How to find number of isomers for any Organic compound?
You have no formula to calculate number of constitutional isomers. You have to make all possibilities and coun
But you have a formula to calculate no. of stereoisomers.
The general formula for calculating stereoisomer is: 2n, where n is the number of chial centers.
A couple of things to keep in mind:
1. Stereoisomers are compounds with the same chemical formula but different spatial arrangement.
2. And chiral centers are carbons that are bonded to 4 different groups.
For example, if we use glucose as our starting compound, the number of chiral carbons that a glucose molecule possesses is 4.
Out of the 6 carbons, the 4 carbons in the middle are chiral. That is to say, the four carbons in the middle are bonded to 4 different groups. If we think about it, the OH groups on each of the four carbons can be either on the left or the right side of the carbon (for Fischer projected images). Thus, since each of the 4 OH groups have two possible sides, the number of stereoisomers is 24, or 16.
This is the simplest formula.
But cannot be used in cases of molecular symmetry.
All formulas are stated below:-
a) When the molecule is unsymmetrical and contains ''n '' chiral carbon atoms, Total no. of stereoisomers = 2n
b) When the molecule is unsymmetrical and has even number of stereogenic centres or chiral carbon atoms,
Total no. of stereoisomers = No. of optical isomers + No. of meso forms = 2(n-1) + 2(n/2-1)
c) When the molecule is symmetrical and has odd no. of stereogenic centres
Total no. of stereoisomers = [ 2(n-1)-2(n/2-1/2)] + 2(n/2-1/2)]The formula for determining the number of stereoisomers is as follows:
a) When the molecule is unsymmetrical and contains ''n '' chiral carbon atoms, Total no. of stereoisomers = 2n
b) When the molecule is unsymmetrical and has even number of stereogenic centres or chiral carbon atoms,
Total no. of stereoisomers = No. of optical isomers + No. of meso forms = 2(n-1) + 2(n/2-1)
c) When the molecule is symmetrical and has odd no. of stereogenic centres
Total no. of stereoisomers = [ 2(n-1)-2(n/2-1/2)] + 2(n/2-1/2)]
You have no formula to calculate number of constitutional isomers. You have to make all possibilities and coun
But you have a formula to calculate no. of stereoisomers.
The general formula for calculating stereoisomer is: 2n, where n is the number of chial centers.
A couple of things to keep in mind:
1. Stereoisomers are compounds with the same chemical formula but different spatial arrangement.
2. And chiral centers are carbons that are bonded to 4 different groups.
For example, if we use glucose as our starting compound, the number of chiral carbons that a glucose molecule possesses is 4.
Out of the 6 carbons, the 4 carbons in the middle are chiral. That is to say, the four carbons in the middle are bonded to 4 different groups. If we think about it, the OH groups on each of the four carbons can be either on the left or the right side of the carbon (for Fischer projected images). Thus, since each of the 4 OH groups have two possible sides, the number of stereoisomers is 24, or 16.
This is the simplest formula.
But cannot be used in cases of molecular symmetry.
All formulas are stated below:-
a) When the molecule is unsymmetrical and contains ''n '' chiral carbon atoms, Total no. of stereoisomers = 2n
b) When the molecule is unsymmetrical and has even number of stereogenic centres or chiral carbon atoms,
Total no. of stereoisomers = No. of optical isomers + No. of meso forms = 2(n-1) + 2(n/2-1)
c) When the molecule is symmetrical and has odd no. of stereogenic centres
Total no. of stereoisomers = [ 2(n-1)-2(n/2-1/2)] + 2(n/2-1/2)]The formula for determining the number of stereoisomers is as follows:
a) When the molecule is unsymmetrical and contains ''n '' chiral carbon atoms, Total no. of stereoisomers = 2n
b) When the molecule is unsymmetrical and has even number of stereogenic centres or chiral carbon atoms,
Total no. of stereoisomers = No. of optical isomers + No. of meso forms = 2(n-1) + 2(n/2-1)
c) When the molecule is symmetrical and has odd no. of stereogenic centres
Total no. of stereoisomers = [ 2(n-1)-2(n/2-1/2)] + 2(n/2-1/2)]
