The hydrocarbon fluorene was treated with potassium t-butoxide in an acid-base reaction

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The hydrocarbon fluorene was treated with potassium t-butoxide in an acid-base reaction, giving the fluorenide anion and t-butyl alcohol. (a) Which way does the equilibrium lie, and by how much? b) What is the proportion of the fluorenide anion to fluorene? (c) Why is fluorene so highly acidic, considering the pKa of an average alkane is above 50?
H
-H
1
+ Но
potassium
t-butoxide
fluorene
fluorenide anion
t-butyl alcohol
pK = 18
pKa
= 18
Transcribed Image Text:H -H 1 + Но potassium t-butoxide fluorene fluorenide anion t-butyl alcohol pK = 18 pKa = 18
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