Q. No.| Column I | Column II | ||
| a. |  |
i. | Primary amine |
| b. |  |
ii. | Secondary amine |
| c. | \(\mathrm{(C_2H_5)_2CHNH_2}\) | iii. | Tertiary amine |
| d. | \(\mathrm{(C_2H_5)_2NH}\) |
1. a-iii; b-iii; c-i; d-ii
2. a-ii; b-iii c-i; d-ii
3. a-i; b-iii c-i; d-ii
4. a-ii; b-i c-i; d-ii
1. 7
2. 6
3. 8
4. 9
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 |
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
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. 5
2. 6
3. 4
4. 3
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
Benzene \(\xrightarrow[conc. HNO_{3}]{conc. H_{2}SO_{4}}\) A \(\xrightarrow[ethanol]{H_{2}, Pd}\) B \(\xrightarrow[CH_{3}Cl]{2 \ moles}\) C
C is-
1. N-methylaniline
2. N,N-dimethylaniline
3. Ethylamine
4. None of the above
Cl−(CH2)4−Cl \(\xrightarrow[]{A}\) \(N\equiv C-(CH_{2})_{4}-C\equiv N\) \(\xrightarrow[]{A}\) hexan-1, 6-diamine
A and B in the given reaction are respectively:
1. Ethanolic AgCN; KMnO4
2. Ethanolic AgCN; H2/ Ni
3. Ethanolic KCN; KMnO4
4. Ethanolic KCN; H2/Ni
© 2024 GoodEd Technologies Pvt. Ltd.