Q11. Arrange the following in decreasing order of their acidic strength. Explain the arrangement. C6H5COOH, FCH2COOH, NO2CH2COOH
Answer: The higher the conjugate base’s stability, the higher the acidity. In C6H5COOH, the phenyl group is insulated, so it exerts only −I effect on the resonance hybrid of the carboxyl group, stabilising it by withdrawing electron density due to higher electronegativity of sp2 hybridised carbon atoms. −NO2- and fluorine are electron-withdrawing, thus decreasing the electron density on the carbon atom and stabilising it. −NO2- is more electronegative than fluorine. Thus NO2CH2COOH would be most acidic, followed by FCH2COOH and C6H5COOH. NO2CH2COOH > FCH2COOH > C6H5COOH.
