‘B’ forms a shining silver mirror on warming with ammoniacal silver nitrate. ‘B’ when treated with an aqueous solution of H₂NCONHNH₂.HCl and sodium acetate gives a product ‘C’. Identify the structure of ‘C’.
(a) CH₃CH₂CH=NNHCONH₂
(b) CH₃–C(CH₃)=NNHCONH₂
(c) CH₃–C(CH₃)=NCONHNH₂
(d) CH₃CH₂CH=NCONHNH₂
Answer:
Ans. (a)
Since (B), M.F. C₃H₆O forms a shining silver mirror on warming with ammonical AgNO₃ (Tollens’ reagent), (B) must be an aldehyde, i.e., CH₃CH₂CHO and ‘A’ with M.F. C₃H₈O must be a 1° alcohol, i.e., CH₃CH₂CH₂OH.
During the reaction of ‘B’ with H₂NCONHNH₂.HCl, the more nucleophilic NH₂ i.e., NH₂ next to NH reacts giving product (C), i.e., option (a) is correct.
CH₃CH₂CH₂OH (A) K₂Cr₂O₇
⟶
H₂SO₄ CH₃CH₂CHO (B) H₂NCONHNH₂.HCl
⟶
Sod. acetate (−HCl, −H₂O) CH₃CH₂CH=NNHCONH₂ (C)

Teacher's Note:
1. In semicarbazide only the terminal NH₂ of the hydrazine part reacts; the CONH₂ nitrogen is deactivated by the C=O group.

Question 10. (CH₃)₂C=CHCOCH₃ can be oxidised to (CH₃)₂C=CHCOOH by :
(a) Chromic acid
(b) NaOI
(c) Cu at 300°C
(d) KMnO₄
Answer:
Ans. (b)
Only NaOI (iodoform reaction) will oxidise COCH₃ to CHI₃ while all other reagents will bring about oxidation both at the double bond as well as the keto group.
(CH₃)₂C=CHCOCH₃ + 3NaOI → (CH₃)₂C=CHCOONa + CHI₃ + 2NaOH; acidification gives (CH₃)₂C=CHCOOH.

Teacher's Note:
1. The haloform reaction is a clean way to turn a methyl ketone into an acid with one carbon less, leaving C=C bonds untouched.

Question 11. In the following reaction sequence, the correct structures of E, F and G are
Ph–CO–*CH₂–COOH Heat
⟶ [E] I₂
⟶
NaOH [F] + [G]
(* implies ¹³C labelled carbon) (IIT Paper II, 2008)
(a) E = Ph–*CO–CH₃, F = Ph–*COO⁻Na⁺, G = CHI₃
(b) E = Ph–CO–*CH₃, F = Ph–COO⁻Na⁺, G = CHI₃
(c) E = Ph–CO–*CH₃, F = Ph–COO⁻Na⁺, G = *CHI₃
(d) E = Ph–CO–*CH₃, F = Ph–*COO⁻Na⁺, G = CH₃I
Answer:
Ans. (c)
Ph–CO–*CH₂–COOH (a β-keto acid) Heat (−CO₂)
⟶
Decarboxylation Ph–CO–*CH₃ (Acetophenone, E)
Ph–CO–*CH₃ I₂/NaOH
⟶
Iodoform reaction Ph–COO⁻Na⁺ (F) + *CHI₃ (G)

Teacher's Note:
1. β-Keto acids lose CO₂ easily on heating; the CO₂ comes from the COOH carbon, so the labelled CH₂ becomes the CH₃ that ends up in iodoform.

