Arrange basicity of given compounds in decreasing order.
(1) `overset(ө)OH`
(2) `CH_(3)-overset(ө)O`
(3) `CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-overset(ө)O`
(4) `CH_(3)-underset(CH_(3))underset(|)CH-overset(ө)O`

Options

A. 1, 2, 3, 4
B. 4, 3, 2, 1
C. 3, 4, 2, 1
D. 3, 4, 1, 2

Answer

To determine the basicity of the given compounds, we need to analyze the electron-donating effects of the substituents around the oxygen atom in each compound. Basicity is related to the ability of a compound to donate a lone pair of electrons, and this ability can be influenced by the presence of electron-donating groups (which increase electron density) or electron-withdrawing groups (which decrease electron density).

### Step-by-Step Solution:

1. **Identify the Compounds**:
- (1) `OH`
- (2) `CH₃-O`
- (3) `CH₃-C(CH₃)₂-O`
- (4) `CH₃-CH(CH₃)-O`

2. **Analyze the Electron-Donating Effects**:
- The basicity of these compounds is influenced by the +I (inductive) effect of the alkyl groups attached to the oxygen atom. Alkyl groups are electron-donating and will increase the electron density on the oxygen, enhancing its ability to donate the lone pair.

3. **Rank the Compounds Based on +I Effect**:
- **Compound (3)**: `CH₃-C(CH₃)₂-O` has the highest number of alkyl groups (3 methyl groups) attached to the oxygen, providing the strongest +I effect. Therefore, it is the most basic.
- **Compound (4)**: `CH₃-CH(CH₃)-O` has 2 methyl groups attached to the oxygen, providing a significant +I effect, making it the second most basic.
- **Compound (2)**: `CH₃-O` has only one methyl group, giving it a moderate +I effect, placing it third in basicity.
- **Compound (1)**: `OH` has no alkyl groups attached, resulting in no +I effect, making it the least basic.

4. **Final Ranking**:
- Based on the analysis, the order of basicity from highest to lowest is:
- (3) > (4) > (2) > (1)

### Conclusion:
The decreasing order of basicity for the given compounds is:
**(3) > (4) > (2) > (1)**