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An organic compound A (molecular formula C_(3)H_(6)O) is resistant is oxidation but forms a compound B (C_(3)H_(8)O) on reduction. B. reacts with HBr to form a bromide C which o treatment with alcoholic KOH forms an alkene D (C_(3)H_(6)). Deduce the structures of A,B,C and D. |
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Answer» SOLUTION :STRUCTURES of A,B, C and D are deduced in the following manner: `CH_(3)-UNDERSET(A)underset("Acetone")underset(O)underset(||)(C)-CH_(3)underset("Reduction")overset(LiAlH_(4))(to)CH_(3)-underset(B)underset("Propane-2-ol")underset(OH)underset(|)(CH)-CH_(3)` `CH_(3)-underset(A)underset("Propan"-2-ol)underset(OH)underset(|)(CH)-CH_(3)+HBR to CH_(3)-underset(C)underset(2-"Bromopropane")underset(Br)underset(|)(CH)-CH_(3)` `CH_(3)-underset(C)underset(2-"Bromopropane")underset(Br)underset(|)(CH)-CH_(3)+KOH(alc)toCH_(3)-underset(D)underset("Propene")(CH)=CH_(2)+KBr+H_(2)O` |
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