Question 12. After completion of reactions (I and II), the organic compound(s) in the reaction mixture is (are)
Reaction I : CH₃COCH₃ (1.0 mole) Br₂ (1.0 mole)
⟶
aqueous NaOH
Reaction II : CH₃COCH₃ (1.0 mole) Br₂ (1.0 mole)
⟶
CH₃COOH
P = CH₃COCH₂Br; Q = CH₃COCBr₃; R = CBr₃COCBr₃; S = BrCH₂COCH₂Br; T = CH₃COONa; U = CHBr₃ (JEE Advanced 2013)
(a) Reaction I : P and Reaction II : P
(b) Reaction I : U, acetone and Reaction II : Q, acetone
(c) Reaction I : T, U, acetone and Reaction II : P
(d) Reaction I : R, acetone and Reaction II : S, acetone
Answer:
Ans. (c)
Under basic conditions, ‘Haloform reaction’ occurs to give CHBr₃ (U) and CH₃COONa (T). Since for completion of haloform reaction, we need 3 moles of Br₂ but only one mole of Br₂ is given, therefore, some of the acetone will remain unused. However, under acidic conditions, the reaction stops at the monobromination stage. Thus,
Reaction I : CH₃COCH₃ (1.0 mole) Br₂ (1.0 mole)
⟶
NaOH (aq), Haloform reaction CHBr₃ (U) + CH₃COONa (T)
Reaction II : CH₃COCH₃ Br₂ (1.0 mole)
⟶
CH₃COOH CH₃COCH₂Br (P) + HBr
So option (c) is correct.

Teacher's Note:
1. In base each Br makes the remaining α-H more acidic, so halogenation runs to CX₃; in acid the first Br slows further enolisation, so it stops at one.

Question 13. The major product in the following reaction is
NaOH
⟶
Δ (JEE Main 2019)
(a)
(b)
(c)
(d)
Answer:
Ans. (c)
Although carbanion can be formed either at CH₃ group or at CH₂ group, since seven-membered rings are difficult to be formed, therefore, carbanion formation occurs at CH₂ group.
Nucleophilic attack of the carbanion (I) on the CH=O group gives cyclized product (II) which upon loss of H₂O gives α,β-unsaturated ketone (III), 1-acetyl-4-methylcyclopentene (option c).

Teacher's Note:
1. In intramolecular aldol reactions the ring size decides the site of the carbanion: five- and six-membered rings are formed preferentially.

Question 14. The cross aldol product formed when propanal acts as the electrophile and butanal as nucleophile is
(a) 3-hydroxy-2-methylpentanal
(b) 3-hydroxy-2-methylhexanal
(c) 2-ethyl-3-hydroxypentanal
(d) 2-ethyl-3-hydroxyhexanal
Answer:
Ans. (c)
The aldehyde which forms the carbanion acts as the nucleophile.
CH₃CH₂CH₂CHO (Butanal) OH⁻
⟶
−H₂O CH₃CH₂C⁻H–CHO (Carbanion, nucleophile)
CH₃CH₂CH=O (Propanal, electrophile) + ⁻CH(CH₂CH₃)–CHO H₂O
⟶
−OH⁻ CH₃CH₂–CH(OH)–CH(CH₂CH₃)–CHO
i.e., 2-Ethyl-3-hydroxypentanal.

Teacher's Note:
1. The nucleophile contributes its α-carbon with the CHO group (C-1, C-2); the electrophile’s carbonyl carbon becomes the CH(OH) carbon.

Question 15. Which one of the following is not formed when acetone reacts with 2-pentanone in the presence of dilute NaOH followed by heating ? (NEET 2022)
(a)
(b)
(c)
(d)
Answer:
Ans. (a)
Self aldol condensation of acetone followed by heating gives product reported in option (d): 2CH₃COCH₃ Dil. NaOH
⟶ (CH₃)₂C(OH)CH₂COCH₃ Δ
⟶
−H₂O (CH₃)₂C=CHCOCH₃
Cross Aldol condensation involving nucleophilic attack by acetone carbanion on 2-pentanone followed by dehydration on heating gives product reported in option (b): CH₃CH₂CH₂C(OH)(CH₃)CH₂COCH₃ Δ
⟶
−H₂O CH₃CH₂CH₂C(CH₃)=CHCOCH₃
Cross aldol condensation involving nucleophilic attack by C₃-carbanion of 2-pentanone on acetone followed by heating gives product reported in option (c): (CH₃)₂C(OH)CH(C₂H₅)COCH₃ Δ
⟶
−H₂O (CH₃)₂C=C(C₂H₅)COCH₃
Product (a) is the self aldol condensation product of 3-pentanone and hence cannot be obtained by self-aldol condensation of 2-pentanone. Thus, option (a) is correct.

