Question

What is $\mathrm{Z}$ in the following sequence of reactions?

$\mathrm{Z}\stackrel{{\mathrm{PCl}}_5}{\to }\mathrm{X}\stackrel{\mathrm{Alc}.\mathrm{KOH}}{\to }\mathrm{Y}\underset{\left(\mathrm{ii}\right){\mathrm{H}}_2\mathrm{O},\mathrm{Boil}}{\overset{\left(\mathrm{i}\right)\mathrm{Conc}.{\mathrm{H}}_2{\mathrm{SO}}_4}{\to }}\mathrm{Z}$

• A. ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{OH}$
• B. ${\mathrm{CH}}_3{\mathrm{CHOHCH}}_3$
• C. ${\left({\mathrm{CH}}_3{\mathrm{CH}}_2\right)}_2\mathrm{CHOH}$
• D. ${\mathrm{CH}}_3\mathrm{CH}={\mathrm{CH}}_2$

Answer 1

The correct option is B

${\mathrm{CH}}_3{\mathrm{CHOHCH}}_3$

Explanation for the correct option:

(B) ${\mathrm{CH}}_3{\mathrm{CHOHCH}}_3$

• When $2$-Propanol (${\mathrm{C}}_3{\mathrm{H}}_8\mathrm{O}$) ($\mathrm{Z}$) reacts with Phosphorus pentachloride (${\mathrm{PCl}}_5$), it forms $2$-Chloropropane (${\mathrm{C}}_3{\mathrm{H}}_7\mathrm{Cl}$) ($\mathrm{X}$) that involves displacement of the hydroxyl group by chlorine.
• $2$-Chloropropane (${\mathrm{C}}_3{\mathrm{H}}_7\mathrm{Cl}$) ($\mathrm{X}$) reacts with alcoholic Potassium hydroxide ($alc\mathit{.}\mathrm{KOH}$) to form $1$-Propene (${\mathrm{C}}_3{\mathrm{H}}_6$) ($\mathrm{Y}$).
• Now, when $1$-Propene (${\mathrm{C}}_3{\mathrm{H}}_6$) ($\mathrm{Y}$) is boiled with $\mathrm{Conc}.$ ${\mathrm{H}}_2{\mathrm{SO}}_4$ and water, it again results in $2$-Propanol (${\mathrm{C}}_3{\mathrm{H}}_8\mathrm{O}$) ($\mathrm{Z}$).

Explanation for the incorrect options:

(A) ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{OH}$

• If $1$-Propanol (${\mathrm{C}}_3{\mathrm{H}}_8\mathrm{O}$) reacts with Phosphorus pentachloride (${\mathrm{PCl}}_5$), it will form $1$-Chloropropane (${\mathrm{C}}_3{\mathrm{H}}_7\mathrm{Cl}$).
• $1$-Chloropropane (${\mathrm{C}}_3{\mathrm{H}}_7\mathrm{Cl}$) will react with alcoholic Potassium hydroxide ($alc\mathit{.}\mathrm{KOH}$) to form $1$-Propene (${\mathrm{C}}_3{\mathrm{H}}_6$).
• Now, when $1$-Propene (${\mathrm{C}}_3{\mathrm{H}}_6$) is boiled with $\mathrm{Conc}.{\mathrm{H}}_2{\mathrm{SO}}_4$ and water, it results $2$-Propanol (${\mathrm{C}}_3{\mathrm{H}}_8\mathrm{O}$) which is different from the initial reactant.

(C) ${\left({\mathrm{CH}}_3{\mathrm{CH}}_2\right)}_2\mathrm{CHOH}$

• If $3$-Pentanol (${\mathrm{C}}_5{\mathrm{H}}_{12}\mathrm{O}$) reacts with Phosphorus pentachloride (${\mathrm{PCl}}_5$), it will form $3$-Chloropentane (${\mathrm{C}}_5{\mathrm{H}}_{11}\mathrm{Cl}$).
• $3$-Chloropentane (${\mathrm{C}}_5{\mathrm{H}}_{11}\mathrm{Cl}$) will react with alcoholic Potassium hydroxide $(\mathrm{alc}.\mathrm{KOH})$ to form $2$-Pentene (${\mathrm{C}}_5{\mathrm{H}}_{10}$).
• Now, when $2$-Pentene (${\mathrm{C}}_5{\mathrm{H}}_{10}$) is boiled with $\mathrm{Conc}.{\mathrm{H}}_2{\mathrm{SO}}_4$ and water, it results in a mixture of $3$-Pentanol and $2$-Pentanol which is different from the initial reactant.

(D) ${\mathrm{CH}}_3\mathrm{CH}={\mathrm{CH}}_2$

• It does not give similar reactions as above.

Hence, option (B) is correct. $\mathrm{Z}$ in the given reaction is ${\mathrm{CH}}_3{\mathrm{CHOHCH}}_3$.