Hydrocarbons Previous Year Questions — JEE Main Chemistry
35 past-year Hydrocarbons questions from JEE Main (Chemistry) — page 2.
Q42 (2025)
Which compound would give 3-methyl-6-oxoheptanal upon ozonolysis?
- Structure (1)
- Structure (2)
- Structure (3)
- Structure (4)
### Related Formula
$$\text{Cyclic Alkene} \xrightarrow{\text{1) } O_3 \text{ 2) } Zn/H_2O} \text{Open-chain Dicarbonyl Compound}$$
### Core Logic
Reductive ozonolysis ($O_3, Zn/H_2O$) cleaves the double bond of 1,4-dimethylcyclohexene and replaces each carbon of the alkene with a carbonyl group ($C=O$).
3-methyl-6-oxoheptanal has a 7-carbon principal chain containing an aldehyde group at C-1, a methyl substituent at C-3, and a ketone group at C-6.
### Step 1: Ozonolysis Verification
Cleaving the endocyclic double bond in structure (2) produces:
$$\mathrm{CH_3-CO-CH_2-CH_2-CH(CH_3)-CH_2-CHO}$$
Numbering from the aldehyde functional group (C-1):
- C-1: $-\mathrm{CHO}$
- C-3: $-\mathrm{CH_3}$ methyl branch
- C-6: $-\mathrm{C=O}$ keto group
- C-7: terminal methyl group
IUPAC Name: 3-methyl-6-oxoheptanal.
### Pattern Recognition
Reverse ozonolysis: Connect aldehyde C-1 and keto C-6 back into a ring double bond.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Q938 (2025)
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:
### Core Logic
Let's map out this complete aromatic synthesis pathway:
1. **Benzene + Oleum:** Sulfonation steps take place to form Benzene Sulfonic Acid ($C_6H_5SO_3H$, Compound X).
2. **Fusion with molten $NaOH$ followed by $H^+$ activation:** The sulfonic group is displaced, passing through a sodium phenoxide intermediate to yield Phenol ($C_6H_5OH$, Compound Y).
3. **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.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Class 12 Chemistry: Alcohols, Phenols and Ethers
Q942 (2025)
Predict the major product of the following reaction sequence:
{{IMG1}}
### Core Logic
Let's analyze the steps of the reaction sequence:
1. **Step 1 ($Br_2 / h\nu$):** Light-induced free radical substitution targeted at the most stable tertiary position, producing 1-bromo-1-methylcyclohexane.
2. **Step 2 (Alcoholic $KOH, \Delta$):** Dehydrohalogenation occurs via an E2 mechanism. Following Saytzeff's rule, elimination favors the formation of the more highly substituted, stable alkene: 1-methylcyclohexene.
3. **Step 3 ($HBr / R-O-O-R, h\nu$):** 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.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Class 12 Chemistry: Haloalkanes and Haloarenes
Q41 (2025)
The number of optically active products obtained from the complete ozonolysis of the given compound is: [cite: 364, 365]
{{IMG1}}
- $2$
- $0$
- $1$
- $4$
### Related Formula
$$\text{R}_1\text{-CH}=\text{CH-R}_2 \xrightarrow[\text{Zn / H}_2\text{O}]{\text{O}_3} \text{R}_1\text{-CHO} + \text{R}_2\text{-CHO}$$
### Core Logic
Complete oxidative cleavage of all double bonds via reductive ozonolysis breaks the molecule into smaller fragments:
$$\text{CH}_3\text{-CH}=\text{CH-CH}(\text{CH}_3)\text{-CH}=\text{CH-CH}(\text{CH}_3)\text{-CH}=\text{CH-CH}_3$$
Let's trace the fragmentation logic mapping visually:
{{SOL_IMG1}}
### Step 1: Tracking Fragment Structures
The chemical reaction outputs two major molecular product types:
1. $\text{CH}3\text{-CHO}$ (Acetaldehyde): Optically inactive as it lacks a chiral carbon.
2. $\text{OHC-CH}(\text{CH}_3)\text{-CHO}$ (2-methylpropanedial): Let's inspect the substituted central carbon. It is bonded to: a hydrogen atom ($-\text{H}$), a methyl group ($-\text{CH}_3$), and two identical formyl groups ($-\text{CHO}$).
Because two of the groups are identical ($-\text{CHO}$), 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.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q50 (2025)
Identify the structure of the final product (D) in the following sequence of the reactions:
{{IMG1}}
Total number of $sp^2$ hybridised carbon atoms in product D is $\dots$.
### Related Formula
$$\text{Terminal Alkyne } (\text{R-C}\equiv\text{CH}) \xrightarrow[2.\,\text{H}_2\text{O}_2 / \text{OH}^-]{1.\,\text{B}_2\text{H}_6} \text{R-CH}_2\text{-CHO} \quad \text{(Anti-Markovnikov Hydroboration-Oxidation)}$$
### Core Logic
Let's track the molecular changes at every intermediate junction:
- Step 1: Acetophenone ($\text{Ph-CO-CH}_3$) reacts with $\text{PCl}_5$ to generate a gem-dichloride intermediate [A]: $\text{Ph-CCl}_2\text{-CH}_3$.
- Step 2: Reaction with $3$ equivalents of the incredibly strong base $\text{NaNH}_2$ triggers dual elimination to form a terminal sodium acetylide salt [B]: $\text{Ph-C}\equiv\text{C}^-\text{Na}^+$.
- Step 3: Acidification yields phenylacetylene [C]: $\text{Ph-C}\equiv\text{CH}$.
- Step 4: Hydroboration-oxidation of phenylacetylene leads to anti-Markovnikov water addition forming an enol structure, which immediately tautomerizes to [D] phenylacetaldehyde: $\text{Ph-CH}_2\text{-CHO}$.
### Step 1: Counting Hybridized Carbons
The step transformations match the sequential tracking map:
{{SOL_IMG1}}
Let's locate all $sp^2$ hybridised carbon environments in product D ($\text{Ph-CH}_2\text{-CHO}$):
1. The aromatic benzene ring contains 6 $sp^2$ carbon atoms.
2. The aldehyde carbonyl carbon ($-\text{CHO}$) is double-bonded to oxygen, adding 1 $sp^2$ carbon atom.
3. The aliphatic link carbon ($-\text{CH}_2-$) is entirely $sp^3$ hybridized.
Total $sp^2$ 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 $sp^2$ site on top of the original aromatic framework.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Class 11 Chemistry: Hydrocarbons
