Question

In the reaction:
 $ Ethanol\xrightarrow{{PC{l_5}}}X\xrightarrow{{Alc.KOH}}Y\xrightarrow[{{H_2}O,\Delta }]{{{H_2}S{O_4},{\text{Room temp}}{\text{.}}}}Z $ , the product Z is:
 $ 1.\;\;\;\;\;\;{C_2}{H_4} \\
  2.\;\;\;\;\;\;C{H_3}C{H_2}OC{H_2}C{H_3} \\
  3.\;\;\;\;\;\;C{H_3}C{H_2}OS{O_3}H \\
  4.\;\;\;\;\;\;C{H_3}C{H_2}OH \\ $

Answer 1

Hint :For solving this question, try to recall first how ethanol reacts with phosphorus (V) chloride $ (PC{l_5}) $ . Then that product will react with Alcoholic $ KOH $ and undergo nucleophilic substitution to give you ‘Y’ finally this will undergo a hydration reaction to give the final product ‘Z’.

Complete Step By Step Answer:
In the above question, we are given a series of continuous reactions taking place and we have to identify the last product ‘Z’ formed at the end of the reaction. But only after finding ‘X’ and ‘Y’ you will get what is ‘Z’.
The reaction given is:
 $ Ethanol\xrightarrow{{PC{l_5}}}X\xrightarrow{{Alc.KOH}}Y\xrightarrow[{{H_2}O,\Delta }]{{{H_2}S{O_4},{\text{Room temp}}{\text{.}}}}Z $
Let us break the reaction series into smaller reactions to arrive at the final product. So first reaction is
 $ Ethanol\xrightarrow{{PC{l_5}}}X $
Ethanol reacts with phosphorus (V) chloride $ (PC{l_5}) $ to give chloroethane. Here nucleophilic substitution takes place. Along with Chloroethane white misty fumes of $ HCl $ are produced in the air. And so our compound ‘X’ is chloroethane.
  $ Ethanol\xrightarrow{{PC{l_5}}}C{H_3}C{H_2}Cl $
Next in the second reaction chloroethane is reacting with Alcoholic $ KOH $ . During the reaction Alcoholic $ KOH $ which is a strong base will abstract hydrogen from the $ \beta $ -carbon of the chloroethane and one molecule of $ HCl $ will be eliminated giving an alkene. So in our case the alkene formed is ethene. Hence the reaction is taking place as follows:
  $ C{H_3}C{H_2}Cl\xrightarrow{{Alc.KOH}}C{H_2} = C{H_2} $
In the last step, ethene reacts with $ {H_2}S{O_4} $ and $ {H_2}O $ at room temperature to give the final product. Ethene on reacting with $ {H_2}S{O_4} $ undergoes hydration reaction and the product formed is ethyl alcohol. So the reaction takes place in following way:
 $ C{H_2} = C{H_2}\xrightarrow[{{H_2}O,\Delta }]{{{H_2}S{O_4},{\text{Room temp}}{\text{.}}}}C{H_3}C{H_2}OH $
Thus we combine all the individual reactions together the complete reaction series is as follows:
 $ Ethanol\xrightarrow{{PC{l_5}}}\mathop {C{H_3}C{H_2}Cl}\limits_{(X)} \xrightarrow{{Alc.KOH}}\mathop {C{H_2} = C{H_2}}\limits_{(Y)} \xrightarrow[{{H_2}O,\Delta }]{{{H_2}S{O_4},{\text{Room temp}}{\text{.}}}}\mathop {C{H_3}C{H_2}OH}\limits_{(Z)} $ Therefore the product ‘Z’ is ethyl alcohol $ C{H_3}C{H_2}OH $
Hence the correct option is $ 4.\;\;\;\;\;\;C{H_3}C{H_2}OH $ .

Note :
For such a problem analyzing the reaction series properly is important. Identify how each reagent is used properly and try to recall in which types of reactions they are used and how they react with the reactant. Sometimes in place of $ PC{l_5} $ , $ PB{r_5} $ may be present since both $ Cl $ and $ Br $ are halogens PBr5 will react in the same manner how $ PC{l_5} $ reacted with ethanol.