The lone pair of amines makes them basic. They react with acids to form acid-base salts. Amines are more basic than alcohols, ethers and water. When an amine is dissolved in water, an equilibrium is established , where water acts as an acid and transfer a proton to the amine. The basic strength of an amine can be measured by basicity constant `K_(b)`.
Arylamines are less basic than alkylamines because the lone pair of nitrogen is delocalised with the aromatic ring and are less available for donation.
Substituted arylamines can be either more basic or less basic than aniline , depending on the substituted . ERG substituents, such as `-CH_(3), -NH_(2) and -OCH_(3)` increases the basicity and EWG substituents , such as `-Cl, -NO_(2) and -CN` decreases basicity. While `sp^(2)-` hybridized nitrogen atom in pyridine is less basic then the `sp^(3)` -hybridized nitrogen in an alkylamine.
The most basic carbanion is :

Options

A.
B. `CH-=overset(Theta)(C)`
C.
D.

Answer

To determine the most basic carbanion among the given options, we need to analyze the stability of each carbanion. The basicity of a carbanion is inversely proportional to its stability; that is, the less stable the carbanion, the more basic it is. Here’s a step-by-step breakdown of the reasoning:

### Step 1: Understand Carbanion Stability
Carbanions are negatively charged species where a carbon atom has an extra electron. The stability of a carbanion is influenced by several factors, including hybridization, resonance, and the presence of electron-withdrawing or electron-donating groups.

### Step 2: Analyze Hybridization
- **sp Hybridization**: Carbanions with sp hybridization (linear geometry) are more stable because the negative charge is held by a carbon atom that is more electronegative due to the higher s-character (50% s-character).
- **sp² Hybridization**: Carbanions with sp² hybridization (trigonal planar geometry) are less stable than sp hybridized ones but more stable than sp³ hybridized ones.
- **sp³ Hybridization**: Carbanions with sp³ hybridization (tetrahedral geometry) are the least stable because they have the lowest s-character (25%).

### Step 3: Consider Resonance Effects
- Carbanions that can delocalize their negative charge through resonance are more stable. For example, if the negative charge can be spread over multiple atoms, it stabilizes the carbanion.
- Conversely, if the negative charge is localized and cannot be delocalized, the carbanion is less stable.

### Step 4: Evaluate the Given Options
Assuming we have four options (A, B, C, D) for carbanions:
- **Option A**: Carbanion with resonance stabilization.
- **Option B**: Carbanion with sp hybridization.
- **Option C**: Carbanion with sp² hybridization.
- **Option D**: Carbanion with sp³ hybridization and no resonance stabilization.

### Step 5: Determine Basicity
- **Option D** is likely the least stable due to its sp³ hybridization and lack of resonance, making it the most basic carbanion.
- **Option A** would be more stable due to resonance, thus less basic.
- **Option B** would be stable due to sp hybridization but still less basic than D.
- **Option C** would be less basic than D but more stable than A.

### Conclusion
The most basic carbanion among the options is **Option D** due to its instability and inability to stabilize the negative charge effectively.