Identity Z in the sequence of reactions, `CH_3CH_2CH=CH_2overset(HBr//H_2O_2)to Y overset(C_2H_5ONa)to Z`

Options

A. `CH_3(CH_2)_4-O-CH_3`
B. `CH_3CH_2-CH(CH_3)-O-CH_2CH_3`
C. `CH_3-(CH_2)_3-O-CH_2CH_3`
D. `(CH_3)_2CH_2-O-CH_2CH_3`

Explanation

To identify compound Z in the sequence of reactions given, we will go through the reactions step by step.

### Step 1: Identify the starting compound
The starting compound is 1-butene, which is represented as \( CH_3CH_2CH=CH_2 \).

### Step 2: Reaction with HBr in the presence of H2O2
When 1-butene reacts with HBr in the presence of hydrogen peroxide (H2O2), it follows the anti-Markovnikov addition rule due to the presence of the peroxide.

- The peroxide generates hydroxyl radicals (•OH), which react with HBr to produce bromine radicals (•Br).
- The bromine radical adds to the less substituted carbon of the alkene (the terminal carbon), leading to the formation of a more stable radical at the more substituted carbon.

The reaction can be summarized as follows:
1. The bromine radical (•Br) adds to the terminal carbon of 1-butene, forming a butyl radical (CH3CH2CH•CH2Br).
2. This radical then reacts with another HBr molecule to yield 1-bromobutane (CH3CH2CH2CH2Br).

### Step 3: Reaction with sodium ethoxide (C2H5ONa)
Next, 1-bromobutane (the product Y) reacts with sodium ethoxide (C2H5ONa), which acts as a nucleophile.

- The ethoxide ion (C2H5O-) will attack the bromine atom in 1-bromobutane, leading to the substitution of the bromine atom with the ethoxy group.

The reaction can be summarized as follows:
1. The ethoxide ion attacks the carbon bonded to bromine, displacing the bromine and forming butyl ethyl ether.

### Final Structure of Z
The final product Z is butyl ethyl ether, which can be represented as:
\[ CH_3CH_2CH_2OCH_2CH_3 \]
This compound has a total of 6 carbon atoms (4 from butyl and 2 from ethyl).

### Conclusion
Thus, the compound Z in the sequence of reactions is butyl ethyl ether, represented as:
\[ CH_3CH_2CH_2OCH_2CH_3 \]