An unknown alkyl halide $[$ X $]$ reacts with alcoholic KOH and produces a hydrocarbon $(C_{4} H_{8})$ as the major product. Ozonolysis of this hydrocarbon affords one mole of propanaldehyde and one mole of formaldehyde. Suggest which organic compound among the following has the correct structure of the above alkyl halide $[$ X $]$ ?

C H_{3} C H_{2} C H_{2} C H_{2} B r

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C H_{3} C H (B r) C H_{2} C H_{3}

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C H_{3} C H (B r) C H (B r) C H_{3}

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B r (C H_{2})_{4} B r

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Solution

The correct option is A $C H_{3} C H_{2} C H_{2} C H_{2} B r$
As the ozonolysis gives propanaldehyde and formaldehyde thus the hydrocarbon is $C H_{3} C H_{2} C H = C H_{2} (C_{4} H_{8}) 1 - b u t e n e$
$1 -$ butene can be obtained by $C H_{3} - C H_{2} - C H_{2} - C H_{2} - B r A l c . K O H \to - H B r C H_{3} C H_{2} C H = C H_{2}$ .

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The correct option is A CH3CH2CH2CH2BrAs the ozonolysis gives propanaldehyde and formaldehyde thus the hydrocarbon is CH3CH2CH=CH2(C4H8)1−butene1−butene can be obtained by CH3−CH2−CH2−CH2−BrAlc.KOH→−HBrCH3CH2CH=CH2.