`CH_(2) = CH_(2) + B_(2) H_(6) underset(H_(2)O_(2))overset(NaOH) to ` Product is :

Options

A. `CH_(3)CH_(2)CHO`
B. `CH_(3)CH_(2)OH`
C. `CH_(3)CHO`
D. None of the above

Answer

To solve the reaction of ethylene (CH₂=CH₂) with diborane (B₂H₆) in the presence of hydrogen peroxide (H₂O₂) and sodium hydroxide (NaOH), we will go through the mechanism of hydroboration-oxidation step by step.

### Step 1: Hydroboration
1. **Formation of the Boron-Alkyl Complex**:
- The alkene (CH₂=CH₂) acts as a nucleophile and donates its π electrons to the boron atom of diborane (B₂H₆).
- This results in the formation of a boron-alkyl complex where one of the hydrogen atoms from boron (BH₃) adds to one of the carbon atoms of the alkene, and the boron atom attaches to the other carbon atom.
- The product of this step is an organoborane intermediate:
\[ \text{CH}_3\text{CH}_2\text{B}H_2 \]

### Step 2: Oxidation
2. **Formation of Alkoxide**:
- In the presence of NaOH and H₂O₂, the hydroxide ion (OH⁻) acts as a nucleophile and attacks the boron atom in the organoborane intermediate.
- This leads to the formation of an alkoxide intermediate:
\[ \text{CH}_3\text{CH}_2\text{O}^- \text{B}H_2 \]

3. **Migration of the Boron Group**:
- The boron group (BH₂) migrates to the oxygen atom, forming a new bond and breaking the bond between boron and carbon.
- This results in the formation of an alcohol:
\[ \text{CH}_3\text{CH}_2\text{OH} \]

### Step 3: Final Product
4. **Protonation**:
- The alkoxide ion (CH₃CH₂O⁻) is then protonated by water (or another proton source) to yield the final product, which is ethanol (C₂H₅OH):
\[ \text{C}_2\text{H}_5\text{OH} \]

### Final Answer:
The product of the reaction is ethanol (C₂H₅OH).

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