Chemical Reaction

The reaction described is an example of the Cannizzaro reaction, which occurs when an aldehyde (lacking an α-hydrogen) is treated with a strong base (like concentrated sodium hydroxide). The reaction leads to disproportionation, where one molecule is reduced to an alcohol, and the other is oxidized to a carboxylate anion. In this case: - p-Methoxy benzaldehyde undergoes the Cannizzaro reaction. - Methanal (formaldehyde) also undergoes the same process. The main products: - The reduction of one aldehyde molecule (in this case, p-Methoxy benzaldehyde) leads to the formation of p-Methoxy benzyl alcohol. - The other aldehyde (formaldehyde) is oxidized to form sodium formate. Therefore, the correct products are p-Methoxy benzyl alcohol and sodium formate.

The reaction described in the question is the Reimer-Tiemann reaction, which is a reaction between aromatic compounds and chloroform in the presence of an alcoholic KOH solution. This reaction leads to the formation of a substituted phenol by introducing a formyl group (-CHO) in the aromatic ring. For this reaction to proceed, the amino group (-NH ) must be in the para position relative to the methyl group in order to get a positive result. In the given options, p-Methylbenzylamine (option C) will provide the required positioning of the amino group to give a positive result when treated with chloroform and alcoholic KOH, leading to a successful Reimer-Tiemann reaction.