To identify $Z$ in the given sequence of reactions, we need to analyze each step.$\\$Step 1: ${CH}_3CH_2=CH_2 \xrightarrow{HBr, H_2O_2} Y \\$• The reaction involves the addition of HBr in the presence of $H_2O_2. \\$• This is an anti-Markovnikov addition due to the presence of $H_2O_2. \\$• The bromine atom will add to the less substituted carbon.$\\$The product $Y$ will be:$\\ {CH}_3CH_2CH_2Br \\$Step 2: $Y \xrightarrow{C_2H_5ONa} Z \\$• The reaction involves the use of sodium ethoxide ($C_2H_5ONa$), which is a strong base.$\\$• This will lead to an elimination reaction, specifically an E2 mechanism.$\\$• The elimination will form a double bond, resulting in an alkene.$\\$The product $Z$ will be:$\\ {CH}_3CH=CH_2 \\$However, the question asks for the ether formed, which suggests a Williamson ether synthesis.$\\$• The sodium ethoxide can also act as a nucleophile, attacking the alkyl halide to form an ether.$\\$The ether formed will be:$\\ {CH}_3CH_2CH_2-O-CH_2CH_3 \\$This corresponds to Option 1: $CH_3 - {(CH_2)}_3 -O- CH_2CH_3 \\$Therefore, the correct answer is Option 1.