In the following reaction sequence, X and Z respectively are :

Core Logic

Step 1: Reaction of Alcohols with PCl₅$PCl_5$ CH₃CH₂CH₂-OH + PCl₅ arrow CH₃CH₂CH₂Cl + POCl₃ + HCl$CH_3CH_2CH_2-OH + PCl_5 \rightarrow CH_3CH_2CH_2Cl + POCl_3 + HCl$ Thus, the by-product X$X$ is POCl₃$POCl_3$.

Step 2: Elimination (Dehydrohalogenation) CH₃CH₂CH₂Cl alc. KOH, Δ CH₃CH=CH₂ (Y)$CH_3CH_2CH_2Cl \xrightarrow{\text{alc. KOH, } \Delta} CH_3CH=CH_2 \quad (Y)$ (Propene)

Step 3: Anti-Markovnikov Addition (Peroxide effect) CH₃CH=CH₂ HBr, (CH₃COO)₂ CH₃CH₂CH₂-Br (Z)$CH_3CH=CH_2 \xrightarrow{HBr, (CH_3COO)_2} CH_3CH_2CH_2-Br \quad (Z)$ (1-Bromopropane)

Step 1: Final Conclusion

Therefore, X = POCl₃$\mathrm{X} = \mathrm{POCl}_3$ and Z = CH₃CH₂CH₂Br$\mathrm{Z} = \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{Br}$.

Pattern Recognition

PCl₅$PCl_5$ yields POCl₃$POCl_3$, whereas PCl₃$PCl_3$ yields H₃PO₃$H_3PO_3$. HBr$HBr$ with peroxide is Kharasch effect (Anti-Markovnikov addition), placing the Bromine on the less hindered terminal carbon.

Chapter Mix

Class 12 Chemistry: Alcohols, Phenols and Ethers Class 12 Chemistry: Haloalkanes and Haloarenes