In the given conversion,the compound A is: | vedclass

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(B) The reaction of $2$-bromophenol with $2$ equivalents of an organolithium reagent (like $n-BuLi$ or $t-BuLi$) involves two steps: $1$. Acid-base re

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Dilithium salt of $2-$bromophenol

Vedclass · Answer

(B) The reaction of $2$-bromophenol with $2$ equivalents of an organolithium reagent (like $n-BuLi$ or $t-BuLi$) involves two steps: $1$. Acid-base reaction: The phenolic $-OH$ proton is acidic and is deprotonated by the first equivalent of the organolithium reagent to form a lithium phenoxide salt. $2$. Halogen-metal exchange: The second equivalent of the organolithium reagent reacts with the aryl bromide to form an aryllithium species. Thus,the intermediate $A$ is the dilithium salt of $2$-bromophenol,which has the structure of a benzene ring with an $-OLi$ group at the ortho position relative to a $-Li$ group. This species then reacts with $CO_2$ followed by acidic workup $(H_3O^+)$ to yield salicylic acid.

In the given conversion,the compound $A$ is:

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