WBJEE 2025 Organic Chemistry Practice Questions with Solutions

WBJEE 2025 Organic Chemistry Practice Questions: Organic Chemistry for WBJEE 2025 covers topics such as Carbon, Bonding, Alkanes, and Haloalkanes. It carries a weightage of approximately 6% and a good 4-5 questions are asked every year from this chapter. Organic Chemistry is one of the most important chapters in the WBJEE Chemistry syllabus 2025. To be able to attempt the maximum number of questions and score high marks in the examination, you should diligently solve WBJEE 2025 Organic Chemistry practice questions from these topics. WBJEE Organic Chemistry sample questions are meant to provide you with a series of probable questions that may be asked in the upcoming WBJEE 2025 exam . Our experts have prepared a set of model questions based on the previous years' question papers and topic weightage. You can find the WBJEE 2025 Organic Chemistry practice questions with solutions on this page.

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List of WBJEE 2025 Organic Chemistry Sub Topics for Exam Preparations

The WBJEE 2025 Chemistry syllabus PDF highlighting the key topics from the Organic Chemistry section is available on the official WBJEE website. The sub-topics contain detailed description, theories and formulae from the Organic Chemistry chapter. It is important to prepare all the sub-topics from Organic Chemistry in order to have a dedicated study plan. Check out the list of Organic Chemistry topics for WBJEE 2025.

The expected marks weightage for WBJEE Organic Chemistry gives you an idea about how many marks the chapter will cover and how many questions may be asked from this section. The Organic Chemistry section includes 4-5 questions and has a weightage of around 6% making it an important part of the WBJEE 2025 syllabus PDF . Let’s have a look at the WBJEE 2025 Organic Chemistry expected marks weightage from the table below.

WBJEE 2025 Organic Chemistry Practice Questions with Solutions

Good exam preparation includes solving WBJEE 2025 Organic Chemistry practice questions. This practice not only helps in building clarity on the Chemistry fundamentals but also gives enough material to practice for the examination. The following are some of the most important WBJEE 2025 Organic Chemistry practice questions with solutions.

a. 3-phenylpropanoic acid

b. Benzylacetic acid

c. Carboxyethylbenzene

d. 2-Phenylpropanioc acid

a. I < II < III

b. I < III <II

c. II < III < I

d. III < II < I

Solution:

To determine the correct acidity order of phenol (I), 4-hydroxybenzaldehyde (II), and 3-hydroxybenzaldehyde (III), we need to consider the electron-withdrawing or electron-donating effects of the substituents on the phenol ring and how these affect the stability of the phenoxide ion after deprotonation.

The presence of the electron-withdrawing aldehyde group influences the acidity of the phenol by stabilizing the negative charge on the oxygen atom of the phenoxide ion. This stabilization is more significant when the aldehyde group is closer (para position) or less steric hindered compared to the hydroxyl group.

1. Phenol (I): Has no additional electron-withdrawing groups to stabilize the phenoxide ion, thus has the least acidity.

4-Hydroxybenzaldehyde (II): The CHO group at the para position significantly withdraws electron density through both inductive and resonance effects, stabilizing the phenoxide ion more effectively than at the meta position.

3-Hydroxybenzaldehyde (III): The CHO group at the meta position primarily withdraws electron density through inductive effects, which is less stabilizing than the para position incorporation.

Therefore, the correct acidity order is:

Phenol (I) < 3-Hydroxybenzaldehyde (III) < 4-Hydroxybenzaldehyde (II)

Or, I < III < II

a. 4-Nitrobenzaldehyde

b. 4-Methoxybenzaldehyde

c. 4-Methylnitrobenzene

d. 4-Nitroacetophenone

a. 6

b. 12

c. 13

d. 15

a. O3

b. CO

c. HCN

d.O3 and HCN

a. MeCN > C5H5N > Me3N

b. C5H5N > Me3N > MeCN

c. MeN > MeCN > C5H5N

d. Electronegativity same in all

a. Y > Z > X

b. Z > Y > X

c. Z > X > Y

d. Y > X > Z

a. 2

b. 1

c. 3

d. 4

a. The rate of reaction increases with increasing nucleophilicity

b. The number 2 denotes a second order reaction

c. Tertiary butyl substrates do not follow this mechanism

d. The optical rotation of substrates always changes from (+) to (-) or from (-) to (+) in the product

Solution:

The correct statements about the SN ₂ reaction mechanism are:

Option A: The rate of reaction increases with increasing nucleophilicity.

This statement is true. In an SN ₂ reaction, the nucleophile attacks the electrophilic carbon in a single, concerted step. Therefore, a stronger (more nucleophilic) nucleophile will generally result in a faster reaction because it can more readily attack the carbon center.

Option B: The number 2 denotes a second-order reaction.

This statement is also true. The term SN ₂ stands for bimolecular nucleophilic substitution and indicates that the rate-determining step involves two reacting species: the nucleophile and the electrophilic substrate. The rate law for an SN ₂ reaction is given by:

Rate = k[Nucleophile][Electrophile]

This indicates that the reaction is second-order overall.

Option C: Tertiary butyl substrates do not follow this mechanism.

This statement is true. Tertiary butyl substrates are typically very sterically hindered, making it difficult for the nucleophile to approach the electrophilic carbon from the backside, which is required for the SN ₂ mechanism. As a result, tertiary substrates usually do not undergo SN2 reactions.

Option D: The optical rotation of substrates always changes from (+) to (-) or from (-) to (+) in the products.

This statement is incorrect because in an SN ₂ reaction inversion does take place but we can never say a dextro substrate will produce a levo product or vice-versa. Because the dextro or levo nature of the reactant or product is purely experimental.

a. ketone

b. alkene

c. ether

d. Xanthate

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