C2H5OH→573KCuA;CH32CHOH→573KCuB;CH33COH→573KCuC;Incorrect statement is

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‘A’ gives positive haloform test and can undergo aldol condensation

‘C’ on reductive ozonolysis gives ‘B’ and formaldehyde

‘B’ does not give haloform test but gives 2,4-DNP test

‘C’ with KMnO4/H+, gives B and CO2

answer is C.

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Detailed Solution

CH3−CH2−OH→Cu/573KCH3−CHOCH32CHOH→Cu/573KCH32C=OCH23COH→Cu/573KCH32C=CH2In fact 30 alcohols undergo dehydration but not dehydrogenation. CH3CHO(A) positive haloform test ……undergo aldol condensation.CH32C=CH2→Zn+H2OO3CH32C=O+HCHO CH3−C|CH3=CH2C→An+H2OO3CH32C=O+HCHO CH3−CO−CH3……..positive haloform test ……undergo aldol condensation. CH32C=CH21→ΔKMnO4/H+CH32C=O+CO2+H2O

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