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# A compound can be divided into two equal halves and contains even $n$ asymmetric carbon atoms .The number of stereoisomers is:A. ${2^n}$B. $2(n + 1)$C. ${2^{(\dfrac{n}{2} - 1)}}$D. ${2^{n - 1}} + {2^{(\dfrac{n}{2} - 1)}}$

Last updated date: 13th Jun 2024
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Hint: A chiral carbon shows optical isomerism by changing the spatial arrangements of the functional groups which are attached to the central carbon atom.
meso isomer is a non-optically active member of a set of stereoisomers, at least two of which are optically active.
The stereoisomers include meso isomers as well as optical isomers, so in order to calculate the number of stereoisomers, the addition of both isomers is required.

A chiral carbon is a type of carbon which has four different functional groups attached to it, for instance a carbon with which a hydrogen along with three different halogens are attached. A diagram is shown below.

Optical isomerism is a type of isomerism where both the isomeric structures have same molecules and connectivity between atoms but different spatial arrangements. Given below is an example of optical isomerism.

For a chiral carbon atom optical isomerism is determined by a compound division suppose if a compound is not divided then optical isomerism shows ${{2}^{n}}$ and meso isomers are zero
Now I am explaining what are known as meso isomers.
meso isomer is a non-optically active member of a set of stereoisomers, at least two of which are optically active. Example of a meso isomer is shown below.

Optical isomers are $2$ compounds which contain the same number and kinds of atoms, bonds are different spatial arrangements of the atoms, but which have non-superimposable mirror images.
Now if a compound is divided then we should check whether they are even or odd.
If they are even then the optical isomer are ${{2}^{n-1}}$ and meso isomers are ${{2}^{(\dfrac{n}{2}-1)}}$.
If they are odd then optical isomers are ${{2}^{n-1}}$ and meso isomers are ${{2}^{(\dfrac{n-1}{2})}}$.
So according to question the compound can be divided into two equal halves so $n$is even according to even it has both optical isomer ${{2}^{n-1}}$ and meso isomers are ${{2}^{(\dfrac{n}{2}-1)}}$

So, the correct answer is Option D.

One stereoisomer, called the cis stereoisomer, has both of the double-bond hydrogens on the identical side of the covalent bond, while the other stereoisomer, called the trans stereoisomer, has the $2$ hydrogens on opposite sides of the covalent bond