Arrange in decreasing order of acidic strengths , `H_(2)O, CH_(3)OH, (CH_(3))_(2)CHOH`.

Answer

To arrange the compounds \( H_2O \), \( CH_3OH \), and \( (CH_3)_2CHOH \) in decreasing order of acidic strength, we will analyze the acidic properties of each compound based on the presence of electron-donating or electron-withdrawing groups.

### Step-by-Step Solution:

1. **Identify the Compounds**:
- The compounds we need to compare are:
- \( H_2O \) (water)
- \( CH_3OH \) (methanol)
- \( (CH_3)_2CHOH \) (isopropanol)

2. **Understand Acidic Strength**:
- Acidic strength is determined by the ability of a compound to donate a proton (\( H^+ \)). The easier it is for a compound to release a proton, the stronger the acid it is.
- Electron-withdrawing groups increase acidity because they stabilize the negative charge that forms when a proton is lost. Conversely, electron-donating groups decrease acidity.

3. **Analyze Each Compound**:
- **Water (\( H_2O \))**:
- There are no electron-donating groups attached to water. It has a neutral structure, making it a relatively strong acid among the three.
- **Methanol (\( CH_3OH \))**:
- The \( CH_3 \) group is an electron-donating group (it exhibits a +I effect). This reduces the acidity compared to water because it makes it harder for the molecule to release a proton.
- **Isopropanol (\( (CH_3)_2CHOH \))**:
- This compound has two \( CH_3 \) groups, both of which are electron-donating. The presence of two electron-donating groups further decreases the acidity compared to both water and methanol.

4. **Rank the Acids**:
- Based on the analysis:
- \( H_2O \) is the most acidic because it has no electron-donating groups.
- \( CH_3OH \) is less acidic than water due to one electron-donating \( CH_3 \) group.
- \( (CH_3)_2CHOH \) is the least acidic because it has two electron-donating \( CH_3 \) groups.

5. **Final Order**:
- The decreasing order of acidic strength is:
\[ H_2O > CH_3OH > (CH_3)_2CHOH \]