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Identify $Z$ in the sequence of reactions: $\newline {CH}_3CH_2=CH_2 \xrightarrow{HBr, H_2O_2} Y \xrightarrow{C_2H_5ONa} Z$

medium

Hydrocarbons

2014

chemistry

$CH_3 - {(CH_2)}_3 -O- CH_2CH_3$

Explanation

To identify $Z$ in the given sequence of reactions, we need to analyze each step. $\\$ Step 1: ${CH}_3CH_2=CH_2 \xrightarrow{HBr, H_2O_2} Y \\$

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${(CH_3)}_2 -O- CH_2CH_3$

$CH_3 {(CH_2)}_4 -O- CH_3$

$CH_3CH_2 - CH(CH_3) - O - CH_2CH_3$

• 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.