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Hint: The isomers which have the same molecular formula but have a difference in the spatial arrangement then they are called stereoisomers and the isomerism is called stereoisomerism. Two kinds of stereoisomers are present enantiomers and diastereomers.
There is a formula to calculate the number of stereoisomers.
Number of stereoisomers = ${{2}^{n}}$ - meso structures.
Where n = number of chiral centres.
Complete step by step solution:
- We have to find the number of stereoisomers for the given compound in the question.
- The structure of the given compound is as follows.
- In the given compound there are six carbons present.
- We can see clearly that there is no plane of symmetry in the given molecule.
- So, there is no chance of formation of mesostructures due to the absence of a plane of symmetry.
- If a plane of symmetry exists then meso compounds are going to form by the given molecule.
- Now the plane of symmetry is ruled out, next we have to see how many numbers of chiral centres are present in the given molecule.
- There is only one chiral centre present in the given molecule at carbon-4.
- Therefore
\[\begin{align}
& Number\text{ }of\text{ }stereoisomers\text{ }=~{{2}^{n}}-\text{ }meso\text{ }structures \\
& Number\text{ }of\text{ }stereoisomers={{2}^{1}}-0 \\
& Number\text{ }of\text{ }stereoisomers=2 \\
\end{align}\]
- Therefore the number of stereoisomers formed by the given molecule is 2.
So, the correct option is (A).
Note: If there is any plane of symmetry existing in a molecule then only meso compounds are going to form. There is no relation between chiral centres and mesostructures. Plane of symmetry means an imaginary plane should divide any molecule into two identical halves then the molecule contains a plane of symmetry.
There is a formula to calculate the number of stereoisomers.
Number of stereoisomers = ${{2}^{n}}$ - meso structures.
Where n = number of chiral centres.
Complete step by step solution:
- We have to find the number of stereoisomers for the given compound in the question.
- The structure of the given compound is as follows.
- In the given compound there are six carbons present.
- We can see clearly that there is no plane of symmetry in the given molecule.
- So, there is no chance of formation of mesostructures due to the absence of a plane of symmetry.
- If a plane of symmetry exists then meso compounds are going to form by the given molecule.
- Now the plane of symmetry is ruled out, next we have to see how many numbers of chiral centres are present in the given molecule.
- There is only one chiral centre present in the given molecule at carbon-4.
- Therefore
\[\begin{align}
& Number\text{ }of\text{ }stereoisomers\text{ }=~{{2}^{n}}-\text{ }meso\text{ }structures \\
& Number\text{ }of\text{ }stereoisomers={{2}^{1}}-0 \\
& Number\text{ }of\text{ }stereoisomers=2 \\
\end{align}\]
- Therefore the number of stereoisomers formed by the given molecule is 2.
So, the correct option is (A).
Note: If there is any plane of symmetry existing in a molecule then only meso compounds are going to form. There is no relation between chiral centres and mesostructures. Plane of symmetry means an imaginary plane should divide any molecule into two identical halves then the molecule contains a plane of symmetry.
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