Arrange the isomers of molecular formula `C_(4)H_(9)CI` in order of decreasing rate of reaction with sodium iodide in acetone.

Answer

To arrange the isomers of the molecular formula \( C_4H_9Cl \) in order of decreasing rate of reaction with sodium iodide in acetone, we need to consider the type of halides and the steric hindrance associated with each isomer. The reaction follows an SN2 mechanism, which is sensitive to steric hindrance.

### Step-by-Step Solution:

1. **Identify the Isomers**: The four isomers of \( C_4H_9Cl \) are:
- n-butyl chloride (1-chlorobutane)
- isobutyl chloride (2-chlorobutane)
- secondary butyl chloride (2-chlorobutane)
- tert-butyl chloride (2-chloro-2-methylpropane)

2. **Classify the Halides**:
- **n-butyl chloride**: This is a primary halide (1°).
- **isobutyl chloride**: This is also a primary halide (1°).
- **secondary butyl chloride**: This is a secondary halide (2°).
- **tert-butyl chloride**: This is a tertiary halide (3°).

3. **Assess Steric Hindrance**:
- Primary halides (n-butyl and isobutyl) have less steric hindrance compared to secondary and tertiary halides.
- Among the primary halides, isobutyl chloride has more steric hindrance due to its branched structure compared to n-butyl chloride.
- The tertiary halide (tert-butyl chloride) has the most steric hindrance.

4. **Determine Reactivity Order**:
- In SN2 reactions, the reactivity decreases with increasing steric hindrance. Therefore, the order of reactivity is:
- n-butyl chloride (most reactive)
- isobutyl chloride
- secondary butyl chloride
- tert-butyl chloride (least reactive)

5. **Final Arrangement**:
- The final order of the isomers of \( C_4H_9Cl \) in terms of decreasing rate of reaction with sodium iodide in acetone is:
1. n-butyl chloride
2. isobutyl chloride
3. secondary butyl chloride
4. tert-butyl chloride