Ethylene oxide when treated with Grignard reagent yields

Options

A. Secondary alcohol
B. Tertiary alcohol
C. Cyclopropyl alcohol
D. Primary alcohol

Explanation

To solve the problem of what ethylene oxide yields when treated with a Grignard reagent, we can follow these steps:

### Step 1: Understand the structure of ethylene oxide
Ethylene oxide is a three-membered cyclic ether with the formula C2H4O. Its structure can be represented as:
\[ \text{CH}_2-\text{CH}_2-\text{O} \]

### Step 2: Identify the Grignard reagent
A Grignard reagent is typically represented as R-MgX, where R is an organic group (alkyl or aryl) and X is a halogen (like Cl, Br, or I).

### Step 3: Mechanism of reaction
When ethylene oxide reacts with a Grignard reagent, the nucleophilic carbon of the Grignard reagent attacks the less hindered carbon of the epoxide. The oxygen in the epoxide has a partial negative charge due to its electronegativity, which makes it a good leaving group.

### Step 4: Breaking the bond
The bond between the oxygen and the carbon atom in the epoxide breaks, leading to the formation of an alkoxide intermediate. The Grignard reagent donates its R group to one of the carbon atoms, resulting in:
\[ \text{CH}_2-\text{O}^-\text{CH}_2R \]

### Step 5: Hydrolysis of the alkoxide
When this intermediate is treated with water (H2O), the alkoxide ion (O^-) is protonated, replacing the MgX group with a hydrogen atom. This results in the formation of an alcohol.

### Step 6: Final product
The final product after hydrolysis is:
\[ \text{RCH}_2-\text{CH}_2-\text{OH} \]
This is a primary alcohol where R is the alkyl group from the Grignard reagent.

### Conclusion
Thus, when ethylene oxide is treated with a Grignard reagent, the final product is:
\[ \text{RCH}_2-\text{CH}_2-\text{OH} \]

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