Statement-1: Hydroboration by oxidation of propene gives anti-Markovnikoff's alcohol.
and
Statement-2: Hydroboration reaction proceeds through Markovnikoff's addition

Answer

To analyze the statements provided in the question, we need to break down the concepts of hydroboration and oxidation, particularly in the context of propene.

### Step 1: Understanding Hydroboration
Hydroboration is a reaction where borane (BH3) adds across a double bond in an alkene. In the case of propene (C3H6), the reaction can be represented as follows:
- Propene reacts with BH3 to form an organoborane intermediate.

### Step 2: Mechanism of Hydroboration
During hydroboration, the addition of BH3 occurs in a syn fashion, meaning both the boron atom and the hydrogen atom add to the same side of the double bond. The boron atom attaches to the less substituted carbon (the carbon with more hydrogen atoms), while the hydrogen atom attaches to the more substituted carbon. This is contrary to the Markovnikov's rule, which states that the more substituted carbon would typically receive the electrophile (in this case, boron).

### Step 3: Oxidation of the Organoborane
After hydroboration, the organoborane intermediate undergoes oxidation, typically using hydrogen peroxide (H2O2) in a basic medium. This step converts the boron atom into a hydroxyl group (-OH), resulting in the formation of an alcohol.

### Step 4: Resulting Alcohol
The final product from the hydroboration-oxidation of propene is an alcohol where the hydroxyl group is attached to the less substituted carbon. This outcome aligns with the anti-Markovnikov's rule, which states that the -OH group will attach to the carbon with more hydrogen atoms.

### Conclusion
- **Statement 1**: Hydroboration by oxidation of propene gives anti-Markovnikov's alcohol. This statement is **true**.
- **Statement 2**: Hydroboration reaction proceeds through Markovnikov's addition. This statement is also **true** because the first step of hydroboration involves the addition of BH3 according to Markovnikov's rule.

### Final Answer
Both statements are true, but they refer to different aspects of the reaction mechanism.

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