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 to`
D. `C_(6)H_(5)I`

Answer

To determine the product of 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 the reaction are:
- Ethoxide ion: \( C_2H_5O^- \)
- Ethyl iodide: \( CH_3CH_2I \)

### Step 2: Understand the Mechanism
The ethoxide ion \( C_2H_5O^- \) acts as a nucleophile due to the negative charge on the oxygen atom. The ethyl iodide \( CH_3CH_2I \) contains a good leaving group, which is the iodide ion \( I^- \).

### Step 3: Nucleophilic Attack
In the reaction, the ethoxide ion will attack the carbon atom that is bonded to the iodide in \( CH_3CH_2I \). This results in the formation of a new bond between the oxygen of the ethoxide and the ethyl group.

### Step 4: Leaving Group Departure
As the ethoxide ion attacks the ethyl iodide, the \( I^- \) leaves, resulting in the formation of a new compound. The leaving group \( I^- \) departs with its electrons, creating a positively charged carbon.

### Step 5: Product Formation
The product formed from this reaction is an ether. The structure of the product can be represented as:
\[ C_2H_5OCH_2CH_3 \]
This can also be written as \( C_2H_5OC_2H_5 \), which is diethyl ether.

### Conclusion
The final product of the reaction between \( C_2H_5O^- \) and \( CH_3CH_2I \) is:
\[ C_2H_5OC_2H_5 \] (diethyl ether).