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Aldehydes, Ketones and Carboxylic Acid ‌–‌ ‌JEE‌ ‌Advanced‌ ‌Previous‌ ‌Year‌ ‌Question With Solutions

Aldehydes, Ketones and Carboxylic Acid ‌JEE‌ ‌Advanced‌ ‌Previous‌ ‌Year‌ ‌Questions‌ ‌with‌ ‌Solutions‌ are provided here to help JEE as[irants get a better idea of the type of questions that can come in the entrance exam. As aspirants go through the list of questions they will further understand the difficulty level of the questions and also find out the important topics to focus on. The solutions that are given along with the questions will also help aspirants to learn the best approach to come up with the right answer. They will also help them get a better grip on the concepts and at the same time have a quick revision of the same. In essence, as aspirants go through the questions and solutions they will develop better skills to solve different types of problems and prepare effectively for the upcoming entrance exam.

Aldehydes, Ketones and Carboxylic Acid JEE Advanced ‌Previous‌ ‌Year‌ ‌Questions‌ ‌with‌ ‌Solutions‌ PDF has been provided below for free download. You can download and refer to these questions at your convenience.

Download Aldehydes, Ketones and Carboxylic Acid Previous Year Solved Questions PDF

JEE Advanced Previous Year Questions on Aldehydes, Ketones and Carboxylic Acid

Question 1. The major product of the following reaction sequence is:\

Solution: (A)

Question 2. Which compound would give 5-keto-2-methyl hexanal upon ozonolysis?

Solution: (B)

Question 3. The major product in the following reaction is

Solution: (D)

Question 4. Among the following the number of reactions that produce benzaldehyde is:

Solution: (All of them)

Benzaldehyde can be obtained from all the four reactions mentioned above.

(A) The first reaction is Gatterman-Koch formylation wherein benzene is treated under high pressure with CO and HCl in the presence of anhydrous AlCl3/CuCl catalyst to form benzaldehyde.

(B) The second reaction involves the hydrolysis of benzyl chloride by heating with water to form benzaldehyde.

(C) In the third reaction there is Rosenmund reduction. Benzoyl chloride is hydrogenated over Pd − BaSO4 catalyst to give benzaldehyde.

(D) In the fourth reaction there is a partial reduction of methyl benzoate with DIBALH in the presence of toluene as a solvent at −78oC and this leads to the formation of benzaldehyde.

Question 5. Consider all possible isomeric ketones, including stereoisomers of MW = 100. All these isomers are independently reacted with NaBH4. (NOTE: stereoisomers are also reacted separately). What is the total number of ketones that give a racemic product(s)?

Solution: (5)

The molecular weight of the ketone is 100

So, molecular formula = C6H12O

[6 × 12 + 12 × 1 + 16].

Degree of unsaturation = (6 + 1) – 12 / 2 = 1

According to the question, the compound contains a ketone group. The compound which contains a chiral centre will form a diastereomer while others produce enantiomers. Various isomers and their possible reduced product are as shown below.

The number of ketones that gives a racemic mixture with NaBH4 is 5 as the ketone with a chiral centre will give diastereoisomer with NaBH4.

Question 6. In the reaction scheme shown below, Q, R, and S are the major products. The correct structure of

Solution: (B and D)

Question 7. Compounds P and R upon ozonolysis produce Q and S, respectively. The molecular formula of Q and S is C8H8O.Q undergoes Cannizzaro reaction but not haloform reaction, whereas S undergoes haloform reaction but not Cannizzaro reaction.

The option(s) with a suitable combination of P and R, respectively, is(are).

Solution: (A and B)

Question 8. The correct statement about the following reaction sequence is(are)

A. R is steam volatile

B. Q gives dark violet colouration with 1% aqueous FeCl3 solution

C. S gives a yellow precipitate with 2,4-dinitrophenylhydrazine

D. S gives dark violet colouration with 1% aqueous FeCl3 solution

Solution: (B and C)

Question 9. Positive Tollens’ test is observed for:

Solution: (A, B and C)

Aldehydes and α-Hydroxyketone show positive Tollen’s test.

Now except for the last option, the rest all of them form the silver mirror.

This is due to the fact that the first two are aldehydes, the third one is α− hydroxy ketone, and the last option is none of the above.

Options A, B and C are correct.

