Aldehydes, Ketones and Carboxylic Acids

Aldehydes, Ketones and Carboxylic Acids

Question

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

  • Propene

  • 1-butene

  • 2-butene

  • ethene

Answer

C.

2-butene

(a) The general formula of aldehyde compound is CnH2nO. First, calculate the value of n. 
(b) Alkene is symmetrical, therefore, only single type of aldehyde is produced as a product.
CnH2nO  = 44
CnH2n = 44-16 = 28
n = 2
Therefore, since, the alkenes is symmetrical, then the structure is
CH3 - CH=CH -CH3
Thus, CH subscript 3 space minus space CH equals CH minus CH subscript 3 space rightwards arrow from left parenthesis ii right parenthesis space Zn divided by straight H subscript 2 straight O to left parenthesis straight i right parenthesis space straight O subscript 3 of space 2 CH subscript 3 minus CH equals straight O

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