Aniline with excess of methyl iodide in the presence of sodium carbonate solution gives:
1. N, N, N−Trimethylanilinium carbonate.
2. N, N-Dimethylaniline.
3. N, N-Diethylaniline.
4. N, N, N−Triethylanilinium carbonate.
Aniline + Benzoyl chloride \(\xrightarrow[]{Base}\) Product
The product produced in the reaction mentioned above is:
1. N-Phenylbenzamide
2. Chlorobenzene
3. N-Chlorobenzamide
4. None of the above
1. a. Br2/H2O; b- NaNO2/HCl; c- H3PO2 /H2O
2. a-NaNO2/HCl; b- H3PO2 /H2O; c-Br2/H2O
3. a-NaNO2/HCl; b-H2O; c-Br2/H2O
4. None of the above
Consider the reactions,
(i) CH3CH2CH2NH2+HCl→ A
(ii) (C2H5)3N+HCl→ B
Products A and B are respectively-
1. CH3CH2CH2NHCl ; (C2H5)3NCl
2. CH3CH2CH2Cl ; (C2H5)3NCl
3. CH3CH2CH2NH3+ ; (C2H5)3NH+
4. None of the above
Benzene \(\xrightarrow[conc. HNO_{3}]{conc. H_{2}SO_{4}}\) A \(\xrightarrow[ethanol]{H_{2}, Pd}\) B
B is-
1. Aniline
2. Benzene
3. Cyclohexylamine
4. 1-Nitrocyclohexane