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`(CH_(3))_(3) N` is basic but `(CF_(3))_(3)N` is not basic. Explain why ? |
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Answer» N has a lone pair of electrons. But due to higher electronegativity of C `(2.5)` than that of `H(2.1)`, `CH_(3)` acts as an electron-donating group. It increases the electron density on N in `(CH_(3))_(3)N`. As a result, the lone pair of electrons on N in `(CH_(3))_(3)N` in easily available for potenation and hence `(CH_(3))_(3)N` acts a base. In contrast, due to much higher electronegativity of `F(4.0)` than that of `(2.5)`, `CF_(3)` acts a strong electron-withdrawing group. It considerably decreases the electron density of N in `(CF_(3))_(3)N`. As a result, the lone pair of electrons on N in `(CF_(3))_(3)N` is not available for protonation and hence `(CF_(3))_(3)N` does not act as a base. |
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