Amines Previous Year Questions — JEE Main Chemistry
40 past-year Amines questions from JEE Main (Chemistry) — page 4.
Q80 (2024)
Given below are two statements:
Statement (I): The $NH_2$ group in Aniline is ortho and para directing and a powerful activating group.
Statement (II): Aniline does not undergo Friedel-Craft's reaction (alkylation and acylation).
In the light of the above statements, choose the most appropriate answer from the options given below:
- $\text{Both Statement I and Statement II are correct}$
- $\text{Both Statement I and Statement II are incorrect}$
- $\text{Statement I is incorrect but Statement II is correct.}$
- $\text{Statement I is correct but Statement II is incorrect}$
### Core Logic
Statement (I): The $-NH_2$ group has a lone pair of electrons on nitrogen, which undergoes resonance with the benzene ring (strong +M effect). This strongly activates the ring towards electrophilic substitution and directs incoming electrophiles to the ortho and para positions.
Statement (II): Friedel-Crafts alkylation and acylation require a Lewis acid catalyst like anhydrous $AlCl_3$. Aniline is a Lewis base (due to the lone pair on N) and reacts with the Lewis acid $AlCl_3$ to form a stable salt/complex ($C_6H_5\overset{+}{N}H_2-AlCl_3^-$). This removes the lone pair from resonance and converts the $-NH_2$ group into a strongly deactivating group, thereby halting the Friedel-Crafts reaction.
### Step 1: Evaluate Statements
Statement I is correct.
Statement II is correct.
### Pattern Recognition
Aniline NEVER undergoes Friedel-Crafts because the base ($NH_2$) reacts with the catalyst ($AlCl_3$) before the reaction can proceed.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
Q1481 (2024)
The product A formed in the following reaction is:
{{IMG1}}
### Related Formula
$$\text{Ar-NH}_2 \xrightarrow{\text{NaNO}_2 + \text{HCl}, 0^\circ\text{C}} \text{Ar-N}_2^+\text{Cl}^- \xrightarrow{\text{Cu}_2\text{Cl}_2} \text{Ar-Cl}$$
### Core Logic
The schematic outlines a classic sequence:
1. **Diazotization Step**: Aniline reacts with nitrous acid generated in situ ($\text{NaNO}_2 + \text{HCl}$) at low temperature ($0-5^\circ\text{C}$) to form benzene diazonium chloride intermediate.
2. **Sandmeyer Replacement Step**: Treating the diazonium salt with cuprous chloride ($\text{Cu}_2\text{Cl}_2$) results in replacement of the diazo group by a chlorine atom.
### Step 1: Visual Verification
Tracing structural shifts through intermediates validates the final outcome:
{{SOL_IMG1}}
### Pattern Recognition
Aniline $\rightarrow$ Diazonium salt $\rightarrow$ Chlorobenzene via standard Sandmeyer synthesis protocols.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Organic Compounds Containing Nitrogen
Q68 (2024)
The arenium ion which is not involved in the bromination of Aniline is.
- ""
- ""
- ""
- ""
### Core Logic
Aniline undergoes electrophilic aromatic substitution (like bromination). The $-NH_2$ group is a strongly activating group and directs incoming electrophiles to the ortho and para positions due to resonance electron donation (+$M$ effect).
### Step 1: Identifying Sigma Complexes (Arenium Ions)
When an electrophile ($Br^+$) attacks the ring, an intermediate arenium ion (sigma complex) is formed.
- If attack occurs at the **ortho** or **para** position, the positive charge is delocalized onto the carbon atom bearing the $-NH_2$ group. The lone pair on nitrogen can then stabilize this positive charge via resonance, forming a highly stable resonance structure (an octet-complete intermediate).
- If attack occurs at the **meta** position, the positive charge delocalizes only over the remaining ring carbons and never rests on the carbon bearing the $-NH_2$ group. Thus, it misses the extra stabilization provided by the nitrogen lone pair.
Because the meta attack intermediate is less stable compared to ortho/para attack, and the $-NH_2$ is strictly o/p directing, the meta-arenium ion is NOT a primary intermediate involved in standard bromination pathways of neutral aniline.
### Step 2: Conclusion
Option 3 displays the arenium ion resulting from a meta-attack (positive charge skips the $-NH_2$ substituted carbon).
{{SOL_IMG1}}
Since $-NH_2$ is ortho/para directing, the meta-arenium ion will not be formed.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
Q1748 (2024)
The products A and B formed in the following reaction scheme are respectively
{{IMG1}}
### Core Logic
Step 1: Nitration of benzene using conc. $HNO_3$ and conc. $H_2SO_4$ gives nitrobenzene.
Step 2: Reduction of nitrobenzene with $Sn/HCl$ yields aniline.
Step 3: Aniline reacts with $NaNO_2/HCl$ at $0-5^\circ C$ to form benzene diazonium chloride (Product A).
Step 4: Benzene diazonium chloride undergoes a coupling reaction with phenol (typically in a mildly alkaline medium) to form p-hydroxyazobenzene, an orange dye (Product B).
{{SOL_IMG1}}
### Pattern Recognition
Nitration $\rightarrow$ Reduction $\rightarrow$ Diazotization $\rightarrow$ Coupling (Azo Dye Test).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
Q1842 (2024)
The final product A, formed in the following multistep reaction sequence is:
{{IMG1}}
### Core Logic
Step 1: Bromobenzene + $Mg, ether \rightarrow Phenylmagnesium bromide$ (Grignard reagent).
Step 2: Grignard + $CO_2$ followed by $H^+ \rightarrow Benzoic acid$ ($C_6H_5COOH$).
Step 3: Benzoic acid + $NH_3, \Delta \rightarrow Benzamide$ ($C_6H_5CONH_2$).
Step 4: Benzamide + $Br_2/NaOH$ (Hoffmann bromamide degradation) $\rightarrow Aniline$ ($C_6H_5NH_2$).
{{SOL_IMG1}}
### Step 1: Tracing the product
The final product 'A' is Aniline.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Class 12 Chemistry: Haloalkanes and Haloarenes
