Consider the reaction of 1-chloro-2-hydroxy-4 | vedclass

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(D) The starting material is $1-chloro-2-hydroxy-4-nitrobenzene$. Treatment with $OH^-$ (base) deprotonates the phenolic $-OH$ group to form a phenoxi

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No reaction

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(D) The starting material is $1-chloro-2-hydroxy-4-nitrobenzene$. Treatment with $OH^-$ (base) deprotonates the phenolic $-OH$ group to form a phenoxide ion. However,the reaction conditions provided $(OH^-, \Delta)$ typically refer to Nucleophilic Aromatic Substitution $(S_NAr)$. In $1-chloro-2-hydroxy-4-nitrobenzene$,the $-OH$ group is ortho to the $-Cl$ group and para to the $-NO_2$ group. Under basic conditions,the phenoxide ion is formed. The electron-withdrawing $-NO_2$ group activates the ring for nucleophilic substitution. However,the $-OH$ group is already present. The reaction with $OH^-$ on a chlorobenzene derivative with an ortho-hydroxyl group does not lead to substitution of the $-Cl$ because the phenoxide is a poor leaving group and the $-OH$ is already present. Actually,this specific substrate is stable under these conditions,or if substitution were to occur,it would be difficult due to the ortho-phenoxide group. Given the options,the reaction does not proceed to replace the $-Cl$ with $-OH$ because the $-OH$ is already present and the phenoxide is electron-donating. Thus,no reaction occurs.

List-$I$ Conversion	List-$II$ Reagents,Conditions used
$A$. Chlorobenzene $\rightarrow$ Phenol	$I$. Warm,$H_2O$
$B$. $1$-Chloro-$4$-nitrobenzene $\rightarrow$ $4$-Nitrophenol	$II$. $(a)$ $NaOH, 368 \ K$; $(b)$ $H_3O^+$
$C$. $1$-Chloro-$2,4$-dinitrobenzene $\rightarrow$ $2,4$-Dinitrophenol	$III$. $(a)$ $NaOH, 443 \ K$; $(b)$ $H_3O^+$
$D$. $1$-Chloro-$2,4,6$-trinitrobenzene $\rightarrow$ $2,4,6$-Trinitrophenol	$IV$. $(a)$ $NaOH, 623 \ K, 300 \ atm$; $(b)$ $H_3O^+$

Consider the reaction of $1-chloro-2-hydroxy-4-nitrobenzene$ with $OH^-$ followed by $H^+$. Based on the above reaction,the major product $P$ would be:

Similar Questions

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In the following reaction,the major product is:

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