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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 Mar 2008 09:08:51 IST
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Which member of the pairs is more basic...
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13 Mar 2008 09:13:18 IST
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are the answers a)aniline and b) 2methyl aniline
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Ajay Antony |
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a) pyridine b) methyl aniline
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13 Mar 2008 11:47:42 IST
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Can u please give reasons...
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13 Mar 2008 12:13:26 IST
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See,in 1st one,add H+ to nitrogen in both 1st one and aniline...
Now,the +ve charge in 1st one is more stabilized due to resonance with pi bonds...but dats not the case with aniline...so 1st is more basic..
In 2nd question,do similar thing...add H+ to nitrogen in both...CH3 group has +I effect..so it will tend to reduce the +ve charge intensity on N and so stabilizes it...so,methyl amine is more basic..
Am i correct??
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MaNuTd RoXxXx..MaNuTd 2 WiN PrEmIeR LeAgUe ThIs SeAsOn ToO AlOnG WiTh ChAmPiOnS LeAgUe.....HaiL RoNaLdO ...HaiL LaMpArD... |
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13 Mar 2008 12:56:54 IST
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The answer for the first one is aniline as the lone pair on nitrogen in the other compound(Its not pyridine) is involved in conjugation and hence not available for donation..... In the 2nd one, methyl increases the electron density on the ring making the lone pair on the nitrogen less welcome..Hence aniline can donate the lone pair more easily than the 2 methyl aniline....
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HOPE U GOT IT... |
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13 Mar 2008 13:19:16 IST
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 >  This is because, in pyrrole the lone pair is a part of the aromatic sextet. Thus, lp is involved in delocalization and is less avialable which makes pyrrole less basic.
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13 Mar 2008 13:23:04 IST
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 > This is because, -CH3 is an electron donating group and it will increase the basicity.
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13 Mar 2008 15:20:56 IST
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Isnt the lone pair on nitrogen in aniline involved in resonance.?so then how do we compare ...the basic strength?
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13 Mar 2008 15:29:05 IST
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LAMPARD's answer luks good to me
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