Teacher's Note:
1. List every possible enolate (from each α-carbon of each ketone) and every electrophile; any product that needs a carbonyl compound not present is impossible.

Question 16. Aldol condensation between which of the following two compounds followed by dehydration gives methyl vinyl ketone ? (AIIMS 2013)
(a) Formaldehyde and acetone
(b) Formaldehyde and acetaldehyde
(c) Two molecules of acetaldehyde
(d) Two molecules of acetone
Answer:
Ans. (a)
CH₂=O (Formaldehyde) + CH₃COCH₃ (Acetone) OH⁻
⟶
Cross aldol condensation HOCH₂–CH₂–COCH₃ H⁺/H₂O, Δ
⟶
−H₂O CH₂=CH–CO–CH₃ (Methyl vinyl ketone)

Teacher's Note:
1. Split the C=C of the product: the CH₂= part comes from the electrophile (HCHO) and the –CH–CO–CH₃ part from the enolate of acetone.

Question 17. Identify the combination of compounds that undergo Aldol condensation followed by dehydration to produce but-2-enal.
(a) methanal and ethanal
(b) two moles of ethanal
(c) methanal and propanone
(d) ethanal and propanone
Answer:
Ans. (b)
2CH₃CHO (Ethanal) Dil. NaOH
⟶
Aldol condensation CH₃–CHOH–CH₂CHO H⁺, Δ
⟶
−H₂O CH₃CH=CHCHO (But-2-enal)

Teacher's Note:
1. But-2-enal has four carbons, the sum of two C₂ aldehydes; crotonaldehyde is the classic aldol product of acetaldehyde.

Question 18. Of the following, which is the product formed when cyclohexanone undergoes aldol condensation followed by heating ? (NEET 2017)
(a)
(b)
(c)
(d)
Answer:
Ans. (a)
OH⁻ removes an α-H of cyclohexanone (−H₂O); the carbanion attacks the C=O group of a second molecule of cyclohexanone.
The alkoxide takes H⁺ from H₂O (−OH⁻) to give the aldol, 2-(1-hydroxycyclohexyl)cyclohexanone, which on heating loses H₂O to give 2-cyclohexylidenecyclohexanone.
Thus, option (a) is correct.

Teacher's Note:
1. In the aldol product the OH is on the carbon that was the C=O of the electrophile molecule; dehydration then puts the C=C exactly there.

Question 19. Which of the following will be dehydrated most readily in alkaline medium ? (AIPMT Mains 2010 ; West Bengal JEE 2016)
(a)
(b)
(c)
(d)
Answer:
Ans. (b)
Aldols, i.e., β-hydroxyaldehydes or β-hydroxyketones (under alkaline or acidic conditions) readily undergo dehydration to form α,β-unsaturated aldehydes or ketones.
OH⁻ removes the acidic α-H (−H₂O) to give a carbanion, which loses OH⁻ (E₁cB) to give CH₃COCH=CHCH₃.

Teacher's Note:
1. Base-promoted dehydration needs an acidic α-H on the carbon next to the C–OH carbon, i.e., a β-hydroxy carbonyl compound.

Question 20. The diketone CH₃–CO–(CH₂)₂–CO–CH₃ on intramolecular aldol condensation gives the final product
(a)
(b)
(c)
(d)
Answer:
Ans. (c)
Aldols are β-hydroxyaldehydes which upon dehydration give α,β-unsaturated aldehydes/ketones.
CH₃–CO–(CH₂)₂–CO–CH₃ OH⁻
⟶
−H₂O CH₃–CO–(CH₂)₂–CO–CH₂⁻; the carbanion attacks the other C=O group to form a five-membered ring.
H₂O (−OH⁻) gives 3-hydroxy-3-methylcyclopentanone, which on loss of H₂O gives 3-methylcyclopent-2-enone.

Teacher's Note:
1. Hexane-2,5-dione can only close a five-membered ring through its CH₃ carbanion; a three-membered ring through the CH₂ carbanion is not favoured.

