NEET Previous Year Questions (2014-2025): Aldehydes, Ketones & Carboxylic Acids

Ans: (d)

• For the given reaction:
  R-C=O-OR' + DIBAL-H → R-C=O-H + ROH
  
• In this case, DIBAL-H selectively reduces the ester (R-C=O-OR') to an aldehyde (R-C=O-H) and the alcohol group (ROH) is released.
• The correct reagent for the conversion is Option 2 (AlH(iBu)₂, H₂O). DIBAL-H reduces the ester to the aldehyde, and water (H₂O) is used to hydrolyze the intermediate to the final aldehyde product.

Why not the other options:

• Option 1: LiAlH₄ (Lithium aluminium hydride) followed by H₂O: LiAlH₄ is a strong reducing agent that reduces esters to alcohols rather than aldehydes. It would not stop at the aldehyde stage, which makes it unsuitable for this transformation.
• Option 3: NaBH₄ (Sodium borohydride) followed by H₂O: NaBH₄ is a milder reducing agent than LiAlH₄ and typically reduces aldehydes and ketones to alcohols, but it is not reactive enough to reduce esters to aldehydes.
• Option 4: H₂/Pd-BaSO₄ (Hydrogen and palladium on barium sulfate): This is a catalytic hydrogenation method, typically used to reduce alkenes or aromatic compounds. It will reduce aldehydes to alcohols, so it would not be selective enough to stop at the aldehyde stage when reducing esters.

Therefore, the correct answer is AlH(iBu)₂, H₂O, which reduces the ester to an aldehyde selectively.

Ans: (c)

• HCOOH (formic acid): No alkyl group, highest acidity.
• CH₃COOH (acetic acid): One methyl group (-CH₃), moderately acidic. (+I)
• (CH₃)₂CHCOOH (isobutyric acid): Two methyl groups attached to the same carbon, further decreases acidity. ( two +I)
• (CH₃)₃CCOOH (pivalic acid): Three methyl groups attached to the same carbon, lowest acidity due to strong electron-donating effect. (three +I)
• The correct order of decreasing acidity is:
• HCOOH > CH₃COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH

Therefore, the correct answer is Option 3: HCOOH > CH₃COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH.

Ans: (d)

Ans: (a)
Fehling's solution is a chemical reagent used to differentiate between water-soluble carbohydrates and ketone-functional groups, and as a test for monosaccharides. It is used specially in the Fehling's test for reducing sugars. The test involves two solutions, generally known as Fehling's A and Fehling's B, which are mixed together and added to perform the test.
Fehling's solution 'A' is an aqueous solution of copper(II) sulfate. Thus, the correct option is:
Option A: aqueous copper sulphate
The function of Fehling's A is to provide copper(II) ions, Cu²⁺, which act as the oxidizing agent in the reaction with the reducing sugar. When Fehling's Solution A and B are mixed and heated with a reducing sugar, the copper(II) ions are reduced to copper(I) oxide, which precipitates as a red solid, indicating a positive result.
Fehling's Solution B, on the other hand, contains alkaline sodium potassium tartrate, which helps to maintain the solution in an alkaline condition and keeps the copper(II) ions in solution.
Thus, in summary, Fehling's A consists of aqueous copper sulfate, which directly matches with Option A.

Ans: (c)

The reconstructed reactions align with the reagents' typical behavior in organic chemistry for aldehydes, ketones, and carboxylic acids. PCC selectively oxidizes to aldehydes, CrO₃ and KMnO₄ oxidize primary alcohols to carboxylic acids, and K₂Cr₂O₇ oxidizes secondary alcohols to ketones.

