Arrange the following compounds in the decreasing order of their boiling points.
(I) Methanol (II) Ethanol (III) propan-1-ol (IV) Butan-1-ol

Options

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

Explanation

To arrange the compounds Methanol (I), Ethanol (II), Propan-1-ol (III), and Butan-1-ol (IV) in decreasing order of their boiling points, we will analyze their molecular weights and structures.

### Step 1: Determine the molecular weights of each compound.
- **Methanol (CH₃OH)**:
- Carbon (C): 12 g/mol
- Hydrogen (H): 4 g/mol (3 from CH₃ and 1 from OH)
- Oxygen (O): 16 g/mol
- Total = 12 + 4 + 16 = **32 g/mol**

- **Ethanol (C₂H₅OH)**:
- Carbon (C): 24 g/mol (2 × 12)
- Hydrogen (H): 6 g/mol (5 from C₂H₅ and 1 from OH)
- Oxygen (O): 16 g/mol
- Total = 24 + 6 + 16 = **46 g/mol**

- **Propan-1-ol (C₃H₇OH)**:
- Carbon (C): 36 g/mol (3 × 12)
- Hydrogen (H): 8 g/mol (7 from C₃H₇ and 1 from OH)
- Oxygen (O): 16 g/mol
- Total = 36 + 8 + 16 = **60 g/mol**

- **Butan-1-ol (C₄H₉OH)**:
- Carbon (C): 48 g/mol (4 × 12)
- Hydrogen (H): 10 g/mol (9 from C₄H₉ and 1 from OH)
- Oxygen (O): 16 g/mol
- Total = 48 + 10 + 16 = **74 g/mol**

### Step 2: Compare the molecular weights.
- Methanol: 32 g/mol
- Ethanol: 46 g/mol
- Propan-1-ol: 60 g/mol
- Butan-1-ol: 74 g/mol

### Step 3: Arrange the compounds in decreasing order of boiling points.
The boiling point is generally higher for compounds with higher molecular weights due to stronger van der Waals forces and increased surface area. Therefore, we can arrange them as follows:

1. **Butan-1-ol (IV)**: 74 g/mol
2. **Propan-1-ol (III)**: 60 g/mol
3. **Ethanol (II)**: 46 g/mol
4. **Methanol (I)**: 32 g/mol

### Final Order:
**Butan-1-ol (IV) > Propan-1-ol (III) > Ethanol (II) > Methanol (I)**