Question #f961a

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Question #f961a

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Answer 1 · 2015-04-26T16:16:33

Ordinarily, we would predict the order to be C > N > O > F.

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C is the least electronegative atom.

It has the least tendency to hold on to its electrons, so it is the best electron donor and the strongest base.

So, why is C out of order in your series?

We are comparing ${CH}_{3}^{-}$ $"CH"_3^-$ and ${CH}_{3} {C\equivC}^{-}$ $"CH"_3"C≡C"^-$ .

The difference is in the hybridization of the C atom.

In ${CH}_{3}^{-}$ $"CH"_3^-$ , the hybridization is $s p^{3}$ $sp^3$ .

In ${CH}_{3} {C\equivC}^{-}$ $"CH"_3"C≡C"^-$ , the hybridization is $s p$ $sp$ .

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An $s p^{3}$ $sp^3$ orbital has 25 % $s$ $s$ character, while an $s p$ $sp$ orbital has 50 % $s$ $s$ character.

$s$ $s$ electrons are held more tightly to the nucleus, so an alkyne C atom is less basic than an alkane carbon.

The order of acidities (with $p K_{a}$ $"p"K_"a"$ values) is

${CH}_{4}$ $"CH"_4$ (48) < ${NH}_{3}$ $"NH"_3$ (38) < ${CH}_{3} C\equivCH$ $"CH"_3"C≡CH"$ (25) < ${CH}_{3} {CH}_{2} OH$ $"CH"_3"CH"_2"OH"$ (16) < $HF$ $"HF"$ (3)

So the order of basicities is

${CH}_{3}^{-}$ $"CH"_3^-$ > ${NH}_{2}^{-}$ $"NH"_2^-$ > ${CH}_{3} {C\equivC}^{-}$ $"CH"_3"C≡C"^-$ > ${CH}_{3} {CH}_{2} O^{-}$ $"CH"_3"CH"_2"O"^-$ > $F^{-}$ $"F"^-$

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