
Organic Chemistry JEE Main Questions 2025 cover almost 30 to 35% of the entire Chemistry section of JEE Mains. Approximately 6-8 questions out of 25 questions are asked from Organic Chemistry which is equivalent to 24 to 32 marks. The important topics from Organic Chemistry in JEE Mains 2025 are General Organic Chemistry (GOC) covering a weight of 2-3 questions, Hydrocarbons, Haloalkanes and Haloarenes, Alcohols, Phenols, and Ethers, Aldehydes and Ketones, Biomolecules and Polymers where at least 1 to 2 questions can be expected.
Organic Chemistry in JEE Mains is one of the most important chapters of the Chemistry section that requires a strategic approach requiring the candidates to understand the entire topic. In this article, we have covered the entire set of Important Practice Organic Chemistry JEE Main Questions with PYQs along with the important topics and their weightage (expected).
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JEE Main 2025 Organic Chemistry Important Questions
Check the important JEE Main 2025 Questions from Organic Chemistry here.
1. The most basic compound among the following is:-
(1) Acetanilide
(2) Benzylamine
(3) p-Nitro aniline
(4) Aniline
Solution:
Basicity is inversely proportional to the resonance of lone pair electrons. Benzylamine is more basic. The electron pairs are not involved in resonance in benzylamine. In other amines, there is delocalization of a lone pair of electrons on the N atom on the ring. In acetanilide, the delocalization of a lone pair of electrons on the N atom is due to the adjacent CO group.
Hence option (2) is the answer.
2. Considering the basic strength of amines in aqueous solution, which one has
the smallest pKb value?
(1) (CH3)3N
(2) C6H5NH2
(3) (CH3)2NH
(4) CH3NH2
Solution:
Because of resonance, arylamines are less basic than alkyl amines. In arylamines, the lone pair of electrons on N is partly shared with the ring and is less available for sharing with a proton. In alkylamines, the electron-releasing alkyl group increases the electron density of the nitrogen atom. Hence increases the protonation ability of amines. So, the more the number of alkyl groups, the higher be basicity of amine. Because of the steric effect, a slight discrepancy occurs in the case of trimethyl amine. So (CH3)2NH has the smallest pKb value.
Hence option (3) is the answer.
3. Among the following the molecule with the lowest dipole moment is:-
(1) CHCl3
(2) CH2Cl2
(3) CCl4
(4) CH3Cl
Solution:
Dipole moment of CCl4 is 0.
The order of dipole moment is CCl4 < CH3Cl < CH2Cl2 < CHCl3.
Hence option (3) is the answer.
4. Which one of the following compounds will not be soluble in sodium bicarbonate?
(1) Benzene sulphonic acid
(2) Benzoic acid
(3) o-Nitrophenol
(4) 2, 4, 6 – Trinitrophenol
Solution:
Benzene sulphonic acid and Benzoic acid are stronger acids and they react with sodium bicarbonate. o-Nitrophenol is a very weak acid and it does not react with sodium bicarbonate. o-Nitrophenol will not be soluble in sodium bicarbonate. 2, 3, 6 – Trinitrophenol is higher in acidity and it reacts with sodium bicarbonate.
Hence option (3) is the answer.
5. The compound formed in the positive test for nitrogen with the Lassaigne solution of an organic compound is
(1) Fe4[Fe(CN)6]3
(2) Na4[Fe(CN)5NOS]
(3) Fe(CN)3
(4) Na3[Fe(CN)6]
Solution:
The compound formed in the positive test for nitrogen with the Lassaigne solution of an organic compound is the Prussian blue-coloured complex compound ferric ferrocyanide.
Hence option (1) is the answer.
6. The general formula CnH2n02 could be for open-chain
(1) carboxylic acids
(2) diols
(3) dialdehydes
(4) diketones
Solution:
The general formula CnH2n02 could be for open-chain carboxylic acid or ester.
Hence option (1) is the answer.
7. Among the following oxoacids, the correct decreasing order of acid strength is :
(1) HClO4 > HClO3 > HClO2 > HOCl
(2) HClO2 > HClO4 > HClO3 > HOCl
(3) HOCl > HClO2 > HClO3 > HClO4
(4) HClO4 > HOCl > HClO2 > HClO3
Solution:
Acidic strength is directly proportional to the oxidation number. Increasing acid strength is because of an increase in the oxidation state of the central atom.
Hence option (1) is the answer.
