Hydrocarbons Previous Year Questions — JEE Main Chemistry
35 past-year Hydrocarbons questions from JEE Main (Chemistry) — page 3.
Q27 (2025)
Following are the four molecules "P", "Q", "R" and "S":
{{IMG1}}
Which one among the four molecules will react with $H-Br(aq)$ at the fastest rate?
- S
- Q
- R
- P
### Related Formula
$$\text{Rate of Electrophilic Addition} \propto \text{Stability of Intermediate Carbocation}$$
### 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. {{SOL_IMG1}}
### Pattern Recognition
Look for conjugated or allylic systems that stabilize the positive charge dynamically via resonance.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Q29 (2025)
Identify product [A], [B] and [C] in the following reaction sequence :
$$CH_{3}-C\equiv CH\frac{Pd/C}{H_{2}}\rightarrow[A]\frac{(i)O_{3}}{(ii)Zn,H_{2}O}\rightarrow[B]+[C]$$
- $[A]:CH_{3}-CH=CH_{2},\ [B]: CH_{3}CHO,\ [C]: HCHO$
- $[A]: CH_2=CH_2,\ [B]: H_3C-CO-CH_3,\ [C]: HCHO$
- $[A]:CH_{3}-CH=CH_{2},\ [B]: CH_{3}CHO,\ [C]: CH_{3}CH_{2}OH$
- $[A]: CH_{3}CH_{2}CH_{3},\ [B]: CH_{3}CHO,\ [C]: HCHO$
### Related Formula
Partial hydrogenation of alkynes using $Pd/C$ yields alkenes:
$$R-C\equiv CH + H_2 \xrightarrow{Pd/C} R-CH=CH_2$$
Ozonolysis cleaves the double bond to form carbonyls:
$$R-CH=CH_2 \xrightarrow{(i)O_3, (ii)Zn/H_2O} R-CHO + HCHO$$
### Core Logic
Step 1: Controlled reduction of propyne gives propene:
$$CH_3-C\equiv CH \xrightarrow{Pd/C, H_2} CH_3-CH=CH_2 \quad [A]$$
Step 2: Reductive ozonolysis of propene ($[A]$) splits the alkene at the $C=C$ bond, creating ethanal ($[B]$) and methanal ($[C]$):
$$CH_3-CH=CH_2 \xrightarrow{O_3, \text{ then } Zn/H_2O} CH_3CHO\ [B] + HCHO\ [C]$$
### Step 1: Final Identification
Hence,
$[A] = CH_3-CH=CH_2$
$[B] = CH_3CHO$
$[C] = HCHO$
### Pattern Recognition
Whenever an alkyne undergoes partial hydrogenation with regular catalysts, count the carbons to trace the matching alkene framework. Cleaving a terminal alkene like propene always results in formaldehyde ($HCHO$) as one of the fragment products.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Q1474 (2024)
Which of the following reaction is correct?
### Related Formula
$$\text{Alkene} + \text{HX} \rightarrow \text{Alkyl Halide} \quad \text{(Markovnikov's Rule)}$$
### Core Logic
The reaction involving the addition of $\text{HI}$ to the double bond follows electrophilic addition mechanism governed by Markovnikov's rule. The proton adds to the less substituted carbon to generate the more stable tertiary carbocation intermediate, which is then attacked by $\text{I}^-$ to yield the corresponding major tertiary alkyl iodide.
### Step 1: Verification of Choices
Option (2) perfectly captures the sound execution of Markovnikov addition mechanics whereas the remaining choices present invalid product distributions or faulty reaction stoichiometry templates.
### Pattern Recognition
Electrophilic addition cleanly forms the most substituted, most stable carbocation before halide attack takes place.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Q1661 (2024)
Identify product A and product B :
{{IMG1}}
### Core Logic
The substrate is cyclohexene, which can undergo two distinct types of reactions with chlorine ($Cl_2$) depending on the reaction conditions.
**Reaction 1: Condition $hv$ (Product A)**
In the presence of light ($h\nu$) or high temperature, halogens undergo homolytic cleavage to generate free radicals. This triggers allylic substitution (a free radical substitution mechanism). The allylic position is targeted because the resulting allylic free radical is resonance stabilized. Thus, substitution occurs at the carbon adjacent to the double bond, yielding 3-chlorocyclohexene as Product A.
**Reaction 2: Condition $CCl_4$ (Product B)**
In the presence of a non-polar solvent like $CCl_4$ and without light/heat, $Cl_2$ undergoes an electrophilic addition reaction across the carbon-carbon double bond. A cyclic chloronium ion intermediate is formed, leading to anti-addition of two chlorine atoms. This yields 1,2-dichlorocyclohexane as Product B.
### Step 1: Structures of A and B
{{SOL_IMG1}}
Product A preserves the double bond and substitutes a Cl at the allylic position.
Product B loses the double bond and adds two Cl atoms adjacently.
### Pattern Recognition
$X_2$ + light ($h\nu$) = Free radical Allylic Substitution.
$X_2$ + dark/solvent ($CCl_4$) = Electrophilic Addition across double bond.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Class 12 Chemistry: Haloalkanes and Haloarenes
Q1680 (2024)
Consider the given reaction.
$$\text{CH}_3-\text{CH}=\text{C}(\text{CH}_3)_2 \xrightarrow[(ii)\ \text{Zn/H}_2\text{O}]{(i)\ \text{O}_3} (\text{P})$$
The total number of oxygen atoms present per molecule of the product (P) is
### Core Logic
The reaction given is the reductive ozonolysis of an alkene, 2-methylbut-2-ene ($CH_3-CH=C(CH_3)_2$).
In reductive ozonolysis ($O_3$ followed by $Zn/H_2O$), the carbon-carbon double bond is completely cleaved. An oxygen atom is placed on each carbon atom of the broken double bond to form carbonyl compounds (aldehydes or ketones).
### Step 1: Identifying the Products
$$CH_3-CH=C(CH_3)_2 \xrightarrow{O_3 / Zn, H_2O} CH_3-CHO + O=C(CH_3)_2$$
The reaction yields two distinct product molecules:
1. Acetaldehyde ($CH_3CHO$) - contains 1 oxygen atom.
2. Acetone ($CH_3COCH_3$) - contains 1 oxygen atom.
The question asks for the number of oxygen atoms present **per molecule** of the product (P). Since any resulting product molecule (either acetaldehyde or acetone) contains exactly 1 oxygen atom, the answer is 1.
### Pattern Recognition
Reductive ozonolysis of a simple alkene without other oxygenated functional groups always creates simple aldehydes or ketones. Each resulting discrete molecule formed from the cleaved double bond will have exactly 1 carbonyl group (1 oxygen atom) unless it's a cyclic alkene opening up (which would have 2).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids
