To determine which amine has the smallest pKb value in an aqueous solution, we need to consider the basicity of each amine. The pKb value is a measure of how well a compound can accept protons (H+) in water; the lower the pKb, the stronger the base. Let's analyze each option provided.
Understanding the Amines
Amines are organic compounds derived from ammonia (NH3) by replacing one or more hydrogen atoms with alkyl or aryl groups. The basicity of amines is influenced by the electron-donating or electron-withdrawing effects of these substituents.
- (CH3)3N (Trimethylamine): This is a tertiary amine. The three methyl groups are electron-donating, which enhances the basicity by stabilizing the positive charge on the nitrogen when it accepts a proton.
- C6H5NH2 (Aniline): This is an aromatic amine. The phenyl group is electron-withdrawing due to resonance, which decreases the basicity of the nitrogen atom, making it less likely to accept a proton.
- (CH3)2NH (Dimethylamine): This is a secondary amine. It has two methyl groups that provide some electron-donating effects, making it more basic than aniline but less than trimethylamine.
- CH3NH2 (Methylamine): This is a primary amine. The single methyl group enhances its basicity, but it is generally less basic than tertiary amines.
Comparing Basicity
To summarize the basicity of these amines:
- Trimethylamine (Tertiary) - Strong base, likely the highest basicity.
- Dimethylamine (Secondary) - Moderate basicity, less than trimethylamine.
- Methylamine (Primary) - Basic, but less than dimethylamine.
- Aniline (Aromatic) - Weak base due to resonance stabilization of the lone pair on nitrogen.
pKb Values
The pKb values generally follow this trend:
- Trimethylamine: Lowest pKb (strongest base)
- Dimethylamine: Higher pKb than trimethylamine
- Methylamine: Higher pKb than dimethylamine
- Aniline: Highest pKb (weakest base)
Based on this analysis, the amine with the smallest pKb value, indicating the strongest basicity, is (CH3)3N (Trimethylamine). Therefore, the answer to your question is A. (CH3)3N.