Arrange the following compounds in order of increasing basicity.
`CH_(3)CH = CH^(-), CH_(3)CH_(2)CH_(2)^(-), CH_(3)C -=C^(-)`

Answer

To arrange the compounds `CH3CH=CH^(-)`, `CH3CH2CH2^(-)`, and `CH3C≡C^(-)` in order of increasing basicity, we need to analyze their conjugate acids and the stability of the negative charge on each compound. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Compounds and Their Conjugate Acids
1. **Compound 1:** `CH3CH=CH^(-)` (Allylic anion)
- Conjugate acid: `CH3CH=CH2` (propene)

2. **Compound 2:** `CH3CH2CH2^(-)` (Alkyl anion)
- Conjugate acid: `CH3CH2CH3` (propane)

3. **Compound 3:** `CH3C≡C^(-)` (Terminal alkyne anion)
- Conjugate acid: `CH3C≡CH` (propyne)

### Step 2: Determine the Hybridization of the Carbons
- In `CH3CH=CH^(-)`, the carbon with the negative charge is **sp² hybridized**.
- In `CH3CH2CH2^(-)`, the carbon with the negative charge is **sp³ hybridized**.
- In `CH3C≡C^(-)`, the carbon with the negative charge is **sp hybridized**.

### Step 3: Analyze the Basicity
Basicity is inversely related to the stability of the conjugate acid. The more stable the conjugate acid, the weaker the base.

1. **sp Hybridization:**
- The `sp` hybridized carbon has a higher electronegativity, making the negative charge more stable. Hence, `CH3C≡C^(-)` is the least basic.

2. **sp² Hybridization:**
- The `sp²` hybridized carbon is less electronegative than `sp`, making `CH3CH=CH^(-)` more basic than `CH3C≡C^(-)`.

3. **sp³ Hybridization:**
- The `sp³` hybridized carbon is the least electronegative, making `CH3CH2CH2^(-)` the most basic.

### Step 4: Order of Increasing Basicity
Based on the analysis:
- Least basic: `CH3C≡C^(-)` (sp)
- Intermediate basic: `CH3CH=CH^(-)` (sp²)
- Most basic: `CH3CH2CH2^(-)` (sp³)

Thus, the order of increasing basicity is:
1. `CH3C≡C^(-)` < `CH3CH=CH^(-)` < `CH3CH2CH2^(-)`

### Final Answer
The order of increasing basicity is:
`CH3C≡C^(-) < CH3CH=CH^(-) < CH3CH2CH2^(-)`

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