If the starting material is 1-methyl-1,2-epox | vedclass

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(B) The reaction of $1-$methyl$-1,2-$epoxycyclopentane with sodium methoxide $(CH_3ONa)$ in methanol $(CH_3OH)$ is a nucleophilic ring-opening reactio

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(B) The reaction of $1-$methyl$-1,2-$epoxycyclopentane with sodium methoxide $(CH_3ONa)$ in methanol $(CH_3OH)$ is a nucleophilic ring-opening reaction of an epoxide under basic conditions.In basic conditions,the nucleophile $(CH_3O^-)$ attacks the less sterically hindered carbon atom of the epoxide ring.In $1-$methyl$-1,2-$epoxycyclopentane,the two carbons of the epoxide are the $C1$ (which is tertiary,attached to a methyl group) and $C2$ (which is secondary).The nucleophile $(CH_3O^-)$ will attack the less hindered $C2$ position.This leads to the opening of the epoxide ring,where the oxygen atom remains attached to the $C1$ position and becomes a hydroxyl group $(-OH)$ after protonation,while the methoxy group $(-OCH_3)$ attaches to the $C2$ position.The resulting product is $2-$methoxy$-2-$methylcyclopentan$-1-$ol,where the $-OCH_3$ and $-OH$ groups are in a trans-configuration relative to each other.

If the starting material is $1-$methyl$-1,2-$epoxycyclopentane,determine which compound correctly represents the product of its reaction with sodium methoxide in methanol.

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