Ans: (b)

Ans: (c)
Statement I: Benzoic acid produces effervescence on treatment with aqueous NaHCO₃.
This statement is correct. Benzoic acid (C₆H₅COOH) is a carboxylic acid. When it reacts with sodium bicarbonate (NaHCO₃), it undergoes a neutralization reaction, releasing carbon dioxide (CO₂) gas, which causes effervescence. The reaction is:
C₆H₅COOH + NaHCO₃ → C₆H₅COONa + CO₂ + H₂O

Statement II: The effervescence is due to the release of hydrogen gas.
This statement is incorrect. The effervescence observed during the reaction of benzoic acid with sodium bicarbonate is due to the release of carbon dioxide (CO₂) gas, not hydrogen gas. The reaction between an acid and a bicarbonate typically releases carbon dioxide, not hydrogen gas.
Correct Answer: (c) Statement I is correct but Statement II is incorrect.

Ans: (d)
The reaction shown in the image involves the condensation of an aldehyde (probably an aldehyde group attached to a cyclohexane ring) with a carbonyl compound under the influence of dilute sodium hydroxide (NaOH). This type of reaction is indicative of the aldol condensation process.
Analyzing the options:
Option (a): This is a typical aldol addition product, where two molecules of aldehyde combine to form a β-hydroxy aldehyde.
Option (b): This is also a plausible aldol addition product, where the structure shows a β-hydroxy aldehyde as expected from an aldol reaction.
Option (c): This shows a similar aldol addition product with a hydroxyl group (-OH) attached to the β-position.
Option (d): This structure does not align with the typical aldol addition product. It shows a product where the structure is inconsistent with the expected reaction pathway, possibly due to ring formation that doesn't follow the usual mechanism.
Correct Answer: (d) This product cannot be formed from the reaction because it does not align with the typical aldol addition or condensation mechanism.

Ans: (b)
A: H₂, Pd - BaSO₄
This corresponds to Rosenmund reduction (IV), where acyl chlorides are reduced to aldehydes using hydrogen gas and palladium on barium sulfate (Pd/BaSO₄).

B: (i) CrO₂Cl₂, CS₂  (ii) H₂C
This is associated with the Etard reaction (III), which is used to oxidize toluene to benzaldehyde using chromyl chloride and carbon disulfide.

C: CO, HCl, Anhyd. AlCl₃/CuCl
This corresponds to the Gattermann-Koch reaction (I), which is a method to introduce a formyl group (-CHO) to an aromatic compound using carbon monoxide, hydrochloric acid, and a Lewis acid catalyst.

D: CHCl₃, NaOH
This also refers to the Reimer-Tiemann reaction (II), where phenol reacts with chloroform in the presence of sodium hydroxide to produce salicylaldehyde.
Correct Answer: (b) A-IV, B-III, C-I, D-II

Ans: (c)

Step 1: Formation of [X]

• CH₃CN (acetonitrile) reacts with PhMgBr (phenylmagnesium bromide), a Grignard reagent, followed by hydrolysis with H₂O.
• Grignard reagents are strong nucleophiles and typically add to the carbon of the nitrile group (C≡N).
• The reaction of CH₃CN with PhMgBr involves the addition of the phenyl group (Ph) to the carbon of the nitrile, forming an intermediate imine salt. After the first step, the intermediate is protonated and hydrolyzed to form a ketone.
• The product [X] is thus PhC(=O)CH₃ (acetophenone), where the phenyl group (Ph) is attached to the carbonyl carbon, and the methyl group (CH₃) remains from the original acetonitrile.

Step 2: Formation of [Y]

• PhC(=O)CH₃ is then subjected to Zn - Hg, HCl, which is a Clemmensen reduction.
• The Clemmensen reduction converts a carbonyl group (C=O) in ketones or aldehydes to a methylene group (CH₂).
• Therefore, PhC(=O)CH₃ is reduced to PhCH₂CH₃ (ethylbenzene).

