The total number of monohalogenated organic p | vedclass

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(C) $1$. The simplest optically active alkene is $3$-methylpent-$1$-ene. $2$. Hydrogenation of $3$-methylpent-$1$-ene with $H_2/Ni/\Delta$ gives $3$-m

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$10$

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(C) $1$. The simplest optically active alkene is $3$-methylpent-$1$-ene. $2$. Hydrogenation of $3$-methylpent-$1$-ene with $H_2/Ni/\Delta$ gives $3$-methylpentane. $3$. Free radical chlorination of $3$-methylpentane $(CH_3-CH_2-CH(CH_3)-CH_2-CH_3)$ produces various monochlorinated isomers: - Substitution at $C_1$: $Cl-CH_2-CH_2-CH(CH_3)-CH_2-CH_3$ ($1$ chiral center,$2$ enantiomers). - Substitution at $C_2$: $CH_3-CHCl-CH(CH_3)-CH_2-CH_3$ ($2$ chiral centers,$4$ stereoisomers). - Substitution at $C_3$: $CH_3-CH_2-CCl(CH_3)-CH_2-CH_3$ ($1$ chiral center,$2$ enantiomers). - Substitution at $C_4$ (methyl group): $CH_3-CH_2-CH(CH_2Cl)-CH_2-CH_3$ ($1$ chiral center,$2$ enantiomers). $4$. Total number of stereoisomers = $2 + 4 + 2 + 2 = 10$.

The total number of monohalogenated organic products (including stereoisomers) formed in the following reaction is:
$A$ (simplest optically active alkene) $\xrightarrow[(ii) X_2/\Delta ]{(i) H_2/Ni/\Delta }$

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The total number of monohalogenated organic products (including stereoisomers) formed in the following reaction is: $A$ (simplest optically active alkene) $\xrightarrow[(ii) X_2/\Delta ]{(i) H_2/Ni/\Delta }$