Complete the missing links
`CH_(3)CHBrCH_(3) overset("alc. KOH")rarr Xoverset("HBr, Peroxide")rarr Y overset(CH_(3)ONa)rarr Z`

Options

A. `{:("X","Y","Z"),(CH_(3)CH=CH_(2),CH_(3)CH(Br)CH_(2)Br,CH_(3)CH(OH)CH_(3)):}`
B. `{:("X","Y","Z"),(CH_(3)CH=CH_(2),CH_(3)CH_(2)CH_(2)Br,CH_(3)CH_(2)OCH_(2)CH_(3)):}`
C. `{:("X","Y","Z"),(CH_(3)CH=CH_(2),CH_(3)CH(Br)CH_(3),CH_(3)CH_(2)OCH_(2)CH_(3)):}`
D. `{:("X","Y","Z"),(CH_(3)CH=CH_(2),CH_(3)CH_(2)CH_(2)Br,CH_(3)CH_(2)CH_(2)OCH_(3)):}`

Answer

To solve the problem, we need to identify the compounds X, Y, and Z in the given reaction sequence. Let's break down the steps:

### Step 1: Identify Compound X
We start with the compound \( CH_3CHBrCH_3 \) and treat it with alcoholic KOH.

- **Reaction Type**: Alcoholic KOH is a strong base and promotes elimination reactions (dehydrohalogenation).
- **Elimination**: The elimination of HBr occurs, where the Br leaves and a hydrogen atom from the adjacent carbon is also removed.

The structure of the starting compound can be represented as:
```
CH3
|
CHBr
|
CH3
```

After elimination, we will form a double bond between the first and second carbon atoms:
```
CH3
|
CH=CH2
```

Thus, compound X is:
\[ X = CH_3CH=CH_2 \]

### Step 2: Identify Compound Y
Next, we treat compound X with HBr in the presence of peroxide.

- **Reaction Type**: The presence of peroxide leads to the anti-Markovnikov addition of HBr.
- **Mechanism**: The bromine atom will add to the less substituted carbon atom (the one with more hydrogen atoms), while the hydrogen will add to the more substituted carbon.

Starting with:
```
CH3
|
CH=CH2
```

The addition of HBr will yield:
```
CH3
|
CH2Br
```

Thus, compound Y is:
\[ Y = CH_3CH_2CH_2Br \]

### Step 3: Identify Compound Z
Finally, we treat compound Y with \( CH_3ONa \) (sodium methoxide).

- **Reaction Type**: This reaction is a Williamson ether synthesis.
- **Mechanism**: The bromine atom is a good leaving group and will leave, allowing the methoxide ion (\( CH_3O^- \)) to attack the carbon atom where the bromine was attached.

Starting with:
```
CH3
|
CH2Br
```

After the reaction, we get:
```
CH3
|
CH2OCH3
```

Thus, compound Z is:
\[ Z = CH_3CH_2CH_2OCH_3 \]

### Summary of Compounds
- **X**: \( CH_3CH=CH_2 \)
- **Y**: \( CH_3CH_2CH_2Br \)
- **Z**: \( CH_3CH_2CH_2OCH_3 \)