The number of isomeric products formed by the | vedclass

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(D) The structure of $2$-methylbutane is $CH_3-CH(CH_3)-CH_2-CH_3$.Monochlorination can occur at four different types of hydrogen atoms:$1$. At $C_1$

Vedclass · Accepted Answer

$6$

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(D) The structure of $2$-methylbutane is $CH_3-CH(CH_3)-CH_2-CH_3$.Monochlorination can occur at four different types of hydrogen atoms:$1$. At $C_1$ (terminal methyl group): $CH_2Cl-CH(CH_3)-CH_2-CH_3$ ($1$-chloro$-2-$methylbutane,achiral).$2$. At $C_2$ (tertiary carbon): $CH_3-CCl(CH_3)-CH_2-CH_3$ ($2$-chloro$-2-$methylbutane,achiral).$3$. At $C_3$ (chiral center): $CH_3-CH(CH_3)-CHCl-CH_3$ ($2$-chloro$-3-$methylbutane). This product has a chiral center,so it exists as a pair of enantiomers ($d$ and $l$ forms).$4$. At $C_4$ (terminal methyl group): $CH_3-CH(CH_3)-CH_2-CH_2Cl$ ($1$-chloro$-3-$methylbutane). This product has a chiral center at $C_2$,so it exists as a pair of enantiomers ($d$ and $l$ forms).Total isomeric products = $1$ (from $C_1$) + $1$ (from $C_2$) + $2$ (from $C_3$) + $2$ (from $C_4$) = $6$.

The number of isomeric products formed by the monochlorination of $2$-methylbutane in the presence of sunlight is . . . . . . .

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