`CH_(3)CH_(2)CH=CH_(2)underset(H_(2)O.H_(2)O_(2).OH^-)overset(B_(2)H_(6))rarrZ`
What is Z?

Options

A. `CH_(3)CH_(2)undersetunderset(OH)(|)(C)HCH_(3)`
B. `CH_(3)CH_(2)CH_(2)CHO`
C. `CH_(3)CH_(2)CH_(2)CH_(3)`
D. `CH_(3)CH_(2)CH_(2)CH_(2)OH`

Explanation

To determine the product Z from the reaction of butene with the reagents B2H6, H2O, H2O2, and OH^-, we will follow the steps of hydroboration-oxidation.

### Step-by-Step Solution:

1. **Identify the Reactant**: The reactant is butene, which can be represented as CH3-CH2-CH=CH2. This is a terminal alkene.

2. **Hydroboration Reaction**: The first step involves hydroboration using B2H6 (diborane). In this reaction, the boron atom from B2H6 will add to the less substituted carbon of the alkene (the terminal carbon), while a hydrogen atom will add to the more substituted carbon. This results in the formation of a trialkyl borane intermediate.

- The boron (B) will bond to the terminal carbon (C1), and a hydrogen (H) will bond to the second carbon (C2). The structure after hydroboration will be:
- CH3-CH2-CH2-CH(BH2)-H

3. **Oxidation Step**: The next step involves oxidation using H2O, H2O2, and OH^-. In this step, the boron atom will be replaced by a hydroxyl group (OH). The reaction will convert the trialkyl borane into alcohols.

- The boron (BH2) will be replaced by OH, leading to the formation of butanol. Since the hydroboration adds the OH group to the terminal carbon, we will get:
- CH3-CH2-CH2-CH2OH (1-butanol)

4. **Final Product (Z)**: The final product Z is 1-butanol (CH3-CH2-CH2-CH2OH).

### Conclusion:
Thus, the product Z from the given reaction is **1-butanol**.