An alkyl halide having molecular formula `C_(4)H_(9)Cl` is optically active. What is its structure ?

Answer

To determine the structure of an alkyl halide with the molecular formula C₄H₉Cl that is optically active, we can follow these steps:

### Step 1: Understand the Molecular Formula
The molecular formula C₄H₉Cl indicates that the compound contains 4 carbon atoms, 9 hydrogen atoms, and 1 chlorine atom.

### Step 2: Identify the Requirement for Optical Activity
For a compound to be optically active, it must have a chiral center (asymmetric carbon). A chiral carbon is one that is bonded to four different groups.

### Step 3: Determine Possible Structures
Given that we need a structure with 4 carbon atoms and one chlorine atom, we can consider various arrangements of these atoms.

### Step 4: Construct the Structure
One possible structure that fits the criteria is 1-chloro-2-methylpropane. In this structure:
- The carbon skeleton consists of 4 carbon atoms.
- The chlorine atom is attached to one of the carbon atoms.
- The carbon to which the chlorine is attached is also bonded to three different groups: a methyl group (−CH₃), an ethyl group (−C₂H₅), and a hydrogen atom (−H).

### Step 5: Verify Chirality
In 1-chloro-2-methylpropane:
- The chiral carbon is the one bonded to Cl, CH₃, C₂H₅, and H.
- Since all four substituents are different, this confirms that the compound is indeed optically active.

### Final Structure
The structure of the optically active alkyl halide C₄H₉Cl is:
```
CH3
|
CH3-CH-CH2-Cl
|
H
```
This represents 1-chloro-2-methylpropane.