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

Year	2026	2025	2024	Total
Questions	14	12	9	35

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Which compound would give 3-methyl-6-oxoheptanal upon ozonolysis ?

A. Structure (1)
Ozonolysis of Alkenes
B. Structure (2)
Ozonolysis of Alkenes
C. Structure (3)
Ozonolysis of Alkenes
D. Structure (4)
Ozonolysis of Alkenes

Solution & Explanation

Core Logic

Let us reconstruct the original alkene from the given ozonolysis product fragments. Write down the line-structure formula for 3-methyl-6-oxoheptanal:

O=CH-CH₂-CH(CH₃)-CH₂-CH₂-C(=O)-CH₃

Remove both carbonyl oxygen atoms and link carbon-1 directly to carbon-6 with a double bond. This cyclizes into a 6-membered ring structure: 1,4-dimethylcyclohexene.

Alkene cyclization path for Q42 - JEE Main 2025 Morning

Step 1: Ozonolysis Verification

Performing reductive ozonolysis ( $O₃ / Zn, H₂O$ ) cleaves the internal endocyclic double bond of 1,4-dimethylcyclohexene, perfectly regenerating the acyclic keto-aldehyde compound.

Pattern Recognition

Shortcut: Count the carbons in the main chain product ( $7 carbons in main chain, 8 total$ ). Ozonolysis of structure (2) creates an open chain containing a terminal aldehyde group on one end and a methyl ketone group on the other.

Evaluation Rubric / Model Answer

Option (B)

Chapter Mix

Class 11 Chemistry: Hydrocarbons

More Hydrocarbons Previous-Year Questions

Q jee_main_2026_21_jan_morning Alkenes Oxidation and Reduction

Identify A in the following reaction.

Alkenes Oxidation and Reduction diagram for Q53 - JEE Main 2026 Morning — A chemical reaction scheme converting an unknown compound A into different cyclic structures using specific reagents.

A.
B.
C.
D.

Solution

Core Logic

Let's analyze the given reaction sequences.

Reaction with $2H₂ / Pt$ $2H_2 / Pt$: Compound A undergoes hydrogenation with 2 moles of hydrogen, suggesting the presence of two double bonds.
Reaction with $KMnO₄ / Δ$ $\mathrm{KMnO_4} / \Delta$: Compound A undergoes oxidative cleavage to give a substituted dicarboxylic acid (cyclohexane-1,2-dicarboxylic acid) and oxalic acid ( $HOOC-COOH$ $HOOC-COOH$).

The structure of A must have a diene system that, when cleaved completely at the double bonds, yields these specific acid fragments.

Alkenes Oxidation and Reduction diagram for Q53 - JEE Main 2026 Morning — A chemical reaction scheme converting an unknown compound A into different cyclic structures using specific reagents.

By matching the cleavage points, compound A is identified as 1,2,3,4,4a,5,8,8a-octahydronaphthalene derivative with two isolated double bonds in the ring.

Alkenes Oxidation and Reduction diagram for Q53 - JEE Main 2026 Morning — A chemical reaction scheme converting an unknown compound A into different cyclic structures using specific reagents.

Pattern Recognition

Hot $KMnO₄$ $KMnO_4$ cleaves double bonds entirely, converting $=C-H$ $=C-H$ into $-COOH$ $-COOH$. The formation of oxalic acid indicates a $-CH=CH-$ $-CH=CH-$ fragment caught between two cleavable double bonds in a ring system.

Chapter Mix

Class 11 Chemistry: Hydrocarbons

Q61 jee_main_2026_22_january_morning Electrophilic Aromatic Substitution

Given below are two statements: Statement I: Benzene is nitrated to give nitrobenzene, which on further treatment with

CH₃COCl / AlCl₃

will give

Acylated nitrobenzene diagram for Q61 - JEE Main 2026 Morning — Image depicts m-nitroacetophenone.

Statement II:

NO₂

group is a m-directing, and deactivating group. In the light of the above statements, choose the most appropriate answer from the options given below.

A. $Statement I is correct but Statement II is incorrect.$
B. $Both Statement I and Statement II are correct.$
C. $Statement I is incorrect but Statement II is correct.$
D. $Both Statement I and Statement II is are incorrect.$

Solution

Related Formula

Nitrobenzene + Friedel Crafts Acylation $arrow$ $\rightarrow$ No Reaction.

