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Nilanjan Bhowmick AIR 3, CSIR NET (Earth Science)
- IIT JAM
- Chemistry (CY)
() 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
() 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
2 Answer(s)
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Ashutosh singh
In (I), there's no Phenyl ring whatsoever so, Lone pair of Nitrogen is localised at one position, easing the attack by any other reagent.
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Sourav
why 2 more basic than 3?
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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, it gets difficult to comment on! So, as soon as you get your all options eliminated, tick the correct option....there's very very little difference between pkb values of ii and iii.
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if you want, I can explain that too.
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ok sir.. do so
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Can you draw the resonating structures of ii and iii?
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ya 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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Sourav
yes sir
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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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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.
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Not every reasonating structure is stable, Sourav.
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Sourav
yes sir
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 Nitrogen won't be delocalised as much as it will get delocalised in iii, because in that case, there's no 3° carbanion.
Sourav, Hope you understand.
got it sir.. ty