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

Ethane

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Propene

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1- butene

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2-butene

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Solution

The correct option is D
2-butene

$R - C H = C H - R (i) O_{3} / C H_{2} C l_{2} - -------- \to (i i) Z n / H_{2} O 2 R - C H O$

$(S y m m . a l k e n e) A l d e h y d e$

Molecular mass of RCHO = 44

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

This is possible, only when R is - $C H_{3}$ group, so the

aldehyde is $C H_{3} C H O$ and the symmetrical alkene is

$C H_{3} - C H = C H - C H_{3} (B u t - 2 - e n e)$

31. $C H_{3} - C H = C H - C H_{3} O z o n o l y s i s - ----- \to 2 C H_{3} C H O$

$B u t - 2 - e n e A c e t a l d e h y d e$

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

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