In the compound $C_6{H}_{5}NHCHNH{C}_6{H}_5$ (benzylideneaniline), which nitrogen atom is more basic?

• A. The nitrogen atom attached to the benzene ring (aniline nitrogen)
• B. Neither nitrogen atom is basic
• C. The nitrogen atom in the imine group (double-bonded to carbon)
• D. Both nitrogen atoms have equal basicity

Answer: (C)

Verified step by step guidance

1. Identify the two nitrogen atoms in the compound benzylideneaniline: one is the aniline nitrogen attached directly to the benzene ring (aniline nitrogen), and the other is the imine nitrogen, which is double-bonded to a carbon atom (C=N).
2. Recall that basicity in nitrogen atoms depends on the availability of the lone pair of electrons to accept a proton (H+). The more available the lone pair, the more basic the nitrogen.
3. Consider the electronic environment of the aniline nitrogen: its lone pair is partially delocalized into the aromatic ring through resonance, which decreases its availability to accept a proton, thus lowering its basicity.
4. Consider the imine nitrogen: its lone pair is involved in the C=N double bond, but it is less delocalized compared to the aniline nitrogen, making its lone pair more available to accept a proton, increasing its basicity relative to the aniline nitrogen.
5. Conclude that the imine nitrogen is more basic than the aniline nitrogen because its lone pair is more available for protonation due to less resonance delocalization.