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Hydrocarbons appeared 35 times across 3 years — 4.1% of Chemistry. This question is from Ozonolysis and Stereochemistry.

Year 2026 2025 2024 Total
Questions 14 12 9 35

The number of optically active products obtained from the complete ozonolysis of the given compound is: [cite: 364, 365]
Ozonolysis and Stereochemistry compound diagram for Q41 - JEE Main 2025 Evening
The image details a symmetrically structured long-chain polyene containing multiple internal alkene bonds and chiral carbon sites.

Solution & Explanation

Related Formula
R₁-CH=CH-R₂ [Zn / H₂O]O₃ R₁-CHO + R₂-CHO
Core Logic

Complete oxidative cleavage of all double bonds via reductive ozonolysis breaks the molecule into smaller fragments:

CH₃-CH=CH-CH(CH₃)-CH=CH-CH(CH₃)-CH=CH-CH₃

Let's trace the fragmentation logic mapping visually:

Ozonolysis and Stereochemistry products diagram for Q41 - JEE Main 2025 Evening
The image details a symmetrically structured long-chain polyene containing multiple internal alkene bonds and chiral carbon sites.

Step 1: Tracking Fragment Structures

The chemical reaction outputs two major molecular product types:

  • CH3-CHO (Acetaldehyde): Optically inactive as it lacks a chiral carbon.
  • OHC-CH(CH₃)-CHO (2-methylpropanedial): Let's inspect the substituted central carbon. It is bonded to: a hydrogen atom (-H), a methyl group (-CH₃), and two identical formyl groups (-CHO).
  • Because two of the groups are identical (-CHO), this molecule does not have a chiral center and is entirely optically inactive.

Step 2: Total Summation

Since every single product formed is achiral, the number of optically active products is zero.

Pattern Recognition

Symmetry check shortcut: When a symmetrical dialkene is cleaved, it yields symmetric fragments. The central carbon is attached to identical flanking aldehyde units post-cleavage, destroying any prior asymmetry and leaving 0 optically active compounds.

Chapter Mix

Class 11 Chemistry: Hydrocarbons Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Reference Study Guides

More Hydrocarbons Previous-Year Questions — Page 5

Q jee_main_2025_04_april_morning Properties of Benzene
Benzene is treated with oleum to produce compound (X) which when further heated with molten sodium hydroxide followed by acidification produces compound (Y). The compound Y is treated with zinc metal to produce compound (Z). Identify the structure of compound (Z) from the following options:
  • A.
  • B.
  • C.
  • D.

Solution

Core Logic

Let's map out this complete aromatic synthesis pathway:

  • Benzene + Oleum: Sulfonation steps take place to form Benzene Sulfonic Acid (C₆H₅SO₃H, Compound X).
  • Fusion with molten NaOH followed by H^+ activation: The sulfonic group is displaced, passing through a sodium phenoxide intermediate to yield Phenol (C₆H₅OH, Compound Y).
  • Phenol + Zinc dust distillation: Phenol undergoes clean deoxygenation reduction when heated with Zinc metal, stripping the hydroxyl group away to reform Benzene (Compound Z).
Pattern Recognition

Zinc dust distillation is a highly reliable reduction tool designed explicitly to strip phenolic hydroxyl groups away, leaving a clean unsubstituted aromatic ring behind.

Chapter Mix

Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Alcohols, Phenols and Ethers

Q jee_main_2025_04_april_morning Free Radical Bromination
Predict the major product of the following reaction sequence:
Alkyl radical halogenation flowchart matrix for Q42 - JEE Main 2025 Morning
The sequence pathways specify Br2/hv free radical substitution, alcoholic KOH elimination, and HBr/peroxide addition.
  • A.
  • B.
  • C.
  • D.

Solution

Core Logic

Let's analyze the steps of the reaction sequence:

  • Step 1 (Br₂ / hν): Light-induced free radical substitution targeted at the most stable tertiary position, producing 1-bromo-1-methylcyclohexane.
  • Step 2 (Alcoholic KOH, Δ): Dehydrohalogenation occurs via an E2 mechanism. Following Saytzeff's rule, elimination favors the formation of the more highly substituted, stable alkene: 1-methylcyclohexene.
  • Step 3 (HBr / R-O-O-R, hν): Radical hydrobromination across the unsymmetrical alkene. The presence of peroxide shifts addition toward the Anti-Markovnikov path, placing the bromine atom cleanly at the less-substituted secondary carbon to yield 1-bromo-2-methylcyclohexane.
Pattern Recognition

Combining Saytzeff elimination with a peroxide-promoted HBr addition allows you to reposition functional groups from highly substituted tertiary carbons to adjacent secondary positions.

