Arrange the compounds in order of decreasing acidity:
(1) `Cl-CH_(2)-CH_(2)-OH`
(2) `Cl_(2)CH-CH_(2)-OH`
(3) `I-CH_(2)-CH_(2)-OH`
(4) `CH_(3)-CH_(2)-OH`

Options

A. `IV gt III gt II gt I`
B. `I gt II gt III gt IV`
C. `I gt II gt IV gt III`
D. `II gt I gt III gt IV`

Answer

To arrange the compounds in order of decreasing acidity, we need to analyze the structure of each compound and the stability of their conjugate bases. The acidity of a compound is directly related to the stability of its conjugate base; a more stable conjugate base corresponds to a stronger acid.

### Step-by-Step Solution:

1. **Identify the Compounds:**
- (1) `Cl-CH2-CH2-OH`
- (2) `Cl2CH-CH2-OH`
- (3) `I-CH2-CH2-OH`
- (4) `CH3-CH2-OH`

2. **Formulate the Conjugate Bases:**
- For (1): The conjugate base is `Cl-CH2-CH2-O^-`
- For (2): The conjugate base is `Cl2CH-CH2-O^-`
- For (3): The conjugate base is `I-CH2-CH2-O^-`
- For (4): The conjugate base is `CH3-CH2-O^-`

3. **Evaluate the Stability of Conjugate Bases:**
- **(1) `Cl-CH2-CH2-O^-`:** The presence of one chlorine atom (an electronegative atom) will stabilize the conjugate base through inductive effect.
- **(2) `Cl2CH-CH2-O^-`:** The presence of two chlorine atoms will provide a stronger inductive effect, pulling more electron density away from the oxygen, thus stabilizing the conjugate base even more than in (1).
- **(3) `I-CH2-CH2-O^-`:** Iodine is less electronegative than chlorine, so it will not stabilize the conjugate base as effectively as chlorine does.
- **(4) `CH3-CH2-O^-`:** There are no electronegative groups to stabilize the conjugate base, making it the least stable and thus the weakest acid.

4. **Rank the Acidity:**
- Based on the stability of the conjugate bases, we can rank the compounds:
- (2) `Cl2CH-CH2-OH` (most acidic)
- (1) `Cl-CH2-CH2-OH`
- (3) `I-CH2-CH2-OH`
- (4) `CH3-CH2-OH` (least acidic)

### Final Order of Decreasing Acidity:
**(2) > (1) > (3) > (4)**