Arrange acididy of given compounds in decreasing order
(I). `CH_(3)-N-CH_(2)-OH`
(II). `CH_(3)-NH-CH_(2)-CH_(2)-CH_(2)-OH`
(III). `(CH_(3))_(3)overset(o+)(N)-CH_(2)-CH_(2)-OH`

Options

A. IIIgtIgtII
B. IIIgtIIgtI
C. IgtIIgtIII
D. IIgtIgtIII

Explanation

To arrange the acidity of the given compounds in decreasing order, we need to evaluate the stability of the conjugate bases formed when each compound loses a proton (H⁺). The more stable the conjugate base, the stronger the acid.

### Step-by-Step Solution:

1. **Identify the Compounds**:
- (I) `CH₃-N-CH₂-OH`
- (II) `CH₃-NH-CH₂-CH₂-CH₂-OH`
- (III) `(CH₃)₃N⁺-CH₂-CH₂-OH`

2. **Determine the Conjugate Bases**:
- For (I): The conjugate base is `CH₃-N-CH₂-O⁻`.
- For (II): The conjugate base is `CH₃-NH-CH₂-CH₂-CH₂-O⁻`.
- For (III): The conjugate base is `(CH₃)₃N⁺-CH₂-CH₂-O⁻`.

3. **Analyze the Stability of the Conjugate Bases**:
- The stability of the conjugate base is influenced by the presence of electron-donating or electron-withdrawing groups.
- In (I) and (II), the nitrogen atom is positively charged, which can stabilize the negative charge on the oxygen when the proton is lost.
- In (III), the presence of three methyl groups (alkyl groups) around the nitrogen increases the electron density, which can destabilize the negative charge on the oxygen.

4. **Compare the Compounds**:
- **Compound (III)** has the most stable conjugate base because the positive charge on nitrogen helps stabilize the negative charge on oxygen.
- **Compound (I)** has fewer alkyl groups compared to (II), making its conjugate base more stable than that of (II).
- **Compound (II)** has the least stability due to the increased number of alkyl groups, which donate electrons and destabilize the negative charge on oxygen.

5. **Arranging in Decreasing Order of Acidity**:
- The order of acidity based on the stability of the conjugate bases is:
- (III) > (I) > (II)

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