NUMBER OF POSSIBLE STEREOISOMERS IN COMPOUNDS CONTAINING DIFFERENT NUMBER OF ASYMMETRIC ATOMS

The number of optical isomers is calculated by the application of the following rules:

The molecule has no symmetry (unsymmetrical) and (n) is the number of asymmetric carbon atoms, then

The number of d- and l- (optically active) forms, a = 2ⁿ

and       The number of meso – forms, m = 0

Total number of optical (stereo) isomers = a + m = 2ⁿ

For example, CH₃—CHOH—COOH (Lactic acid)

Where, n = 1, a 2¹ = 2, m = 0, r = (2/2) = 1

So, total optical isomers = 2 + 0 = 2.

(ii) When the molecule can be divided into equal halves, i.e., the molecule has symmetry and the number (n) of asymmetric carbon atoms is even, then

The number of d- and l- (optically active) forms, a = 2^(n-1)

and       The number of meso – forms, m = 2^(n/2)-1

Total number of optical isomers = a + m

= 2^(n-1) + 2^(n/2)-1

For example, HOOC – *CHOH – *CHOH – COOH

Where, n=2, a=2^n-1=2^2-1=2¹=2

m=2^(n/2) – 1 = 2^(2/2) – 1 = 2¹ – 1 = 1    

So, total optical isomers = 2 + 1 = 3.

(iii) When the molecule can be divided into two equal halves (Symmetrical) and the number (n) of asymmetric carbon atoms is odd, then 

The number of d-and l-forms, a = 2^(n-1) – 2^(n-1)/2

And   The number of meso-forms, m=2^(n-1)/2

Total number of optical isomers = a + m = 2^(n-1)

For example,

CH₂OH – *CHOH – *CHOH – *CHOH – CH₂OH

Where, n=3, a=2^(n-1) – 2^(n-1)/2 = 2² – 2¹ = 2

m = 2^(n-1)/2 = 2^(3-1)/2 = 2¹ = 2

So, total optical isomers = 2 + 2 = 4

In all the above three cases, the number of racemic forms will be = a/2.

I) HOCH₂ – (CHOH)₄ – CHO

II) HOCH₂ – (CHOH)₄ – CH₂OH