Arrange the following acid in decreasing order of their acidic strength :
(I) `CH_(3)-OH` (II) `CH_(3)CH_(2)-OH`
(III) `CH_(3)-underset(CH_(3))underset(|)(CH)-OH` (IV) `CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-OH`

Options

A. (a)`II gt III gt IV gt I`
B. (b)`I gt II gt III gt IV`
C. (c)`I gt III gt II gt IV`
D. (d)`II gt IV gt III gt I`

Answer

To determine the decreasing order of acidic strength for the given alcohols, we need to analyze the stability of the corresponding alkoxide ions formed when each alcohol donates a proton (H⁺). The more stable the alkoxide ion, the stronger the acid.

### Step-by-Step Solution:

1. **Identify the Alcohols and Their Alkoxide Ions**:
- (I) `CH₃OH` → Alkoxide: `CH₃O⁻`
- (II) `CH₃CH₂OH` → Alkoxide: `CH₃CH₂O⁻`
- (III) `CH₃C(CH₃)₂OH` → Alkoxide: `CH₃C(CH₃)₂O⁻`
- (IV) `CH₃C(CH₃)₃OH` → Alkoxide: `CH₃C(CH₃)₃O⁻`

2. **Analyze the Alkoxide Stability**:
- The stability of the alkoxide ion is influenced by the electron-donating or electron-withdrawing effects of the substituents attached to the carbon atom bearing the negative charge.
- Alkyl groups (like CH₃) are electron-donating due to the +I (inductive) effect, which increases the negative charge on the alkoxide ion, thus destabilizing it.

3. **Stability Comparison**:
- For `CH₃O⁻` (from (I)), there is one CH₃ group donating electrons.
- For `CH₃CH₂O⁻` (from (II)), there are two CH₃ groups donating electrons, making it less stable than (I).
- For `CH₃C(CH₃)₂O⁻` (from (III)), there are three CH₃ groups, which further destabilizes the ion due to increased negative charge.
- For `CH₃C(CH₃)₃O⁻` (from (IV)), there are four CH₃ groups, leading to the highest destabilization of the alkoxide ion.

4. **Order of Stability**:
- The order of stability of the alkoxide ions (and thus the order of acidic strength) is:
- (I) `CH₃O⁻` > (II) `CH₃CH₂O⁻` > (III) `CH₃C(CH₃)₂O⁻` > (IV) `CH₃C(CH₃)₃O⁻`

5. **Final Order of Acidic Strength**:
- Therefore, the decreasing order of acidic strength is:
- (I) > (II) > (III) > (IV)
- In terms of the original compounds, the order is:
- `CH₃OH` > `CH₃CH₂OH` > `CH₃C(CH₃)₂OH` > `CH₃C(CH₃)₃OH`

### Conclusion:
The final answer is:
1. `CH₃OH` (I)
2. `CH₃CH₂OH` (II)
3. `CH₃C(CH₃)₂OH` (III)
4. `CH₃C(CH₃)₃OH` (IV)