Question 21. Cyclohexene on ozonolysis followed by reaction with zinc dust and water gives compound E. Compound E on further treatment with aqueous KOH yields compound F. Compound F is (IIT Paper II, 2007)
(a)
(b)
(c)
(d) HO₂C–CH₂CH₂CH₂CH₂–CO₂H
Answer:
Ans. (a)
Cyclohexene (i) Ozonolysis
⟶
(ii) Zn/H₂O OHC–CH₂CH₂CH₂CH₂–CHO (Hexane-1,6-dial, E)
With aq. KOH, the α-carbanion of one CHO group attacks the other CHO group (intramolecular aldol condensation) giving 2-hydroxycyclopentanecarbaldehyde, which loses H₂O to form cyclopent-1-enecarbaldehyde (F).

Teacher's Note:
1. The aldol product dehydrates so that the C=C is conjugated with the CHO group, i.e., the double bond is at C-1 of the ring.

Question 22. In the following reaction, the product E is
OHC–CHO 1. NaOH
⟶
2. H⁺ E
(a) CH₂OH–CHO
(b) CHO–CO₂H
(c) CH₂OH–CO₂H
(d) CO₂H–CO₂H
Answer:
Ans. (c)
OHC–CHO 1. NaOH
⟶ CH₂OH–COONa 2. H⁺
⟶ CH₂OH–COOH
Glyoxal has no α-H and undergoes intramolecular Cannizzaro reaction: one CHO is oxidised and the other reduced.

Teacher's Note:
1. In an intramolecular Cannizzaro reaction one CHO becomes CH₂OH and the other COO⁻ in the same molecule.

Question 23. The product of the following reaction is
(i) Conc. NaOH
⟶
(ii) H⁺/H₂O
(a)
(b)
(c)
(d)
Answer:
Ans. (b)
Phthalaldehyde has no α-H; with conc. NaOH it undergoes intramolecular Cannizzaro reaction: one CHO is oxidised to COO⁻ and the other is reduced to CH₂OH.
On acidification the 2-(hydroxymethyl)benzoic acid formed loses water (intramolecular esterification) to give the lactone, phthalide.

Teacher's Note:
1. A γ-hydroxy acid (OH and COOH placed to form a five-membered ring) lactonises spontaneously in acid.

Question 24. In the following species, the one which is likely to be intermediate during benzoin condensation of benzaldehyde, is
(a) Ph–C≡O⁺
(b) Ph–C⁺(OH)(CN)
(c) Ph–C⁻(OH)(CN)
(d) Ph–C⁻=O
Answer:
Ans. (c)
Ph–CH=O + CN⁻ → Ph–CH(O⁻)–CN (Less stable) Proton transfer
⟶ Ph–C⁻(OH)–CN (More stable)
This carbanion then attacks the C=O group of a second benzaldehyde molecule.

Teacher's Note:
1. The cyanide ion makes the aldehyde carbon nucleophilic: its carbanion is stabilised by both the CN group and the phenyl ring.

Question 25. In the Cannizzaro reaction given below,
2 Ph–CHO OH⁻
⟶ Ph–CH₂OH + PhCO₂⁻,
the slowest step is (AIEEE 2009)
(a) the attack of OH⁻ at the carbonyl group
(b) the transfer of hydride ion to the carbonyl group
(c) the abstraction of a proton from the carboxylic acid
(d) the deprotonation of Ph–CH₂OH.
Answer:
Ans. (b)
OH⁻ adds rapidly to the C=O group of one molecule of benzaldehyde to give Ph–CH(OH)O⁻.
This anion transfers a hydride ion to the C=O group of a second molecule; this hydride transfer is the slow (rate-determining) step. The proton exchanges that follow are fast.

Teacher's Note:
1. Proton transfers are always fast; in Cannizzaro the slow step is the intermolecular hydride shift.

Question 26. Which of the following combination of aldehydes gives cross Cannizzaro reaction ?
(a) CH₃CHO, HCHO
(b) C₆H₅CHO, CH₃CHO
(c) C₆H₅CHO, HCHO
(d) all of these
Answer:
Ans. (c)
Both the aldehydes should not have α-hydrogens, i.e., C₆H₅CHO and HCHO.

