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:

Similar Questions

The major product formed in the following reaction is :
$CH_3CH=CHCH(CH_3)_2 \xrightarrow{HBr}$

Taking into account the stability of various cycloalkanes and carbocations,as well as the rules governing mechanisms of carbocation rearrangements,what is the most likely product of this reaction?

In the following sequence of reactions,the alkene affords the compound $B$.
$CH_3-CH=CH-CH_3$ $\xrightarrow{O_3} A$ $\xrightarrow[Zn]{H_2O} B$
The compound $B$ is:

The major product formed when $but-2-ene$ is reacted with $O_3$ followed by treatment with $Zn / H_2O$ is:

For the given reaction sequence,identify the products $(C)$ and $(D)$ formed after ozonolysis of the major product $(B)$:
$(CH_3)_2CH-CHCl-CH_3$ $\xrightarrow{C_2H_5ONa / C_2H_5OH} B$ $\xrightarrow{O_3 / H_2O_2} C + D$

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