The chemical reaction is shown below:C_6H_5-C | vedclass

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(C) The reaction follows Markovnikov's addition of $HCl$ to the alkene.$1$. The $H^+$ ion attacks the terminal carbon of the double bond to form the m

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(C) The reaction follows Markovnikov's addition of $HCl$ to the alkene.$1$. The $H^+$ ion attacks the terminal carbon of the double bond to form the most stable carbocation.$2$. The intermediate formed is a $3^{\circ}$ benzylic carbocation $(C_6H_5-CH^+-CH_2-CH_3)$,which is highly stabilized by resonance with the phenyl ring.$3$. The $Cl^-$ ion then attacks this carbocation to form the final product: $1$-chloro-$1$-phenylpropane $(C_6H_5-CH(Cl)-CH_2-CH_3)$.

The chemical reaction is shown below:
$C_6H_5-CH_2-CH=CH_2 + HCl \rightarrow X$,$X$ is:

Similar Questions

How many products will be formed in the reaction shown below?
$CH_3-CH_2-C(CH_3)=C(CH_3)-CH_2-CH_3 + HBr \xrightarrow{R_2O_2} \text{Products}$

Arrange $HI$,$HCl$,$HBr$,and $HF$ in increasing order of rate of reaction with alkene.

Identify $X, Y$ and $Z$ in the following reactions:

Reactant $(A)$ in this reaction is

$Ph-CH=C(CH_3)_2 \xleftarrow{HBr} A$ and $Ph-CH=C(CH_3)_2 \xrightarrow[\text{Peroxide}]{HBr} B$. Identify $A$ and $B$.

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The chemical reaction is shown below:$C_6H_5-CH_2-CH=CH_2 + HCl \rightarrow X$,$X$ is: