
Hydrocarbons JEE Main Questions 2025
are available here. Candidates preparing for the JEE Mains 2025 Exam for January and April Sessions are advised to practice the Important Practice Questions with PYQs of Hydrocarbons in JEE Mains. Hydrocarbons in
JEE Mains 2025
is one of the most important chapters holding a weight of 2 to 3% of the total Chemistry section. The important chapters of Hydrocarbons that candidates should emphasise are alcohols, aldehydes, and ketones. Questions asked from Hydrocarbons are conceptual questions on reaction mechanisms, numerical questions on quantitative aspects of reactions, and name reactions and product prediction. The expected number of Hydrocarbons JEE Main Questions is 1 to 2 questions which is approximately 4 to 8 marks in JEE Mains. In this article, we have provided the Expected Hydrocarbons JEE Main Questions 2025.
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JEE Main 2025 Hydrocarbons Important Questions
1. The gas liberated by the electrolysis of the Dipotassium succinate solution is :
(1) Ethyne
(2) Ethene
(3) Propene
(4) Ethane
Solution:
(CH2COO–)2 → CH2 = CH2 + 2CO2 (g) + 2e–
2H2O + 2e– → 2OH– + H2 (g)
So the gas generated during electrolysis of Dipotassium succinate solution is ethene.
Hence option (2) is the answer.
2. One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44 u. The alkene is:-
(1) Ethene
(2) Propene
(3) 1-Butene
(4) 2-Butene
Solution:
2-Butene on ozonolysis gives 2 moles of acetaldehyde with a molecular mass of 44gm/mol.
Hence option (4) is the answer.
3. 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.
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4. 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.
5. Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of:-
(1) an isopropyl group
(2) an acetylenic triple bond
(3) two ethylenic double bonds
(4) a vinyl group
Solution:
In Ozonolysis, the presence of the vinyl group gives formaldehyde as one of the products.
Hence option (3) is the answer.
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6. In the given transformation, which of the following is the most appropriate reagent?
(1) NaBH4
(2) NH2-NH2, OH
(3) Zn – Hg / HCl
(4) Na, Liq.NH3
Solution:
In NH2-NH2, OH, the medium is basic so it will not have any effect on the OH group in the reagent.
Hence option (2) is the answer.
7. In the hydroboration – oxidation reaction of propene with diborane, H2O2 and NaOH, the organic compound formed is :
(1) CH3CH2CH2OH
( 2) (CH3)3COH
(3) CH3CHOHCH3
(4) CH3CH2OH
Solution:
The hydroboration–oxidation reaction is a two-step hydration reaction that converts an alkene into alcohol. It is an anti-Markovnikov reaction. The organic compound formed in the hydroboration–oxidation reaction of propene with diborane, H2O2 and NaOH is CH3CH2CH2OH.
Hence option (1) is the answer.
8. The number and type of bonds in C22- ion in CaC2 is:
(1) Two σ bonds and one π – bond
(2) Two σ bonds and two π – bonds
(3) One σ bond and two π – bonds
(4) One σ bond and one π bond
Solution:
Ca+2 [C≡C] -2
One σ bond and two π – bonds are there in C22- ion in CaC2.
Hence option (3) is the answer.
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9. 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 does 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.
10. The gas evolved on heating CH3MgBr in methanol is
(1) methane
(2) ethane
(3) propane
(4) HBr
Solution:
The gas evolved by heating Methyl magnesium bromide in methanol is methane.
CH3-O-H + CH3MgBr → CH3-H + Ch3-OMgBr
Hence option (1) is the answer.
11. The reagent needed for converting
is
(1) H2 / lindlar Cat.
(2) Catalytic Hydrogenation
(3) LiAlH4
(4) Li/ NH3
Solution:
Li/NH3 is the birch reagent. It reduces the alkyne to trans alkene.
Hence option (4) is the answer.
12. 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.
13. The hydrocarbon which can react with sodium in liquid ammonia is
(1) CH3CH2C ≡ CCH2CH3
(2) CH3CH2CH2C ≡ CCH2CH2CH3
(3) CH3CH2C ≡ CH
(4) CH3CH ≡ CHCH3
Solution:
Terminal alkynes have acidic hydrogen. Terminal alkynes react with sodium in liquid ammonia to give ionic compounds.
Hence option (3) is the answer.
14. Which one of the following is reduced with zinc and hydrochloric acid to give the corresponding hydrocarbon?
(1) Ethyl acetate
(2) Acetic acid
(3) Acetamide
(4) Butan-2-one
Solution:
Butan-2-one will get reduced into butane when treated with zinc and hydrochloric acid following Clemmensen reaction whereas Zn/HCl do not reduce ester, acid and amide.
Hence option (4) is the answer.
15. Polysubstitution is a major drawback in
(a) Reimer-Tiemann reaction
(b) Friedel-Crafts acylation
(c) Friedel-Crafts alkylation
(d) Acetylation of aniline.
Solution:
The product obtained is more activated in Friedel-Crafts alkylation and thus, polysubstitution will take place.
Hence option (3) is the answer.
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Important Topics From Hydrocarbons in JEE Mains
Check the table below to learn the important topics of Hydrocarbons in JEE Main 2025 Chemistry.
Topics | Sub-Topics from Hydrocarbons |
---|---|
Alkanes | Wurtz Reaction, Kolbe's Electrolytic Method), Reactions (Halogenation, Combustion |
Alkenes | Dehydration Of Alcohols, Dehydrohalogenation Of Alkyl Halides), Markovnikov's And Anti-Markovnikov's Rule, Oxidation (Baeyer's Test), And Addition Reactions |
Alkynes | Dehydrohalogenation, Dehalogenation), Acidic Nature, And Reactions Like Ozonolysis |
Aromatic Hydrocarbons | Benzene Structure, Electrophilic Substitution Reactions (Nitration, Halogenation, Friedel-Crafts Alkylation/Acylation), And Resonance |
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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.
Hydrocarbons are commonly utilized as sources of energy, such as LPG (Liquefied Petroleum Gas) and CNG (Compressed Natural Gas). They are utilized in the production of polymers like polyethene and polystyrene, among others. These organic compounds are used as foundational materials in the production of pharmaceuticals and colourants. They act as lubricating oils and greases.
Hydrocarbons can be classified into several categories based on their bonding and structure. Saturated hydrocarbons, known as alkanes, consist solely of single bonds between carbon atoms, having the general formula CnH2n+2 and featuring sp³ hybridized carbon atoms. Unsaturated hydrocarbons include alkenes and alkynes, which contain double and triple bonds respectively, with general formulas of CnH2n for alkenes and CnH2n-2 for alkynes. Cycloalkanes are characterized by one or more carbon rings with hydrogen atoms attached. Aromatic hydrocarbons, or arenes, contain at least one aromatic ring, while aliphatic hydrocarbons feature straight-chain structures without rings. Alicyclic hydrocarbons, on the other hand, possess a ring structure and can include carbon atoms that are sp, sp², or sp³ hybridized.
A hydrocarbon is an organic compound consisting of hydrogen and carbon found in crude oil, natural gas, and coal. Hydrocarbons are highly combustible and the main energy source of the world. Its uses consist of gasoline, jet fuel, propane, kerosene, and diesel, to name just a few.
The important chapters of Hydrocarbons that candidates should emphasise are alcohols, aldehydes, and ketones.
Hydrocarbons in JEE Mains 2025 is one of the most important chapters holding a weight of 2 to 3% of the total Chemistry section.
JEE Main Previous Year Question Paper
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