Question #6b37a

1 Answer
Truong-Son N.
Jan 16, 2018

Alcoholic #"KOH"# just means it was dissolved in an alcohol, e.g. methanol, ethanol, propanol, etc.

The longer the alkyl chain attached to the hydroxyl head, usually the more basic the conjugate base is (and the less nucleophilic).

This is seen in the #"pK"_a# of the Lewis acids:

  • #"H"_2"O"#, #"pK"_a = 15.7#
  • #"CH"_3"OH"#, #"pK"_a = 15.5#
  • #"CH"_3"CH"_2"OH"#, #"pK"_a = 15.9#
  • #"CH"_3("CH"_2)_2"OH"#, #"pK"_a = 16.0#
  • #"CH"_3("CH"_2)_3"OH"#, #"pK"_a = 16.1#

The higher the #"pK"_a#, the weaker the acid. As a result, the stronger the conjugate base.

Therefore, in alcohols with more than one carbon, the alkoxide is a stronger base than the hydroxide. And with more steric hindrance, the tendency to act as a base increases, and the tendency to act as a nucleophile decreases...

Consider tert-butanol, #("CH"_3)_3"C"-"OH"#, compared to butanol:

  • #"CH"_3("CH"_2)_3"OH"#, #"pK"_a = 16.1#
  • #("CH"_3)_3"C"-"OH"#, #"pK"_a = 16.54#

Big things are slow. Slow things are bad nucleophiles. But tert-butoxide is still a better base than butoxide...

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