8. Ortho-nitrophenol is less soluble in water than p– and m– Nitrophenols because:
(1) The melting point of o–Nitrophenol is lower than those of m– and p– isomers
(2) o–Nitrophenol is more volatile in steam than those of m– and p– isomers
(3) o–Nitrophenol shows Intramolecular H–bonding
(4) o–Nitrophenol shows Intermolecular H–bonding
Solution:
Intramolecular H-bonding is present in o-nitrophenol. So solubility in water is decreased.
Hence option (3) is the answer.
9. The order of basicity of amines in the gaseous state is :
(1) 30 > 20 > NH3 > 10
(2) 10 > 20 > 30 > NH3
(3) NH3> 10 > 20 > 30
(4) 30 > 20 > 10 > NH3
Solution:
The basicity is proportional to the +I effect. The presence of an electron-donating group increases the basicity of amines. The presence of a withdrawing group decreases the basicity of amines.
Hence option (4) is the answer.
10. Which one of the following conformations of cyclohexane is chiral?
(1) Twist boat
(2) Rigid
(3) Chair
(4) Boat
Solution:
Twist boat is chiral because it does not have a plane of symmetry.
Hence option (1) is the answer.
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11. Increasing order of stability among the three main conformations (i.e. Eclipse, Anti, Gauche) of 2-fluoroethanol is
(1)Eclipse, Gauche, Anti
(2)Gauche, Eclipse, Anti
(3)Eclipse, Anti, Gauche
(4)Anti, Gauche, Eclipse
Solution:
Eclipse is the least stable and Gauche is the most stable.
Hence option (3) is the answer.
12. The conjugate base of hydrazoic acid is:-
(1) NH3–
(2) N3–
(3) N2–
(4) N-3
Solution:
Hydraulic acid is HN3.
HN3 → H+ + N3–
Hence option (2) is the answer.
13. The correct order of increasing basicity of the given conjugate base (R =CH3) is:
(1) RCOO– < HC ≡ C– < NH2– < R–
(2) RCOO– < HC ≡ C– < R– < NH2
(3) R– < HC ≡ C– < RCOO– < NH2–
(4) RCOO < NH2 < HC ≡ C– < R–
Solution:
The basic strength is inversely proportional to the stability of the conjugate base. In basicity, if the availability of the electrons is greater, then more readily they can be donated to form a new bond to the proton and, hence the stronger base. If bronsted acid is a strong acid then its conjugate base is weak.
The correct order of increasing basic strength is RCOO– < HC ≡ C– < NH2– < R–
Hence option (1) is the answer.
14. Which types of isomerism are shown by 2,3-dichlorobutane?
(1) Diastereo
(2) Optical
(3) Geometric
(4) Structural
Solution:
2,3- dichlorobutane shows optical isomerism.
Hence option (2) is the answer.
15. Due to the presence of an unpaired electron, free radicals are
(1)Chemically reactive
(2) Chemically inactive
(3) Anions
(4) Cations
Solution:
Due to the presence of an unpaired electron, the free radicals are chemically active.
Hence option (1) is the answer.
16. Which one of the following does not have sp2 hybridized carbon?
(1) Acetone
(2) Acetamide
(3) Acetonitrile
(4) Acetic acid
Solution:
Acetonitrile has only sp3 and sp hybridized carbon atoms.
Hence option (3) is the answer.
17. Which one of the following classes of compounds is obtained by polymerization of acetylene?
(1) Poly-ene
(2) Poly-yne
(3) Poly-amide
(4) Poly-ester
Solution:
nHC ≡ CH → ( CH =CH)n poly-yne
Hence option (2) is the answer.
18. Which branched chain isomer of the hydrocarbon with molecular mass 72 u gives only one isomer of mono-substituted alkyl halide?
(1) Neopentane
(2) Isohexane
(3) Neohexane
(4) Tertiary butyl chloride
Solution:
Molecular mass indicates that it is pentane. Neopentane can only form one mono-substituted alkyl halide as all the hydrogens are equivalent in neopentane.
Hence option (1) is the answer.
19. Which one of the following has the minimum boiling point?
(1) n-Butane
(2) 1-Butyne
(3) 1-Butene
(4) Isobutene
Solution:
Among the isomeric alkanes, the normal isomer has a higher boiling point than the branched-chain isomer. The higher the branching of the chain, the lower the boiling point. The n-alkanes have more surface area in comparison to branched-chain isomers. So, intermolecular forces are weaker in branched-chain isomers. Hence they have lower boiling points in comparison to straight-chain isomers.
Hence option (4) is the answer.
