Question

Calculate the total number of open chain isomeric carbonyl compounds of molecular formula \[{C_5}{H_8}O\] which can’t show geometric isomerism.

Answer 1

Hint: The open-chain compounds are the acyclic compounds that have a linear structure, rather than a cyclic one. Geometric isomerism is a type of isomerism also known as ‘cis-trans’ isomerism. For the molecule, \[{C_5}{H_8}O\] we will not count this isomer.

Complete step by step answer:
First, we will understand the isomers. Isomers are the molecules of the compounds which possess a similar molecular formula which means that the number of individual atoms present in the parent compound or molecule is the same as the isomer formed. Isomer possesses a different chemical structure. Now we will discuss one of the types of isomerism is geometrical isomerism. So, geometrical isomerism is the in which directional arrangement is considered. In geometrical isomerism, the rotation of the attached atoms or groups of atoms around the double-bonded atoms helps to produce a new compound that possesses similar molecular formula but a different chemical structure. Now we will consider the compound given to us which is \[{C_5}{H_8}O\]. We will try to frame only pen chain isomeric carbonyl compounds. So the possible isomers for \[{C_5}{H_8}O\] as follows.

From the above isomers, we can observe that all the possible isomers possess a similar molecular formula which is \[{C_5}{H_8}O\] but the chemical structure is different. The position of the carbonyl group and the double bond is changed to form an isomer of a similar molecular formula.
Therefore, the total number of open chain isomeric carbonyl compounds of molecular formula \[{C_5}{H_8}O\] which can’t show geometric isomerism are \[8\].

Note:
The total number of possible open-chain isomers of the compound \[{C_5}{H_8}O\] are \[43\]. The isomers contain \[15\] alcohols, \[16\] ether, \[8\] aldehydes, and \[4\] ketones. This optical isomerism and the geometrical isomerism are also included.