Question

What is Z in the following sequence of reaction?
$Z\xrightarrow{{PC{l_5}}}X\xrightarrow{{Alc.KOH}}Y\xrightarrow[{2){H_2}O;boil}]{{1)Conc.{H_2}S{O_4}}}Z$
A) $C{H_3}C{H_2}C{H_2}OH$
B) $C{H_3}CHOHC{H_3}$
C) ${(C{H_3}C{H_2})_2}CHOH$
D) $C{H_3}CH = C{H_2}$

Answer 1

In the given reaction Alc. KOH is used, which is basically used for elimination reactions in Halide Substituted compounds or alcohols. $PC{l_5}$ is majorly used for chlorination purposes. And ${H_2}O/{H_2}S{O_4}$ is used for formation of alcohols.

Complete answer:
In the first reaction $PC{l_5}$ is added to Z. Solid Phosphorus (V) Chlorine when added to alcohols, reacts violently at room temperature, leading to formation of Hydrogen Chloride gas. It was probably used for making Chloroalkanes. It is also an indicator test used to detect the presence of alcohol. Hence, we can conclude that Z compound is a 3-carbon containing alcohol only. Compound Z could be a secondary alcohol; 2-propanol. The first reaction can be given as:
$C{H_3}CHOHC{H_3}\xrightarrow{{PC{l_5}}}C{H_3}CHClC{H_3} + POC{l_3} + HCl$
Compound X is hence 2-Chloropropane. Some side reactions of $POC{l_3}$ with alcohol also occur.
Alkyl Halide on reaction with Alc. KOH leads to an elimination reaction by the formation of alkoxide ions. This alkoxide ion is basic in nature and it abstracts the Beta-Hydrogen from the saturated alkyl halide. Hence on addition of Alc. KOH to 2-Chloropropane will form a double bonded compound, 1-propene. The reaction can be given as:
$C{H_3}CHClC{H_3}\xrightarrow{{Alc.KOH}}C{H_3}CH = C{H_2}$
Compound Y is 1-propene.
Hydrolysis of alkene is usually carried out in the presence of acid. The addition of H-OH occurs according to the Markovnikov Rule. This leads to the formation of Secondary Alcohol as the Major product.
$C{H_3}CH = C{H_2}\xrightarrow[{{H_2}O;Boil}]{{{H_2}S{O_4}}}C{H_3} - CH(OH) - C{H_3}$
The product Z is obtained and is similar to the starting product.

The correct answer is Option (B).

Note:
Markovnikov rules state that the negative part of the substrate will get attached to that carbon atom having less number of Hydrogen Atoms. In this case the negative part of the substrate (water) is $- O{H^ - }$ which gets attached to the second carbon atom. 1-propanol is also formed but as a minor product.