Observe the following reactions:1. C_6H_5-CH= | vedclass

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(C) The products formed are:$X$: $C_6H_5-CH(Br)-CH_3$ ($1$-phenylethyl bromide,secondary benzylic halide)$Y$: $C_6H_5-C(Br)(CH_3)-CH_2-CH_3$ (Wait,the

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$Y > X > Z$

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(C) The products formed are:$X$: $C_6H_5-CH(Br)-CH_3$ ($1$-phenylethyl bromide,secondary benzylic halide)$Y$: $C_6H_5-C(Br)(CH_3)-CH_2-CH_3$ (Wait,the structure is $C_6H_5-C(CH_3)=CH_2$,so $Y$ is $C_6H_5-C(Br)(CH_3)-CH_3$,a tertiary benzylic halide)$Z$: $C_6H_5-CH_2-CH_2-Br$ ($2$-phenylethyl bromide,primary alkyl halide)Reactivity towards $S_N1$ depends on the stability of the carbocation intermediate formed.For $Y$,the carbocation is $C_6H_5-C^+(CH_3)_2$,which is tertiary and benzylic (highly stable).For $X$,the carbocation is $C_6H_5-CH^+-CH_3$,which is secondary and benzylic (stable).For $Z$,the carbocation is $C_6H_5-CH_2-CH_2^+$,which is primary (least stable).Therefore,the order of reactivity is $Y > X > Z$.

Observe the following reactions:
$1$. $C_6H_5-CH=CH_2 + HBr \rightarrow X$
$2$. $C_6H_5-C(CH_3)=CH_2 + HBr \rightarrow Y$
$3$. $C_6H_5-CH=CH_2 + HBr \xrightarrow{(C_6H_5COO)_2} Z$
The correct order of reactivity of $X, Y, Z$ towards $S_N1$ reaction is:

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Observe the following reactions:$1$. $C_6H_5-CH=CH_2 + HBr \rightarrow X$$2$. $C_6H_5-C(CH_3)=CH_2 + HBr \rightarrow Y$$3$. $C_6H_5-CH=CH_2 + HBr \xrightarrow{(C_6H_5COO)_2} Z$The correct order of reactivity of $X, Y, Z$ towards $S_N1$ reaction is: