When phenol reacts with CHCL3 and KoH , the product obtained would be salicylaldehyde, p-hydroxy benzaldehyde , Both a and b, chloretone.
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When phenol treated with chloroform in the presence of aqueous alkali base at 340 K temperature followed by hydrolysis yields 2-hydroxybenzaldehyde also called as salicylaldehyde and reaction called as Reimer-Tiemann reaction.
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Answer: It will form Salicylalydehde.
Explanation:
It is an example of Riemer-Tiemer Reaction. First, base will help to form phenoxide. Also, the base will remove Hydrogen and one chlorine atom from CHCl3 by taking H+ , will finally leads to leaving of Cl-, whivh forms CCl2(carbene). Carbene attaches to Phenoxide on Ortho position. Then the intermediate formed does SN2 with the base. The new intermediate reacts with Hydronium to form Salicylalydehde.
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