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Question
The basicites of
(I)o−toluidine
(II) m−toluidine
(III) p−toluidine
(IV)aniline follow the order:
The basicites of
(I)o−toluidine
(II) m−toluidine
(III) p−toluidine
(IV)aniline follow the order:
(II) m−toluidine
(III) p−toluidine
(IV)aniline follow the order:
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Solution
The correct option is D III>II>IV>I
The m-toluidine the CH3 group is closer to N atom and also in O- toluidine. So it should not increase the electron density on N- atom. NH2 group in O-tuluidine accepting a proton becomes −NH+3 which moves out of the plane due to steric repulsion and hence the positive charge is not delocalised making the compound less stable so it became less basic. Aniline is more basic than or the toulidine because of steric hindrance of protonation of cation formed at NH2. On the other hand, in meta touluidine only +I of ME is operation whereas in para - toluidine both hyper conjugation effect and +I effect of Me. So -P- toluidine is more basic than m-toluidine.Hence, order is,P-toluidine > m-toludine> aniline > 0-toluidine.
The m-toluidine the CH3 group is closer to N atom and also in O- toluidine. So it should not increase the electron density on N- atom. NH2 group in O-tuluidine accepting a proton becomes −NH+3 which moves out of the plane due to steric repulsion and hence the positive charge is not delocalised making the compound less stable so it became less basic. Aniline is more basic than or the toulidine because of steric hindrance of protonation of cation formed at NH2. On the other hand, in meta touluidine only +I of ME is operation whereas in para - toluidine both hyper conjugation effect and +I effect of Me. So -P- toluidine is more basic than m-toluidine.
Hence, order is,
P-toluidine > m-toludine> aniline > 0-toluidine.
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