Which of the following best explains why diisopropylamine is a stronger base than aniline ($C_6{H}_5{\mathrm{NH}}_2$)?

• A. Diisopropylamine is less basic because its bulky groups hinder protonation.
• B. The lone pair on the nitrogen atom in aniline is delocalized into the aromatic ring, making it less available for protonation, whereas in diisopropylamine the lone pair is localized and more available to accept a proton.
• C. Aniline is more basic because the aromatic ring increases electron density on the nitrogen atom.
• D. Both compounds have similar basicity because they both contain a nitrogen atom with a lone pair.

Answer: (B)

Verified step by step guidance

1. Step 1: Understand the concept of basicity in amines, which depends on the availability of the nitrogen's lone pair to accept a proton (H\+). The more available the lone pair, the stronger the base.
2. Step 2: Analyze the structure of aniline (C\_6H\_5NH\_2). The nitrogen's lone pair in aniline is conjugated with the aromatic ring, meaning it can delocalize into the pi system of the benzene ring. This delocalization reduces the electron density on nitrogen, making the lone pair less available for protonation.
3. Step 3: Examine diisopropylamine, a secondary amine with bulky alkyl groups attached to nitrogen. Alkyl groups are electron-donating via the inductive effect, which increases electron density on nitrogen and makes the lone pair more available for protonation.
4. Step 4: Consider steric effects: although diisopropylamine has bulky groups, these do not significantly hinder protonation compared to the resonance effect in aniline. The resonance delocalization in aniline has a stronger impact on reducing basicity.
5. Step 5: Conclude that diisopropylamine is a stronger base than aniline because its nitrogen lone pair is localized and more available to accept a proton, whereas aniline's lone pair is delocalized into the aromatic ring, decreasing its basicity.