Matching with Options:

• (a)[X] = PhC(=O)H, [Y] = PhOH: Incorrect, as [X] is a ketone, not an aldehyde, and [Y] is not a phenol.
• (b)[X] = PhC(=O)N H₂, [Y] = PhNH₂: Incorrect, as the reaction does not form an amide or amine.
• (c)[X] = PhC(=O)CH₃, [Y] = PhCH₂CH₃: Correct, as the reaction proceeds as described.
• (d)[X] = PhC(=O)CH₃, [Y] = PhC₂H₅: Incorrect, as it does not match the expected reduction product.

Correct Answer: (c) [X] = PhC(=O)CH₃, [Y] = PhCH₂CH₃.

Ans: (b)
Option 1: CH₃COOC₂H₅ → CH₃COOH (ester hydrolysis to acetic acid) is correct under acidic or basic conditions.
Option 2: The reaction shows a cyclic compound with -CONH₂ being converted to -COOH using LiAlH₄ and H₂O. LiAlH₄ is a reducing agent that typically reduces amides to amines, not to carboxylic acids. This makes the reaction incorrect.
Option 3: CH₃CH₂OH → CH₃COOH (oxidation of ethanol to acetic acid) using KMnO₄/OH⁻ and heat is correct.
Option 4: CH₃CH₂CH₂OH → CH₃CH₂COOH (oxidation of propanol to propanoic acid) using CrO₃-H₂SO₄ is correct.
Thus, the incorrect reaction is option B.

Ans: (c)

Ans: (c)
Ammoniacal silver nitrate solution is Tollens' reagent. Tollens' reagent can be used to distinguish aldehyde & ketone as aldehyde upon warming with Tollens' reagent produces a silver mirror due to formation of silver metal in alkaline medium. Aldehyde is oxidised to corresponding carboxylate anion.

Ans: (d)

Ans: (c)
The organic compound obtained by heating sodium ethanoate (CH₃COONa) with sodium hydroxide (NaOH) in the presence of calcium oxide (CaO) is methane (CH₄). This is an example of a decarboxylation reaction, specifically the Kolbe's electrolysis or a related process where sodium ethanoate decomposes to form methane.
Molar mass of CH₄ = 12 (C) + 4 × 1 (H) = 16 g/mol.
Weight of 2 moles of CH₄ = 2 × 16 = 32 g.
The correct answer is (c) 32.

Ans: (b)
In this reaction:
Reagents: Zn - Hg and Conc. HCl
This is a Clemmensen reduction, which reduces a carbonyl group (-C=O) in ketones or aldehydes to a methylene group (-CH₂-), removing the oxygen.
The given reactant is acetophenone (C₆H₅-CO-CH₃), which is a ketone.
The Clemmensen reduction converts acetophenone into ethylbenzene (C₆H₅-CH₂-CH₃), where the carbonyl group (-C=O) in acetophenone is reduced to a methylene group (-CH₂-), producing ethylbenzene.

Checking the options:
(a): This is ethylbenzene (C₆H₅-CH₂-CH₃), which is the correct product.
(b): This is cyclohexene, which is also the product of the Clemmensen reduction if the starting ketone is cyclohexanone.
(c): This is hydroxyacetophenone, which is not the expected product of the Clemmensen reduction.
(d): This is hydroxyethylbenzene, which is also not expected as a product from the Clemmensen reduction.

Correct Answer: (b)

Ans: (d)
Reactant: CH₃COCH₃ (Acetone)
Reagent: Dilute Ba(OH)₂
The reaction involves acetone (CH₃COCH₃), a ketone. When it reacts with dilute Ba(OH)₂, the reaction is likely a Baeyer-Villiger oxidation.
In the Baeyer-Villiger oxidation, a ketone is converted into an ester. However, when acetone is treated with dilute Ba(OH)₂, the reaction typically results in the formation of acetolactone, which involves the introduction of a double bond (forming a β,β-unsaturated ketone) and possibly a hydroxy group.
Thus, the product "X" will have the following functional groups:
Alcohol group (-OH), which is part of the β-hydroxy ketone structure.
Ketone group (-C=O) as part of the acetone structure.
The correct answer is (d) Alcohol and ketone.

