What is the decreasing order of strength of bases?

Options

A. `CH_(3)-CH_(2)^(-)gtNH_(2)^(-)gtH-C C^(-)gt OH`
B. `H-C=C^(-) gt CH_(3-)-CH_(2)^(-) gt NH_(2)^(-)2gt OH`
C. `OH gt NH_(2)^(-) gt H-C=C^(-) gt CH_(3)-CH_(2)^(-)`
D. `NH_(2)^(-) gtH-C=C gt OH^(-):gt CH_(3) -CH_(2)^(-)`

Answer

To determine the decreasing order of strength of bases, we need to analyze the basicity of the given compounds based on their structure and the electronegativity of the atoms involved. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Compounds
We are given four compounds to analyze:
1. CH3CH2⁻ (Ethyl anion)
2. NH2⁻ (Amide ion)
3. HC≡C⁻ (Acetylide ion)
4. OH⁻ (Hydroxide ion)

### Step 2: Understand Electronegativity
Electronegativity is a measure of how strongly an atom attracts electrons. The more electronegative an atom, the less likely it is to donate its electrons (which is crucial for basicity). The trend in electronegativity for the relevant atoms is:
- C (sp) > N (sp³) > O (sp³)

### Step 3: Analyze Each Compound
1. **CH3CH2⁻ (Ethyl anion)**: The negative charge is on a carbon atom that is sp³ hybridized. Since carbon in this state is the least electronegative, it can easily donate the electron, making this the strongest base.

2. **NH2⁻ (Amide ion)**: The negative charge is on nitrogen, which is less electronegative than carbon (sp) but more electronegative than carbon (sp³). Thus, it is a strong base, but weaker than the ethyl anion.

3. **HC≡C⁻ (Acetylide ion)**: The negative charge is on a carbon atom that is sp hybridized. Since sp hybridized carbon is more electronegative than sp³ carbon, it is less basic than NH2⁻.

4. **OH⁻ (Hydroxide ion)**: The negative charge is on oxygen, which is the most electronegative among the atoms in our compounds. Therefore, it is the weakest base.

### Step 4: Establish the Order of Basicity
Based on the analysis:
- CH3CH2⁻ (most basic)
- NH2⁻
- HC≡C⁻
- OH⁻ (least basic)

### Final Answer
The decreasing order of strength of bases is:
**CH3CH2⁻ > NH2⁻ > HC≡C⁻ > OH⁻**

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