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Aldehydes, Ketones And Carboxylic Acids

Question
CBSEENCH12010734
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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

Solution

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

Some More Questions From Aldehydes, Ketones and Carboxylic Acids Chapter

Write the structures of the following compounds.
Di-sec-butyl ketone

Write the structures of the following compounds.
4-fluoro acetophenone.

Write the structures of the following reactions:

Write the structures of products of the following reactions:


    

Arrange the following compounds in increasing order of their boiling points:
CH3CHO, CH3CH2OH, CH3OCH3, CH3CH2CH3.

Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions:    
Ethanal, Propanal, Propanone, Butanone.

Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions:    
Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone.

Predict the products of the following reactions:

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