Top 50 Most Repeated AMINES PYQs | JEE MAINS
A curated collection of the most important questions from AMINES, fully solved with step-by-step concepts to prepare for JEE MAINS.
A curated collection of the most important questions from AMINES, fully solved with step-by-step concepts to prepare for JEE MAINS.
In Gabriel synthesis, the alkyl halide first reacts with potassium phthalimide to give N‑alkylphthalimide. Subsequent hydrolysis liberates the primary...
Read Full Step-by-Step Solution →The Sandmeyer reaction is a type of electrophilic aromatic substitution reaction that produces phenols from diazonium salts....
Read Full Step-by-Step Solution →Exhaustive methylation of aniline gives quaternary ammonium salt; Hoffmann elimination yields least substituted alkene — styrene....
Read Full Step-by-Step Solution →Diazotisation requires nitrous acid (generated from NaNO₂ + HCl) at low temperature (0–5 °C) to keep the unstable diazonium ion from decomposing....
Read Full Step-by-Step Solution →Electron-donating groups increase basicity; aniline is weakest due to resonance....
Read Full Step-by-Step Solution →In Hofmann bromamide reaction, R-CONH₂ → R-NH₂ with loss of CO₂. The alkyl group migrates, so carbon count decreases by one. Propanamide (C₂H₅CONH₂) g...
Read Full Step-by-Step Solution →Diazotisation follows a 1:1 stoichiometry between aniline and NaNO₂. Hence, 0.10 mol of NaNO₂ is needed....
Read Full Step-by-Step Solution →The Hofmann bromamide reaction proceeds via bromination of the amide followed by base‑induced elimination, resulting in decarboxylation and formation ...
Read Full Step-by-Step Solution →Gabriel synthesis produces only primary aliphatic amines via alkylation of phthalimide....
Read Full Step-by-Step Solution →NH₂ group in aniline donates electrons via resonance (+M), activating the ring for electrophilic attack....
Read Full Step-by-Step Solution →Coupling reaction in acidic medium forms orange azo dye....
Read Full Step-by-Step Solution →In Gabriel synthesis, the alkyl halide first reacts with potassium phthalimide to give N‑alkylphthalimide. Subsequent hydrolysis liberates the primary...
Read Full Step-by-Step Solution →The Sandmeyer reaction is a type of electrophilic aromatic substitution reaction that produces phenols from diazonium salts....
Read Full Step-by-Step Solution →Exhaustive methylation of aniline gives quaternary ammonium salt; Hoffmann elimination yields least substituted alkene — styrene....
Read Full Step-by-Step Solution →Diazotisation requires nitrous acid (generated from NaNO₂ + HCl) at low temperature (0–5 °C) to keep the unstable diazonium ion from decomposing....
Read Full Step-by-Step Solution →Electron-donating groups increase basicity; aniline is weakest due to resonance....
Read Full Step-by-Step Solution →In Hofmann bromamide reaction, R-CONH₂ → R-NH₂ with loss of CO₂. The alkyl group migrates, so carbon count decreases by one. Propanamide (C₂H₅CONH₂) g...
Read Full Step-by-Step Solution →Diazotisation follows a 1:1 stoichiometry between aniline and NaNO₂. Hence, 0.10 mol of NaNO₂ is needed....
Read Full Step-by-Step Solution →The Hofmann bromamide reaction proceeds via bromination of the amide followed by base‑induced elimination, resulting in decarboxylation and formation ...
Read Full Step-by-Step Solution →Gabriel synthesis produces only primary aliphatic amines via alkylation of phthalimide....
Read Full Step-by-Step Solution →NH₂ group in aniline donates electrons via resonance (+M), activating the ring for electrophilic attack....
Read Full Step-by-Step Solution →Coupling reaction in acidic medium forms orange azo dye....
Read Full Step-by-Step Solution →In Gabriel synthesis, the alkyl halide first reacts with potassium phthalimide to give N‑alkylphthalimide. Subsequent hydrolysis liberates the primary...
Read Full Step-by-Step Solution →The Sandmeyer reaction is a type of electrophilic aromatic substitution reaction that produces phenols from diazonium salts....
Read Full Step-by-Step Solution →Exhaustive methylation of aniline gives quaternary ammonium salt; Hoffmann elimination yields least substituted alkene — styrene....
Read Full Step-by-Step Solution →Diazotisation requires nitrous acid (generated from NaNO₂ + HCl) at low temperature (0–5 °C) to keep the unstable diazonium ion from decomposing....
Read Full Step-by-Step Solution →Electron-donating groups increase basicity; aniline is weakest due to resonance....
Read Full Step-by-Step Solution →In Hofmann bromamide reaction, R-CONH₂ → R-NH₂ with loss of CO₂. The alkyl group migrates, so carbon count decreases by one. Propanamide (C₂H₅CONH₂) g...
Read Full Step-by-Step Solution →Diazotisation follows a 1:1 stoichiometry between aniline and NaNO₂. Hence, 0.10 mol of NaNO₂ is needed....
Read Full Step-by-Step Solution →The Hofmann bromamide reaction proceeds via bromination of the amide followed by base‑induced elimination, resulting in decarboxylation and formation ...
Read Full Step-by-Step Solution →Gabriel synthesis produces only primary aliphatic amines via alkylation of phthalimide....
Read Full Step-by-Step Solution →NH₂ group in aniline donates electrons via resonance (+M), activating the ring for electrophilic attack....
Read Full Step-by-Step Solution →Coupling reaction in acidic medium forms orange azo dye....
Read Full Step-by-Step Solution →In Gabriel synthesis, the alkyl halide first reacts with potassium phthalimide to give N‑alkylphthalimide. Subsequent hydrolysis liberates the primary...
Read Full Step-by-Step Solution →The Sandmeyer reaction is a type of electrophilic aromatic substitution reaction that produces phenols from diazonium salts....
Read Full Step-by-Step Solution →Exhaustive methylation of aniline gives quaternary ammonium salt; Hoffmann elimination yields least substituted alkene — styrene....
Read Full Step-by-Step Solution →Diazotisation requires nitrous acid (generated from NaNO₂ + HCl) at low temperature (0–5 °C) to keep the unstable diazonium ion from decomposing....
Read Full Step-by-Step Solution →Electron-donating groups increase basicity; aniline is weakest due to resonance....
Read Full Step-by-Step Solution →In Hofmann bromamide reaction, R-CONH₂ → R-NH₂ with loss of CO₂. The alkyl group migrates, so carbon count decreases by one. Propanamide (C₂H₅CONH₂) g...
Read Full Step-by-Step Solution →Diazotisation follows a 1:1 stoichiometry between aniline and NaNO₂. Hence, 0.10 mol of NaNO₂ is needed....
Read Full Step-by-Step Solution →The Hofmann bromamide reaction proceeds via bromination of the amide followed by base‑induced elimination, resulting in decarboxylation and formation ...
Read Full Step-by-Step Solution →Gabriel synthesis produces only primary aliphatic amines via alkylation of phthalimide....
Read Full Step-by-Step Solution →NH₂ group in aniline donates electrons via resonance (+M), activating the ring for electrophilic attack....
Read Full Step-by-Step Solution →Coupling reaction in acidic medium forms orange azo dye....
Read Full Step-by-Step Solution →In Gabriel synthesis, the alkyl halide first reacts with potassium phthalimide to give N‑alkylphthalimide. Subsequent hydrolysis liberates the primary...
Read Full Step-by-Step Solution →The Sandmeyer reaction is a type of electrophilic aromatic substitution reaction that produces phenols from diazonium salts....
Read Full Step-by-Step Solution →Exhaustive methylation of aniline gives quaternary ammonium salt; Hoffmann elimination yields least substituted alkene — styrene....
Read Full Step-by-Step Solution →Diazotisation requires nitrous acid (generated from NaNO₂ + HCl) at low temperature (0–5 °C) to keep the unstable diazonium ion from decomposing....
Read Full Step-by-Step Solution →Electron-donating groups increase basicity; aniline is weakest due to resonance....
Read Full Step-by-Step Solution →In Hofmann bromamide reaction, R-CONH₂ → R-NH₂ with loss of CO₂. The alkyl group migrates, so carbon count decreases by one. Propanamide (C₂H₅CONH₂) g...
Read Full Step-by-Step Solution →