Question

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

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

Answer 1

The correct option is C (iv)<(ii)<(iii)<(i)

Basicity is inversely proportional to electronegativity. Here $C{N}^{-}$ is most stable anion and hence least basic because negative charge is present on highly electronegative $N$ atom.

Negative charge in (i) is most destabilised because of electron donating (methyl) groups on $C$. So it is most basic in nature.

Comparing (ii) and (iii), as the $\%s\text{-character}$ in hybridisation of $C$ is more, more is the electronegativity, more is the stability of anion and hence lesser is the basicity. Hybridisation of $C$ in alkyne and alkane is $spands{p}^3$ respectively. So negative charge on alkyne; with maximum $s-\text{character}$, its anion is more stabilised and is less basic than anion of alkane for the same reason.

Hence, (c) is the correct option.