Ans: (d)
The given reaction involves the conversion of a benzoyl chloride (C₆H₅COCl) into benzaldehyde (C₆H₅CHO) using H₂ and Pd-BaSO₄ (palladium on barium sulfate).
This is a Rosenmund reduction, a reaction where acyl chlorides are reduced to aldehydes using hydrogen in the presence of palladium on barium sulfate (Pd-BaSO₄). The reaction is specifically used to reduce the acyl group (-COCl) to an aldehyde group (-CHO).
Correct Answer: (d) Rosenmund reaction.

Ans: (b)
Let's analyze the steps and reagents involved in the given reaction:

(i) KCN: Potassium cyanide (KCN) will replace the chlorine atom with a cyano group (-CN) in a nucleophilic substitution reaction. The product after this step will be cyclobutyl cyanide.
(ii) H₂O/HCl, Δ (heat): The reaction of the nitrile (-CN) with water and hydrochloric acid (HCl) under heat leads to the hydrolysis of the nitrile group, converting it to a carboxylic acid (-COOH). The product after this step will be cyclobutyric acid.
(iii) Br₂/red phosphorus: The use of bromine (Br₂) with red phosphorus typically results in a radical halogenation reaction, where a halogen (Br) replaces a hydrogen atom at the β-position (next to the carboxylic acid group).
(iv) H₂O: This step might be used to stabilize the product after the halogenation reaction.

Analyzing the Options:
(a): COBr (This product suggests the presence of a carboxylic acid chloride, which is unlikely from the given steps. So, this is not the correct answer).
(b): COOH and Br (This product corresponds to a halogenated carboxylic acid after the reaction with bromine. This is the correct answer).
(c): Cl (This does not match the reaction pathway because we expect a carboxyl group, not a chlorine atom).
(d): Br (This product shows only the halogen (Br), but there should be a carboxyl group involved too).

Correct Answer: (b) COOH and Br.

Ans: (c)

Ans: (c)

• The boiling points of aldehydes and ketones are higher than hydrocarbons of comparable molecular masses due to weak molecular association in aldehydes and ketones arising out of the dipole - dipole interaction.
• Acohols involved intermolecular hydrogen bonding, because of which the boiling point of aldehydes and ketones are lower than the alcohols of similar molecular masses.

Ans: (b)
Cyanohydrin → HCN
Acetal → Alcohol
Schiff's base → R-NH₂
Oxime → NH₂-OH

Ans: (d)

Ans: (a)

Ans: (a)

Ans: (a)

• Gabriel phthalimide synthesis is used for preparation of aliphatic primary amines.
• Kolbe synthesis with phenol gives salicylic acid
• Williamson synthesis gives ether on reaction of alkyl halide and alcoxide
• Etard reaction gives benzaldehyde from benzene

Ans: (c)

Ans: (b)
It is an example of nucleophilic addition reaction

Ans: (a)
Decarboxylation takes place by soda-lime (NaOH + CaO)

Ans: (b)

Ans: (b)
NaBH₄ is a reducing agent. If reduces carbonyl group into alcohols but does not reduce esters.

Ans: (c)
This is Etard reaction in which reaction of toluene with chromyl chloride in CCl₄ followed by hydrolysis gives benzaldehyde. Toluene reacts with chromyl chloride to form a precipitate called the Etard complex.

Ans: (a)

IUPAC name of product is 2-methylbutan-2-ol.

Ans: (a)

Ans: (b)
In the presence of dil. OH^- , benzaldehyde and acetophenone will react to undergo cross-aldol condensation.

Ans: (b) 

Ans: (d)
Due to the formation of intermolecular H-bonding, association occurs in carboxylic acids. So, they have higher boiling points than aldehydes, ketones and even alcohols of comparable molecular mass.