Phenol can be converted into salicylic acid b | vedclass

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(D) The conversion of phenol to salicylic acid can be achieved through several methods:$1$. Kolbe-Schmitt reaction: Phenol reacts with $CO_2$ under pr

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All of the above

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(D) The conversion of phenol to salicylic acid can be achieved through several methods:$1$. Kolbe-Schmitt reaction: Phenol reacts with $CO_2$ under pressure in the presence of an alkali $(NaOH)$ to form sodium salicylate,which upon acidification yields salicylic acid.$2$. Reimer-Tiemann reaction with $CCl_4$: Phenol reacts with $CCl_4$ in the presence of an alkali $(NaOH)$ to form salicylic acid directly.$3$. Reimer-Tiemann reaction with $CHCl_3$: Phenol reacts with $CHCl_3$ and alkali to form salicylaldehyde,which can then be oxidized to salicylic acid.Since all these methods can lead to the formation of salicylic acid,the correct answer is $D$.

Column-$I$ (Reaction)	Column-$II$ (Electrophile)
$A$. $C_6H_6 + CH_3Cl \xrightarrow{AlCl_3} C_6H_5CH_3$	$P$. $CO_2$
$B$. $C_6H_5OH + CO_2 \xrightarrow[H^+]{NaOH} \text{Salicylic acid}$	$Q$. $:CCl_2$
$C$. $C_6H_5OH + CHCl_3 \xrightarrow{KOH} \text{Salicylaldehyde}$	$R$. $\stackrel{+}{C}H_3$
$D$. $C_6H_6 + SO_3 \xrightarrow{H_2SO_4} C_6H_5SO_3H$	$S$. $\stackrel{+}{S}O_3H$
	$T$. $SO_3$

Column-$I$ (Reaction)	Column-$II$ (Bond broken)
$(A)$ $RCH_2OH \xrightarrow{CrO_3} RCHO$	$(i)$ $C-O$ bond breaks.
$(B)$ Phenol $\xrightarrow{HNO_3}$ $p$-Nitrophenol + $o$-Nitrophenol	(ii) $O-H$ bond breaks.
$(C)$ Phenol + $NaOH \rightarrow$ Sodium phenoxide + $H_2O$	(iii) $O-H$ and $C-H$ bonds break.
$(D)$ $CH_3CH_2CH_2OH \xrightarrow{H_2SO_4, \Delta} CH_3CH=CH_2$	(iv) $C-H$ bond breaks.

Phenol can be converted into salicylic acid by heating with

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