In the Reimer-Tiemann formylation reaction,by | vedclass

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(A) In the Reimer-Tiemann reaction,phenol $(C_6H_5OH)$ reacts with chloroform $(CHCl_3)$ in the presence of aqueous $NaOH$ to form salicylaldehyde $(C

Vedclass · Accepted Answer

$28$

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(A) In the Reimer-Tiemann reaction,phenol $(C_6H_5OH)$ reacts with chloroform $(CHCl_3)$ in the presence of aqueous $NaOH$ to form salicylaldehyde $(C_6H_4(OH)CHO)$.The chemical equation is: $C_6H_5OH + CHCl_3 + 3NaOH ightarrow C_6H_4(OH)CHO + 3NaCl + 2H_2O$.The molecular weight of phenol $(C_6H_5OH)$ is $6 	imes 12 + 6 	imes 1 + 16 = 94 \ g/mol$.The molecular weight of salicylaldehyde $(C_6H_4(OH)CHO)$ is $7 	imes 12 + 6 	imes 1 + 2 	imes 16 = 84 + 6 + 32 = 122 \ g/mol$.The increase in molecular weight is $122 - 94 = 28$.This corresponds to the replacement of one $H$ atom (mass $1$) with one $CHO$ group (mass $29$),resulting in a net increase of $29 - 1 = 28$.

In the Reimer-Tiemann formylation reaction,by how much does the molecular weight of phenol increase?

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