Choose the correct answer among the alternatives given,
Complete the missing links:
$C H_{3} C H B r C H_{3} a l c . K O H - ---- \to X H B r - ---- \to P e r o x i d e Y C H_{3} O N a - ---- \to Z$ :

C H_{3} C H = C H_{2}, C H_{3} C H (B r) C H_{2} B r, C H_{3} C H (O H) C H_{3}

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C H_{3} C H = C H_{2}, C H_{3} C H_{2} C H_{2} B r, C H_{3} C H_{2} O C H_{2} C H_{3}

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C H_{3} C H = C H_{2}, C H_{3} C H (B r) C H_{3}, C H_{3} C H_{2} O C H_{2} C H_{3}

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C H_{3} C H = C H_{2}, C H_{3} C H_{2} C H_{2} B r, C H_{3} O C H_{2} C H_{2} C H_{3}

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Solution

The correct option is D $C H_{3} C H = C H_{2}, C H_{3} C H_{2} C H_{2} B r, C H_{3} O C H_{2} C H_{2} C H_{3}$
Given: $C H_{3} C H B r C H_{3} a l c . K O H - ---- \to X H B r - ---- \to P e r o x i d e Y C H_{3} O N a - ---- \to Z$

Alcoholic KOH forms a strong base as alkoxide which on reaction with alkyl halide cause elimination product as alkene. The alkene on treatment with HBr in the presence of peroxide undergoes addition reaction across double bond according to anti-Markovnikov's rule and give an alkyl halide as the product. The alkyl halide reacts with sodium alkoxide according to Williamson's synthesis and gives ether. The reaction sequence can be represented as:

$C H_{3} C H B r C H_{3} a l c . K O H - ---- \to C H_{3} C H = C H_{2} H B r - ---- \to P e r o x i d e C H_{3} C H_{2} C H_{2} B r C H_{3} O N a - ---- \to C H_{3} - O - C H_{2} C H_{2} C H_{3}$

thus $X = C H_{3} C H = C H_{2}, Y = C H_{3} C H_{2} C H_{2} B r, Z = C H_{3} - O - C H_{2} C H_{2} C H_{3}$

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Similar questions

I . C H_{3} C H_{2} C H_{2} B r + K O H \to C H_{3} C H = C H_{2} + K B r + H_{2} O

Which of the following statements is correct?

C H_{3} C H = C H_{2} + H_{2} S O_{4} \to

+ H_{2} O - --- \to B o i l

+ H_{2} S O_{4}

C H_{3} C H = C H_{2} + H B r

$C H_{3} C H = C H_{2} + B . r \to C H_{3} . C H_{2} B r$

C H_{3} C H = C H_{2}

C H_{3} C H = C H C H_{2} C H_{3}

C H_{3} C H = C H C H_{3}

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