A halogen compound X (C_4 H_9 Br) on hydrolys | vedclass

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Question

(A) The reaction sequence is as follows:$1$. Hydrolysis of $C_4 H_9 Br$ $(X)$ gives an alcohol $Y$ $(C_4 H_9 OH)$.$2$. The alcohol $Y$ undergoes dehyd

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$(CH_3)_3 CBr$

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(A) The reaction sequence is as follows:$1$. Hydrolysis of $C_4 H_9 Br$ $(X)$ gives an alcohol $Y$ $(C_4 H_9 OH)$.$2$. The alcohol $Y$ undergoes dehydration with $20 \% H_3 PO_4$ at $358 \ K$ to form an alkene.$3$. Tertiary butyl bromide,$(CH_3)_3 CBr$,on hydrolysis gives tertiary butyl alcohol,$(CH_3)_3 COH$.$4$. Tertiary butyl alcohol undergoes easy dehydration with $20 \% H_3 PO_4$ at $358 \ K$ to form isobutylene ($2$-methylpropene),which is consistent with the provided reaction conditions.$5$. Therefore,$X$ is $(CH_3)_3 CBr$.

$A$ halogen compound $X$ $(C_4 H_9 Br)$ on hydrolysis gives an alcohol $Y$. The alcohol $Y$ undergoes dehydration with $20 \% H_3 PO_4$ at $358 \ K$. What is $X$?

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Consider the following anions:
$(I)$ $CF_3SO_3^-$
$(II)$ $C_6H_5SO_3^-$
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$(i) \, (CH_3)_2CH-CH_2Br \xrightarrow{C_2H_5OH} (CH_3)_2CH-CH_2OC_2H_5 + HBr$
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The reaction mechanisms for $(i)$ and $(ii)$ are respectively:

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