Answer
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Hint: Try to remember the general formula of alkene and the mechanism of ozonolysis reaction. The general formula for alkenes is ${C_n}{H_{2n}}$. With the general formula try to draw the structure of alkene.
Complete step by step answer:
In the question, they gave the products in reaction with ozone and alkene. The structures of propanal and pentan$ - 3 - $ one are $C{H_3}C{H_2}CHO$ and $C{H_3}C{H_2}COC{H_2}C{H_3}$ respectively.
Ozonolysis of alkenes-When alkene is reacted with ozone then alkenes are oxidized with ozone to form alcohols, aldehydes, or ketones or carboxylic acids. From the question, they gave one aldehyde and one ketone.
In the first step of the reaction mechanism, the double bond breaks, and the oxygen atom is added on both sides. And in the second step, an easier molecule loses its oxygen atom and combines with hydrogen to form a water molecule. So from the products, we can easily say that the double bond is between the third carbon atom in pent$ - 3 - $one and the first carbon atom in propanal.
The required structure of alkene is
$C{H_3} - C{H_2} - C(C{H_2} - C{H_3}) = CH - C{H_2} - C{H_3}$
The IUPAC name of the above alkene is $4 - ethyl - hex - 3 - ene$.
The double bond between the third and fourth carbon atom breaks and the right part of the double bond gives propanal and the left part gives pentan $ - 3 - $ one.
Note:
The reaction mechanism is more important for all the named reactions. All reactions of ozonolysis with alkenes don’t give ketones and aldehydes as products for example ozonolysis of ethene gives carbon dioxide and water molecules. Don’t get confused with the general formula of alkenes, alkanes, and alkynes.
Complete step by step answer:
In the question, they gave the products in reaction with ozone and alkene. The structures of propanal and pentan$ - 3 - $ one are $C{H_3}C{H_2}CHO$ and $C{H_3}C{H_2}COC{H_2}C{H_3}$ respectively.
Ozonolysis of alkenes-When alkene is reacted with ozone then alkenes are oxidized with ozone to form alcohols, aldehydes, or ketones or carboxylic acids. From the question, they gave one aldehyde and one ketone.
In the first step of the reaction mechanism, the double bond breaks, and the oxygen atom is added on both sides. And in the second step, an easier molecule loses its oxygen atom and combines with hydrogen to form a water molecule. So from the products, we can easily say that the double bond is between the third carbon atom in pent$ - 3 - $one and the first carbon atom in propanal.
The required structure of alkene is
$C{H_3} - C{H_2} - C(C{H_2} - C{H_3}) = CH - C{H_2} - C{H_3}$
The IUPAC name of the above alkene is $4 - ethyl - hex - 3 - ene$.
The double bond between the third and fourth carbon atom breaks and the right part of the double bond gives propanal and the left part gives pentan $ - 3 - $ one.
Note:
The reaction mechanism is more important for all the named reactions. All reactions of ozonolysis with alkenes don’t give ketones and aldehydes as products for example ozonolysis of ethene gives carbon dioxide and water molecules. Don’t get confused with the general formula of alkenes, alkanes, and alkynes.
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