Question

The increasing order of basicity for the following intermediates is (from weak to strong)

(i) \((\mathrm{CH_3})_3\mathrm{C}^{\ominus}\)
(ii) \(\mathrm{H_2C=CH-\overset{\ominus}{C}H_2}\)
(iii) \(\mathrm{HC\equiv C^{\ominus}}\)
(iv) \(\overset{\ominus}{\mathrm{C}}\mathrm{H_3}\)
(v) \(\mathrm{{}^{\ominus}CN}\)

• A. (iii) < (i) < (ii) < (iv) < (v)
• B. (iii) < (iv) < (ii) < (i) < (v)
• C. (v) < (i) < (iv) < (ii) < (iii)
• D. (v) < (iii) < (ii) < (iv) < (i)

Hint:

Basicity of an anion is the reverse of the acidity of its conjugate acid: compare the pKa of HCN, HC\(\equiv\)CH, propene, CH₄ and isobutane.

Answer 1

The Correct Option is D

The basicity of an anion increases as the acidity of its conjugate acid decreases.
Conjugate acids and approximate pKa: HCN (9.2) for (v); HC\(\equiv\)CH (25) for (iii); CH₂=CH-CH₃ (about 43) for (ii); CH₄ (about 48) for (iv); (CH₃)₃CH (about 53) for (i).
Reasons: CN^- has the negative charge on an sp carbon and is also delocalised towards the electronegative N, so it is the weakest base; the acetylide has 50% s-character; the allyl anion is resonance stabilised; CH₃^- has no stabilising or destabilising group; the tert-butyl anion is destabilised by three +I methyl groups and is the strongest base.
Increasing basicity: (v) < (iii) < (ii) < (iv) < (i).
Answer: (4) (v) < (iii) < (ii) < (iv) < (i)