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Question

One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44 amu. The alkene is :


A

Ethane

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B

Propene

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C

1- butene

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D

2-butene

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Solution

The correct option is D

2-butene


RCH=CHR(i) O3/CH2Cl2−−−−−−−−(ii) Zn/H2O2RCHO

(Symm.alkene) Aldehyde

Molecular mass of RCHO = 44

i.e., R + 12 + 1 + 16 = 44 or R = 15

This is possible, only when R is - CH3 group, so the

aldehyde is CH3CHO and the symmetrical alkene is

CH3CH=CHCH3(But2ene)

31. CH3CH=CHCH3Ozonolysis−−−−−2CH3CHO

But2ene Acetaldehyde


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