Answer :
Certainly! Let's analyze the given chemical reaction:
The reaction is:
[tex]\[ CH_3-CH_2-CH_2-O-CH_3 + HBr \][/tex]
This is a classic example of an ether cleavage reaction with hydrogen bromide (HBr). Ethers can be cleaved by strong acids like HBr into alcohols and alkyl halides.
The ether in question is [tex]\(CH_3-CH_2-CH_2-O-CH_3\)[/tex]. This structure consists of a propyl group [tex]\(CH_3-CH_2-CH_2-\)[/tex] and a methyl group [tex]\(CH_3-\)[/tex] connected via an oxygen atom.
Upon reacting with HBr, the ether will be cleaved:
1. The bond between the oxygen and the carbon attached to the smaller alkyl group (methyl group, [tex]\(CH_3\)[/tex]) will break.
2. Hydrogen from HBr will combine with the methyl group ([tex]\(CH_3\)[/tex]) to form methanol ([tex]\(CH_3OH\)[/tex]).
3. The bromine atom from HBr will attach to the propyl group ([tex]\(CH_3-CH_2-CH_2-\)[/tex]) to form 1-bromopropane ([tex]\(CH_3-CH_2-CH_2Br\)[/tex]).
So, the products of the reaction are:
[tex]\[ CH_3OH \quad \text{(methanol)} \quad \text{and} \quad CH_3-CH_2-CH_2Br \quad \text{(1-bromopropane)} \][/tex]
To summarize, the cleavage of the ether [tex]\(CH_3-CH_2-CH_2-O-CH_3\)[/tex] by HBr produces:
[tex]\[ CH_3OH \quad \text{and} \quad CH_3-CH_2-CH_2Br \][/tex]
These are methanol and 1-bromopropane, respectively.
The reaction is:
[tex]\[ CH_3-CH_2-CH_2-O-CH_3 + HBr \][/tex]
This is a classic example of an ether cleavage reaction with hydrogen bromide (HBr). Ethers can be cleaved by strong acids like HBr into alcohols and alkyl halides.
The ether in question is [tex]\(CH_3-CH_2-CH_2-O-CH_3\)[/tex]. This structure consists of a propyl group [tex]\(CH_3-CH_2-CH_2-\)[/tex] and a methyl group [tex]\(CH_3-\)[/tex] connected via an oxygen atom.
Upon reacting with HBr, the ether will be cleaved:
1. The bond between the oxygen and the carbon attached to the smaller alkyl group (methyl group, [tex]\(CH_3\)[/tex]) will break.
2. Hydrogen from HBr will combine with the methyl group ([tex]\(CH_3\)[/tex]) to form methanol ([tex]\(CH_3OH\)[/tex]).
3. The bromine atom from HBr will attach to the propyl group ([tex]\(CH_3-CH_2-CH_2-\)[/tex]) to form 1-bromopropane ([tex]\(CH_3-CH_2-CH_2Br\)[/tex]).
So, the products of the reaction are:
[tex]\[ CH_3OH \quad \text{(methanol)} \quad \text{and} \quad CH_3-CH_2-CH_2Br \quad \text{(1-bromopropane)} \][/tex]
To summarize, the cleavage of the ether [tex]\(CH_3-CH_2-CH_2-O-CH_3\)[/tex] by HBr produces:
[tex]\[ CH_3OH \quad \text{and} \quad CH_3-CH_2-CH_2Br \][/tex]
These are methanol and 1-bromopropane, respectively.
