Answer:

When sodium ethoxide reacts with an organic compound, it undergoes a nucleophilic substitution reaction, also known as the Williamson ether synthesis. In this reaction, the sodium ethoxide acts as a nucleophile, attacking the carbon atom of the organic compound and replacing the leaving group with an alkoxide ion (-OCH2CH3) to form an ether.

In order to determine which of the given compounds would react most rapidly with sodium ethoxide, we need to consider the reactivity of the leaving group. The leaving group should be a good leaving group, meaning it should be able to depart easily and stabilize the resulting negative charge.

Reactivity of the leaving groups:
a) Chlorobenzene: Chlorobenzene has a poor leaving group, as the chloride ion is a weak base and does not stabilize the negative charge well. Therefore, it would not react rapidly with sodium ethoxide.
b) 2-nitrotoluene: The nitro group (-NO2) is a poor leaving group, as it is highly electron-withdrawing and does not stabilize the negative charge well. Therefore, it would not react rapidly with sodium ethoxide.
c) p-nitrochlorobenzene: The nitro group (-NO2) is a poor leaving group, as mentioned above, and the chloride ion is also a weak base. Therefore, it would not react rapidly with sodium ethoxide.
d) m-(chloromethyl)-toluene: The leaving group in this compound is a chloride ion, which is a good leaving group and stabilizes the resulting negative charge. Therefore, it would react most rapidly with sodium ethoxide.

Conclusion:
Among the given options, (d) m-(chloromethyl)-toluene would react most rapidly with sodium ethoxide to produce an ether. The presence of a good leaving group (chloride ion) enhances the reactivity of the compound in the nucleophilic substitution reaction.