End product S of the reaction sequence is
`CH_3-CH_2Broverset(KCN)(rarr)P overset(H_2O //H^+)(rarr)Q overset(SOCl_2)(rarr) R overset((C_2H_5)_(2)Cd)(rarr)S`

Options

A. `CH_3CH_2-CH_2-O-C_2H_5`
B. `C_2H_5COOCH_2-CH_3`
C. `CH_3-CH_2-COC_2H_5`
D. `CH_3COOC_2H_5`

Answer

To determine the end product S of the given reaction sequence, let's break down the steps involved in the transformation of the starting material through each reaction.

### Step-by-Step Solution:

1. **Starting Material**: The starting compound is `CH3-CH2Br` (Bromoethane).

2. **Reaction with KCN**:
- Bromoethane reacts with potassium cyanide (KCN).
- The cyanide ion (CN⁻) acts as a nucleophile and attacks the electrophilic carbon in bromoethane, leading to the substitution of bromine.
- **Product after this step**: `CH3-CH2CN` (Propionitrile).

3. **Hydrolysis**:
- The next step involves hydrolysis of the nitrile group (CN) in the presence of water and an acid (H⁺).
- During hydrolysis, the nitrile is converted to a carboxylic acid.
- **Product after this step**: `CH3-CH2COOH` (Propanoic acid).

4. **Reaction with SOCl2**:
- The carboxylic acid reacts with thionyl chloride (SOCl2) to form an acyl chloride.
- The hydroxyl group (OH) is replaced by a chlorine atom (Cl).
- **Product after this step**: `CH3-CH2COCl` (Propionyl chloride).

5. **Reaction with (C2H5)2Cd**:
- The acyl chloride reacts with diethylcadmium ((C2H5)2Cd), where cadmium acts as a nucleophile.
- The nucleophile attacks the electrophilic carbonyl carbon, leading to the formation of a new carbon-carbon bond.
- The leaving group (Cl) is expelled.
- **Final Product S**: `CH3-CH2C(=O)C2H5` (Butyric acid derivative).

### Final Product:
The end product S of the reaction sequence is `CH3-CH2C(=O)C2H5`.