The product of following reaction is-
`C_2H_5O^(-) + CH_(3)CH_(2)I to`

Options

A. `C_(6)H_(5)OC_(2)H_(5)`
B. `C_(2)H_(5)OC_(2)H_(5)`
C. `C_(6)H_(5)O^(-) + CH_(3)CH_(2)I `
D. `C_(6)H_(5)I`

Answer

To solve the reaction between \( C_2H_5O^- \) (ethoxide ion) and \( CH_3CH_2I \) (ethyl iodide), we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- Ethoxide ion \( C_2H_5O^- \)
- Ethyl iodide \( CH_3CH_2I \)

### Step 2: Understand the Nature of the Reactants
The ethoxide ion \( C_2H_5O^- \) is a strong nucleophile because it has a negative charge on the oxygen atom, making it electron-rich. Ethyl iodide \( CH_3CH_2I \) is an alkyl halide, where iodine is a good leaving group due to its ability to stabilize the negative charge after leaving.

### Step 3: Nucleophilic Attack
In the reaction, the ethoxide ion will perform a nucleophilic attack on the carbon atom of ethyl iodide that is bonded to the iodine atom. This carbon is electron-deficient due to the presence of the electronegative iodine atom.

### Step 4: Formation of the Product
When the ethoxide ion attacks the carbon atom, the bond between the carbon and iodine breaks, resulting in the formation of a new bond between the ethoxide ion and the carbon atom. The product formed is:
\[ C_2H_5OCH_2CH_3 \]
This compound is known as diethyl ether or ethoxyethane.

### Step 5: Write the Final Equation
The overall reaction can be summarized as:
\[ C_2H_5O^- + CH_3CH_2I \rightarrow C_2H_5OCH_2CH_3 + I^- \]

### Conclusion
The product of the reaction is diethyl ether \( C_2H_5OCH_2CH_3 \).

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