Teacher's Note:
1. If either aldehyde has an α-H, aldol condensation takes place instead of a Cannizzaro reaction.

Question 27. The correct structure of product ‘H’ formed in the following reaction is
PhCHO NaOD
⟶
in D₂O H + Ph–CO–O⁻ (JEE Main 2022)
(a) Ph–CH₂–OD
(b) Ph–CH₂–OH
(c) Ph–CHD–OD
(d) Ph–CD₂–OH
Answer:
Ans. (a)
OD⁻ adds to the C=O group of one PhCHO molecule to give Ph–CH(OD)O⁻, which transfers its hydride ion (H, not D) to the C=O group of a second molecule.
This gives Ph–CO–OD + Ph–CH₂–O⁻; the alkoxide takes D⁺ from D₂O to give Ph–CH₂–OD and the acid gives Ph–CO–O⁻. Thus, option (a) is correct.

Teacher's Note:
1. The H that moves in Cannizzaro is the aldehyde’s own C–H; only the O–H position picks up deuterium from the solvent.

Question 28. Which of the following reagents reacts differently with HCHO, CH₃CHO and CH₃COCH₃ ?
(a) HCN
(b) NH₂NH₂
(c) NH₂OH
(d) NH₃
Answer:
Ans. (d)
NH₃ reacts differently. With HCHO, it gives urotropine ; with acetaldehyde, it gives CH₃CH=NH and with acetone, it gives diacetone amine.

Teacher's Note:
1. HCN, NH₂NH₂ and NH₂OH give the same type of product (cyanohydrin, hydrazone, oxime) with all three compounds.

One or More Correct Answers

Question 29. Which of the following reagents are used for detecting the presence of carbonyl group ?
(a) NH₂OH
(b) NH₂NH₂
(c) H₂NCONHNH₂.HCl
(d) C₆H₅NHNH₂.HCl
Answer:
Ans. (a, b, c, d)
All these ammonia derivatives react with the carbonyl group (addition followed by elimination of water) to give oximes, hydrazones, semicarbazones and phenylhydrazones respectively, which are crystalline solids with sharp melting points.

Teacher's Note:
1. The salts (hydrochlorides) are used with sodium acetate, which frees the base and keeps the pH mildly acidic.

Question 30. Positive Tollens’ test is observed for (JEE Advanced 2016)
(a) CH₂=CH–CHO
(b) C₆H₅CHO
(c) Ph–CH(OH)–CO–Ph
(d) Ph–CH=CH–CO–Ph
Answer:
Ans. (a, b, c)
α,β-Unsaturated aldehydes, aromatic aldehydes and α-hydroxyketones all give Tollens’ test but ketones (including α,β-unsaturated) do not give this test. Thus, options (a, b and c) are correct.

Teacher's Note:
1. α-Hydroxy ketones such as benzoin are easily oxidised to 1,2-diketones, so they reduce Tollens’ reagent even though they are ketones.

Passage-Based Questions

Question 31. Comprehension 1: In chemical analysis, the presence of a methyl ketonic group is tested with I₂ in presence of an alkali like NaOH or KOH. In another chemical analysis, aldehyde group is tested either by Tollens’ reagent (AgNO₃ + NH₄OH) which gives silver mirror or by Fehling’s solution (CuSO₄ + NaOH + Roschelle salt) which gives red ppt. of copper (I) oxide.
Fehling solution will oxidise
I. C₆H₅CHO   II. C₆H₅CH(OH)CH₃   III. C₆H₅CH₂CHO   IV. HCHO
(a) All
(b) Only I and IV
(c) Only II and IV
(d) Only III and IV
Answer:
Ans. (d)
Fehling’s solution oxidises only aliphatic aldehydes, III and IV.
C₆H₅CH₂CHO has its CHO group on an aliphatic carbon, so it behaves as an aliphatic aldehyde; C₆H₅CHO is aromatic and C₆H₅CH(OH)CH₃ is an alcohol.

Teacher's Note:
1. Phenylacetaldehyde is an aliphatic aldehyde (CHO not attached to the ring) and does give the Fehling’s test.

Matrix Match Questions

Teacher's Note:
1. Remember the reagent fingerprints: CO + HCl for Gattermann–Koch, NaOX for haloform, X₂/red P for HVZ.

Integer and Numerical Answer Questions

Question 33. How many of the compounds listed below undergo Cannizzaro reaction ?
Methanal, 2-methylpentanal, benzaldehyde, benzophenone, cyclohexanone, 1-phenyl-propanone, phenylacetaldehyde, butan-1-ol and 2, 2-dimethylbutanal.
Answer:
Ans. 3
Three aldehydes, i.e., methanal, benzaldehyde and 2, 2-dimethylbutanal do not contain α-hydrogens and hence undergo Cannizzaro reaction.

Teacher's Note:
1. Cannizzaro needs an aldehyde without α-H; benzophenone has no α-H but is a ketone, so it does not react.

Question 34. How many of the isomeric ketones having the molecular formula C₆H₁₂O undergo iodoform test ?
Answer:
Ans. 4
Four isomeric ketones, i.e., CH₃COCH₂CH₂CH₂CH₃, CH₃COCH(CH₃)CH₂CH₃, CH₃COCH₂CH(CH₃)₂ and CH₃COC(CH₃)₃ undergo iodoform test.

Teacher's Note:
1. Only methyl ketones (CH₃CO–) give the iodoform test, so count the C₄H₉ groups that can be attached to CH₃CO–.

Question 35. Consider all possible isomeric ketones, including stereoisomers of MW = 100. All these isomers are independently reacted with NaBH₄ (NOTE : stereoisomers are also reacted). The total number of ketones that give a racemic product(s) is/are (JEE Advanced 2014)
Answer:
Ans. 5
General formula of ketones = C_nH_2nO; MW of the ketone = 100
∴ 12 × n + 1 × 2n + 1 × 16 = 100 or 14n = 84 or n = 6. ∴ M.F. of the ketone = C₆H₁₂O.
The total number of isomeric ketones having the M.F. C₆H₁₂O is six: (i) CH₃CH₂CH₂CH₂COCH₃ (ii) CH₃CH₂*CH(CH₃)COCH₃ (iii) (CH₃)₂CHCH₂COCH₃ (iv) (CH₃)₃CCOCH₃ (v) CH₃CH₂CH₂COCH₂CH₃ (vi) (CH₃)₂CHCOCH₂CH₃
Whenever a new chiral centre is created by reaction of achiral compounds with achiral reagents, a racemic mixture of products is always produced. But chiral compounds under these conditions will produce diastereomeric optically active products.
Since there are five achiral ketones, therefore, on reduction with NaBH₄ will produce five racemic mixture of products.

Teacher's Note:
1. Reduction of an achiral ketone R–CO–R′ creates a new stereocentre equally from both faces, giving a racemic alcohol.

Question 36. How many different carboxylic acids are obtained when all the isomeric arenes having the molecular formula C₈H₁₀ are oxidised with alk. KMnO₄ followed by acidification ?
Answer:
Ans. 4
Four isomeric carboxylic acids are formed from four different arenes : ethylbenzene → benzoic acid, o-xylene → phthalic acid, m-xylene → isophthalic acid and p-xylene → terephthalic acid.

Teacher's Note:
1. Any alkyl side chain with a benzylic H is oxidised to a single COOH, so ethylbenzene gives benzoic acid, not phenylacetic acid.

Assertion–Reason Questions

Question 37. Statement-1: Esters on reduction with LiAlH₄ give alcohols while amides give primary amines.
Statement-2: Alkoxide ion is a better leaving group than amide ion.
(a) Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1
(b) Statement-1 is True, Statement-2 is True ; Statement-2 is not a correct explanation for Statement-1
(c) Statement-1 is True, Statement-2 is False
(d) Statement-1 is False, Statement-2 is True
Answer:
Ans. (a)
With an ester, H⁻ adds to C=O and the tetrahedral intermediate expels the alkoxide ion (a good leaving group) to give an aldehyde, which is further reduced to a 1° alcohol.
With an amide, the amide ion (NH₂⁻) is a very poor leaving group, so oxygen (as O–Al) is lost instead, giving an iminium ion that is reduced to the amine, RCH₂NH₂. Statement-2 explains Statement-1.

