Reimer Tiemann Reaction of Para Cresol Products

To determine the product(s) formed in the Reimer-Tiemann reaction of $p$-cresol ($4$-methylphenol) with chloroform ($\text{CHCl}_3$) in the presence of a strong base ($\text{OH}^-$), we analyze the step-by-step reaction mechanism:

1. Generation of Electrophile

Chloroform reacts with hydroxide base via deprotonation followed by loss of a chloride ion to generate dichlorocarbene ($:\text{CCl}_2$), an electron-deficient, neutral electrophile: $\text{CHCl}_3 + \text{OH}^- \rightleftharpoons :\text{CCl}_3^- + \text{H}_2\text{O}$ $:\text{CCl}_3^- \xrightarrow{-\text{Cl}^-} :\text{CCl}_2 \quad (\text{dichlorocarbene})$

Simultaneously, $p$-cresol is deprotonated by $\text{OH}^-$ to form the more reactive nucleophile, $p$-methylphenoxide ion.

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2. Attack of Electrophile

Case I: Ortho Attack (Formation of Major Product $S$)

The $p$-methylphenoxide ion undergoes electrophilic aromatic substitution at the unoccupied ortho-position:

1. The phenoxide ion attacks $:\text{CCl}_2$ at the ortho-position to give a cyclohexadienyl anion intermediate.
2. Proton shift restores aromaticity, yielding the $o$-dichloromethyl substituted intermediate.
3. Alkaline hydrolysis of the $-\text{CHCl}_2$ group converts it into a formyl group ($-\text{CHO}$): $-\text{CHCl}_2 \xrightarrow{\text{OH}^-} -\text{CH(OH)}_2 \xrightarrow{-\text{H}_2\text{O}} -\text{CHO}$

This leads to 2-hydroxy-5-methylbenzaldehyde ($S$). Because aromaticity is fully restored during this path, $S$ is the major product.

Case II: Para Attack / Ipso Attack (Formation of Minor Product $Q$)

The phenoxide ion can also attack the electrophile $:\text{CCl}_2$ at the para-position (ipso to the methyl group):

1. The electrophilic attack forms a intermediate where the para-carbon carries both the methyl ($-\text{CH}_3$) group and the $-\text{CHCl}_2^-$ group.
2. Proton abstraction from this para-position is not possible because the para-carbon is quaternary (it lacks a hydrogen atom).
3. As a result, the intermediate cannot undergo tautomerization to restore aromaticity.
4. Protonation of the negative charge on dichloromethyl carbon gives the stable non-aromatic dienone structure, 4-methyl-4-(dichloromethyl)cyclohexa-2,5-dien-1-one ($Q$).

Because loss of aromatic stability makes this pathway less favorable, $Q$ is formed as a minor (abnormal Reimer-Tiemann) product.

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Conclusion

• $S$ is formed as the major product (normal Reimer-Tiemann product).
• $Q$ is formed as the minor product (abnormal Reimer-Tiemann product).

Therefore, the correct choices are (B) and (D).