() arrange these compounds in basic strength order: (a) i>ii> iii (i) (iii) (b) ii>i> iii (c) iii>i>ii (d) ii> iii>i

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Ashutosh singh Best Answer
You'll find that while drawing resonating structures of ii, you'll get one 3° carbanion i.e tertiary carbanion...but, tertiary carbanions are very unstable so, the lone pair of Nit...
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Ashutosh singh
Sourav, Hope you understand.
Ashutosh singh
Sourav...basicity is basically how easily the lone pair of Nitrogen is available?
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Ashutosh singh
do you agree?
S
Sourav
why 2 more basic than 3?
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Ashutosh singh
Dear Sourav, listen! In questions like these, there's always one option which stands out in particular because in (ii) and (iii) there's not much difference between pkb values so, ...
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Ashutosh singh
In options (II) and (III), the lone pair of Nitrogen is involved in the resonance with the Phenyl group.
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Ashutosh singh
There will be 3 resonating structures each for ii and iii, for ii, 1 out of these 3 will be really unstable because of 3°...rest will be of the similar stability as that of iii.
Ashutosh singh
Option A is correct. (I) will be most basic.
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Sourav
how sir
Ashutosh singh
Is this clear now, Sourav?
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Sourav
sir why 2 more basic than 3?
S
Sourav
yes sir
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Sourav
yes sir
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Sourav
what about 2 and 3?
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Ashutosh singh
Therefore, The lone pair of Nitrogen is not easily available to get attacked because it gets delocalised to multiple sites...therefore they are less basic than (I).
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