Animated Solution for Chemistry - Organic Chemistry: The major product obtained in the following reaction is

Imagine you are looking at a molecular battlefield. On one side, we have our reactant, $o$-aminophenol. This molecule is special because it carries two functional groups: a hydroxyl group ($-OH$) and an amino group ($-N{H}_2$).

On the other side, we have our reagent, acetic anhydride ($({\text{CH}}_3\text{CO}{)}_2\text{O}$), waiting to be attacked. We are also given exactly $1$ equivalent of this reagent, which means there is only enough for one of the functional groups to react.

The question is: which group will strike first?

Electronegativity and Nucleophilicity

To determine the winner, we need to evaluate the nucleophilicity of both groups. A nucleophile is an electron-rich species that loves to donate its electrons. Both the nitrogen in the $-N{H}_2$ group and the oxygen in the $-OH$ group have lone pairs available for donation.

However, not all lone pairs are created equal. The key lies in electronegativity. Oxygen is more electronegative ($\chi \approx 3.5$) than nitrogen ($\chi \approx 3.0$).

Because oxygen is highly electronegative, it holds onto its lone pairs very tightly. It is greedy and reluctant to share. Nitrogen, being less electronegative, is much more generous. It holds its lone pair loosely, making it far more available for donation.

Therefore, the $-N{H}_2$ group is a much stronger nucleophile than the $-OH$ group.

Now that we know the $-N{H}_2$ group is the superior nucleophile, let's watch the reaction unfold.

The lone pair on the nitrogen atom launches a nucleophilic attack on the highly electrophilic carbonyl carbon of the acetic anhydride. This attack pushes the pi electrons of the $C=O$ bond up onto the oxygen, forming a temporary tetrahedral intermediate.

This intermediate is unstable. The electrons on the oxygen swing back down to reform the double bond, and in doing so, they kick out the acetate ion (${\text{CH}}_3{\text{COO}}^{-}$) as a leaving group. This entire process is a classic nucleophilic acyl substitution.

Finally, the reaction mixture contains pyridine, which acts as a mild base. The pyridine swoops in and removes the extra proton from the positively charged nitrogen, neutralizing the molecule.

Because we only had $1$ equivalent of acetic anhydride, the reaction stops here. The $-OH$ group remains completely untouched.

The final major product is N-(2-hydroxyphenyl)acetamide, where only the amine group has been acetylated. This perfectly matches option (b).