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 :

Options

A. Optically active `1^(@)-`alcohol
B. Optically active `2^(@)-`alcohol
C. Optically inactive `1^(@)-`alchohol
D. Optically inactive `3^(@)-`alcohol

Answer

To solve the problem, we will follow the steps involved in the hydroboration-oxidation reaction of the given alkene with borane (BH₃) and hydrogen peroxide (H₂O₂) in an alkaline medium.

### Step-by-Step Solution:

1. **Identify the Alkene Structure**:
The given alkene is `CH₃-CH(CH₃)-C=CH₂`. This structure has a double bond between the last two carbon atoms (C=C) and is substituted with methyl groups.

2. **Hydroboration Reaction**:
- When BH₃ is added to the alkene, it undergoes hydroboration.
- In this step, the boron atom from BH₃ adds to the less substituted carbon of the double bond (the terminal carbon in this case), while a hydrogen atom adds to the more substituted carbon.
- This results in the formation of trialkyl borane, where the boron atom is attached to the less substituted carbon.

3. **Formation of Trialkyl Borane**:
- After the addition of BH₃, the product can be represented as:
```
CH₃-CH(CH₃)-C(BR)-CH₂
```
- Here, BR represents the boron atom attached to the less substituted carbon.

4. **Oxidation Step**:
- The next step involves the oxidation of the trialkyl borane using hydrogen peroxide (H₂O₂) in the presence of a base (OH⁻).
- During this oxidation, the boron atom is replaced by a hydroxyl group (OH) at the same position where boron was attached.

5. **Final Product**:
- The final product after oxidation will be:
```
CH₃-CH(CH₃)-C(OH)-CH₂
```
- This compound is a primary alcohol because the carbon bearing the hydroxyl group is attached to only one other carbon.

6. **Stereochemistry**:
- The reaction is stereospecific and results in syn addition, meaning that the hydroxyl group and the hydrogen atom are added to the same side of the double bond.
- Since there are no chiral centers in the final product, it is optically inactive.

### Conclusion:
The final product Z is a primary alcohol, specifically `CH₃-CH(CH₃)-C(OH)-CH₂`, which is optically inactive.