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

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Question

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

Vedclass · Accepted Answer

No reaction

Vedclass · Answer

(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.

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:

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