Animated Solution for Chemistry - Organic Chemistry: The correct order of stability of given carbocation is

Decoding Carbocation Stability

When you are faced with a lineup of carbocations and asked to rank their stability, you are essentially being asked to evaluate how well each molecule can handle an electron deficiency. A carbocation features a carbon atom with only six electrons in its valence shell, bearing a formal positive charge. Because it is electron-starved, any structural feature that can donate electron density toward that carbon will stabilize the entire system.

In this classic problem, we are given four distinct carbocations: - (A) Benzyl carbocation ($C_6{H}_{5}C{H}_2^{+}$) - (B) Vinyl carbocation ($C{H}_2=C{H}^{+}$) - (C) Ethyl carbocation ($C{H}_3-C{H}_2^{+}$) - (D) Ethynyl carbocation ($HC\equiv C^{+}$)

To solve this, we must rely on the golden rules of electronic effects: Resonance is the most powerful stabilizing force, followed by Hyperconjugation, and finally, the Inductive Effect. Additionally, we must carefully consider the hybridization of the carbon atom bearing the positive charge.

The Power of Resonance

Let's begin with structure (A), the benzyl carbocation. Notice that the positively charged $C{H}_2^{+}$ group is attached directly to a benzene ring. This is a highly privileged position. The adjacent $\pi$ electrons of the aromatic ring can easily delocalize, flowing out of the ring to form a double bond with the external carbon, thereby shifting the positive charge into the ring itself.

This phenomenon is known as resonance. By spreading the positive charge over multiple atoms (the ortho and para positions of the ring), the burden of the electron deficiency is shared. Because resonance is the most dominant stabilizing effect, the benzyl carbocation is exceptionally stable. It easily takes the crown as the most stable carbocation in our lineup.

The Magic of Hyperconjugation

Next, we examine structure (C), the ethyl carbocation ($C{H}_3-C{H}_2^{+}$). Here, we do not have any $\pi$ bonds adjacent to the positive charge, so resonance is off the table. However, we do have an adjacent methyl group ($C{H}_3$).

This methyl group provides three $\alpha$-hydrogens. The electrons in the $C-H$ $\sigma$ bonds can partially overlap with the empty $p$-orbital of the positively charged carbon. This stabilizing interaction is called hyperconjugation (often referred to as "no-bond resonance"). Furthermore, the alkyl group exerts a positive inductive effect ($+I$), pushing electron density through the $\sigma$ framework toward the positive center. While hyperconjugation is a strong stabilizing force, it is generally weaker than full $\pi$-electron resonance. Therefore, (C) is highly stable, but it ranks strictly below (A).

The Hybridization Trap

Things take a dramatic turn when we look at structures (B) and (D). In both cases, the positive charge is located directly on an unsaturated carbon atom.

In (B), the vinyl carbocation ($C{H}_2=C{H}^{+}$), the positive charge resides on an $s{p}^2$ hybridized carbon. In (D), the ethynyl carbocation ($HC\equiv C^{+}$), the charge is on an $sp$ hybridized carbon. To understand their stability, we must look at the s-character of these hybrid orbitals.

An $s{p}^2$ orbital has $33\%$ s-character, while an $sp$ orbital has a massive $50\%$ s-character. Because s-orbitals are spherical and closer to the nucleus than p-orbitals, a higher s-character means the electrons are held more tightly by the nucleus. Consequently, an atom's effective electronegativity increases as its s-character increases.

Putting a positive charge on a highly electronegative atom is chemically disastrous—it is akin to forcing a starving person to give away their last meal! Because the $sp$ carbon in (D) is more electronegative than the $s{p}^2$ carbon in (B), the ethynyl carbocation is significantly more unstable than the vinyl carbocation.

By synthesizing all these principles, the hierarchy of stability becomes crystal clear: 1. (A) is the most stable due to powerful resonance delocalization. 2. (C) follows, stabilized by hyperconjugation and the $+I$ effect. 3. (B) is highly unstable because the charge is on an electronegative $s{p}^2$ carbon. 4. (D) is the least stable of all, as the charge is trapped on an extremely electronegative $sp$ carbon.

Thus, the correct decreasing order of stability is A > C > B > D, which corresponds perfectly to option (a).