Teacher's Note:
1. LiAlH₄ converts RCOOR′ into RCH₂OH (+ R′OH) but RCONH₂ into RCH₂NH₂, i.e., the C=O becomes CH₂ in amides.

Question 38. Statement-1: The esterification of a carboxylic acid with an alcohol is a nucleophilic acyl substitution.
Statement-2: Electron-withdrawing groups in the carboxylic acid will increase the rate of esterification reaction.
(a) Both statement-1 and statement-2 are correct.
(b) Both statement-1 and statement-2 are incorrect.
(c) Statement-1 is correct but statement-2 is incorrect.
(d) Statement-1 is incorrect but statement-2 is correct.
Answer:
Ans. (a)
The alcohol (nucleophile) attacks the protonated C=O group and OH of the acid is replaced by OR (nucleophilic acyl substitution), so statement-1 is correct.
Electron-withdrawing groups make the carbonyl carbon more electrophilic and so increase the rate of nucleophilic attack; statement-2 is also correct.

Teacher's Note:
1. In Fischer esterification the OH lost as water comes from the acid and the OR group from the alcohol.

Question 39. Assertion: Benzaldehyde is less reactive than ethanal towards nucleophilic attack.
Reason: All the carbon atoms of benzaldehyde are sp²-hybridized. (AIIMS 2015)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (b)
Correct explanation. Due to electron-donating effect of the benzene ring, the carbonyl carbon of benzaldehyde is less electrophilic.
Reason is true but it does not explain the assertion.

Teacher's Note:
1. Resonance of the ring with C=O reduces the positive charge on the carbonyl carbon of aromatic aldehydes.

Question 40. Assertion: Both Grignard reagent and dialkylcadmium react with acid chlorides to form tert-alcohols.
Reason: Grignard reagents are as reactive as dialkylcadmium. (AIIMS 2011)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (d)
Correct assertion. Grignard reagents react with acid chlorides to give tert-alcohols but dialkylcadmium gives ketones.
Correct reason. Grignard reagents are more reactive than dialkylcadmium.

Teacher's Note:
1. Dialkylcadmium (R₂Cd) is used to make ketones from acid chlorides precisely because it is too weak to attack the ketone formed.

Question 41. Assertion: 2-Butenal lacks enolisable H-atom, α- to carbonyl group, still it has sufficient acidic character.
Reason: The conjugate base of 2-butenal is stabilised by resonance. (AIIMS 2009)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
Reason is the correct explanation of the assertion.
[H–CH₂–CH=CH–CH=O OH⁻
⟶
−H₂O ⁻CH₂–CH=CH–CH=O ↔ CH₂=CH–CH=CH–O⁻]; the conjugate base is stabilized by resonance.

Teacher's Note:
1. The acidity of a γ-H in an α,β-unsaturated carbonyl compound is called vinylogy: the C=C relays the effect of C=O.

Question 42. Assertion: 2, 2-Dimethylpropanal gives Cannizzaro reaction with concentrated NaOH.
Reason: Cannizzaro is a disproportionation reaction. (AIIMS 2010)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (b)
Correct explanation. 2, 2-Dimethylpropanal does not contain α-H and hence in presence of conc. alkali, it undergoes Cannizzaro reaction.
Reason is true but is not the explanation.

Teacher's Note:
1. Ask ‘does it have an α-H?’ first: no α-H means Cannizzaro; α-H means aldol.

Question 43. Assertion: Aldol condensation is usually carried out in dilute solution of a strong base.
Reason: Concentrated solution of strong base involves Cannizzaro reaction. (AIIMS 2014)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (c)
Correct reason. Only aldehydes which do not contain α-hydrogens on treatment with a concentrated solution of a base undergo Cannizzaro reaction.

Teacher's Note:
1. Aldehydes with α-H give resinous products, not Cannizzaro products, with concentrated alkali.