Which of the following arrangements with respect to their reactivity in nucleophilic addition reaction is correct?

Nucleophilic addition to a carbonyl group occurs at the electrophilic carbon of the $$C = O$$ bond. The rate depends on two main factors:
  • Electronic effects: groups that withdraw electrons ( $$-I$$ and/or $$-R$$) increase the partial positive charge on the carbonyl carbon and enhance reactivity. Groups that donate electrons ( $$+I$$ and/or $$+R$$) decrease the charge and slow the reaction.
  • Steric effects: less‐hindered carbonyls are more accessible to the nucleophile.

First compare aldehydes and ketones attached to an aromatic ring.
  • Aldehydes have only one alkyl/aryl group attached to the carbonyl carbon, so they are less sterically hindered and receive less $$+I$$ donation than ketones. Hence aromatic aldehydes are more reactive than aromatic ketones.
  • Therefore $$\text{acetophenone (Ar-CO-CH}_3)$$ is expected to be the least reactive of the four compounds.

Now compare the three aromatic aldehydes:
  1. $$\text{p-nitrobenzaldehyde}$$ contains a $$NO_2$$ group. $$NO_2$$ is strongly electron-withdrawing by both $$-I$$ (inductive) and $$-R$$ (resonance) effects, so it makes the carbonyl carbon highly electrophilic. This gives the HIGHEST reactivity.
  2. $$\text{benzaldehyde}$$ has no extra substituent on the ring, so its reactivity is taken as the reference among aldehydes.
  3. $$\text{p-tolualdehyde}$$ (para-methylbenzaldehyde) contains a $$CH_3$$ group. $$CH_3$$ is electron-donating by $$+I$$ and weakly $$+R$$, so it reduces the electrophilicity of the carbonyl carbon, lowering the reactivity below that of unsubstituted benzaldehyde.

Putting the steric and electronic arguments together:

$$\text{Least reactive} \;\; \longrightarrow \;\; \text{acetophenone} \lt \text{p-tolualdehyde} \lt \text{benzaldehyde} \lt \text{p-nitrobenzaldehyde} \;\; \longrightarrow \;\; \text{Most reactive}$$

This sequence matches Option D.

Answer: Option D  $$\left(\text{acetophenone} \lt \text{p-tolualdehyde} \lt \text{benzaldehyde} \lt \text{p-nitrobenzaldehyde}\right)$$