Step 1: Identify the main factor influencing basicity

• The basicity of amines and related compounds depends on the availability of the lone pair on the nitrogen atom for donation.
• Anything that decreases the availability of this lone pair decreases basicity.

Step 2: Examine all options

Let us write the names and analyse:

Option A: Aniline ($\ce{C6H5NH2}$)

Option B: p-Nitroaniline ($\ce{p-NO2-C6H4NH2}$)

Option C: Benzylamine ($\ce{C6H5CH2NH2}$)

Option D: Ammonia ($\ce{NH3}$)

Step 3: Analyse effects

1. Aniline ($\ce{C6H5NH2}$):

• The lone pair on nitrogen in aniline is partially delocalized into the aromatic ring (resonance), reducing its availability for protonation $\rightarrow$ decreases basicity.

1. p-Nitroaniline ($\ce{p-NO2-C6H4NH2}$):

• Same as aniline, but now a highly electron-withdrawing nitro group ($\ce{NO2}$) is attached at the para position.
• The nitro group (–M effect and –I effect) pulls electrons even more strongly from the ring, further reducing the electron density on the nitrogen.
• This makes the lone pair least available for donation, so very low basicity.

1. Benzylamine ($\ce{C6H5CH2NH2}$):

• The $\ce{NH2}$ group is not directly attached to the aromatic ring.
• Thus, no resonance with the aromatic ring — lone pair is fully available.
• More basic than both aniline and p-nitroaniline.

1. Ammonia ($\ce{NH3}$):

• A simple molecule; only weakly basic, but nitrogen lone pair is completely available.
• Comparable to benzylamine but more basic than aniline and p-nitroaniline.

Step 4: Order of Basicity

Based on above:

$$ \text{Benzylamine} > \text{Ammonia} > \text{Aniline} > \text{p-Nitroaniline} $$

Final Step: Answer

The least basic compound is Option B: p-Nitroaniline.

Reason: The electron-withdrawing nitro group at the para position, along with resonance of the aniline system, greatly decreases the electron density on the nitrogen and makes its lone pair least available for bonding with a proton.