(R)-3-bromocyclopentene (shown below) reacts | vedclass

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(A) The reaction of $(R)-3-$bromocyclopentene with $Br_2/CCl_4$ involves the anti-addition of bromine across the double bond. Since the starting mater

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$1,2,3-$tribromocyclopentane (all trans)

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(A) The reaction of $(R)-3-$bromocyclopentene with $Br_2/CCl_4$ involves the anti-addition of bromine across the double bond. Since the starting material has a bromine atom at the $3-$position,the addition of $Br_2$ results in the formation of $1,2,3-$tribromocyclopentane. Anti-addition means the two new bromine atoms will be added on opposite faces of the cyclopentene ring. For the $(R)-3-$bromocyclopentene isomer,one of the resulting diastereomers will possess an internal plane of symmetry,making it a meso compound. $A$ meso compound is achiral and therefore optically inactive. Based on the stereochemistry of anti-addition to the $(R)-$enantiomer,the product $Y$ that is optically inactive (meso) has the structure where the bromine atoms are arranged such that a plane of symmetry exists.

$(R)-3-$bromocyclopentene (shown below) reacts with $Br_2/CCl_4$ to form two products,$Y$ and $Z$. $Y$ is not optically active (does not rotate plane-polarized light). What is the structure of $Y$?

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