20. In the presence of peroxide, HCl and HI do not give anti-Markovnikov’s addition to alkenes because
(1) All the steps are exothermic in HCl and HI
(2) One of the steps is endothermic in HCl and HI
(3) HCl is oxidizing and the HI is reducing
(4) Both HCl and HI are strong acids
Solution:
Bromides will take part in radical-based reactions in the presence of organic peroxides. Fission of the peroxide O-O linkage causes a Br radical which behaves differently from Bromide, by adding to the less substituted side of the alkene (anti–Markovnikov). HI, and HCl do not do this for energetic reasons. The addition of Cl and I radicals to the alkene in an anti-Markovnikov fashion is an endothermic reaction. Hence it is unfavourable. All hydrogen halides will be added according to the Markovnikov rule, without the presence of peroxide,
Hence option (2) is the answer.
Organic Chemistry JEE Mains Weightage 2025
Check the table to get a detailed idea of the JEE Mains expected weightage for the organic chemistry section.
Topics
|
Subtopics
|
Question Allocation
|
Marks Allocation
|
---|---|---|---|
General Organic Chemistry (GOC) | Resonance, inductive effects, hyperconjugation, acidity/basicity | 2 to 3 questions | 8 to 12 marks |
Hydrocarbons | Reactions of alkanes, alkenes, and alkynes; free radical mechanisms | 1 to 2 questions | 4 to 8 marks |
Haloalkanes and Haloarenes | SN1 and SN2 mechanisms, stereochemistry | 1 to 2 questions | 4 to 8 marks |
Alcohols, Phenols, and Ethers | Reactions of alcohols and phenols, preparation of ethers | 1 to 2 questions | 4 to 8 marks |
Aldehydes and Ketones | Aldol condensation, nucleophilic addition reactions | 1 to 2 questions | 4 to 8 marks |
Carboxylic Acids and Derivatives | Decarboxylation, acid derivatives’ reactions | 1 question | 4 marks |
Amines | Hofmann bromamide reaction, Gabriel synthesis, diazonium salts | 1 question | 4 marks |
Biomolecules and Polymers | Structure of glucose, properties of amino acids, types of polymers | 1 to 2 questions | 4 to 8 marks |
Environmental Chemistry | Reactions involving environmental pollutants and their effects | 1 question | 4 marks |
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FAQs
The JEE Main exam incorporates various question types, comprising multiple-choice questions (MCQs), numerical value-based questions, as well as drawing-based questions. With MCQs, candidates must select the correct answer from the given options.
No. The JEE Mains exam alone is not enough for securing admission into an IIT. JEE Advanced is the crucial examination required for IIT admissions, and candidates are required to qualify in the JEE Main exam to be eligible for the JEE Advanced exam.
Yes, 90 is considered to be a good percentile in the JEE Main 2025 entrance exam. Candidates can secure admission into NITs, IITs or IISc Bangalore if they score a rank between 1,00,000 and 1,50,000 with a 90 percentile in the JEE Main exam.
Mathematics is often considered to be the most challenging subject in the JEE Main exam, but its difficulty level varies from person to person. Some students may find other subjects more challenging based upon their strengths and weaknesses.
NCERT books are very important to prepare for the JEE Main 2025 exam. The JEE Main exam follows the CBSE syllabus for 11th and 12th classes, so NCERT books are considered to be the most reliable and necessary study materials. NCERT books explain basic fundamental concepts in a simple and clear way.
To score a 99+ percentile in the JEE Main 2025 exam, candidates require more than just knowing the JEE Main syllabus, as it required consistent practice, strategic revision, time management, and the ability to solve complex problems under pressure. However, with the appropriate resources, one can achieve these requirements.
All sections namely Mathematics, Physics, and Chemistry are mandatory in the JEE Main exam, with questions covering the entire JEE Main syllabus. To be eligible for admission, candidates must achieve the minimum qualifying marks in each subject and the overall aggregate.
Candidates can study from the NCERT textbooks of class 11 and 12 Physics, Chemistry and Mathematics to cover the JEE Mains syllabus. No changes have been made to the JEE Main Syllabus 2025. NTA removed several topics of Physics, Chemistry and Mathematics last year from the JEE Main syllabus.
Candidates must keep in mind that there is no requirement of 75% marks in class 12 for appearing in the JEE Mains exam. The JEE Main eligibility criteria of minimum 75% marks in class 12 is required at the time of securing admission across NITs, IIITs and GFTIs. Candidates can apply and appear in the JEE Mains 2025 exam irrespective of their class 12 marks.
Previous year question papers are an important source for the JEE Main preparations. However, relying solely upon PYQs is not enough to fully prepare for the JEE Mains exam. Candidates must also focus on covering the entire syllabus, developing strong conceptual understanding, practising time management, and attempting mock tests.
JEE Main Previous Year Question Paper
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