Hydrocarbons Previous Year Questions — JEE Main Chemistry
35 past-year Hydrocarbons questions from JEE Main (Chemistry) — page 7.
Q70 (2026)
Given below are two statements :
Statement I : There are several conformers for n-butane. Out of those conformers,
{{IMG1}}
(X)
the least stable and most stable conformer is
{{IMG2}}
(Y)
Statement II : As the dihedral angle increases, torsional strain decreases from (X) to (Y).
In the light of the above statements, choose the correct answer from the options given below :
- Both Statement I and Statement II are false
- Statement I is false but Statement II is true
- Statement I is true but Statement II is false
- Both Statement I and Statement II are true
### Core Logic
Statement I: Conformer (X) represents the fully eclipsed conformation where the two bulky methyl groups are at a dihedral angle of $0^{\circ}$, making it the least stable due to maximum torsional and steric strain. Conformer (Y) is the anti-staggered conformation where the methyl groups are at $180^{\circ}$ apart, rendering it the most stable. Thus, Statement I is true.
Statement II: Moving from (X) to (Y) increases the dihedral angle from $0^{\circ}$ to $180^{\circ}$. As the bonds move farther apart, the repulsion between bonding electron pairs minimizes, meaning torsional strain indeed decreases. Thus, Statement II is true.
### Step 1: Final Conclusion
Both Statement I and Statement II are true.
### Pattern Recognition
Anti (dihedral = $180^{\circ}$) is the stability zenith for n-butane. Fully eclipsed (dihedral = $0^{\circ}$) is the energetic peak (least stable). Strain scales inversely with spatial separation of bulky/repulsive groups.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Q58 (2026)
Given below are two statements for the following reaction sequence.
{{IMG1}}
Statement I: Compound 'Z' will give yellow precipitate with NaOI.
Statement II: Compound 'Q' has two different types of 'H' atoms (aromatic : aliphatic) in the ratio 1 : 3.
In the light of the above statements, choose the correct answer from the option given below:
- Statement I is true but Statement II is false
- Both Statement I and Statement II are true
- Statement I is false but Statement II is true
- Both Statement I and Statement II are false
### Core Logic
From the sequence provided:
$C_3H_6Cl_2$ (X) on treating with excess $\mathrm{NaNH}_2$ undergoes double dehydrohalogenation to give propyne, $CH_3-C\equiv CH$ (Compound Y).
{{SOL_IMG1}}
Compound Y on passing through a red-hot iron tube trimerizes to form 1,3,5-trimethylbenzene (Mesitylene), which is Compound Q ($C_9H_{12}$).
Compound Y on hydration with dil. $\mathrm{H_2SO_4}/\mathrm{HgSO_4}$ gives acetone, $CH_3-CO-CH_3$ (Compound Z).
### Step 1: Analyze Statement I
Compound Z is acetone ($CH_3-CO-CH_3$). It contains a methyl ketone group ($CH_3-CO-$), so it will give a positive Iodoform test (yellow precipitate with $\mathrm{NaOI}$). Thus, Statement I is true.
### Step 2: Analyze Statement II
Compound Q (Mesitylene) has a symmetrical structure with 3 identical aromatic protons and 9 identical aliphatic protons (from the three methyl groups).
Ratio of aromatic H : aliphatic H = 3 : 9 = 1 : 3.
Thus, Statement II is true.
### Final Conclusion
Both Statement I and Statement II are correct.
### Pattern Recognition
Propyne trimerization always yields 1,3,5-trimethylbenzene. Acetone is the classic positive target for the iodoform reaction.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids
Q65 (2026)
$\text{Ph-CH=CH}_{2} \xrightarrow[\text{HBr}]{\text{(PhCOO)}_{2}} \text{Product}$
Consider the above reaction
A. The reaction proceeds through a more stable radical intermediate.
B. The role of peroxide is to generate $\dot{H}$ (Hydrogen radical).
C. During this reaction, benzene is formed as a biproduct.
D. 1-Bromo-2-phenylethane is fanned as the minor product.
E. The same reaction in absence of peroxide proceeds via carbocation intermediate.
Identify the correct statements. Choose the correct answer from the options given below:
- A & E Only
- A, B & D Only
- C, D & E Only
- A, C & E Only
### Core Logic
This is the classic Peroxide Effect (Kharasch effect) showing Anti-Markovnikov addition of $\mathrm{HBr}$ across an alkene.
$\mathrm{Ph-CH=CH_2 + HBr} \xrightarrow{\text{Peroxide}} \mathrm{Ph-CH_2-CH_2-Br}$ (1-Bromo-2-phenylethane is the MAJOR product).
