`CH_(2) = CH_(2) + B_(2) H_(6) underset(H_(2)SO_(4))overset(NaOH) to ` Pr oduct. 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 given reaction step by step, we will analyze the reaction of ethylene (CH₂=CH₂) with diborane (B₂H₆) in the presence of sulfuric acid (H₂SO₄) and sodium hydroxide (NaOH). This process is known as Hydroboration-Oxidation.

### Step 1: Hydroboration
- Ethylene (CH₂=CH₂) reacts with diborane (B₂H₆).
- In this reaction, borane adds across the double bond of ethylene.
- The boron atom from diborane forms a bond with one of the carbon atoms, while the hydrogen from borane attaches to the other carbon atom.
- The product formed at this stage is CH₃CH₂BH₂ (ethylborane).

### Step 2: Oxidation
- The next step involves the oxidation of the alkylborane (CH₃CH₂BH₂) using sulfuric acid (H₂SO₄).
- The boron atom (BH₂) is replaced by a hydroxyl group (OH) through a proton exchange mechanism.
- This reaction leads to the formation of ethanol (CH₃CH₂OH).

### Final Product
- The final product of the reaction is ethanol (C₂H₅OH).

### Summary of the Reaction
1. Ethylene (CH₂=CH₂) reacts with diborane (B₂H₆) to form ethylborane (CH₃CH₂BH₂).
2. Ethylborane is then oxidized by sulfuric acid (H₂SO₄) to yield ethanol (C₂H₅OH).

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

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