Propanal and pentan-3-one are the ozonolysis products of an alkene. What is the structural formula of the alkene:
(a)
(b)
(c)
(d)
Answer
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Hint: Alkenes on reaction with ozone gives aldehydes and ketones. They give symmetrical or asymmetrical aldehydes and ketones depending on the alkene. Ozone cleaves the unsaturated bonds of alkenes, alkynes and azo compounds.
Complete step by step solution:
Let us consider the given alkene as A. So we can write the ozonolysis reaction as
\[\underset{alkene}{\mathop{A}}\,+{{O}_{3}}\to \underset{ozonide}{\mathop{X}}\,\]
-First the alkene A is reacted with\[{{O}_{3}}\] which gives an ozonide X.
-X in the presence of Zn and \[{{H}_{2}}O\] gives the desired product, that is propanal and pentan-3-one.
The ozonide X has a structure
The products are obtained by the cleavage of the ozonide X, as shown in the image.
-By the addition of \[{{O}_{3}}\] to alkene we get the ozonide X which cleaves in the presence of Zn and\[{{H}_{2}}O\] to give the products. So the alkene must be option (a).
-There is another method by which we can solve this problem.
-Both products are written in such a way that the doubly bonded O of the products are facing each other.
-By removing the O atoms and the fragments obtained are bonded by a double bond, we get the desired product.
Ozonolysis reaction is used as a preparation of desired aldehydes and ketones from alkenes. They mainly undergo reductive ozonolysis. If symmetrical alkenes are used we get only 1 aldehyde or ketone (depending on the alkene), using asymmetrical alkenes we get a mixture of aldehyde and ketones.
Thus, the correct answer to the question is (b).
Note: The possible confusion that can occur in solving this question is how the ozonide is formed and which all positions it is cleaved to get the product. Therefore, the second method shown above could be an easy way to solve questions of this type.
Complete step by step solution:
Let us consider the given alkene as A. So we can write the ozonolysis reaction as
\[\underset{alkene}{\mathop{A}}\,+{{O}_{3}}\to \underset{ozonide}{\mathop{X}}\,\]
-First the alkene A is reacted with\[{{O}_{3}}\] which gives an ozonide X.
-X in the presence of Zn and \[{{H}_{2}}O\] gives the desired product, that is propanal and pentan-3-one.
The ozonide X has a structure
The products are obtained by the cleavage of the ozonide X, as shown in the image.
-By the addition of \[{{O}_{3}}\] to alkene we get the ozonide X which cleaves in the presence of Zn and\[{{H}_{2}}O\] to give the products. So the alkene must be option (a).
-There is another method by which we can solve this problem.
-Both products are written in such a way that the doubly bonded O of the products are facing each other.
-By removing the O atoms and the fragments obtained are bonded by a double bond, we get the desired product.
Ozonolysis reaction is used as a preparation of desired aldehydes and ketones from alkenes. They mainly undergo reductive ozonolysis. If symmetrical alkenes are used we get only 1 aldehyde or ketone (depending on the alkene), using asymmetrical alkenes we get a mixture of aldehyde and ketones.
Thus, the correct answer to the question is (b).
Note: The possible confusion that can occur in solving this question is how the ozonide is formed and which all positions it is cleaved to get the product. Therefore, the second method shown above could be an easy way to solve questions of this type.
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