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 arrange the compounds in order of decreasing basicity, we need to analyze the stability of the negative charges in each compound. The more stable the negative charge, the less basic the compound will be.

Let's break down the compounds:

1. **(I) CH₃CH₂MgBr**:
- Here, the negative charge is on the carbon atom (CH₃CH₂⁻). Carbon is less electronegative than oxygen, which means the negative charge is less stable compared to when it is on oxygen.

2. **(II) HC≡CMgBr**:
- In this compound, the negative charge is on a carbon atom that is part of a triple bond (HC≡C⁻). The sp-hybridized carbon has more s-character (50% s-character) compared to sp² or sp³, which makes it more electronegative and thus stabilizes the negative charge better than in (I). However, it is still less stable than a negative charge on oxygen.

3. **(III) CH₃CH₂OMgBr**:
- Here, the negative charge is on the oxygen atom (CH₃CH₂O⁻). Oxygen is more electronegative than carbon, which means the negative charge is more stable on oxygen. Therefore, this compound will be the least basic.

Now, let's summarize the stability of the negative charges:
- (III) CH₃CH₂OMgBr: Negative charge on oxygen (most stable, least basic)
- (II) HC≡CMgBr: Negative charge on sp-hybridized carbon (more stable than sp³, but less than oxygen)
- (I) CH₃CH₂MgBr: Negative charge on sp³ carbon (least stable, most basic)

### Order of Decreasing Basicity:
1. **(I) CH₃CH₂MgBr** (most basic)
2. **(II) HC≡CMgBr**
3. **(III) CH₃CH₂OMgBr** (least basic)

### Final Answer:
The order of decreasing basicity is: **(I) > (II) > (III)**.