Isobutyl magneisum bromide with dry ether and absolute alcohol gives

Options

A. `CH_(3)-underset(CH_(3))underset(|)(CH)-CH_(2)OH` and `CH_(3)CH_(2)MgBr`
B. `CH_(3)-underset(CH_(3))underset(|)(CH)-CH_(2)-CH_(2)-CH_(3)` and `Mg(OH)Br`
C. `CH_(3)-underset(CH_(3))underset(|)(CH)-CH_(3),CH_(2)=CH_(2)` and `Mg(OH)Br`
D. `CH_(3)-underset(CH_(3))underset(|)(CH)-CH_(3)` and `CH_(3)CH_(2)OMgBr`

Answer

To solve the question regarding the reaction of isobutyl magnesium bromide with dry ether and absolute alcohol, we will follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- Isobutyl magnesium bromide (a Grignard reagent), which can be represented as \( \text{(CH}_3)_2\text{CHCH}_2\text{MgBr} \).
- Absolute alcohol, which is ethyl alcohol or ethanol, represented as \( \text{CH}_3\text{CH}_2\text{OH} \).

### Step 2: Understand the Reaction Type
This is a Grignard reaction. Grignard reagents react with alcohols to form alkanes and alkoxides. The Grignard reagent acts as a nucleophile, while the alcohol acts as an electrophile.

### Step 3: Write the Reaction
The reaction can be represented as:
\[ \text{(CH}_3)_2\text{CHCH}_2\text{MgBr} + \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{dry ether}} \text{(CH}_3)_2\text{CHCH}_3 + \text{CH}_3\text{CH}_2\text{O}^- \text{MgBr}^+ \]

### Step 4: Analyze the Products
From the reaction:
- The isobutyl group (\( \text{(CH}_3)_2\text{CHCH}_2 \)) will form isobutane (\( \text{(CH}_3)_2\text{CHCH}_3 \)).
- The alcohol will form an alkoxide (\( \text{CH}_3\text{CH}_2\text{O}^- \text{MgBr}^+ \)).

### Step 5: Conclusion
The products of the reaction are:
1. Isobutane (\( \text{(CH}_3)_2\text{CHCH}_3 \))
2. Ethyl magnesium bromide alkoxide (\( \text{CH}_3\text{CH}_2\text{O}^- \text{MgBr}^+ \))

### Final Answer
The reaction of isobutyl magnesium bromide with dry ether and absolute alcohol gives isobutane and ethyl magnesium bromide alkoxide.

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