2 is the least basic as the lone pair is involved in aromatizing the ring 1 is the maximum basic as one of the lone pairs is used to aromatize the ring and the other lone pair is available but how do we distinguish between 3 and 4 the options are:





by the above discussion we can make out that the answer is a. but how to distinguish between option 3 and 4. how can we say that 3 is more basic than 4.

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    $\begingroup$ All your answers are wrong, and that's particularly bad question. 3 and 4 are more basic than 2. $\endgroup$ – Mithoron Dec 18 '15 at 17:08
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    $\begingroup$ As Mithoron points out this is a bad question. The basicities of 3 and 4 are close and they flip-flop depending upon whether we are discussing gas phase or solution basicity. In solution (you can google the pKas) the correct answer would be 3>4>1>2. You are correct that 2 is least basic because the lone pair is involved in aromaticity. 2 is more basic because the most available of the 2 lone pairs is in an sp2 orbital. Finally 3 and 4 are the most basic because their lone pairs are in an sp3 orbital. $\endgroup$ – ron Dec 18 '15 at 17:30

I think 4 should me more basic than 3 since $ -CH_3 $ has $ +I $ effect due to which electron density on $N$ increases.

  • $\begingroup$ but according to the answer it is 3>4 $\endgroup$ – Prakhar Dec 18 '15 at 10:07
  • $\begingroup$ How can you say 1 is most basic? both 3 and 4 too have localized lone pair and since in 1 Nitrogen is $sp^2$ hybridized its basic strength decreases with respect to 3 and 4. $\endgroup$ – Vaibhav Dec 18 '15 at 14:37
  • $\begingroup$ 1 is the most basic because the lone pair on nitrogen(the one that is at the bottom)will delocalise to make the ring aromatic, and the other lone pair is always available(a stronger base is one which can donate lone pairs). $\endgroup$ – Prakhar Dec 18 '15 at 18:29
  • $\begingroup$ In both 3 and 4 lone pairs are available too and 3°N is more basic than 2°N $\endgroup$ – Vaibhav Dec 19 '15 at 12:24

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