MHTCET 11 May 2024 Shift 1 Paper

Concept:The Reimer-Tiemann reaction introduces a formyl group at the ortho position of a phenol using chloroform and a base.Explanation:The reagent used is chloroform ($\mathrm{CHCl}_3$) in the presence of aqueous sodium hydroxide ($\mathrm{NaOH}$), followed by acid hydrolysis with $\mathrm{H}_3\mathrm{O}^{+}$.Under basic conditions, $\mathrm{CHCl}_3$ undergoes deprotonation and loses $\mathrm{HCl}$ to form dichlorocarbene ($\mathrm{CCl}_2$), a highly reactive intermediate.This electrophilic carbene then attacks the electron-rich ortho position of the phenoxide ion.The resulting dichloromethyl intermediate is hydrolyzed by $\mathrm{H}_3\mathrm{O}^{+}$ to yield the aldehyde group, forming ortho-hydroxybenzaldehyde as the major product.Therefore, the correct reagent combination for the Reimer-Tiemann reaction is $\mathrm{CHCl}_3$, aq. $\mathrm{NaOH}$, and $\mathrm{H}_3\mathrm{O}^{+}$.Answer:Option B: $\mathrm{CHCl}_3$, aq. $\mathrm{NaOH}$, $\mathrm{H}_3\mathrm{O}^{+}$