In the given reactions:
(i). `CH_(3)Br overset(Na,ether)to X underset(hupsilon)overset(Br_(2))toY overset(Na,ether)Z`
(ii) `CH_(3)COOH overset(NaOH)to X underset(CaO)overset(NaOH)to Y underset(h upsilon)overset(Br_(2))toZ`
Identify, X,Y and Z

Options

A. X-(i) `CH_(4)` (ii) `CH_(3)COONa`
Y-`CH_(3)Br, CH_(3)CH_(3)`
Z-`CH_(3)CH_(3), CH_(3)CH_(2)Br`
B. X-(i) `CH_(3)CH_(3)` (ii) `CH_(3)COONa`
Y-`CH_(4), CH_(4)`, Z-`CH_(3)Br,CH_(3)CH_(2)CH_(3)`
C. X-(i) `CH_(3)CH_(3)` (ii)- `CH_(3)COONa`, Y-`CH_(4),CH_(3)CH_(3), Z-CH_(3)CH_(3)CH_(3), CH_(3)Br`
D. X-(i) `CH_(3)CH_(3)` (ii) `CH_(3)COONa`, Y-`CH_(3)CH_(2)Br,CH_(4)` Z-`CH_(3)CH_(2)CH_(2)CH_(3), CH_(3)Br`

Answer

To solve the given reactions and identify the compounds X, Y, and Z, we will break down each reaction step by step.

### Reaction (i):
1. **Starting Compound**: The first compound is `CH3Br` (bromomethane).
2. **Reaction with Sodium in Ether**:
- When `CH3Br` reacts with sodium in the presence of ether, it undergoes a reaction known as Wood's reaction.
- This reaction leads to the formation of an alkane by eliminating sodium bromide (NaBr).
- The product formed is `C2H6` (ethane).
- **Identified Product X**: `X = C2H6` (ethane).

3. **Bromination of Ethane**:
- The product `X` (ethane) then reacts with `Br2` in the presence of light (hν).
- This results in the bromination of ethane, following a free radical mechanism.
- The product formed is `C2H5Br` (bromoethane).
- **Identified Product Y**: `Y = C2H5Br` (bromoethane).

4. **Wood's Reaction Again**:
- The product `Y` (bromoethane) reacts again with sodium in ether.
- This leads to another Wood's reaction, resulting in the formation of a symmetrical alkane.
- The product formed is `C4H10` (butane).
- **Identified Product Z**: `Z = C4H10` (butane).

### Summary of Reaction (i):
- **X**: Ethane (`C2H6`)
- **Y**: Bromoethane (`C2H5Br`)
- **Z**: Butane (`C4H10`)

### Reaction (ii):
1. **Starting Compound**: The first compound is `CH3COOH` (acetic acid).
2. **Reaction with Sodium Hydroxide**:
- When `CH3COOH` reacts with sodium hydroxide (NaOH), it forms sodium acetate (`CH3COONa`).
- **Identified Product X**: `X = CH3COONa` (sodium acetate).

3. **Decarboxylation Reaction**:
- The sodium acetate then reacts with soda lime (NaOH and CaO), which leads to a decarboxylation reaction.
- In this reaction, carbon dioxide (CO2) is eliminated, and methane (`CH4`) is formed.
- **Identified Product Y**: `Y = CH4` (methane).

4. **Bromination of Methane**:
- The product `Y` (methane) reacts with `Br2` in the presence of light (hν).
- This results in the formation of bromomethane (`CH3Br`), following a free radical mechanism.
- **Identified Product Z**: `Z = CH3Br` (bromomethane).

### Summary of Reaction (ii):
- **X**: Sodium acetate (`CH3COONa`)
- **Y**: Methane (`CH4`)
- **Z**: Bromomethane (`CH3Br`)

### Final Identification:
- For Reaction (i):
- X = Ethane (`C2H6`)
- Y = Bromoethane (`C2H5Br`)
- Z = Butane (`C4H10`)

- For Reaction (ii):
- X = Sodium acetate (`CH3COONa`)
- Y = Methane (`CH4`)
- Z = Bromomethane (`CH3Br`)