In the following reaction 'A' is :
`C_(2)H_(5)MgBr+"epoxyethane" overset(H_(2)O)rarr A`

Options

A. `C_(2)H_(5)CH_(2)CHO`
B. `C_(2)H_(5)CH_(2)CH_(2)OH`
C. `C_(2)H_(5)CH_(2)OH`
D. `C_(2)H_(5)CHO`

Answer

To determine the compound 'A' formed in the reaction between C₂H₅MgBr (ethylmagnesium bromide) and epoxyethane upon hydrolysis with water, we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- C₂H₅MgBr (ethylmagnesium bromide), which is a Grignard reagent.
- Epoxyethane (also known as oxirane), which is a three-membered cyclic ether.

### Step 2: Understand the Grignard Reaction
Grignard reagents are nucleophiles and can attack electrophilic centers. In epoxyethane, the carbon atoms are electrophilic due to the strain in the three-membered ring. The nucleophilic carbon from the Grignard reagent will attack one of the carbon atoms in the epoxyethane.

### Step 3: Nucleophilic Attack
When C₂H₅MgBr attacks epoxyethane, the nucleophilic ethyl group (C₂H₅⁻) will attack one of the carbon atoms in the epoxy ring. This will lead to the opening of the ring and the formation of an alkoxide intermediate.

The reaction can be represented as:
\[ \text{C}_2\text{H}_5\text{MgBr} + \text{epoxyethane} \rightarrow \text{C}_2\text{H}_5\text{CH}_2\text{CH}_2\text{O}^- \]

### Step 4: Hydrolysis of the Alkoxide
The alkoxide formed will then undergo hydrolysis when treated with water (H₂O). This step converts the alkoxide into an alcohol.

The hydrolysis reaction can be represented as:
\[ \text{C}_2\text{H}_5\text{CH}_2\text{CH}_2\text{O}^- + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{CH}_2\text{CH}_2\text{OH} + \text{OH}^- \]

### Step 5: Identify the Product
The final product 'A' after hydrolysis is:
\[ \text{C}_2\text{H}_5\text{CH}_2\text{CH}_2\text{OH} \]
This can also be written as:
\[ \text{C}_5\text{H}_{12}\text{O} \]
which is 1-butanol.

### Conclusion
Thus, the compound 'A' formed in the reaction is 1-butanol (C₄H₉OH).

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