Which is decreasing order of acidity in,
`HCOOH(I), CH_(3)COOH(II), CH_(3)CH_(2)COOH(III) and C_(6)H_(5)COOH(IV)`?

Options

A. `IgtIIgtIIIgtIV`
B. `IVgtIIIgtIIgtI`
C. `IVgtIgtIIgtIII`
D. `IgtIVgtIIgtIII`

Answer

To determine the decreasing order of acidity among the compounds HCOOH (formic acid), CH₃COOH (acetic acid), CH₃CH₂COOH (propanoic acid), and C₆H₅COOH (benzoic acid), we need to analyze the influence of substituents on the acidity of these carboxylic acids.

### Step-by-Step Solution:

1. **Identify the Compounds:**
- HCOOH (I): Formic acid
- CH₃COOH (II): Acetic acid
- CH₃CH₂COOH (III): Propanoic acid
- C₆H₅COOH (IV): Benzoic acid

2. **Understand Acidity in Carboxylic Acids:**
- The acidity of carboxylic acids is primarily determined by the stability of the conjugate base formed after deprotonation. A more stable conjugate base corresponds to a stronger acid.

3. **Analyze Electron Withdrawing and Donating Effects:**
- Electron-withdrawing groups (EWGs) increase acidity by stabilizing the negative charge on the conjugate base.
- Electron-donating groups (EDGs) decrease acidity by destabilizing the conjugate base.

4. **Evaluate Each Compound:**
- **HCOOH (I)**: No substituents, so it has the highest acidity. The conjugate base is stable as there are no destabilizing effects.
- **C₆H₅COOH (IV)**: The phenyl group (C₆H₅) can act as a mild electron-withdrawing group due to resonance. This stabilizes the conjugate base, making it more acidic than the alkyl-substituted acids.
- **CH₃COOH (II)**: The methyl group (CH₃) is an electron-donating group, which slightly decreases acidity compared to HCOOH and C₆H₅COOH.
- **CH₃CH₂COOH (III)**: The ethyl group (CH₂CH₃) is also an electron-donating group, making it the least acidic among the four compounds.

5. **Determine the Order of Acidity:**
- Based on the analysis, the decreasing order of acidity is:
1. HCOOH (I)
2. C₆H₅COOH (IV)
3. CH₃COOH (II)
4. CH₃CH₂COOH (III)

### Final Order of Acidity:
**HCOOH > C₆H₅COOH > CH₃COOH > CH₃CH₂COOH**