Which of the following aromatic amines has lower pK_b value than ammonia?

Benzylamine

The correct answer is option 4. Benzylamine.

Let us go through each aromatic amine and compare their basicity with ammonia to explain why benzylamine has a lower \( pK_b \) value than ammonia.

The basicity of a compound is inversely related to its \( pK_b \) value: the lower the \( pK_b \), the stronger the base. Basicity is determined by the ability of the nitrogen atom in the amine to donate its lone pair of electrons to accept a proton (H\(^+\)).

Aromatic Amines

1. N,N-Dimethylbenzenamine (N,N-Dimethylaniline)

The nitrogen is bonded to an aromatic ring and two methyl groups. The electron-donating methyl groups slightly increase the electron density on nitrogen, making it more basic than aniline but still less basic than ammonia due to the electron-withdrawing effect of the aromatic ring.

2. N-Methylbenzenamine (N-Methylaniline)

The nitrogen is bonded to an aromatic ring and one methyl group. The methyl group increases the electron density on nitrogen slightly, making it more basic than aniline but still less basic than ammonia.

3. Benzenamine (Aniline)

The nitrogen is directly bonded to an aromatic ring. The electron-withdrawing effect of the aromatic ring reduces the availability of the lone pair on nitrogen, making aniline less basic than ammonia.

4. Benzylamine

The nitrogen is bonded to a benzyl group, which is a \( \text{C}_6\text{H}_5\text{CH}_2- \) group. The benzyl group has a slight electron-donating effect through the -CH\(_2\)- group, and it does not withdraw electron density as strongly as a directly bonded aromatic ring. This makes benzylamine more basic than aniline and similar in basicity to ammonia, but actually, it is more basic than ammonia.

Comparison with Ammonia

Ammonia (\( \text{NH}_3 \))

Ammonia is a simple amine with no electron-withdrawing groups, making its lone pair readily available for protonation.

Basicity Order

Based on the above analysis, the basicity order for the given compounds is:
\(\text{Benzylamine} > \text{Ammonia} > \text{N-Methylaniline} > \text{N,N-Dimethylaniline} > \text{Aniline}\)

Conclusion

Benzylamine has a lower \( pK_b \) value (indicating higher basicity) than ammonia. The presence of the benzyl group makes the lone pair on the nitrogen more available for protonation compared to the nitrogen in ammonia.

Therefore, the correct answer is: Benzylamine.