Reductive ozonolysis of Y yields :

2 moles of CH3CHO

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C H_{3} C H_{2} C H O

& HCHO

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C H_{3} C | | O - C H_{3}

& HCHO

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C H_{3} C H_{2} C O O H

& HCOOH

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Solution

The correct option is B $C H_{3} C H_{2} C H O$ & HCHO
$C H_{3} C H_{2} C H_{2} C H_{2} - B r X, (C_{4} H_{9} B r) a l c . K O H \to C H_{3} C H_{2} C H = C H_{2} Y B r_{2} \to C H_{3} C H_{2} C H B r C H_{2} B r Z (D i b r o m i d e)$
$N a N H_{2} \to C H_{3} C H_{2} C \equiv C H W (g a s) A g N O_{3} \to S o l^{n} A g B r$
Reductive ozonolysis of Y yields $C H_{3} C H_{2} C H O$ and HCHO.
$C H_{3} C H_{2} C H = C H_{2} r e d u c t i v e o z o n o l y s i s - ----------- \to C H_{3} C H_{2} C H O$ .

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Reductive ozonolysis of Y yields :

The correct option is B CH3CH2CHO & HCHOCH3CH2CH2CH2−BrX,(C4H9Br)alc.KOH→CH3CH2CH=CH2YBr2→CH3CH2CHBrCH2BrZ(Dibromide)NaNH2→CH3CH2C≡CHW(gas)AgNO3→SolnAgBrReductive ozonolysis of Y yields CH3CH2CHO and HCHO.CH3CH2CH=CH2reductiveozonolysis−−−−−−−−−−−−→CH3CH2CHO.