Identify Z in the sequence of reaction, $C H_{3} C H_{2} C H = C H_{2} H B r / H_{2} O_{2} \to Y C_{2} H_{5} O N a \to Z$

(C H_{3})_{2} C H_{2} - O - C H_{2} C H_{3}

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C H_{3} (C H_{2})_{4} - O - C H_{3}

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C H_{3} C H_{2} - C H (C H_{3}) - O - C H_{2} C H_{3}

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C H_{3} - (C H_{2})_{3} - O - C H_{2} C H_{3}

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Solution

The correct option is D $C H_{3} - (C H_{2})_{3} - O - C H_{2} C H_{3}$
When HBr is added to unsymmetrical alkenes in the presence of peroxide, 1-bromobutane is formed contrary to 2-bromopropane (according to Markovnikov’s rule). This reaction is better known as Kharash effect or peroxide effect.
$C H_{3} C H_{2} C H = C H_{2} H B r / H_{2} O_{2} \to C H_{3} C H_{2} C H_{2} C H_{2} B r$
Williamson's ether synthesis involve an alkoxide that reacts with a primary haloalkane to produce ether. Alkoxides consist of the conjugate base of an alcohol and are comprised of an R group bonded to an oxygen atom. They are often written as $R O^{-}$ , where R is the organic substituent.
It is used for the preparation of symmetrical as well as unsymmetrical ethers.
$C H_{3} C H_{2} C H_{2} C H_{2} B r C_{2} H_{5} O N a \to C H_{3} C H_{2} C H_{2} C H_{2} O C_{2} H_{5}$ . Thus Z is Ethyl butylether. Hence option (d) is correct.

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