Q.Explain the following:

Both statements are true, and dichlorocarbene formation is exactly the mechanistic reason salicylaldehyde forms.

Assertion: In the Reimer–Tiemann reaction, phenol is treated with $CHCl_3$ and $NaOH$ at about 340 K; the product (after hydrolysis) is predominantly the ortho-hydroxybenzaldehyde, i.e. salicylaldehyde. True.

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This is the classic Reimer-Tiemann reaction, in which phenol reacts with chloroform and KOH to introduce a -CHO group ortho to the -OH, giving salicylaldehyde.

$\text{C}_6\text{H}_5\text{OH} + \text{CHCl}_3 + \text{KOH} \xrightarrow{\Delta} \text{o-HOC}_6\text{H}_4\text{CHO (salicylaldehyde)} + \text{KCl} + \text{H}_2\text{O}$

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This is the Reimer–Tiemann reaction: phenol reacts with chloroform under strongly basic conditions to install a formyl ($-CHO$) group at the ortho position, giving salicylaldehyde after hydrolysis.

Mechanism outline

1. NaOH deprotonates $CHCl_3$ to form the trichloromethyl carbanion $:CCl_3^-$, which rapidly loses $Cl^-$ to generate the highly reactive electrophile dichlorocarbene ($:CCl_2$).
2. The phenoxide ion (from phenol + NaOH) attacks $:CCl_2$ preferentially at the ortho position (directed by the strongly activating $-O^-$ group), giving a dichloromethyl-substituted intermediate.
3. Hydrolysis of the $-CHCl_2$ group under the alkaline reaction conditions converts it to $-CHO$.

Overall:

$C_6H_5OH \xrightarrow[\text{NaOH}]{CHCl_3} \text{o-hydroxybenzaldehyde (salicylaldehyde)}$

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Dichlorocarbene, generated from CHCl3+NaOH, formylates phenol's activated ortho ring position, and hydrolysis of the resulting dichloromethyl intermediate reveals the aldehyde.

In the Reimer–Tiemann reaction, phenol is treated with chloroform and concentrated NaOH. Base generates dichlorocarbene (:CCl₂) from CHCl₃ (by α-elimination), which is attacked by the electron-rich phenoxide ring (ortho position preferred), forming a ortho-substituted dichloromethyl-phenol intermediate; hydrolysis of this gem-dihalide under the basic conditions (then acidification) re …

Phenol reacts with CHCl3/NaOH via a dichlorocarbene intermediate to install a -CHO group ortho to the -OH.

In the Reimer-Tiemann reaction, phenol is treated with chloroform (CHCl3) and concentrated aqueous sodium hydroxide, heated at around 340 K. NaOH first generates the electrophilic species dichlorocarbene (:CCl2) from CHCl3, which attacks the electron-rich phenoxide ring (mainly at the position ortho to -OH). After hydrolysis of the resulting intermediate, the final product is salicylaldehyde (2-hydroxyb …

Phenol + CHCl3 + NaOH (Reimer-Tiemann) introduces a -CHO group ortho to -OH, giving salicylaldehyde.

In the Reimer-Tiemann reaction, phenol is treated with chloroform (CHCl3) in the presence of aqueous NaOH, followed by acidification. Dichlorocarbene (:CCl2), generated in situ, attacks the ring mainly at the ortho position, and after hydrolysis a -CHO group is introduced:

Phenol --CHCl3 / NaOH, then H+--> salicylaldehyde (2-hydroxybenzaldehyde)

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Treating phenol with chloroform and aqueous $KOH$, then hydrolysing, introduces an $-CHO$ group ortho to $-OH$ — this is the Reimer–Tiemann reaction.

In the Reimer–Tiemann reaction, phenol is treated with chloroform in the presence of aqueous sodium/potassium hydroxide; the reagent generates dichlorocarbene ($:CCl_2$) in situ, which attacks the electron-rich phenoxide ring (predominantly at the ortho position), and subsequent hydrolysis converts the resulting dichloromethyl intermediate into an aldehyde group …