An alkene $A$ contains three $C - C$, eight $C - H$ $\sigma - bonds$ and one $C - C$ $\pi - bond$. $A$ on ozonolysis gives two moles of an aldehyde of molar mass ${\text{44 u}}$. ${\text{IUPAC}}$ name of $A$ is:
A) But$ - 1 - $ene
B) But$ - 2 - $ene
C) $2 - $methylpropane
D) None of these
Answer
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Hint: In the question, the molar mass and the functional group name is given itself. One can guess the products by that data. In the ozonolysis reaction, the breaking of the carbon double bond happens and the carbon atom gets oxidized. One can guess the alkene by this data.
Complete step by step answer:
1) First of all let's find out the structure of $A$ on the basis of given data. The aldehyde is formed due to an ozonolysis reaction on the compound which has four carbon atoms and one carbon-carbon double bond which shows the structure is identical on both sides.
2) The formation of two moles of an aldehyde as a product shows the presence of identical structural units on both sides of the carbon double bond.
3) The structure $A$ has three $C - C$, eight $C - H$ $\sigma - bonds$ , and one $C - C$ $\pi - bond$ i.e. double bond. By using this data the structure $A$ can be shown as, ${H_3}C - CH = CH - C{H_3}$
4) Now, let's see the ozonolysis reaction,
$C{H_3} - CH = CH - C{H_3}\xrightarrow[{Zn + {H_2}O}]{{{O_3}}}2C{H_3} - CHO$
Let's see if the aldehyde we got as the product has a molar mass equal to ${\text{44 u}}$,
$C{H_3} - CHO = 2\left( C \right) + 4\left( H \right) + \left( O \right) = 2 \times 12 + 1 \times 4 + 16 = 44u$
The molar mass of aldehyde is found equal to ${\text{44 u}}$. Therefore, the starting reactant alkene $A$ is ${H_3}C - CH = CH - C{H_3}$
5) Now let's find out the ${\text{IUPAC}}$ name of alkene, the structure has four carbon atoms in it and shows the presence of a double bond at second carbon in the structure. Hence, the ${\text{IUPAC}}$ name of the structure is But$ - 2 - $ene
So, the correct answer is Option B .
Note:
In the ozonolysis reaction there is always breaking of an unsymmetrical bond of alkenes, alkynes by using ozone. This reaction is also used for identifying the position of the double bond in an alkene. Ozonolysis is a type of oxidation reaction where the oxygen-containing carbonyl group is formed on carbon.
Complete step by step answer:
1) First of all let's find out the structure of $A$ on the basis of given data. The aldehyde is formed due to an ozonolysis reaction on the compound which has four carbon atoms and one carbon-carbon double bond which shows the structure is identical on both sides.
2) The formation of two moles of an aldehyde as a product shows the presence of identical structural units on both sides of the carbon double bond.
3) The structure $A$ has three $C - C$, eight $C - H$ $\sigma - bonds$ , and one $C - C$ $\pi - bond$ i.e. double bond. By using this data the structure $A$ can be shown as, ${H_3}C - CH = CH - C{H_3}$
4) Now, let's see the ozonolysis reaction,
$C{H_3} - CH = CH - C{H_3}\xrightarrow[{Zn + {H_2}O}]{{{O_3}}}2C{H_3} - CHO$
Let's see if the aldehyde we got as the product has a molar mass equal to ${\text{44 u}}$,
$C{H_3} - CHO = 2\left( C \right) + 4\left( H \right) + \left( O \right) = 2 \times 12 + 1 \times 4 + 16 = 44u$
The molar mass of aldehyde is found equal to ${\text{44 u}}$. Therefore, the starting reactant alkene $A$ is ${H_3}C - CH = CH - C{H_3}$
5) Now let's find out the ${\text{IUPAC}}$ name of alkene, the structure has four carbon atoms in it and shows the presence of a double bond at second carbon in the structure. Hence, the ${\text{IUPAC}}$ name of the structure is But$ - 2 - $ene
So, the correct answer is Option B .
Note:
In the ozonolysis reaction there is always breaking of an unsymmetrical bond of alkenes, alkynes by using ozone. This reaction is also used for identifying the position of the double bond in an alkene. Ozonolysis is a type of oxidation reaction where the oxygen-containing carbonyl group is formed on carbon.
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