Answer :
To solve the given chemical reaction, we need to follow the specific conditions provided:
Reaction:
[tex]\[ \text{CH}_3\text{-CH}_2\text{-Br} + \text{Mg} \xrightarrow{\text{Dry ether}} \][/tex]
Step-by-step solution:
1. Identify the reactants:
- The first reactant is an alkyl halide: [tex]\(\text{CH}_3\text{-CH}_2\text{-Br}\)[/tex] (ethyl bromide).
- The second reactant is magnesium: [tex]\(\text{Mg}\)[/tex].
2. Reaction conditions:
- The reaction occurs in the presence of dry ether. This is crucial because dry ether prevents the formation of unwanted byproducts that could occur if moisture were present.
3. Reaction type:
- This is a Grignard reaction where an alkyl halide reacts with magnesium in dry ether to form a Grignard reagent.
4. Understanding the Grignard reagent formation:
- In general, when an alkyl halide (R-X) reacts with magnesium (Mg) in an anhydrous (dry) ether, a Grignard reagent (R-Mg-X) is formed.
- The general chemical equation for the reaction is:
[tex]\[ \text{R-X} + \text{Mg} \rightarrow \text{R-Mg-X} \][/tex]
5. Apply this to our specific reactants:
- Here, the alkyl halide is ethyl bromide ([tex]\(\text{CH}_3\text{-CH}_2\text{-Br}\)[/tex]).
- After reacting with magnesium ([tex]\(\text{Mg}\)[/tex]) in dry ether, the product will be ethyl magnesium bromide ([tex]\(\text{CH}_3\text{-CH}_2\text{-Mg-Br}\)[/tex]).
Thus, the detailed step-by-step solution to the reaction is:
[tex]\[ \text{CH}_3\text{-CH}_2\text{-Br} + \text{Mg} \xrightarrow{\text{Dry ether}} \text{CH}_3\text{-CH}_2\text{-Mg-Br} \][/tex]
So, the final product of this reaction is ethyl magnesium bromide ([tex]\(\text{CH}_3\text{-CH}_2\text{-Mg-Br}\)[/tex]).
Reaction:
[tex]\[ \text{CH}_3\text{-CH}_2\text{-Br} + \text{Mg} \xrightarrow{\text{Dry ether}} \][/tex]
Step-by-step solution:
1. Identify the reactants:
- The first reactant is an alkyl halide: [tex]\(\text{CH}_3\text{-CH}_2\text{-Br}\)[/tex] (ethyl bromide).
- The second reactant is magnesium: [tex]\(\text{Mg}\)[/tex].
2. Reaction conditions:
- The reaction occurs in the presence of dry ether. This is crucial because dry ether prevents the formation of unwanted byproducts that could occur if moisture were present.
3. Reaction type:
- This is a Grignard reaction where an alkyl halide reacts with magnesium in dry ether to form a Grignard reagent.
4. Understanding the Grignard reagent formation:
- In general, when an alkyl halide (R-X) reacts with magnesium (Mg) in an anhydrous (dry) ether, a Grignard reagent (R-Mg-X) is formed.
- The general chemical equation for the reaction is:
[tex]\[ \text{R-X} + \text{Mg} \rightarrow \text{R-Mg-X} \][/tex]
5. Apply this to our specific reactants:
- Here, the alkyl halide is ethyl bromide ([tex]\(\text{CH}_3\text{-CH}_2\text{-Br}\)[/tex]).
- After reacting with magnesium ([tex]\(\text{Mg}\)[/tex]) in dry ether, the product will be ethyl magnesium bromide ([tex]\(\text{CH}_3\text{-CH}_2\text{-Mg-Br}\)[/tex]).
Thus, the detailed step-by-step solution to the reaction is:
[tex]\[ \text{CH}_3\text{-CH}_2\text{-Br} + \text{Mg} \xrightarrow{\text{Dry ether}} \text{CH}_3\text{-CH}_2\text{-Mg-Br} \][/tex]
So, the final product of this reaction is ethyl magnesium bromide ([tex]\(\text{CH}_3\text{-CH}_2\text{-Mg-Br}\)[/tex]).
