Hydrocarbons Previous Year Questions — JEE Main Chemistry
35 past-year Hydrocarbons questions from JEE Main (Chemistry).
Q403 (2025)
Given below are two statements:
Statement (I): Neopentane forms only one monosubstituted derivative.
Statement (II): Melting point of neopentane is higher than n-pentane
In the light of the above statements, choose the most appropriate answer from the options given below:
- $\text{Statement I is correct but Statement II is incorrect}$
- $\text{Both Statement I and Statement II are correct}$
- $\text{Both Statement I and Statement II are incorrect}$
- $\text{Statement I is incorrect but Statement II is correct}$
### Related Formula
$$\text{Symmetric Structure} \propto \text{Melting Point (Packing efficiency)}$$
### Core Logic
**Statement (I) is correct**: Neopentane (2,2-dimethylpropane) contains a quaternary carbon bonded to four methyl groups. There are 12 hydrogen atoms, all of which are primary and chemically equivalent. Halogenation yields exactly one monosubstituted derivative:
$$\mathrm{(CH_3)_4C + X_2 \xrightarrow{h\nu} (CH_3)_3C-CH_2X + HX}$$
Statement (II) is correct: Neopentane has a compact, symmetrical, nearly spherical molecular structure. In the solid crystal lattice, these spherical molecules pack much more efficiently compared to the floppy, linear n-pentane. This robust crystalline packing dramatically increases its melting point ($256.4~\mathrm{K}$) compared to that of n-pentane ($143.4~\mathrm{K}$).
### Step 1: Final Verification
Since both statements are theoretically and experimentally correct, the correct option is (2).
{{SOL_IMG1}}
### Pattern Recognition
While branching *decreases* the boiling point (due to decreased surface area and weaker van der Waals forces), branching that creates a highly symmetric structure *increases* the melting point because of close-packing efficiency in the solid phase.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Q183 (2025)
In the following series of reactions identify the major products A & B respectively:
{{IMG1}}
### Related Formula
Orienting effects in electrophilic aromatic substitution:
- Bromine ($-Br$) is ortho/para-directing (para-dominated due to steric hindrance).
- Sulfonic acid group ($-SO_3H$) is a strong deactivating, meta-directing group.
{{SOL_IMG1}}
### Core Logic
Analyzing the first step:
- Sulfonation of bromobenzene with $\mathrm{SO}_3/\mathrm{H}_2\mathrm{SO}_4$ yields 4-bromobenzenesulfonic acid as the major product (A) due to steric hindrance at the ortho-position.
### Step 1: Determine orientation for the second step
In 4-bromobenzenesulfonic acid, we have two substituents:
- $-Br$ (ortho/para director)
- $-SO_3H$ (meta director)
The positions meta to the deactivating $-SO_3H$ group correspond to the positions ortho to the $-Br$ group. Both directing effects align on the same position (carbon-3/carbon-5). Since $-Br$ is activating relative to $-SO_3H$, it controls the orientation.
### Step 2: Identify Product B
Halogenation with $\mathrm{Br}_2/\mathrm{Fe}$ introduces a bromine atom ortho to the existing bromine atom (meta to $-SO_3H$):
$$\text{Product B} = \text{3,4-dibromobenzenesulfonic acid}$$
This matches Option (2).
### Pattern Recognition
When an activating group ($-Br$) and a deactivating group ($-SO_3H$) compete on a benzene ring, the orienting influence of the activating group wins. Position ortho to the bromine atom is favored over meta positions of the sulfonic acid group.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Class 12 Chemistry: Haloalkanes and Haloarenes
Q26 (2025)
Given below are two statements:
Statement I: Ozonolysis followed by treatment with $\mathrm{Zn}$, $\mathrm{H}_2\mathrm{O}$ of cis-2-butene gives ethanal.
Statement II: The product obtained by ozonolysis followed by treatment with $\mathrm{Zn}$, $\mathrm{H}_2\mathrm{O}$ of 3,6-dimethyloct-4-ene has no chiral carbon atom.
In the light of the above statements, choose the correct answer from the options given below:
- $\text{Both Statement I and Statement II are true}$
- $\text{Statement I is false but Statement II is true}$
- $\text{Statement I is true but Statement II is false}$
- $\text{Both Statement I and Statement II are false}$
### Related Formula
$$\text{R}_1-\text{CH}=\text{CH}-\text{R}_2 \xrightarrow[\text{(ii) Zn, H}_2\text{O}]{\text{(i) O}_3} \text{R}_1-\text{CHO} + \text{R}_2-\text{CHO}$$
### Core Logic
Statement I: Cis-2-butene ($CH_3-CH=CH-CH_3$) undergoes reductive ozonolysis to cleave the double bond and yield two molecules of acetaldehyde (ethanal, $CH_3CHO$). This statement is true.
Statement II: 3,6-dimethyloct-4-ene ($CH_3-CH_2-CH(CH_3)-CH=CH-CH(CH_3)-CH_2-CH_3$) undergoes reductive ozonolysis to yield 2-methylbutanal ($CH_3-CH_2-CH(CH_3)-CHO$). This product contains a chiral carbon atom (the $C2$ carbon bonded to $-H$, $-CH_3$, $-CHO$, and $-C_2H_5$). Thus, the statement that it has no chiral carbon atom is false.
### Pattern Recognition
To check chirality after ozonolysis, draw the cleaved fragment structure first. Any carbon with four different groups is chiral. 2-methylbutanal has four unique groups attached to $C2$, making Statement II clearly false.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Class 12 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q33 (2025)
Given below are two statements :
Statement (I): On nitration of m-xylene with $HNO_{3}$, $H_{2}SO_{4}$ followed by oxidation, 4-nitrobenzene-1, 3-dicarboxylic acid is obtained as the major product.
Statement (II) : $CH_{3}$ group is o/p-directing while $-NO_{2}$ group is m-directing group.
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
- Both Statement I and Statement II are true
- Statement I is true but Statement II is false
### Core Logic
Statement I is true: In m-xylene, both methyl groups direct incoming electrophiles to position 4 (synergistic ortho/para effect). Nitration gives 4-nitro-m-xylene. Subsequent strong oxidation of both $-CH_3$ groups yields 4-nitrobenzene-1,3-dicarboxylic acid.
{{SOL_IMG33}}
Statement II is true: Alkyl groups ($-CH_3$) act as ortho/para directors via hyperconjugation, whereas nitro groups ($-NO_2$) are strongly deactivating meta directors.
### Pattern Recognition
When evaluating multi-substituted benzenes, identify whether directors reinforce the same positions. In m-xylene, position 4 is ortho to one methyl and para to the other.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Q47 (2025)
Isomeric hydrocarbons giving negative Baeyer's test have the molecular formula $C_{9}H_{12}$.
The total number of isomers from above with exactly four different non-aliphatic substitution sites is ________.
### Core Logic
A negative Baeyer's test confirms that the hydrocarbon structural isomers contain no aliphatic alkene or alkyne unsaturations, establishing that they are purely aromatic benzene derivatives with side alkyl chains.
{{SOL_IMG47}}
To find structures possessing four distinct ring positions available for electrophilic substitution, we examine the symmetries of specific tri-substituted configurations. There are exactly $2$ such structural isomers satisfying these spatial conditions.
### Pattern Recognition
Negative test = aromatic ring constraint. Calculate positional substitution patterns meticulously to ensure ring symmetry matches the required counts.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Hydrocarbons
