The ^1 H NMR spectrum of phenol (C_6 H_5 OH) shows three absorptions in the aromatic region: 6.70

The ^1 H NMR spectrum of phenol (C_6 H_5 OH) shows three absorptions in the aromatic region: 6.70 (2 ortho H #x27​;s), 7.14 ( 2 meta H #x27​;s), and 6.80 ( 1 para H) ppm. Explain why the ortho and para absorptions occur at lower chemical shift than the meta absorption.
solutions;
"The $^1\ce{H}$ NMR spectrum of phenol ($\ce{C_6H_5OH}$) shows different absorptions in the aromatic region due to the electronic environment around the hydrogen atoms. Let's break down why the ortho and para hydrogens have lower chemical shifts compared to the meta hydrogens:

1. *Resonance Effect of the Hydroxyl Group ($\ce{OH}$):*
The $\ce{OH}$ group on the phenol molecule is an electron-donating group through resonance. This enhances the electron density particularly at the ortho and para positions more than the meta position.
The resonance structures of phenol show that the ortho and para positions have increased electron density due to resonance donation by the oxygen atom.

2. *Shielding Effect:*
The increased electron density at the ortho and para positions due to the resonance effect leads to these hydrogens experiencing greater electronic shielding.
This increased shielding results in the ortho and para hydrogens appearing at lower chemical shifts (6.70 and 6.80 ppm) as they have a reduced effective magnetic field applied on them.

3. *Reduced Shielding at the Meta Position:*
At the meta positions, the resonance effect of the $\ce{OH}$ group does not increase the electron density as significantly.
Consequently, the meta hydrogens experience less shielding compared to ortho and para hydrogens, leading to their absorption at a higher chemical shift (7.14 ppm).

In summary, the ortho and para hydrogens appear at lower chemical shifts because they are more shielded due to the resonance electron-donating effect of the $\ce{OH}$ group, increasing electron density specifically at those positions, unlike the meta position."