### Step 1: Analyze Radical Mechanism (A, B, C)
$\bullet$ The reaction proceeds via the most stable free radical intermediate (a secondary benzylic radical). Statement A is correct.
$\bullet$ Peroxide (Benzoyl peroxide) undergoes homolytic cleavage to form benzoyloxy radicals, which lose $CO_2$ to form phenyl radicals ($\dot{Ph}$). Phenyl radical abstracts H from HBr to form a Bromine radical ($\dot{Br}$), not a Hydrogen radical ($\dot{H}$). Statement B is false.
$\bullet$ The abstraction of H from HBr by the phenyl radical ($\dot{Ph}$) yields Benzene ($Ph-H$) as a byproduct. Statement C is correct.
### Step 2: Product Check (D, E)
$\bullet$ 1-Bromo-2-phenylethane is the *major* product, not the minor product. Statement D is false.
$\bullet$ In the absence of peroxide, standard electrophilic addition occurs via a stable carbocation intermediate (Markovnikov addition). Statement E is correct.
### Final Conclusion
Statements A, C, and E are correct.
### Pattern Recognition
Peroxides generate $\dot{Br}$, never $\dot{H}$. The initiating radical pulls H from H-Br since H-Br is weaker than forming C-H bond, leaving $\dot{Br}$ to attack the alkene.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Q51 (2026)
Identify the correct statements:
The presence of $-NO_{2}$ group in benzene ring
(A) activates the ring towards electrophilic substitutions.
(B) deactivates the ring towards electrophilic substitutions.
(C) activates the ring towards nucleophilic substitutions.
(D) deactivates the ring towards nucleophilic substitutions.
- $(1)\text{ B and D Only}$
- $(2)\text{ C and A Only}$
- $(3)\text{ A and D Only}$
- $(4)\text{ B and C Only}$
### Core Logic
Presence of $-NO_{2}$ group in Benzene ring strongly deactivates the ring towards electrophilic substitution reaction due to $-M$ effect.
Conversely, it activates the ring towards nucleophilic substitution reactions by stabilizing the intermediate carbanion.
### Step 1: Final Conclusion
Statements B and C are correct.
### Pattern Recognition
Electron-withdrawing groups (EWG) like $-NO_2$ withdraw electron density, raising the activation energy for electrophilic attacks while lowering it for nucleophilic attacks.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Class 12 Chemistry: Haloalkanes and Haloarenes
Q56 (2026)
The reactions which produce alcohol as the product area:
(A) $CH_{4} + O_{2} \xrightarrow[\Delta]{Mo_{2}O_{3}}$
(B) $2CH_{3}CH_{3} + 3O_{2} \xrightarrow[\Delta]{(CH_{3}COO)_{2}Mn}$
(C) $(CH_{3})_{3}CH \xrightarrow{KMnO_{4}}$
(D) $2CH_{4} + O_{2} \xrightarrow{Cu/523K/100atm.}$
(E) $CH_{3}-CH=CH-CH_{3} \xrightarrow{KMnO_{4}/H^{+}}$
Choose the correct answer from the options given below:
- $(1)\text{ A and D Only}$
- $(2)\text{ A, C and E Only}$
- $(3)\text{ C and D Only}$
- $(4)\text{ B, D and E Only}$
### Core Logic
Analyzing each oxidation reaction:
(A) $CH_{4} + O_{2} \xrightarrow[\Delta]{Mo_{2}O_{3}} HCHO + H_2O$
(Yields formaldehyde, an aldehyde)
(B) $2CH_{3}CH_{3} + 3O_{2} \xrightarrow[\Delta]{(CH_{3}COO)_{2}Mn} 2CH_{3}COOH + 2H_2O$
(Yields acetic acid, a carboxylic acid)
(C) $(CH_{3})_{3}CH \xrightarrow{KMnO_{4}} (CH_3)_3C-OH$
(Oxidation of $3^{\circ}$ alkane yields tert-butyl alcohol)
{{SOL_IMG1}}
(D) $2CH_{4} + O_{2} \xrightarrow[523\text{K, } 100\text{ atm}]{Cu} 2CH_{3}OH$
(Controlled oxidation yields methanol)
(E) $CH_{3}-CH=CH-CH_{3} \xrightarrow{KMnO_{4}/H^{+}} 2CH_{3}COOH$
(Strong oxidative cleavage yields acetic acid)
### Step 1: Final Conclusion
Reactions (C) and (D) yield alcohols as products. Therefore, C and D only.
### Pattern Recognition
Molybdenum catalyst yields aldehydes. Copper catalyst yields alcohols. $KMnO_4$ attacks tertiary hydrogens to give tertiary alcohols, while vigorously cleaving alkenes to carboxylic acids.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