Core Logic

Statement I claims that nitrobenzene undergoes Friedel-Crafts acylation with $CH₃COCl/AlCl₃$ $CH_3COCl/AlCl_3$ to yield a product. This is strictly false. The $-NO₂$ $-NO_2$ group is highly electron-withdrawing, meaning it severely deactivates the benzene ring toward electrophilic aromatic substitution. Nitrobenzene does not undergo Friedel-Crafts alkylation or acylation.

Statement II correctly states that the $NO₂$ $NO_2$ group is a meta-directing and deactivating group. Because it withdraws electron density, the ring is less reactive than benzene itself, and any substitution that does occur (under harsh conditions) happens at the meta position.

Step 1: Final Conclusion

Statement I is incorrect but Statement II is correct.

Pattern Recognition

Highly deactivated rings (e.g., nitrobenzene) act as a solvent in Friedel-Crafts reactions precisely because they are completely unreactive to the alkylating/acylating conditions.

Chapter Mix

Class 11 Chemistry: Hydrocarbons

Q71 jee_main_2026_22_january_morning Alkenes and Alkynes

The cycloalkane (X) on bromination consumes one mole of bromine per mole of (X) and gives the product (Y) in which C:Br ratio is 3 : 1. The percentage of bromine in the product (Y) is ____%. (Nearest integer) (Given: Molar mass in

g mol⁻¹

H: 1, C: 12, O: 16, Br: 80)

Numerical Answer. Answer: 66 to 66

Solution

Related Formula

Mass percentage of Br = Mass of BrMolar mass of compound \times 100

Core Logic

The given compound is a cycloalkane. Since it consumes one mole of bromine and undergoes addition/substitution giving a C:Br ratio of 3:1, let's analyze the formula. Bromination of an alkane normally requires UV light and is a substitution reaction, yielding mono or dibromo products. But the problem says it "consumes one mole of bromine" to give a product. If it's a cyclopropane or cyclobutane ring, it might undergo ring-opening addition. Let's assume the product has 2 bromine atoms since $Br₂$ $Br_2$ is consumed entirely per mole. If $Br = 2$ $Br = 2$, then from the $C:Br = 3:1$ $C:Br = 3:1$ ratio, Carbon atoms $= 3 \times 2 = 6$ $= 3 \times 2 = 6$. So the cycloalkane (X) has 6 carbon atoms. Being a cycloalkane, its formula is $C₆H₁₂$ $C_6H_{12}$. Wait, the solution provided states $C₆H₁₀ Br₂ C₆H₁₀Br₂$ . This implies X is cyclohexene (a cycloalkene), not a cycloalkane! Since we must follow the PDF's logic precisely: The PDF assumes $X$ $X$ is $C₆H₁₀$ $C_6H_{10}$ (cyclohexene) despite the text saying 'cycloalkane'. It undergoes addition of $Br₂$ $Br_2$ to form $C₆H₁₀Br₂$ $C_6H_{10}Br_2$. For $C₆H₁₀Br₂$ $C_6H_{10}Br_2$: Number of C = 6, Number of Br = 2. C:Br ratio = 6:2 = 3:1. This perfectly fits the given condition.

Step 1: Calculate Molar Mass

Molar mass of product (Y) $C₆H₁₀Br₂$ $C_6H_{10}Br_2$: $M = (12 \times 6) + (1 \times 10) + (80 \times 2)$ $M = 72 + 10 + 160 = 242 g mol⁻¹$

Step 2: Calculate Percentage

% of Br = (160)/(242) \times 100 \approx 66.11%

Step 3: Rounding

Nearest integer is 66.

Pattern Recognition

Be prepared to spot exam typos (like 'cycloalkane' instead of 'cycloalkene') when a fixed atomic ratio constraint immediately points to a specific molecular formula.

Chapter Mix

Class 11 Chemistry: Hydrocarbons Class 11 Chemistry: Some Basic Concepts of Chemistry

Q52 jee_main_2026_22_january_evening Alkyne Synthesis and Alkylation

Consider the following reaction:

Alkyne reaction scheme for Q52 - JEE Main 2026 Evening — Displays a vicinal dibromide reacting with NaNH2 to form terminal alkyne intermediate X, followed by alkylation to yield product Y.

