Arrange the ortho (I), meta (II) and para (III) methoxyphenols in decreasing order of acidity.
(I)
(II)
(III)

Question

Arrange the ortho (I), meta (II) and para (III) methoxyphenols in decreasing order of acidity. (I) (II) (III)

TestHub · Accepted Answer

OCH₃ is a +R donor from ortho and para, so it destabilises the phenoxide ion there. At meta only its -I effect acts, which stabilises the ion, so the meta isomer is the most acidic.

The -I effect is stronger at ortho than at para, so ortho $>$ para.

Order : II $>$ I $>$ III (pKa 9.65, 9.98, 10.21).

Chemistry - GOC Question with Solution | TestHub

Options:

Doubts & Discussion