General Organic Chemistry

Salicylaldehyde has - OH and - CHO ortho to each other, enabling intramolecular H-bonding.

$m$-hydroxy and benzoic acid form intermolecular H-bonds.

In electrophilic aromatic substitution (EAS) reactions, the nitro group ($ \text{NO}_2 $) is known to be a strong electron-withdrawing group due to its resonance and inductive effects. This property significantly influences the electron distribution in the aromatic ring, thereby affecting the position where incoming electrophiles are most likely to attack.

Let's analyze the options provided in this context:

• Option A: Decreases electron density at ortho-and para-positions
• Option B: Decreases electron density at meta-position
• Option C: Increases electron density at meta-position
• Option D: Increases electron density at ortho-and para-positions

The nitro group, being electron-withdrawing, pulls electron density away from the aromatic ring through two main mechanisms:

• Inductive Effect: The nitro group withdraws electron density through the sigma bonds due to its high electronegativity.
• Resonance Effect: The nitro group can participate in resonance, pulling electron density away from the ring, particularly affecting the ortho and para positions relative to the nitro group.

Due to the electron-withdrawing nature of the nitro group, electron density is significantly reduced at the ortho and para positions more than the meta positions. This makes the ortho and para positions less reactive towards electrophiles since electrophiles are attracted to electron-rich areas.

As a result, the meta position remains relatively electron-rich in comparison to the ortho and para positions. Thus, the electrophilic substitution is more likely to occur at the meta position.

Based on this understanding, the correct answer is:

Option A: Decreases electron density at ortho-and para-positions

If acid is weak, its conjugate base (nucleophile) is strong and vice-versa.

I. is conjugate base of

II. CH$_3$O$^-$ is a conjugate base of CH$_3$OH

III. CN$^-$ is a conjugate base of HCN

IV. is a conjugate base of

Order of acidic nature is IV > I > III > II.

So, nucleophilicity order is II > III > I > IV.

In option (c), the hydrogen of $-$CH group is

surrounded by three groups as:

Here, O of CO attracts, the electron density towards itself. Thus, (c) has most acidic hydrogen among given compounds.

Or

The anion formed after removal of

acidic proton in is stablised by $-R$-effect of three $\mathrm{C}=\mathrm{O}$ groups.

Thus, (c) has most acidic hydrogen among given compounds.