JEE Chemistry 2025 Question with Solution

Given: The question asks for the relative reactivity of benzaldehyde, acetophenone, p-tolualdehyde and p-nitrobenzaldehyde in nucleophilic addition reactions.

Find: The correct increasing order of reactivity.

In nucleophilic addition reactions, electron-withdrawing groups increase the electrophilicity of the carbonyl carbon and increase reactivity, while electron-donating groups decrease electrophilicity and reduce reactivity. Steric hindrance also lowers reactivity.

• Acetophenone is the least reactive because it is a ketone and also has greater steric hindrance.
• p-Tolualdehyde contains a methyl group, which is electron-donating, so it is less reactive than benzaldehyde.
• Benzaldehyde is more reactive than p-tolualdehyde because it does not have the electron-donating methyl group.
• p-Nitrobenzaldehyde is the most reactive because the nitro group is strongly electron-withdrawing and increases the electrophilicity of the carbonyl carbon.

Therefore, the reactivity order is:

$$\text{acetophenone} < \text{p-tolualdehyde} < \text{benzaldehyde} < \text{p-nitrobenzaldehyde}$$

So, the correct option is C.

Given: Reactivity in nucleophilic addition depends mainly on electrophilicity of the carbonyl carbon and steric accessibility.

Find: Which option matches the correct order.

The general trend is:

• Electron-withdrawing groups increase reactivity.
• Electron-donating groups decrease reactivity.
• Aldehydes are generally more reactive than ketones because they are less hindered.

Now analyze each compound:

1. Acetophenone: the carbonyl group belongs to a ketone, so it is less reactive than aromatic aldehydes. The phenyl group also contributes to reduced electrophilicity and steric hindrance.
2. p-Tolualdehyde: it is an aldehyde, so it is more reactive than acetophenone, but the para methyl group donates electron density and slightly decreases reactivity.
3. Benzaldehyde: it has no para electron-donating substituent, so it is more reactive than p-tolualdehyde.
4. p-Nitrobenzaldehyde: the para nitro group strongly withdraws electrons, making the carbonyl carbon most electrophilic.

Hence the correct order is:

$$\text{acetophenone} < \text{p-tolualdehyde} < \text{benzaldehyde} < \text{p-nitrobenzaldehyde}$$

This matches option C.

Note: the second solution incorrectly states that option $\text{(3)}$ corresponds to p-Tolualdehyde, but the worked order clearly matches option C.