The correct increasing order of basic strength is
Show Hint
- \(NH_3<C_2H_5NH_2<C_6H_5NH_2<C_6H_5CH_2NH_2\)
- \(C_6H_5NH_2<NH_3<C_6H_5CH_2NH_2<C_2H_5NH_2\)
- \(C_6H_5NH_2<C_6H_5CH_2NH_2<NH_3<C_2H_5NH_2\)
- \(C_6H_5CH_2NH_2<NH_3<C_2H_5NH_2<C_6H_5NH_2\)
\(C_6H_5NH_2<NH_3<C_2H_5NH_2<C_6H_5CH_2NH_2\)
The Correct Option is B
Solution and Explanation
Step 1: Aniline (\(C_6H_5NH_2\)) is less basic than ammonia (\(NH_3\)) due to the electron-withdrawing nature of the phenyl group via resonance.
Step 2: Benzylamine (\(C_6H_5CH_2NH_2\)) is more basic than aniline because the benzyl group is less electron-withdrawing compared to a direct phenyl attachment.
Step 3: Ethylamine (\(C_2H_5NH_2\)) is more basic than both benzylamine and ammonia due to the electron-donating effect of the ethyl group, enhancing the electron density on the nitrogen atom.
Step 4: The correct order of increasing basicity, considering the electronic effects, is therefore aniline, ammonia, benzylamine, and ethylamine.
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