The product Y formed is:

A. 2-methylhex-2-yne
B. 5-methylhex-2-yne
C. 2-methylhex-3-yne
D. Isopropylbut-1-yne

Solution

Related Formula

Vicinal Dibromide 2NaNH₂ Terminal Alkyne (X) [alkyl halide]NaNH₂ Internal Alkyne (Y)

Core Logic

Step 1: Dehydrohalogenation of dibromo compound using $NaNH₂$ $\text{NaNH}_2$ yields terminal alkyne sodium acetylide intermediate $(X)$ $(X)$.

Step 2: Nucleophilic substitution ( $SN2$ $S_N2$) of acetylide ion with isopropyl bromide yields 2-methylhex-3-yne as final product $(Y)$ $(Y)$.

Detailed mechanism of alkyne formation for Q52 - JEE Main 2026 Evening — Displays a vicinal dibromide reacting with NaNH2 to form terminal alkyne intermediate X, followed by alkylation to yield product Y.

Detailed mechanism of alkyne formation for Q52 - JEE Main 2026 Evening — Displays a vicinal dibromide reacting with NaNH2 to form terminal alkyne intermediate X, followed by alkylation to yield product Y.

Pattern Recognition

Sees: Double elimination followed by alkylation of acetylide ion. Shortcut: Count main chain carbon skeleton including the added isopropyl fragment to find IUPAC name: 2-methylhex-3-yne.

Chapter Mix

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

Q60 jee_main_2026_23_january_morning Alkenes Preparation from Alkynes

But-2-yne and hydrogen (one mole each) are separately treated with (i) Pd/C and (ii)

Na/liq. NH₃

to give the products X and Y respectively.

Alkenes Preparation from Alkynes diagram for Q60 - JEE Main 2026 Morning — The image shows the partial reduction pathways of But-2-yne using Lindlar-type conditions and dissolving metal conditions.

Identify the incorrect statements. A. X and Y are stereoisomers. B. Dipole moment of X is zero C. Boiling point of X is higher than Y. D. X and Y react with

O₃/Zn + H₂O

to give different products. Choose the correct answer from the options given below :

A. $B and C only$ $\text{B and C only}$
B. $B and D only$ $\text{B and D only}$
C. $A and B only$ $\text{A and B only}$
D. $A and C only$ $\text{A and C only}$

Solution

Core Logic

Identify the geometrical isomers formed by different hydrogenation conditions. $Pd/C$ $Pd/C$ (or Lindlar's catalyst) provides syn-addition forming the cis-alkene (X). $Na/liq. NH₃$ $Na/liq. NH_3$ (Birch reduction) provides anti-addition forming the trans-alkene (Y).

Step 1: Structural Analysis

X = cis-But-2-ene. Y = trans-But-2-ene.

Alkenes Preparation from Alkynes diagram for Q60 - JEE Main 2026 Morning — The image shows the partial reduction pathways of But-2-yne using Lindlar-type conditions and dissolving metal conditions.

Step 2: Checking Statements

A. X and Y are geometrical isomers, a subset of stereoisomers. (Correct statement) B. Dipole moment of X (cis) is non-zero because the dipole moments of the two $CH₃$ $CH_3$ groups reinforce each other, whereas Y (trans) has a zero dipole moment due to symmetry cancellation. (Incorrect statement) C. Cis-isomers generally have higher boiling points than trans-isomers because they are more polar, leading to stronger intermolecular dipole-dipole forces. (Correct statement) D. Both cis and trans-But-2-ene yield the exact same product upon reductive ozonolysis ( $O₃ / Zn + H₂O$ $O_3 / Zn + H_2O$): 2 moles of ethanal (acetaldehyde). (Incorrect statement)

Step 3: Final Conclusion

Statements B and D are incorrect.

Pattern Recognition

Lindlar = cis (polar, higher BP), Birch = trans (non-polar symmetric, zero dipole). Ozonolysis breaks the double bond structurally and ignores original stereochemistry.

Chapter Mix

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

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