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(B) The reaction of phenol with chloroform $(CHCl_3)$ in the presence of aqueous $NaOH$ is the Reimer-Tiemann reaction,which yields salicylaldehyde ($

Vedclass · Accepted Answer

$68.85$

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(B) The reaction of phenol with chloroform $(CHCl_3)$ in the presence of aqueous $NaOH$ is the Reimer-Tiemann reaction,which yields salicylaldehyde ($o$-hydroxybenzaldehyde) as the major product $P$.The molecular formula of salicylaldehyde $(P)$ is $C_7H_6O_2$.The molar mass of $C_7H_6O_2 = (7 	imes 12) + (6 	imes 1) + (2 	imes 16) = 84 + 6 + 32 = 122 \ g/mol$.The mass percentage of carbon in $P$ is given by:$	ext{Mass } \% C = \frac{	ext{Total mass of Carbon}}{	ext{Molar mass of } P} 	imes 100$$	ext{Mass } \% C = \frac{7 	imes 12}{122} 	imes 100 = \frac{84}{122} 	imes 100 \approx 68.85\%$.Rounding to the nearest integer,we get $69\%$.

$A$ solution of phenol in chloroform when treated with aqueous $NaOH$ gives compound $P$ as a major product. The mass percentage of carbon in $P$ is $..............$ (to the nearest integer)
(Atomic mass : $C = 12$; $H = 1$; $O = 16$)

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$A$ solution of phenol in chloroform when treated with aqueous $NaOH$ gives compound $P$ as a major product. The mass percentage of carbon in $P$ is $..............$ (to the nearest integer)(Atomic mass : $C = 12$; $H = 1$; $O = 16$)