During the conversion :
`C_(6)H_(5)CH_(2)CH_(3) overset((a))to X overset((b))to C_(6)H_(5)CH=CH_(2)`
the reagents (a) and (b) are respectively

Options

A. `SOCl_(2)`, alc. KOH
B. `Cl_(2) // hv, H_(2)O`
C. `SO_(2)Cl_(2)`, aq. KOH
D. `SO_(2)Cl_(2)`, alc. KOH

Explanation

To solve the question regarding the conversion of `C6H5CH2CH3` to `C6H5CH=CH2`, we need to identify the reagents (a) and (b) used in the process. Let's break down the steps:

### Step 1: Identify the starting material and the final product
- The starting material is `C6H5CH2CH3`, which is a primary alkyl chain with a benzene ring (ethylbenzene).
- The final product is `C6H5CH=CH2`, which is an alkene (specifically, styrene).

### Step 2: Determine the first reagent (a)
- To convert an alkane (or a primary alkyl chain) to an alkyl halide, we can use reagents that will introduce a halogen (like chlorine) to the molecule.
- A suitable reagent for this conversion is `SO2Cl2` (thionyl chloride), which can convert the alkane to an alkyl chloride.

### Step 3: Determine the second reagent (b)
- After obtaining the alkyl halide (let's say `C6H5CH2CH2Cl`), we need a reagent that can facilitate the elimination of HCl to form the alkene.
- A common reagent for this elimination reaction is `alcoholic KOH`, which acts as a strong base to promote the elimination reaction.

### Conclusion
Thus, the reagents (a) and (b) are:
- (a) `SO2Cl2`
- (b) `alcoholic KOH`

### Final Answer
- Reagent (a): `SO2Cl2`
- Reagent (b): `alcoholic KOH`