MCQ
In the given pair identify most acidic compound in $(A)$ and $(B)$. Most basic in $(C)$ and $(D)$.
  • A
    $A - I, B - II, C - I, D - II$
  • $A - II, B - I, C - I, D - II$
  • C
    $A - II, B - II, C - II, D - II$
  • D
    $A - I, B - II, C - I, D - I$

Answer

Correct option: B.
$A - II, B - I, C - I, D - II$
b
$(b)$ The ortho-effect: This is a special effect that is shown by $o$ -sbstituents of  benzene and its derivatives, but is not necessarily just $a'$ steric effect, e.g, the  basicities of some $o$ -substituted anilines were explained in terms of steric effects  and differences in crowding round the nitrogen atom. - This ortho-effect also operates  with the benzoic acids. Irrespective of the polar type, nearly all $o$-substituted benzoic acids are stronger than benzoic acid. As w.e have seen, benzoic  acid is a resonance hybrid, and so the carboxyl groups is coplanar with the ring. An $o$ -substituent tends to prevent this coplanarity. Thus resonance is diminished (or  prevented), and so the $o$ -atom of the $OH$ groups has a greater positive charge,  resulting increased acid strength. It follows from this that the greater the steric  inhibition to resonance, the stronger is the acid. Support for this is the fo llowing order  of strengths of substituted benzoic acids. 

$2,6 - di - Me (pK_a \,3.21) > 2 - t - Bu \,(pK_a \,3.46) > 2 - Me \,(pK_a\, 3.91)$.

Here again, if we consider the stability of the anion, steric inhibition of resonance  prevents the $+ R$ effect of the ring coming into operation (see above), and since this  weakens acid strength, its absence results in increased acid strength. 

For options $B$ and $D$, no ortho effect is valid and order of acidity and basicity is  calculated by nearly examining the inductive effect.

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