Q28 / 180 · COMEDK 2026 (9 May Shift 2) · Chemistry
Choose the incorrect statement.
- AAniline cannot be prepared by Phthalimide reaction
- BPropan–2–amine can be obtained by reacting acetoxime with Na / C$_2$H$_5$OH
- CThe decreasing order of basic strength of amines in aqueous solution is Ethanamine > N,N–Dimethylaniline > Benzenamine
- Correct answerDFluorobenzene cannot be prepared from Benzenediazonium chloride by Sandmeyer's reaction because Fluorination of the Diazonium salt is highly endothermic in nature
Solution
Gabriel phthalimide synthesis is used for the preparation of primary aliphatic amines. Aniline cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide. Statement A is correct.
Acetoxime is $(CH_3)_2C=NOH$. Reduction of acetoxime with $Na / C_2H_5OH$ yields propan-2-amine, $(CH_3)_2CHNH_2$. Statement B is correct.
Aliphatic amines are more basic than aromatic amines. Among aromatic amines, the $+I$ effect of two methyl groups in N,N-dimethylaniline increases the electron density on nitrogen, making it more basic than benzenamine. Thus, the basic strength order is Ethanamine $>$ N,N-Dimethylaniline $>$ Benzenamine. Statement C is correct.
Fluorobenzene cannot be prepared by Sandmeyer's reaction, but it is prepared by the Balz-Schiemann reaction. The reason it cannot be prepared by Sandmeyer's reaction is not because the fluorination of the diazonium salt is highly endothermic. In fact, direct fluorination is highly exothermic and violent, and $CuF$ is not effective for the Sandmeyer reaction. Therefore, the reason provided in statement D is incorrect.
Answer: Fluorobenzene cannot be prepared from Benzenediazonium chloride by Sandmeyer's reaction because Fluorination of the Diazonium salt is highly endothermic in nature
Acetoxime is $(CH_3)_2C=NOH$. Reduction of acetoxime with $Na / C_2H_5OH$ yields propan-2-amine, $(CH_3)_2CHNH_2$. Statement B is correct.
Aliphatic amines are more basic than aromatic amines. Among aromatic amines, the $+I$ effect of two methyl groups in N,N-dimethylaniline increases the electron density on nitrogen, making it more basic than benzenamine. Thus, the basic strength order is Ethanamine $>$ N,N-Dimethylaniline $>$ Benzenamine. Statement C is correct.
Fluorobenzene cannot be prepared by Sandmeyer's reaction, but it is prepared by the Balz-Schiemann reaction. The reason it cannot be prepared by Sandmeyer's reaction is not because the fluorination of the diazonium salt is highly endothermic. In fact, direct fluorination is highly exothermic and violent, and $CuF$ is not effective for the Sandmeyer reaction. Therefore, the reason provided in statement D is incorrect.
Answer: Fluorobenzene cannot be prepared from Benzenediazonium chloride by Sandmeyer's reaction because Fluorination of the Diazonium salt is highly endothermic in nature
Hence, the correct answer is option: 'D'.