Answer :
Certainly! Let's analyze each reaction to determine whether it is an elimination reaction.
1. Reaction: [tex]\( CH_3OH + CH_3COOH \rightarrow CH_3CO_2CH_3 + H_2O \)[/tex]
This reaction involves methanol ([tex]\(CH_3OH\)[/tex]) and acetic acid ([tex]\(CH_3COOH\)[/tex]) combining to form methyl acetate ([tex]\(CH_3CO_2CH_3\)[/tex]) and water ([tex]\(H_2O\)[/tex]). This type of reaction is known as an esterification reaction. In esterification, an ester and water are formed from an alcohol and a carboxylic acid.
[tex]\[ CH_3OH + CH_3COOH \rightarrow CH_3CO_2CH_3 + H_2O \][/tex]
This process does not involve the elimination of atoms or groups from a single molecule resulting in the formation of a double bond, which is the hallmark of an elimination reaction. Therefore, this is not an elimination reaction.
2. Reaction: [tex]\( C_3H_7OH \rightarrow C_3H_6 + H_2O \)[/tex]
Here, [tex]\( C_3H_7OH \)[/tex] represents propanol. In this reaction, propanol is converted to propene ([tex]\(C_3H_6\)[/tex]) and water ([tex]\(H_2O\)[/tex]). This fits the general form of an elimination reaction where a single reactant loses atoms or groups, resulting in the formation of a double bond and a separate small molecule (such as water).
[tex]\[ C_3H_7OH \rightarrow C_3H_6 + H_2O \][/tex]
This is known as a dehydration reaction, which is a type of elimination reaction. Hence, this is an elimination reaction.
3. Reaction: [tex]\( C_2H_5Br + NaOH \rightarrow C_2H_4 + NaBr + H_2O \)[/tex]
In this reaction, an alkyl halide ([tex]\(C_2H_5Br\)[/tex]) reacts with sodium hydroxide ([tex]\(NaOH\)[/tex]) and forms ethene ([tex]\(C_2H_4\)[/tex]), sodium bromide ([tex]\(NaBr\)[/tex]), and water ([tex]\(H_2O\)[/tex]). In this reaction, the hydroxide ion ([tex]\(OH^-\)[/tex]) facilitates the removal of a hydrogen atom and a bromine atom from the ethyl group ([tex]\(C_2H_5Br\)[/tex]), leading to the formation of a double bond between the carbon atoms.
[tex]\[ C_2H_5Br + NaOH \rightarrow C_2H_4 + NaBr + H_2O \][/tex]
This type of reaction is another example of an elimination reaction, specifically a dehydrohalogenation, where a hydrogen halide is eliminated from the alkyl halide, resulting in the formation of an alkene.
Summary
- Reaction 1: Not an elimination reaction.
- Reaction 2: Elimination reaction.
- Reaction 3: Elimination reaction.
Therefore, the elimination reactions among the given options are:
[tex]\[ C_3H_7OH \rightarrow C_3H_6 + H_2O \][/tex]
[tex]\[ C_2H_5Br + NaOH \rightarrow C_2H_4 + NaBr + H_2O \][/tex]
So, the reactions corresponding to elimination reactions are:
[tex]\[ \boxed{2 \text{ and } 3} \][/tex]
1. Reaction: [tex]\( CH_3OH + CH_3COOH \rightarrow CH_3CO_2CH_3 + H_2O \)[/tex]
This reaction involves methanol ([tex]\(CH_3OH\)[/tex]) and acetic acid ([tex]\(CH_3COOH\)[/tex]) combining to form methyl acetate ([tex]\(CH_3CO_2CH_3\)[/tex]) and water ([tex]\(H_2O\)[/tex]). This type of reaction is known as an esterification reaction. In esterification, an ester and water are formed from an alcohol and a carboxylic acid.
[tex]\[ CH_3OH + CH_3COOH \rightarrow CH_3CO_2CH_3 + H_2O \][/tex]
This process does not involve the elimination of atoms or groups from a single molecule resulting in the formation of a double bond, which is the hallmark of an elimination reaction. Therefore, this is not an elimination reaction.
2. Reaction: [tex]\( C_3H_7OH \rightarrow C_3H_6 + H_2O \)[/tex]
Here, [tex]\( C_3H_7OH \)[/tex] represents propanol. In this reaction, propanol is converted to propene ([tex]\(C_3H_6\)[/tex]) and water ([tex]\(H_2O\)[/tex]). This fits the general form of an elimination reaction where a single reactant loses atoms or groups, resulting in the formation of a double bond and a separate small molecule (such as water).
[tex]\[ C_3H_7OH \rightarrow C_3H_6 + H_2O \][/tex]
This is known as a dehydration reaction, which is a type of elimination reaction. Hence, this is an elimination reaction.
3. Reaction: [tex]\( C_2H_5Br + NaOH \rightarrow C_2H_4 + NaBr + H_2O \)[/tex]
In this reaction, an alkyl halide ([tex]\(C_2H_5Br\)[/tex]) reacts with sodium hydroxide ([tex]\(NaOH\)[/tex]) and forms ethene ([tex]\(C_2H_4\)[/tex]), sodium bromide ([tex]\(NaBr\)[/tex]), and water ([tex]\(H_2O\)[/tex]). In this reaction, the hydroxide ion ([tex]\(OH^-\)[/tex]) facilitates the removal of a hydrogen atom and a bromine atom from the ethyl group ([tex]\(C_2H_5Br\)[/tex]), leading to the formation of a double bond between the carbon atoms.
[tex]\[ C_2H_5Br + NaOH \rightarrow C_2H_4 + NaBr + H_2O \][/tex]
This type of reaction is another example of an elimination reaction, specifically a dehydrohalogenation, where a hydrogen halide is eliminated from the alkyl halide, resulting in the formation of an alkene.
Summary
- Reaction 1: Not an elimination reaction.
- Reaction 2: Elimination reaction.
- Reaction 3: Elimination reaction.
Therefore, the elimination reactions among the given options are:
[tex]\[ C_3H_7OH \rightarrow C_3H_6 + H_2O \][/tex]
[tex]\[ C_2H_5Br + NaOH \rightarrow C_2H_4 + NaBr + H_2O \][/tex]
So, the reactions corresponding to elimination reactions are:
[tex]\[ \boxed{2 \text{ and } 3} \][/tex]
