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In my worksheet there is a question that asks for the comparison of basicities of the three anisidine isomers and aniline:

enter image description here

After my analysis using:

$\text{Basicity} \propto \text{+M effect (except for meta annisidine)} \propto \dfrac{1}{\text{-I effect}}$,

I could make the following conclusions:

  1. m-anisidine should be the least basic
  2. p-anisidine should be more basic than aniline.

But the main problem is comparing the basicity of aniline and o-anisidine and comparing them with p-anisidine. I doubt the presence of ortho effect in ortho isomer after seeing the ball and stick model which makes the interaction of lone pair of nitrogen with hydrogen atoms of $\ce{-OCH3}$ improbable.

The answer given is:

p-anisidine> aniline > o-anisidine > m-anisidine (I have verified it with $pKa$ values from PubChem)

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  • $\begingroup$ I would say SIP. $\endgroup$
    – Avyansh Katiyar
    Commented Mar 21, 2018 at 1:16
  • $\begingroup$ But I want to think of it in terms of "tendency to donate electrons" @AvnishKabaj $\endgroup$
    – Archer
    Commented Mar 21, 2018 at 1:20
  • $\begingroup$ Steric inhibition of protonation reduces the "tendency to donate electrons" $\endgroup$
    – Avyansh Katiyar
    Commented Mar 21, 2018 at 1:27

1 Answer 1

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Why you considering the methyl group on $\ce{-OCH3}$ to be "locked" in position, facing away from the $\ce{-NH2}$ group? It is free to rotate in all directions, and it will come at a position next to the $\ce{-NH2}$ group, where the ortho effect will apply.

For a visualization, here is a similar image:

(taken from Martin's excellent post)

PS: I believe that Steric Inhibition of Resonance should not be invoked here, because both sides of the $\ce{-NH2}$ group are not blocked by the $\ce{-OCH3}$ group, and $\ce{-OCH3}$ is certainly not as bulky a group as t-butyl.

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  • $\begingroup$ Nice! So basically in ortho effect here, hydrogen bond will be formed between the hydrogen of NH2 and O? $\endgroup$
    – Archer
    Commented Mar 21, 2018 at 1:42
  • $\begingroup$ Nope, that way the lone pair of nitrogen is still free. The interaction will be between N of NH2 and H of OCH3 (also see) $\endgroup$
    – Gaurang Tandon
    Commented Mar 21, 2018 at 1:45

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