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Arrange the following compounds in decreasing order of their basicity: (i) H2C=CH-Na (vinyl sodium), (ii) CH3CH2Na (ethyl sodium), (iii) CH3CH2ONa (sodium ethoxide), (iv) HC=-C-Na (sodium acetylide).

1. H2C=CH-Na
2. CH3CH2Na
3. CH3CH2ONa
4. HC=-C-Na

Correct answer: CH3CH2Na

Solution

The basicity order of carbanions/anions is determined by their conjugate acid pKa: ethane (sp3, pKa~50) > ethylene (sp2, pKa~44) > acetylene (sp, pKa~25) > ethanol (O-H, pKa~16). So basicity order is: CH3CH2Na > H2C=CHNa > HC=-CNa > CH3CH2ONa. The most basic single species is CH3CH2Na (ethyl sodium), so the answer to 'which is most basic' is CH3CH2Na.

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