The sequence of reactions occurs in two steps:

1. Step 1 (Formation of Y): But-1-ene ($\text{CH}_3\text{CH}_2\text{CH=CH}_2$) reacts with $\text{HBr}$ in the presence of a peroxide ($\text{H}_2\text{O}_2$). This addition follows the anti-Markovnikov rule (Kharash or peroxide effect), where the free radical mechanism leads to the bromide atom attaching to the terminal (less substituted) carbon atom . This yields 1-bromobutane as intermediate $\text{Y}$: $\text{CH}_3\text{CH}_2\text{CH=CH}_2 + \text{HBr} \xrightarrow{\text{Peroxide}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}$ (Compound Y)

2. Step 2 (Formation of Z): 1-bromobutane undergoes an $S_N2$ nucleophilic substitution reaction (Williamson ether synthesis) with sodium ethoxide ($\text{C}_2\text{H}_5\text{O}^-\text{Na}^+$). The ethoxide ion attacks the primary carbon of the alkyl halide, displacing the bromide ion to form ethyl butyl ether as product $\text{Z}$ : $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br} + \text{C}_2\text{H}_5\text{O}^-\text{Na}^+ \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{-O-CH}_2\text{CH}_3 + \text{NaBr}$

The product $\text{Z}$ is $\text{CH}_3\text{-(CH}_2)_3\text{-O-CH}_2\text{CH}_3$.