The reaction of HBr with the following compou | vedclass

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Question

(A) The reaction of $HBr$ with $4-hydroxystilbene$ $(Ph-CH=CH-C_6H_4-OH)$ follows Markovnikov's rule. $1$. The double bond attacks the $H^+$ from $HBr

Vedclass · Accepted Answer

$1-$($4$-hydroxyphenyl)$-1-$bromo$-2-$phenylethane

Vedclass · Answer

(A) The reaction of $HBr$ with $4-hydroxystilbene$ $(Ph-CH=CH-C_6H_4-OH)$ follows Markovnikov's rule. $1$. The double bond attacks the $H^+$ from $HBr$ to form the most stable carbocation. $2$. The carbocation formed at the benzylic position adjacent to the $4-hydroxyphenyl$ ring is more stable due to resonance stabilization from both the phenyl ring and the lone pair on the oxygen atom of the $-OH$ group. $3$. The bromide ion $(Br^-)$ then attacks this stable carbocation to form $1-(4-hydroxyphenyl)-1-bromo-2-phenylethane$.

The reaction of $HBr$ with the following compound would produce:

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