The arenium ion which is not involved in the bromination of Aniline is.
A.""
Electrophilic Substitution in Amines
B.""
Electrophilic Substitution in Amines
C.""
Electrophilic Substitution in Amines
D.""
Electrophilic Substitution in Amines
Solution & Explanation
Core Logic
Aniline undergoes electrophilic aromatic substitution (like bromination). The -NH₂$-NH_2$ group is a strongly activating group and directs incoming electrophiles to the ortho and para positions due to resonance electron donation (+M$M$ effect).
When an electrophile (Br^+$Br^+$) 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 -NH₂$-NH_2$ 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 -NH₂$-NH_2$ 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 -NH₂$-NH_2$ 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 -NH₂$-NH_2$ substituted carbon).
Electrophilic Substitution in Amines diagram for Q68 - JEE Main 2024 Morning
Since -NH₂$-NH_2$ is ortho/para directing, the meta-arenium ion will not be formed.
The image outlines a synthesis pathway converting bromobenzene through several steps finally using Hoffmann bromamide reaction.
Step 1: Tracing the product
The final product 'A' is Aniline.
Chapter Mix
Class 12 Chemistry: Amines
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Class 12 Chemistry: Haloalkanes and Haloarenes
Q78jee_main_2024_30_jan_morningChemical Reactions of Amines
Following is a confirmatory test for aromatic primary amines. Identify reagent (A) and (B)
Ph-NH₂ A Ph-N₂^+Cl^- B Scarlet red dye$Ph-NH_2 \xrightarrow{A} Ph-N_2^+Cl^- \xrightarrow{B} \text{Scarlet red dye}$
The reaction sequence represents the classic dye test for aromatic primary amines.
Step 1 (Diazotization): Aniline (Ph-NH₂$Ph-NH_2$) reacts with nitrous acid (generated in situ from NaNO₂ + HCl$NaNO_2 + HCl$) at low temperature (0-5^° C$0-5^\circ C$) to form benzene diazonium chloride (Ph-N₂^+Cl^-$Ph-N_2^+Cl^-$).
Thus, Reagent A is NaNO₂ + HCl$NaNO_2 + HCl$ at 0-5^° C$0-5^\circ C$.
Step 2: Coupling Reaction
Step 2: The diazonium salt undergoes an electrophilic substitution (coupling reaction) with an electron-rich aromatic ring to form an azo dye.
The formation of a 'scarlet red dye' is specifically the result of coupling benzene diazonium chloride with β$\beta$-naphthol in a weakly basic medium (NaOH).
Chemical Reactions of Amines solution diagram for Q78 - JEE Main 2024 Morning
Chapter Mix
Class 12 Chemistry: Amines
Qjee_main_2024_31_jan_eveningReactions of Diazonium Salts
The azo-dye (Y) formed in the following reactions is
Sulphanilic acid + NaNO₂ + CH₃COOH arrow X$\text{Sulphanilic acid } + NaNO_2 + CH_3COOH \rightarrow X$The image shows the reaction of intermediate X with N,N-dimethylaniline to form an azo dye (Y).
A.
B.
C.
D.
Solution
Core Logic
Sulphanilic acid reacts with NaNO₂$NaNO_2$ and CH₃COOH$CH_3COOH$ to form a diazonium salt (X).
The diazonium salt (X) then reacts with N,N-dimethylaniline (given in the coupling step image). The coupling takes place at the para position of the highly activated N,N-dimethylaniline ring.
This coupling yields Methyl Orange, an azo dye. Its structure is p$p$-dimethylaminoazobenzenesulphonic acid.
The image shows the reaction of intermediate X with N,N-dimethylaniline to form an azo dye (Y).
Step 1: Final Identification
The final product (Y) matches option (4) structurally, containing the sulphonic acid group on one ring, the azo linkage, and the N,N-dimethylamine group on the para position of the other ring.
Chapter Mix
Class 12 Chemistry: Amines
Q70jee_main_2024_31_jan_eveningChemical Reactions of Amines
Given below are two statements:
Statement I: Aniline reacts with con. H₂SO₄$H_2SO_4$ followed by heating at 453-473 K gives p-aminobenzene sulphonic acid, which gives blood red colour in the 'Lassaigne's test'.
Statement II: In Friedel-Crafts alkylation and acylation reactions, aniline forms salt with the AlCl₃$AlCl_3$ catalyst. Due to this, nitrogen of aniline acquires a positive charge and acts as deactivating group.
In the light of the above statements, choose the correct answer from the options given below:
A.(1) Statement I is false but statement II is true$\text{(1) Statement I is false but statement II is true}$
B.(2) Both statement I and statement II are false$\text{(2) Both statement I and statement II are false}$
C.(3) Statement I is true but statement II is false$\text{(3) Statement I is true but statement II is false}$
D.(4) Both statement I and statement II are true$\text{(4) Both statement I and statement II are true}$
Solution
Core Logic
Statement I: Aniline reacting with concentrated H₂SO₄$H_2SO_4$ gives anilinium hydrogensulphate, which on heating at 453-473 K produces sulphanilic acid (p-aminobenzene sulphonic acid). Because sulphanilic acid contains both Nitrogen and Sulphur, it gives a blood-red colouration in Lassaigne's test due to the formation of thiocyanate ion SCN^-$SCN^-$ which reacts with Fe³⁺$Fe^{3+}$ to form [Fe(SCN)]²⁺$[Fe(SCN)]^{2+}$. Thus, Statement I is true.
Statement II: In Friedel-Crafts reactions, the Lewis acid catalyst AlCl₃$AlCl_3$ reacts with the lone pair on the nitrogen atom of aniline to form a salt. This generates a positive charge on the nitrogen, transforming the -NH₂$-NH_2$ group from a strong activating group into a strong deactivating group, thus preventing the Friedel-Crafts reaction from occurring. Thus, Statement II is true.
Chemical Reactions of Amines diagram for Q70 - JEE Main 2024 Evening
Step 1: Final Conclusion
Both Statement I and Statement II are true. Option (4) is correct.
Chapter Mix
Class 12 Chemistry: Amines
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q83jee_main_2024_31_jan_eveningAcylation of Amines
A compound (x) with molar mass 108 ~g mol⁻¹$108\mathrm{~g\,mol^{-1}}$ undergoes acetylation to give product with molar mass 192 ~g mol⁻¹$192\mathrm{~g\,mol^{-1}}$. The number of amino groups in the compound (x) is ________.
During the acetylation of an amino group, one hydrogen atom (mass = 1 g/mol$1\text{ g/mol}$) is replaced by an acetyl group (-COCH₃$-COCH_3$, mass = 43 g/mol$43\text{ g/mol}$).
Gain in molecular weight for every one -NH₂$-NH_2$ group acetylated = 43 - 1 = 42 g/mol$43 - 1 = 42\text{ g/mol}$.
Step 1: Calculating Number of Groups
Total increase in molecular weight = Final mass - Initial mass = 192 - 108 = 84 g/mol$192 - 108 = 84\text{ g/mol}$.
Number of amino groups = Total mass increaseMass increase per group = (84)/(42) = 2$$\text{Number of amino groups} = \frac{\text{Total mass increase}}{\text{Mass increase per group}} = \frac{84}{42} = 2$$
Chapter Mix
Class 12 Chemistry: Amines
More Amines Questions — jee_main_2024_29_jan_morning
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
