Select the correct statement.

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(D) Option $(A)$ is correct because the carbocation formed from $(CH_3)_2C=CH-CH_2-Cl$ is resonance-stabilized,making it much more reactive in solvoly

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All of these

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(D) Option $(A)$ is correct because the carbocation formed from $(CH_3)_2C=CH-CH_2-Cl$ is resonance-stabilized,making it much more reactive in solvolysis compared to a simple alkyl chloride.Option $(B)$ is correct because the reaction of $CH_3-CH=CH-CH_2-OH$ with $HBr$ involves the formation of an allylic carbocation intermediate,which can be attacked by $Br^-$ at two different positions,leading to a mixture of $1$-bromo-$2$-butene and $3$-bromo-$1$-butene.Option $(C)$ is correct because $3$-buten-$2$-ol in aqueous acid undergoes protonation and loss of water to form an allylic carbocation,which can be attacked by water at either the $C-1$ or $C-3$ position,leading to an equilibrium mixture of $3$-buten-$2$-ol and $2$-buten-$1$-ol.Since all statements are correct,the correct answer is $(D)$.

List-$I$	List-$II$
$A$. Phenol $\rightarrow$ Salicylaldehyde	$I$. $Br_2$ in $CS_2$
$B$. Phenol $\rightarrow$ Benzene	$II$. $Na_2Cr_2O_7/H_2SO_4$
$C$. Phenol $\rightarrow$ $p$-Benzoquinone	$III$. $Zn$
$D$. Phenol $\rightarrow$ $p$-Bromophenol	$IV$. $CHCl_3/NaOH$

Column-$I$ (Reaction)	Column-$II$ (Bond broken)
$(A)$ $RCH_2OH \xrightarrow{CrO_3} RCHO$	$(i)$ $C-O$ bond breaks.
$(B)$ Phenol $\xrightarrow{HNO_3}$ $p$-Nitrophenol + $o$-Nitrophenol	(ii) $O-H$ bond breaks.
$(C)$ Phenol + $NaOH \rightarrow$ Sodium phenoxide + $H_2O$	(iii) $O-H$ and $C-H$ bonds break.
$(D)$ $CH_3CH_2CH_2OH \xrightarrow{H_2SO_4, \Delta} CH_3CH=CH_2$	(iv) $C-H$ bond breaks.

Column-$I$	Column-$II$
$(i)$ $2-$methylphenol	$(a)$ Hydroquinone
$(ii)$ Benzene$-1,2-$diol	$(b)$ Phenetole
$(iii)$ Benzene$-1,3-$diol	$(c)$ Catechol
$(iv)$ Benzene$-1,4-$diol	$(d)$ $o$-Cresol
$(v)$ Methoxybenzene	$(e)$ Quinone
$(vi)$ Ethoxybenzene	$(f)$ Resorcinol
	$(g)$ Anisole

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