Calculate the `DU` of following compounds :
(i) `C_(6)H_(6)ClBrO` , (ii) `C_(5)H_(9)N`

Explanation

To calculate the Degree of Unsaturation (DU) for the given compounds, we will use the formula:

\[ DU = \frac{C - H + X + N}{2} + 1 \]

Where:
- \( C \) = number of carbon atoms
- \( H \) = number of hydrogen atoms
- \( X \) = number of halogens (F, Cl, Br, I)
- \( N \) = number of nitrogen atoms

### Step-by-Step Solution:

**(i) For the compound \( C_6H_6ClBrO \):**

1. **Identify the number of atoms:**
- Carbon (C) = 6
- Hydrogen (H) = 6
- Halogens (Cl + Br) = 2 (1 Cl and 1 Br)
- Nitrogen (N) = 0 (there are no nitrogen atoms)

2. **Plug the values into the formula:**
\[ DU = \frac{6 - 6 + 2 + 0}{2} + 1 \]

3. **Calculate:**
\[ DU = \frac{6 - 6 + 2 + 0}{2} + 1 = \frac{2}{2} + 1 = 1 + 1 = 2 \]

4. **Final result for the first compound:**
- Degree of Unsaturation (DU) = 2

**(ii) For the compound \( C_5H_9N \):**

1. **Identify the number of atoms:**
- Carbon (C) = 5
- Hydrogen (H) = 9
- Halogens = 0
- Nitrogen (N) = 1

2. **Plug the values into the formula:**
\[ DU = \frac{5 - 9 + 0 + 1}{2} + 1 \]

3. **Calculate:**
\[ DU = \frac{5 - 9 + 0 + 1}{2} + 1 = \frac{-3}{2} + 1 = -1.5 + 1 = -0.5 \]

Since the degree of unsaturation cannot be negative, we need to check our calculations. The correct calculation should be:
\[ DU = \frac{5 - 9 + 0 + 1}{2} + 1 = \frac{-3 + 2}{2} = \frac{-1}{2} + 1 = 0.5 \]
This indicates that the compound is saturated or has a very low degree of unsaturation.

4. **Final result for the second compound:**
- Degree of Unsaturation (DU) = 0.5 (but practically it indicates saturation)

### Summary of Results:
- For \( C_6H_6ClBrO \), DU = 2
- For \( C_5H_9N \), DU = 0.5