Q.Carboxylic acids are more acidic than phenols. Give reasons.

Resonance Stabilization Effect: Why It Works

The Resonance Stabilization Effect explains why certain molecules or ions are more stable than a single Lewis structure would suggest. Let's build the reasoning from the ground up.

1. The Core Problem: Localized vs. Delocalized Electrons

In a simple Lewis structure, we draw localized bonds — electrons are assigned to specific atoms or bonds. But in reality, for molecules like benzene ($\text{C}_6\text{H}_6$) or the carboxylate ion ($\text{RCOO}^-$), the electrons are delocalized over multiple atoms.

• Localized picture: One double bond, one single bond — but this doesn't match experimental bond lengths or stability.
• Delocalized reality: All bonds are identical (e.g., benzene's C–C bonds are all 1.39 Å, between single and double).

Key insight: Delocalization lowers the energy of the system. This energy lowering is the resonance stabilization energy.

2. The Mathematical Foundation: Linear Combination of Atomic Orbitals (LCAO)

Resonance is best understood through Molecular Orbital Theory. For a system with $n$ atomic orbitals (AOs) that can overlap, we form $n$ molecular orbitals (MOs) as linear combinations:

where:

• $\psi_j$ = $j$-th molecular orbital
• $\phi_i$ = $i$-th atomic orbital
• $c_{ji}$ = coefficient (contribution of $\phi_i$ to $\psi_j$)

The energy of each MO is found by solving the secular determinant:

where $H_{ij} = \langle \phi_i | \hat{H} | \phi_j \rangle$ (resonance integral) and $S_{ij} = \langle \phi_i | \phi_j \rangle$ (overlap integral).

3. The Simplest Case: The Allyl System (3 Carbon Atoms)

Consider the allyl radical ($\text{CH}_2=\text{CH}-\text{CH}_2^\bullet$) or allyl cation/anion. Three $p$ orbitals (one per carbon) combine.

Step 1: Set up the Hückel approximation

• Assume all $S_{ij} = 0$ for $i \neq j$ (zero overlap approximation)
• $H_{ii} = \alpha$ (Coulomb integral, same for all carbons)
• $H_{ij} = \beta$ for adjacent carbons, $0$ otherwise

Step 2: The secular determinant

For three atoms in a line (1–2–3):

Step 3: Solve for energies

Let $x = \frac{\alpha - E}{\beta}$. Then:

Expanding: $x(x^2 - 1) - 1(x) = 0 \implies x^3 - 2x = 0 \implies x(x^2 - 2) = 0$

So $x = 0$ or $x = \pm \sqrt{2}$.

Thus the three MO energies are:

(Since $\beta < 0$, $E_1$ is lowest, $E_3$ highest.)

4. Why Stabilization Occurs: The Energy Lowering

For the allyl cation (2 π electrons):

• Electrons fill the lowest MO: $E_1 = \alpha + \sqrt{2}\beta$
• Total energy = $2(\alpha + \sqrt{2}\beta) = 2\alpha + 2\sqrt{2}\beta$

Compare to localized picture (one isolated double bond):

• One double bond = 2 electrons in a bonding MO of energy $\alpha + \beta$
• Total energy = $2(\alpha + \beta) = 2\alpha + 2\beta$

Resonance stabilization energy:

Since $\beta$ is negative, $\Delta E$ is negative → stabilization.

General formula for a linear conjugated system with $n$ atoms:

The Hückel energy levels are:

The total π-electron energy for $N$ electrons (filling from lowest up) is:

The resonance stabilization energy is the difference between $E_\pi$ and the energy of the best localized structure.

5. The Key Formula: Resonance Energy

For a cyclic conjugated system (like benzene, $n=6$):

For benzene ($n=6$):

• $k=0$: $E = \alpha + 2\beta$
• $k=\pm1$: $E = \alpha + \beta$
• $k=\pm2$: $E = \alpha - \beta$
• $k=3$: $E = \alpha - 2\beta$ …