Question

What is Z in the following sequence reactions?
\[Z\xrightarrow{{PC{l_5}}}X\xrightarrow{{Alc.KOH}}Y\xrightarrow[{\left( {ii} \right){H_2}O,boil}]{{\left( i \right)conc.{H_2}S{O_4}}}Z\]
A.\[C{H_3}C{H_2}C{H_2}OH\]
B.\[C{H_3}CHOHC{H_3}\]
C.\[{\left( {C{H_3}C{H_2}} \right)_2}CHOH\]
D.\[C{H_3} - CH = C{H_2}\]

Answer 1

Hint: We need to know that the phosphorus pentachloride is a solid reagent having the formula, \[PC{l_5}\]. This reagent is widely used in the manufacturing process of electrolytes of lithium ion batteries and used as a chlorinating agent in the preparation of antibiotics. When the phosphorus pentachloride is reacting with alcohol, it will produce hydrogen chloride gas. This reagent violently reacts with alcohol at room temperature.

Complete answer:
We have to remember that when the ethanol is reacted with phosphorus pentachloride will get ethyl chloride. This will react with alcoholic potassium hydroxide, there is a formation of ethene. And this ethene is reacting with there will form ethyl hydrogen sulphate. Thus, we can say that it is not equal to ‘Z’. Hence, option (A) is incorrect.
The reactant (Z) in the given reaction is isopropyl alcohol. When it is mixed with phosphorus pentachloride, it will form \[2 - \]chloropropane, (X) by the replacement of hydroxyl group with chlorine group. This \[2 - \]chloropropane is reacted with alcoholic potassium hydroxide, this is converted into an alkene, and here it is propene (Y). When the ethene is boiled with concentrated sulphuric acid and water, again there is a formation of isopropyl alcohol. Let’s see the reaction;
\[C{H_3}CHOHC{H_3}\xrightarrow{{PC{l_5}}}C{H_3}CHClC{H_3}\xrightarrow{{Alc.KOH}}C{H_3} - CH = C{H_2}\xrightarrow[{\left( {ii} \right){H_2}O,boil}]{{\left( i \right)conc.{H_2}S{O_4}}}C{H_3}CHOHC{H_3}\]Hence, option (B) is correct.
In the given sequence reaction, \[{\left( {C{H_3}C{H_2}} \right)_2}CHOH\] is not ‘Z’.
So, the correct answer is “Option C”.

Note:
We have to know that when phosphorus pentachloride reacts with alcohol, it will replace the hydroxyl group present in the alcohol by chlorine and there is a formation of hydrocarbons which is alkane. When the hydrocarbon reacts with alcohol, the alkane is converted to alkene by the formation of a double bond. When the alkene and concentrated acid react together, the alkene again reforms into an alcohol.