Question 10. The major product of the following reaction is:

Solution: (A)

Question 11. After completion of the reactions (I and II), the organic compound(s) in the reaction mixtures is(are):

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

Solution: (C)

When acetone reacts with Br2 in a basic medium, bromoform is formed.

When CH3COCH3 and Br2 are in equimolar quantity, all the Br2 (limiting reactant) is converted into desired products and 2/3 mole of CH3COCH3 remains unreacted, being in excess. When acetone reacts with Br2 in an acidic medium, there is monobromination of acetone.

CH3COCH3 and Br2 react in a 1:1 mole ratio and (P) are formed.

In reaction, I, (U) and (T) are formed and acetone (reactant) remains unreacted. In reaction II, (P) is formed.

Question 12. The correct order of acid strength of the following carboxylic acids is:

A. III > II > I > IV

B. I > II > III > IV

C. I > III > II > IV

D. II > I > IV > III

Solution: (C)

Stronger will be the acid when its conjugate base is highly stabilised. (I) is the most acidic and (IV) is the least since CO2 group is attached with an sp hybridised carbon in (I) and with an sp3 hybridised carbon in (IV).

(III) is less acidic than (II) as the electron-donating group (−OMe) destabilises the carboxylate anion by increasing the electron density in the ring.

Hence the order is: I > II > III > IV

Question 14. The sodium salt of an organic acid ‘X’ produces effervescence with conc. H2SO4. ‘X’ reacts with the acidified aqueous CaCl2 solution to give a white precipitate that decolourises the acidic solution of KMnO4. ‘X’ is




D. Na2C2O4

Solution: (D)

Question 15. The correct order of acidity for the following compound is:

A. I > II > III > IV

B. III > I > II > IV

C. III > IV > II > I

D. I > III > IV > II

Solution: (A)

Due to ortho-effect, (I) and (II) are stronger acids than (III) and (IV). Due to two ortho hydroxyl groups in (I), it is a stronger acid than (II). (Ill) is a stronger acid than (IV) because at m-position, −OH group cannot exert its +R effect but can only exert its −I effect while at p− position, −OH group exerts its strong +R effect. Thus, the correct order of acidity is I > II > III > IV.

Stable the conjugate base, stronger the acid.

o-hydroxybenzoic acid is more acidic than the corresponding para isomer because the carboxylate ion formed from ortho hydroxybenzoic acid attains stability due to intermolecular hydrogen bonding. But in the case of the para isomer, the -OH group shows a +R effect, which increases the electron density and hence the stability of the corresponding carboxylate ion is less compared to the ortho and the meta hydroxybenzoic acids.

Question 16. The compound that does not liberate CO2, on treatment with aqueous NaHCO3 solution, is:

A. Benzoic acid

B. Benzenesulphonic acid

C. Salicylic acid

D. Carbolic acid

Solution: (D)

NaHCO3 ⇌ Na+ + HCO-3

HCO-3 is decomposed by acid releasing CO2

HCO-3 + H+ ⟶ H2O + CO2

If acid is stronger than HCO-3 then CO2 is released. Phenol is less acidic and thus does not liberate CO2 with NaHCO3.

Question 17. The total number of carboxylic acid groups in the product ‘p’ is (numeral) ___.

Solution: (2)

The first step involves acid hydrolysis of cyclic anhydride.

The second one is the decarboxylation of beta keto acids.

Oxidative ozonolysis is the third step.

As per the reaction, the total number of carboxylic acid groups in the product P is 2.

Question 18. In the following reaction, compound Q is obtained from compound P via an ionic intermediate.

What is the degree of unsaturation of Q?

Solution: (18)

The total degree of unsaturation = 18

Question 19. Reagent(s) which can be used to bring about the following transformation is/are:

A. LiAlH4 in (C2H5)2O

B. BH3 in THF

C. NaBH4 in C2H5OH

D. Raney Ni/H2 in THF

Solution: (C)

NaBH4 in C2H5OH

NaBH4/C2H5OH do not reduce acid (COOH), ester (COOR) and epoxide, as it is a mild oxidising agent. It reduces only aldehyde to alcohol.

While the LiAlH4 is a strong oxidising agent it reduces the ester, acid, aldehyde and ketone.

Question 20. The reaction(s) leading to the formation of 1,3,5-trimethyl benzene is (are):

Solution: (A, B and D)


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