What is the product of the reaction of ethylbromide with the nucleophiles `CH_(3)OH?`

Options

A. `CH_(3)CH_(2)OCH_(3)`
B. `CH_(3)CH_(2)NH_(2)`
C. `CH_(3)CH_(2)N^(+)(CH_(3))_(2)Br^(+)`
D. None of these

Explanation

To determine the product of the reaction between ethyl bromide (C2H5Br) and methanol (CH3OH), we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- Ethyl bromide (C2H5Br)
- Methanol (CH3OH)

### Step 2: Understand the Reaction Type
This reaction is a nucleophilic substitution reaction. In this case, methanol acts as the nucleophile, and it will attack the carbon atom bonded to the bromine in ethyl bromide.

### Step 3: Identify the Leaving Group
In ethyl bromide, bromine (Br) is the leaving group. When the nucleophile (methanol) attacks, the bromine will leave as bromide ion (Br⁻).

### Step 4: Mechanism of the Reaction
1. The oxygen atom in methanol has a lone pair of electrons, which allows it to attack the electrophilic carbon atom in ethyl bromide.
2. As the oxygen atom forms a bond with the carbon atom, the C-Br bond breaks, leading to the formation of bromide ion (Br⁻).
3. Simultaneously, the hydrogen atom from the hydroxyl group of methanol is lost, allowing the oxygen to bond with the ethyl group.

### Step 5: Write the Product
The product of this reaction will be:
- Ethyl alcohol (ethanol) or C2H5OH

### Step 6: Final Product Name
The IUPAC name for the product is ethanol.

### Conclusion
The product of the reaction between ethyl bromide and methanol is ethanol (C2H5OH).

### Summary of the Reaction:
C2H5Br + CH3OH → C2H5OH + HBr