Find the number of optical isomers for CH_3-C | vedclass

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(D) The given molecule is $CH_3-CH(OH)-CH(OH)-CH(OH)-CH_3$. It has $n = 3$ chiral centers. The molecule is unsymmetrical. The number of optically acti

Vedclass · Accepted Answer

$4$

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(D) The given molecule is $CH_3-CH(OH)-CH(OH)-CH(OH)-CH_3$. It has $n = 3$ chiral centers. The molecule is unsymmetrical. The number of optically active isomers $(a)$ is given by $a = 2^{n-1} = 2^{3-1} = 2^2 = 4$. The number of meso isomers $(m)$ is given by $m = 2^{(n-2)/2} = 2^{(3-1)/2} = 2^1 = 2$. Wait,for an unsymmetrical molecule with $n$ chiral centers,the total number of stereoisomers is $2^n = 2^3 = 8$. However,the question asks for optical isomers. For $CH_3-CH(OH)-CH(OH)-CH(OH)-CH_3$,the number of stereoisomers is $2^n = 2^3 = 8$. These consist of $4$ pairs of enantiomers (optically active). Thus,the total number of optical isomers is $8$.

Find the number of optical isomers for $CH_3-CH(OH)-CH(OH)-CH(OH)-CH_3$.

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