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Question: An alkene A contains three \(C - C\), eight \(C - H\) \(\sigma - \)bonds and one \(C - C{\text{ }}\p...

An alkene A contains three C−CC - C, eight C−HC - H σ−\sigma - bonds and one C−C πC - C{\text{ }}\pi bond. A on ozonolysis gives two moles of an aldehyde of molar mass 44 u44{\text{ u}}. Write IUPAC name of A.

Explanation

Solution

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 answer:

  1. First of all let's find out the structure of AA 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.
    - 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.
    -The structure AA has three C−CC - C, eight C−HC - H σ−bonds\sigma - bonds , and one C−CC - C π−bond\pi - {\text{bond}} i.e. double bond. By using this data the structure AA can be shown as, H3C−CH=CH−CH3{H_3}C - CH = CH - C{H_3}

-Now, let's see the ozonolysis reaction,
CH3−CH=CH−CH3→Zn+H2OO32CH3−CHOC{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 44 u44{\text{ u}},
CH3−CHO=2(C)+4(H)+(O)=2×12+1×4+16=44 uC{H_3} - CHO = 2\left( C \right) + 4\left( H \right) + \left( O \right) = 2 \times 12 + 1 \times 4 + 16 = 44{\text{ u}}

The molar mass of aldehyde is found equal to 44 u44{\text{ u}}. Therefore, the starting reactant alkene AA is H3C−CH=CH−CH3{H_3}C - CH = CH - C{H_3}
-Now let's find out the IUPAC{\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 IUPAC{\text{IUPAC}} name of the structure is But−2−- 2 -ene which shows option B as a correct choice.
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.