If we use pyrene (CCl_4) in the Riemer-Tieman | vedclass

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(C) When phenol reacts with $CCl_4$ in the presence of aqueous $NaOH$,the reaction is a variation of the Riemer-Tiemann reaction. Instead of forming a

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Salicylic acid

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(C) When phenol reacts with $CCl_4$ in the presence of aqueous $NaOH$,the reaction is a variation of the Riemer-Tiemann reaction. Instead of forming an aldehyde group (as with $CHCl_3$),the use of $CCl_4$ results in the introduction of a carboxylic acid group $(-COOH)$ at the ortho position of the phenol. The reaction is: $C_6H_5OH + CCl_4 + 4NaOH \rightarrow C_6H_4(OH)(COOH) + 4NaCl + 2H_2O$. The product formed is Salicylic acid.

If we use pyrene $(CCl_4)$ in the Riemer-Tiemann reaction in place of chloroform,the product formed is

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