Arrange in order of decreasing basicity
`(I) CH_3CH_2MgBr`
`(II) HC = CMgBr " " (III) CH_3CH_3OMgBr`

Options

A. I, II, III
B. I, III, II
C. III, II, I
D. II, I, III

Answer

To determine the order of decreasing basicity among the given compounds, we need to analyze the stability of the corresponding anions formed when these compounds act as bases. The more stable the anion, the less basic the compound will be. Here's the step-by-step solution:

### Step 1: Identify the Compounds
We have three compounds:
1. \( \text{(I) CH}_3\text{CH}_2\text{MgBr} \)
2. \( \text{(II) HC} \equiv \text{CMgBr} \)
3. \( \text{(III) CH}_3\text{CH}_2\text{OMgBr} \)

### Step 2: Determine the Anions
When these compounds act as bases, they will generate the following anions:
1. From \( \text{(I) CH}_3\text{CH}_2\text{MgBr} \): \( \text{CH}_3\text{CH}_2^- \) (ethyl anion)
2. From \( \text{(II) HC} \equiv \text{CMgBr} \): \( \text{C} \equiv \text{C}^- \) (acetylide anion)
3. From \( \text{(III) CH}_3\text{CH}_2\text{OMgBr} \): \( \text{CH}_3\text{CH}_2\text{O}^- \) (alkoxide anion)

### Step 3: Analyze the Stability of the Anions
- **Ethyl Anion \( \text{CH}_3\text{CH}_2^- \)**: This anion is relatively stable due to the inductive effect of the ethyl group, but it is still a carbanion which is less stable than an anion stabilized by resonance or electronegativity.

- **Acetylide Anion \( \text{C} \equiv \text{C}^- \)**: This anion is more stable due to the sp-hybridization of the carbon, which has more s-character (50% s-character). This makes the negative charge more stable, thus less basic.

- **Alkoxide Anion \( \text{CH}_3\text{CH}_2\text{O}^- \)**: The negative charge is on oxygen, which is highly electronegative and can stabilize the negative charge effectively. Therefore, this anion is the most stable and least basic.

### Step 4: Order of Basicity
Since the stability of the anions is inversely related to their basicity, we can arrange the compounds in order of decreasing basicity as follows:
1. \( \text{(I) CH}_3\text{CH}_2\text{MgBr} \) - most basic
2. \( \text{(II) HC} \equiv \text{CMgBr} \)
3. \( \text{(III) CH}_3\text{CH}_2\text{OMgBr} \) - least basic

### Final Answer
The order of decreasing basicity is:
\[ \text{(I) > (II) > (III)} \]