In the reaction:
\[Ethanol\xrightarrow{PC{{l}_{5}}}X\xrightarrow{alc.KOH}Y\xrightarrow[{{H}_{2}}O,\Delta ]{{{H}_{2}}S{{O}_{4}},room\text{ }temperature}Z\]
The product Z is:
A. ${{C}_{2}}{{H}_{4}}$
B. $C{{H}_{3}}C{{H}_{2}}OC{{H}_{2}}C{{H}_{3}}$
C. $C{{H}_{3}}C{{H}_{2}}OS{{O}_{3}}H$
D. $C{{H}_{3}}C{{H}_{2}}OH$
Answer
608.7k+ views
Hint: Phosphorus pentachloride acts as a chlorinating agent on alcohols. Alcoholic KOH reagent acts as a strong base and used in elimination reactions. Sulphuric acid reacts with alkenes and forms respective sulphonates as the products.
Complete step by step solution:
- In the given question there is a chemical reaction, we have to find the product Z in the given reaction.
- The given chemical reaction is as follows.
\[Ethanol\xrightarrow{PC{{l}_{5}}}X\xrightarrow{alc.KOH}Y\xrightarrow[{{H}_{2}}O,\Delta ]{{{H}_{2}}S{{O}_{4}},room\text{ }temperature}Z\]
- In the above chemical reaction there are three products X, Y and Z.
- In the first step ethanol reacts with Phosphorus pentachloride and forms an alkyl halide called ethyl chloride (X) as the product.
- The chemical reaction of ethanol with Phosphorus pentachloride is as follows.
\[Ethanol\xrightarrow{PC{{l}_{5}}}{{C}_{2}}{{H}_{5}}Cl\]
- The formed ethyl chloride in the above reaction reacts with alc. KOH (Strong base) and forms an alkene (Y) as the product.
- The chemical reaction of ethyl chloride with alc. KOH is as follows.
\[{{C}_{2}}{{H}_{5}}Cl\xrightarrow{alc.KOH}{{C}_{2}}{{H}_{4}}\]
- The product formed in the above reaction is ethylene when it reacts with sulphuric acid in the presence of water and heat produces ethyl alcohol as the product.
- The chemical reaction of ethylene with sulphuric acid at room temperature in the presence of water and heat is as follows.
\[{{C}_{2}}{{H}_{4}}\xrightarrow[{{H}_{2}}O,\Delta ]{{{H}_{2}}S{{O}_{4}},room\text{ }temperature}{{C}_{2}}{{H}_{5}}OH\]
- Therefore the product Z is ethyl alcohol.
So, the correct answer is “Option D”.
Note: Ethyl alcohol can be prepared from ethylene by treating the ethylene with sulphuric acid in the presence of water at higher temperature. But in industries this step is not followed due to the utilization of higher temperature.
Complete step by step solution:
- In the given question there is a chemical reaction, we have to find the product Z in the given reaction.
- The given chemical reaction is as follows.
\[Ethanol\xrightarrow{PC{{l}_{5}}}X\xrightarrow{alc.KOH}Y\xrightarrow[{{H}_{2}}O,\Delta ]{{{H}_{2}}S{{O}_{4}},room\text{ }temperature}Z\]
- In the above chemical reaction there are three products X, Y and Z.
- In the first step ethanol reacts with Phosphorus pentachloride and forms an alkyl halide called ethyl chloride (X) as the product.
- The chemical reaction of ethanol with Phosphorus pentachloride is as follows.
\[Ethanol\xrightarrow{PC{{l}_{5}}}{{C}_{2}}{{H}_{5}}Cl\]
- The formed ethyl chloride in the above reaction reacts with alc. KOH (Strong base) and forms an alkene (Y) as the product.
- The chemical reaction of ethyl chloride with alc. KOH is as follows.
\[{{C}_{2}}{{H}_{5}}Cl\xrightarrow{alc.KOH}{{C}_{2}}{{H}_{4}}\]
- The product formed in the above reaction is ethylene when it reacts with sulphuric acid in the presence of water and heat produces ethyl alcohol as the product.
- The chemical reaction of ethylene with sulphuric acid at room temperature in the presence of water and heat is as follows.
\[{{C}_{2}}{{H}_{4}}\xrightarrow[{{H}_{2}}O,\Delta ]{{{H}_{2}}S{{O}_{4}},room\text{ }temperature}{{C}_{2}}{{H}_{5}}OH\]
- Therefore the product Z is ethyl alcohol.
So, the correct answer is “Option D”.
Note: Ethyl alcohol can be prepared from ethylene by treating the ethylene with sulphuric acid in the presence of water at higher temperature. But in industries this step is not followed due to the utilization of higher temperature.
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