The structure of E-(S)-1-chloro-1-bromo-2-met | vedclass

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(D) $1$. Analyze the $E$ configuration: In $E-(S)-1$-chloro-$1$-bromo-$2$-methyl-$2$-hexene,the double bond is between $C_2$ and $C_3$. For the $E$ is

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(D) $1$. Analyze the $E$ configuration: In $E-(S)-1$-chloro-$1$-bromo-$2$-methyl-$2$-hexene,the double bond is between $C_2$ and $C_3$. For the $E$ isomer,the higher priority groups on each carbon of the double bond must be on opposite sides. On $C_2$,the groups are $-CH_3$ and the chiral group $-CH(Cl)Br$. On $C_3$,the groups are $-H$ and $-CH_2CH_2CH_3$. The higher priority group on $C_2$ is the chiral group (due to $Br/Cl$ vs $C$),and on $C_3$ is the propyl group. Thus,the propyl group and the chiral group must be on opposite sides.$2$. Analyze the $S$ configuration: The chiral center is at $C_1$ (attached to $Cl, Br, H$ and the $C_2$ double bond). Assigning priorities: $Br (1) > Cl (2) > C_2=C_3 (3) > H (4)$. For $S$ configuration,with $H$ on a dashed bond,the sequence $1$ $\rightarrow 2$ $\rightarrow 3$ must be counter-clockwise.$3$. Evaluating the structures: Structure $D$ shows the propyl group and the chiral group on opposite sides ($E$ configuration) and the $S$ configuration at the chiral center ($Br$ is wedge,$Cl$ is in-plane,$H$ is dash,$1$ $\rightarrow 2$ $\rightarrow 3$ is counter-clockwise).

The structure of $E-(S)-1$-chloro-$1$-bromo-$2$-methyl-$2$-hexene is best represented by

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