Hydrocarbons Previous Year Questions — JEE Main Chemistry
35 past-year Hydrocarbons questions from JEE Main (Chemistry) — page 5.
Q87 (2024)
Number of alkanes obtained on electrolysis of a mixture of $CH_3COONa$ and $C_2H_5COONa$ is
### Core Logic
Kolbe's electrolytic method generates free radicals at the anode, which then combine to form alkanes.
The given mixture yields two types of carboxylate radicals which decarboxylate to form alkyl radicals:
$CH_3COONa \rightarrow \dot{C}H_3$
$C_2H_5COONa \rightarrow \dot{C}_2H_5$
These radicals can couple in three different ways:
1. Cross coupling: $\dot{C}H_3 + \dot{C}_2H_5 \rightarrow CH_3-CH_2-CH_3$ (Propane)
2. Self-coupling 1: $\dot{C}H_3 + \dot{C}H_3 \rightarrow CH_3-CH_3$ (Ethane)
3. Self-coupling 2: $\dot{C}_2H_5 + \dot{C}_2H_5 \rightarrow CH_3-CH_2-CH_2-CH_3$ (Butane)
Thus, a total of 3 different alkanes are formed.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Q2098 (2026)
Identify A in the following reaction.
{{IMG1}}
### Core Logic
Let's analyze the given reaction sequences.
1. Reaction with $2H_2 / Pt$: Compound A undergoes hydrogenation with 2 moles of hydrogen, suggesting the presence of two double bonds.
2. Reaction with $\mathrm{KMnO_4} / \Delta$: Compound A undergoes oxidative cleavage to give a substituted dicarboxylic acid (cyclohexane-1,2-dicarboxylic acid) and oxalic acid ($HOOC-COOH$).
The structure of A must have a diene system that, when cleaved completely at the double bonds, yields these specific acid fragments.
{{SOL_IMG1}}
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.
{{SOL_IMG2}}
### Pattern Recognition
Hot $KMnO_4$ cleaves double bonds entirely, converting $=C-H$ into $-COOH$. The formation of oxalic acid indicates a $-CH=CH-$ fragment caught between two cleavable double bonds in a ring system.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Q61 (2026)
Given below are two statements:
Statement I: Benzene is nitrated to give nitrobenzene, which on further treatment with $CH_{3}COCl / AlCl_{3}$ will give
{{IMG1}}
Statement II: $NO_{2}$ 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.
- $\text{Statement I is correct but Statement II is incorrect.}$
- $\text{Both Statement I and Statement II are correct.}$
- $\text{Statement I is incorrect but Statement II is correct.}$
- $\text{Both Statement I and Statement II is are incorrect.}$
### Related Formula
Nitrobenzene + Friedel Crafts Acylation $\rightarrow$ No Reaction.
### Core Logic
Statement I claims that nitrobenzene undergoes Friedel-Crafts acylation with $CH_3COCl/AlCl_3$ to yield a product. This is strictly false. The $-NO_2$ 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 $NO_2$ 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.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Q71 (2026)
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 $g\text{ mol}^{-1}$ H: 1, C: 12, O: 16, Br: 80)
### Related Formula
$$\text{Mass percentage of Br} = \frac{\text{Mass of Br}}{\text{Molar mass of compound}} \times 100$$
### 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 $Br_2$ is consumed entirely per mole.
If $Br = 2$, then from the $C:Br = 3:1$ ratio, Carbon atoms $= 3 \times 2 = 6$.
So the cycloalkane (X) has 6 carbon atoms. Being a cycloalkane, its formula is $C_6H_{12}$.
Wait, the solution provided states $C_6H_{10} \xrightarrow{Br_2} C_6H_{10}Br_2$. This implies X is cyclohexene (a cycloalkene), not a cycloalkane!
Since we must follow the PDF's logic precisely: The PDF assumes $X$ is $C_6H_{10}$ (cyclohexene) despite the text saying 'cycloalkane'. It undergoes addition of $Br_2$ to form $C_6H_{10}Br_2$.
For $C_6H_{10}Br_2$:
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) $C_6H_{10}Br_2$:
$M = (12 \times 6) + (1 \times 10) + (80 \times 2)$
$M = 72 + 10 + 160 = 242\text{ g mol}^{-1}$
### Step 2: Calculate Percentage
$$\% \text{ of Br} = \frac{160}{242} \times 100 \approx 66.11\%$$
### 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.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q52 (2026)
Consider the following reaction:
{{IMG1}}
The product Y formed is:
- 2-methylhex-2-yne
- 5-methylhex-2-yne
- 2-methylhex-3-yne
- Isopropylbut-1-yne
### Related Formula
$$\text{Vicinal Dibromide} \xrightarrow{2\text{NaNH}_2} \text{Terminal Alkyne} (X) \xrightarrow[\text{alkyl halide}]{\text{NaNH}_2} \text{Internal Alkyne} (Y)$$
### Core Logic
Step 1: Dehydrohalogenation of dibromo compound using $\text{NaNH}_2$ yields terminal alkyne sodium acetylide intermediate $(X)$.
Step 2: Nucleophilic substitution ($S_N2$) of acetylide ion with isopropyl bromide yields 2-methylhex-3-yne as final product $(Y)$.
{{SOL_IMG1}}
{{SOL_IMG2}}
### 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.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Class 12 Chemistry: Haloalkanes and Haloarenes
