A mixture of all isomers possible from the mo | vedclass

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(B) The structure provided is $n$-pentane $(CH_3-CH_2-CH_2-CH_2-CH_3)$.Mono-chlorination of $n$-pentane can occur at three non-equivalent carbon posit

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$3$

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(B) The structure provided is $n$-pentane $(CH_3-CH_2-CH_2-CH_2-CH_3)$.Mono-chlorination of $n$-pentane can occur at three non-equivalent carbon positions:$1$. At the $C_1$ position (terminal carbon): $CH_3-CH_2-CH_2-CH_2-CH_2Cl$ ($1$-chloropentane).$2$. At the $C_2$ position: $CH_3-CH_2-CH_2-CHCl-CH_3$ ($2$-chloropentane).$3$. At the $C_3$ position: $CH_3-CH_2-CHCl-CH_2-CH_3$ ($3$-chloropentane).Note that $2$-chloropentane and $3$-chloropentane are chiral molecules,meaning each exists as a pair of enantiomers ($R$ and $S$ forms).However,fractional distillation separates compounds based on their boiling points. Enantiomers have identical boiling points and cannot be separated by fractional distillation.Therefore,the distinct products with different boiling points are:- $1$-chloropentane- $2$-chloropentane (racemic mixture)- $3$-chloropentane (racemic mixture)This results in $3$ distinct fractions.

$A$ mixture of all isomers possible from the mono-chlorination of the structure $n$-pentane $(CH_3-CH_2-CH_2-CH_2-CH_3)$ is subjected to fractional distillation. How many fractions will be obtained?

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