In the reaction$C{{H}_{3}}COOH\xrightarrow{LiAl{{H}_{4}}}A\xrightarrow{PC{{l}_{5}}}B\xrightarrow{Alc.KOH}C$, the product C is:
(A) Ethylene
(B) Acetyl Chloride
(C) Acetaldehyde
(D) Acetylene
Answer
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Hint: In the first step acid is being reduced into alcohol .In the second step this alcohol is converted into corresponding alkyl chloride and in the last step this alkyl chloride is being converted into an alkene and this is a basic method for the preparation of alkenes.
Complete step by step answer:
-In the first step the acetic acid is treated with lithium aluminium hydride. As we know, the carboxylic acids can be transformed to primary alcohols using Lithium aluminium hydride$LiAl{{H}_{4}}$.The reaction is written below,
\[C{{H}_{3}}COOH\xrightarrow{LiAl{{H}_{4}}}C{{H}_{3}}C{{H}_{2}}OH\]
Thus when acetic acid is treated with $LiAl{{H}_{4}}$ ethanol (A) is being formed. The reaction mechanism involves deprotonation, nucleophilic attack by hydride ion, removal of leaving group, nucleophilic attack by hydride anion and finally the protonation of alkoxide
- In the second step A (ethanol) is treated with Phosphorus penta-cholride$PC{{l}_{5}}$.When alcohols are treated with Phosphorus penta-cholride corresponding alkyl chloride and hydrogen chloride gas is formed. The conversion of ethanol to form ethyl chloride is given below
\[{{C}_{2}}{{H}_{5}}OH\xrightarrow{PC{{l}_{5}}}C{{H}_{3}}C{{H}_{2}}Cl+HCl+POC{{l}_{3}}\]
Thus B is ethyl chloride or chloroethane.
- In the third step ethyl chloride (B) is treated with alcoholic KOH. When alcoholic potassium hydroxide is used elimination reaction takes place, corresponding alkene is formed. In the given question ethylene is formed along with HCl. This reaction can be written as follows
\[C{{H}_{3}}C{{H}_{2}}Cl\xrightarrow{Alc.KOH}{{H}_{2}}C=C{{H}_{2}}+HCl\]
Thus C is ethylene.
So, the correct answer is “Option A”.
Note: Note that $NaB{{H}_{4}}$ is not strong enough to convert the carboxylic acids or esters to alcohols. An aldehyde is produced as an intermediate during the first reaction in which acid is converted into alcohol, but this aldehyde cannot be isolated because it is more reactive than the original carboxylic acid.
Complete step by step answer:
-In the first step the acetic acid is treated with lithium aluminium hydride. As we know, the carboxylic acids can be transformed to primary alcohols using Lithium aluminium hydride$LiAl{{H}_{4}}$.The reaction is written below,
\[C{{H}_{3}}COOH\xrightarrow{LiAl{{H}_{4}}}C{{H}_{3}}C{{H}_{2}}OH\]
Thus when acetic acid is treated with $LiAl{{H}_{4}}$ ethanol (A) is being formed. The reaction mechanism involves deprotonation, nucleophilic attack by hydride ion, removal of leaving group, nucleophilic attack by hydride anion and finally the protonation of alkoxide
- In the second step A (ethanol) is treated with Phosphorus penta-cholride$PC{{l}_{5}}$.When alcohols are treated with Phosphorus penta-cholride corresponding alkyl chloride and hydrogen chloride gas is formed. The conversion of ethanol to form ethyl chloride is given below
\[{{C}_{2}}{{H}_{5}}OH\xrightarrow{PC{{l}_{5}}}C{{H}_{3}}C{{H}_{2}}Cl+HCl+POC{{l}_{3}}\]
Thus B is ethyl chloride or chloroethane.
- In the third step ethyl chloride (B) is treated with alcoholic KOH. When alcoholic potassium hydroxide is used elimination reaction takes place, corresponding alkene is formed. In the given question ethylene is formed along with HCl. This reaction can be written as follows
\[C{{H}_{3}}C{{H}_{2}}Cl\xrightarrow{Alc.KOH}{{H}_{2}}C=C{{H}_{2}}+HCl\]
Thus C is ethylene.
So, the correct answer is “Option A”.
Note: Note that $NaB{{H}_{4}}$ is not strong enough to convert the carboxylic acids or esters to alcohols. An aldehyde is produced as an intermediate during the first reaction in which acid is converted into alcohol, but this aldehyde cannot be isolated because it is more reactive than the original carboxylic acid.
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