Which of the following is least alkaline ?

The basicity of substituted anilines depends on the nature of the substituent present on the benzene ring. Electron-donating groups (EDG) increase the electron density on the nitrogen atom, thereby increasing basicity. Electron-withdrawing groups (EWG) decrease the electron density on the nitrogen atom, thereby decreasing basicity.

Among the given options:
$\text{-NO}_2$ is a strong electron-withdrawing group due to its strong $-I$ and $-M$ effects. It strongly pulls the electron density away from the amine group, making the lone pair on nitrogen least available for protonation.
$\text{-OCH}_3$ is an electron-donating group due to its $+M$ effect, which increases basicity.
$\text{-C}_6\text{H}_5$ has a weak $-I$ effect, making it slightly less basic than aniline but much more basic than nitroaniline.
Aniline ($\text{C}_6\text{H}_5\text{NH}_2$) has no substituents.

Therefore, due to the strong electron-withdrawing nature of the nitro group, $\text{O}_2\text{N-C}_6\text{H}_4\text{-NH}_2$ is the least alkaline.

Answer: $\text{O}_2\text{N-C}_6\text{H}_4\text{-NH}_2$