Hydrocarbons Previous Year Questions — JEE Main Chemistry
35 past-year Hydrocarbons questions from JEE Main (Chemistry) — page 4.
Q77 (2024)
Compound A formed in the following reaction reacts with B gives the product C. Find out A and B.
$$CH_3-C\equiv CH + Na \rightarrow A \xrightarrow{B} CH_3-C\equiv C-CH_2-CH_2-CH_3 + NaBr$$
- $A=CH_3-C\equiv C^-Na^+, B=CH_3-CH_2-CH_2-Br$
- $A=CH_3-CH=CH_2, B=CH_3-CH_2-CH_2-Br$
- $A=CH_3-CH_2-CH_3, B=CH_3-C\equiv CH$
- $A=CH_3-C\equiv C^-Na^+, B=CH_3-CH_2-CH_3$
### Core Logic
Terminal alkynes possess acidic hydrogen. When treated with a strong base or active metal like Sodium ($Na$), they form sodium acetylide salts.
$CH_3-C\equiv C-H + Na \rightarrow CH_3-C\equiv C^-Na^+ + \frac{1}{2}H_2$
So, A is Sodium propynide ($CH_3-C\equiv C^-Na^+$).
### Step 1: Analyzing the second step
The product is $CH_3-C\equiv C-CH_2-CH_2-CH_3$. This indicates an $S_N2$ substitution reaction between the acetylide ion (nucleophile) and an alkyl halide (electrophile). Since $NaBr$ is a byproduct, B must be a propyl bromide.
$CH_3-C\equiv C^-Na^+ + CH_3-CH_2-CH_2-Br \rightarrow CH_3-C\equiv C-CH_2-CH_2-CH_3 + NaBr$
### Step 2: Conclusion
A is $CH_3-C\equiv C^-Na^+$ and B is $CH_3-CH_2-CH_2-Br$ (1-Bromopropane).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Class 12 Chemistry: Haloalkanes and Haloarenes
Q1925 (2024)
Identify major product 'P' formed in the following reaction.
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- $\text{(1) Product A}$
- $\text{(2) Product B}$
- $\text{(3) Product C}$
- $\text{(4) Product D}$
### Core Logic
The given reaction is an intramolecular Friedel-Crafts alkylation.
1) The alkyl chloride reacts with anhydrous $AlCl_3$ to form a carbocation intermediate.
2) The carbocation generated will act as an electrophile and attack the adjacent phenyl ring.
3) The intermediate carbocation will undergo electrophilic aromatic substitution to form a new six-membered ring, as a 6-membered ring is highly stable and preferred over other ring sizes.
{{SOL_IMG1}}
### Pattern Recognition
Intramolecular Friedel-Crafts usually prefers forming 5 or 6 membered rings due to lesser angle strain. Here, the tether length is perfect for closing into a 6-membered tetralin-like system.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Q1926 (2024)
Major product of the following reaction is -
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### Core Logic
Addition of $D-Cl$ across the double bond takes place via an electrophilic addition mechanism following Markovnikov's rule.
1) The electrophile $D^+$ attacks the double bond to generate the most stable carbocation. The tertiary carbocation formed at the methyl-substituted carbon is more stable than the secondary carbocation.
2) The nucleophile $Cl^-$ then attacks the planar tertiary carbocation from either face (top or bottom), resulting in a racemic mixture if a new chiral center is fully free, but here the stereochemistry depends on the relative anti/syn addition logic or thermodynamic stability. A mixture of diastereomers can be formed, but standard electrophilic additions often yield predominantly the trans-product due to steric reasons or via a bridged intermediate depending on conditions, though pure $HCl/DCl$ addition is non-stereospecific.
However, based on the official answer key (Option 3), we identify the correct stereochemical representation provided by the examining body.
{{SOL_IMG1}}
### Note on Discrepancy
According to the official NTA key, Option 3 is correct. According to our experts, both options 3 and 4 can be formed as a mixture since the carbocation is planar and $Cl^-$ can attack from both sides. We proceed with the officially accepted answer (3).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Q1939 (2024)
The correct order of reactivity in electrophilic substitution reaction of the following compounds is:
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- $\text{}B > C > A > D$
- $\text{}D > C > B > A$
- $\text{}A > B > C > D$
- $\text{}B > A > C > D$
### Core Logic
Electrophilic substitution reactivity depends on the electron density of the aromatic ring, which is influenced by the inductive ($I$) and mesomeric ($M$) effects of the substituents.
- Compound A (Benzene): Standard reference.
- Compound B (Toluene): The $-CH_3$ group shows $+I$ and hyperconjugation effects, activating the ring. Most reactive.
- Compound C (Chlorobenzene): The $-Cl$ group shows $+M$ and $-I$ effects, but the $-I$ effect dominates, mildly deactivating the ring.
- Compound D (Nitrobenzene): The $-NO_2$ group shows strong $-M$ and $-I$ effects, heavily deactivating the ring. Least reactive.
Order of reactivity: Toluene (B) > Benzene (A) > Chlorobenzene (C) > Nitrobenzene (D).
### Step 1: Final Order
Reactivity: $B > A > C > D$. This matches option (4).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Class 12 Chemistry: Haloalkanes and Haloarenes
Q84 (2024)
Number of isomeric products formed by mono-chlorination of 2-methylbutane in presence of sunlight is ________
### Core Logic
The structure of 2-methylbutane is $CH_3-CH(CH_3)-CH_2-CH_3$. It has four different types of hydrogen atoms, which can be substituted to form structural isomers:
1) 1-chloro-2-methylbutane: Chlorination at terminal $CH_3$ near branch. Yields a chiral center at C2 (2 enantiomers).
2) 2-chloro-2-methylbutane: Chlorination at the tertiary carbon (1 achiral product).
3) 2-chloro-3-methylbutane: Chlorination at the $CH_2$ group. Yields a chiral center at C2 (2 enantiomers).
4) 1-chloro-3-methylbutane: Chlorination at the far terminal $CH_3$. No chiral center (1 achiral product).
{{SOL_IMG1}}
### Step 1: Counting Stereoisomers
Total isomeric products = $2 (from 1st) + 1 (from 2nd) + 2 (from 3rd) + 1 (from 4th) = 6$.
### Pattern Recognition
When asked for "isomeric products" in halogenation without specifying "structural isomers", you must count stereoisomers (enantiomers) as well.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Class 12 Chemistry: Haloalkanes and Haloarenes
