Animated Solution for Chemistry - Organic Chemistry: The increasing order of basicity for the following intermediates is (from weak to strong)

Stability vs Basicity

Welcome to a classic showdown in General Organic Chemistry! We are tasked with arranging five different carbanions in increasing order of their basicity. To conquer this, we must first understand the fundamental relationship between basicity and stability.

The Core Philosophy: A base is a species that donates an electron pair. If a carbanion is highly unstable, it is desperate to donate its extra electrons and form a bond, making it a strong base. Conversely, if a carbanion is highly stable, it is perfectly happy holding onto its electrons, making it a weak base. Therefore, the order of basicity is exactly the reverse of the order of stability.

Let's evaluate our contenders. We start with the most stable ones: $C{N}^{\ominus }$ (v) and $HC\equiv C^{\ominus }$ (iii). In both of these species, the negative charge resides on an $sp$ hybridized carbon atom.

Why does this matter? An $sp$ hybridized orbital has 50% s-character, which means the orbital is physically closer to the positively charged nucleus. This makes the carbon atom highly electronegative, allowing it to stabilize the negative charge exceptionally well.

Between the two, $C{N}^{\ominus }$ is even more stable because the adjacent nitrogen atom is highly electronegative and further pulls electron density towards itself. Thus, $C{N}^{\ominus }$ is the most stable, followed by $HC\equiv C^{\ominus }$.

Resonance Stabilization

Next, we look at the allyl carbanion, $H_{2}C=CH-C{H}_2^{\ominus }$ (ii). Here, the negative charge is not stuck on a single atom; it is delocalized over two carbon atoms via resonance.

While resonance provides significant stability, it is generally not as powerful as the extreme electronegativity of an $sp$ hybridized carbon. Therefore, the allyl carbanion sits comfortably in the middle of our stability ranking.

Inductive Effect in $s{p}^3$ Carbons

Finally, we examine the localized $s{p}^3$ carbanions: the methyl carbanion $C{H}_3^{\ominus }$ (iv) and the tertiary carbanion $(C{H}_3{)}_3{C}^{\ominus }$ (i).

Carbanions are already electron-rich. Alkyl groups (like methyls) are electron-donating through the +I (Inductive) effect. Pumping more electron density into an already negative center creates intense electrostatic repulsion, drastically destabilizing the species.

The tertiary carbanion has three methyl groups pushing electrons onto the central carbon, making it the least stable of the bunch. The methyl carbanion, lacking these +I groups, is relatively more stable.

Putting it all together, our decreasing order of stability is:

Since basicity is the exact opposite of stability, we flip the signs to get our increasing order of basicity:

This perfectly matches option (a). You can also verify this by looking at the $p{K}_a$ values of their conjugate acids. $HCN$ is a much stronger acid ($p{K}_a\approx 9.2$) than isobutane ($p{K}_a\approx 50$), confirming that $C{N}^{\ominus }$ is a much weaker base than the tertiary carbanion. Master this inverse relationship, and you will never miss a basicity question again!