Chapter Mix

Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Haloalkanes and Haloarenes

Q50 jee_main_2025_07_april_evening Reaction Mechanisms and Hybridization
Identify the structure of the final product (D) in the following sequence of the reactions:
Reaction Mechanisms scheme diagram for Q50 - JEE Main 2025 Evening
The image maps out a sequential multi-step chemical reaction scheme moving from acetophenone via gem-dichloride to an alkyne, hydroboration, and product D.
Total number of sp² hybridised carbon atoms in product D is .
Numerical Answer. Answer: 6.5 to 7.5

Solution

Related Formula
Terminal Alkyne (R-C ) [2. H₂O₂ / OH^-]1. B₂H₆ R-CH₂-CHO (Anti-Markovnikov Hydroboration-Oxidation)
Core Logic

Let's track the molecular changes at every intermediate junction:

  • Step 1: Acetophenone (Ph-CO-CH₃) reacts with PCl₅ to generate a gem-dichloride intermediate [A]: Ph-CCl₂-CH₃.
  • Step 2: Reaction with 3 equivalents of the incredibly strong base NaNH₂ triggers dual elimination to form a terminal sodium acetylide salt [B]: Ph-C ^-Na^+.
  • Step 3: Acidification yields phenylacetylene [C]: Ph-C.
  • Step 4: Hydroboration-oxidation of phenylacetylene leads to anti-Markovnikov water addition forming an enol structure, which immediately tautomerizes to [D] phenylacetaldehyde: Ph-CH₂-CHO.
Step 1: Counting Hybridized Carbons

The step transformations match the sequential tracking map:

Structural analysis product diagram for Q50 - JEE Main 2025 Evening
The image maps out a sequential multi-step chemical reaction scheme moving from acetophenone via gem-dichloride to an alkyne, hydroboration, and product D.

Let's locate all sp² hybridised carbon environments in product D (Ph-CH₂-CHO):

  • The aromatic benzene ring contains 6 sp² carbon atoms.
  • The aldehyde carbonyl carbon (-CHO) is double-bonded to oxygen, adding 1 sp² carbon atom.
  • The aliphatic link carbon (-CH₂-) is entirely sp³ hybridized.
  • Total sp² carbon count = 6 + 1 = 7.

Pattern Recognition

Alkyne oxidation mapping: Hydroboration-oxidation transforms a terminal alkyne into an aldehyde carbonyl group, while oxymercuration-demercuration yields a ketone carbonyl. Both introduce precisely one extra carbonyl sp² site on top of the original aromatic framework.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids Class 11 Chemistry: Hydrocarbons

Q27 jee_main_2025_24_jan_morning Electrophilic Addition to Alkenes
Following are the four molecules "P", "Q", "R" and "S":
Electrophilic Addition to Alkenes diagram for Q27 - JEE Main 2025 Morning
The image shows four cyclic and acyclic alkene molecules labelled P, Q, R, and S.
Which one among the four molecules will react with H-Br(aq) at the fastest rate?
  • A. S
  • B. Q
  • C. R
  • D. P

Solution

Related Formula
Rate of Electrophilic Addition ∝ Stability of Intermediate Carbocation
Core Logic

Addition of H-Br(aq) follows an electrophilic addition pathway where a carbocation intermediate is formed in the rate-determining step. Among the given structures, compound Q forms a resonance-stabilized allylic/benzylic carbocation, rendering it highly stable compared to the others.

Electrophilic Addition to Alkenes solution diagram for Q27 - JEE Main 2025 Morning
The image shows four cyclic and acyclic alkene molecules labelled P, Q, R, and S.

Pattern Recognition

Look for conjugated or allylic systems that stabilize the positive charge dynamically via resonance.

Chapter Mix

Class 11 Chemistry: Hydrocarbons

More Hydrocarbons Questions — jee_main_2025_07_april_evening

Practice all Hydrocarbons previous-year questions →

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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)