1-methyl ethylene oxide (propylene oxide) whe | vedclass

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Question

(B) The reaction of $1$-methyl ethylene oxide (propylene oxide) with an excess of $HBr$ proceeds as follows:$1$. The oxygen atom of the epoxide ring g

Vedclass · Accepted Answer

$CH_3-CH(Br)-CH_2Br$

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(B) The reaction of $1$-methyl ethylene oxide (propylene oxide) with an excess of $HBr$ proceeds as follows:$1$. The oxygen atom of the epoxide ring gets protonated by $H^+$,forming a cyclic oxonium ion.$2$. The bromide ion $(Br^-)$ attacks the more substituted carbon atom of the epoxide ring due to the development of partial positive charge character,which is more stable at the more substituted carbon.$3$. This leads to the opening of the ring,forming a bromohydrin intermediate $(CH_3-CH(OH)-CH_2Br)$.$4$. Since $HBr$ is in excess,the hydroxyl group $(-OH)$ is further protonated and replaced by another bromide ion $(Br^-)$ via an $S_N2$ mechanism,resulting in a vicinal dibromide.$5$. The final product is $1,2$-dibromopropane,which is $CH_3-CH(Br)-CH_2Br$.

$1$-methyl ethylene oxide (propylene oxide) when treated with an excess of $HBr$ produces:

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