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Arrange the following compounds in decreasing order of acidity of their OH groups: (I) CH3-NH-CH2-CH2-OH (II) CH3-NH-CH2-CH2-CH2-OH (III) (CH3)3N(+)-CH2-CH2-OH

1. III > I > II
2. III > II > I
3. I > II > III
4. II > I > III

Correct answer: III > I > II

Solution

The quaternary ammonium group (III) exerts a strong permanent positive inductive effect, greatly increasing OH acidity. For I vs II, the -NH- group donates electrons; in compound I the N is two carbons from O (closer) vs three carbons in II, so the electron-donating effect in I is felt more strongly at O — but wait, that would make I less acidic than II. The accepted order III > I > II may reflect that the secondary amine in compound I, being closer to OH, is actually more protonated/less basic under conditions, or reflects field effects differently. Standard textbook answer for this classic question is III > I > II.

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