Borane is an electron deficient compound. It has only six valence eletons, so the boron atom lacks an octet. Acquiring an octet is the driving force for the unusual bonding structure found in boron compounds. As an electron deficient compound, `BH_(3)` is a strong electrophile, capable of adding to a double bond. This hydroboration of double bond is though to oC Cur in one step, with the boron atom adding to the less highly substituted end of the double bond. In transition state, the boron atom withdraws electrons from the pi bond and the carbon at theother end of the double bond acquires a partial positive charge. This positive charge is more stable on the more highly subsituted carbon atom. The second step is the oxidation of boron atom, removing it from carbon and replacing it with hydroxyl group by using `H_(2)O_(2)//OH^(bar(..))`.
The simultaneous addition of boron and hydrogen to the double bond leads to a syn addition. Oxidation of the trialkyl borane replaces boron with a hydroxyl group in the same stereochemical position. Thus, hydroboration of alkenen is an example of steropecific reaction, in which different steroisomers of starting compounds react to give different steroisomers of the product.
`CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-CH=CH_(2)` `underset((ii)H_(2)O_(2)//OH^(bar(..)))overset((i)BH_(3)//THF)rarrZ`.
`Z` is :

Answer

To solve the problem regarding the hydroboration of an alkene using borane (BH3), we will follow these steps:

### Step 1: Identify the Alkene and the Reagents
The alkene given is `CH3-CH(CH3)-CH=CH2`, and the reagents are `BH3` in `THF` (tetrahydrofuran) followed by `H2O2` and `OH-`.

### Step 2: Hydroboration Reaction
In the hydroboration step, borane (BH3) acts as an electrophile and adds to the double bond of the alkene. The addition occurs in a syn fashion, meaning that both boron and hydrogen will add to the same side of the double bond.

- The boron atom will preferentially add to the less substituted carbon of the double bond (the terminal carbon in this case).
- The hydrogen from BH3 will add to the more substituted carbon.

### Step 3: Formation of Trialkylborane
After the hydroboration, we will form a trialkylborane. In this case, the product after the first addition of BH3 will be:

- Boron (B) will be attached to the less substituted carbon (the terminal carbon).
- The more substituted carbon will have a hydrogen atom from BH3.

This leads to the formation of a trialkylborane structure, which can be represented as:
```
CH3
|
CH3-CH-CH2-BH2
```

### Step 4: Oxidation Step
In the second step, we perform oxidation using `H2O2` and `OH-`. This step replaces the boron atom with a hydroxyl group (OH) while maintaining the stereochemistry of the addition.

- The boron atom is removed, and a hydroxyl group is added to the same carbon where boron was attached.
- The reaction will yield the final alcohol product.

### Step 5: Final Product
The final product after the oxidation step will be:
```
CH3
|
CH3-CH-CH2-OH
```
This product will be an alcohol, specifically 2-butanol, following the anti-Markovnikov rule where the hydroxyl group ends up on the more substituted carbon.

### Summary of the Reaction
The overall reaction can be summarized as:
```
CH3-CH(CH3)-CH=CH2 + BH3 → CH3-CH(CH3)-CH2-BH2 (trialkylborane)
CH3-CH(CH3)-CH2-BH2 + H2O2/OH- → CH3-CH(CH3)-CH2-OH (2-butanol)
```