The reaction of 50% aq KOH on an equimolar mixture of 4-methylbenzaldehyde and formaldehyde followed by acidification gives.

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Answer

To solve the problem of the reaction of 50% aqueous KOH on an equimolar mixture of 4-methylbenzaldehyde and formaldehyde followed by acidification, we will follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- 4-methylbenzaldehyde (C6H5CHO with a methyl group at the para position)
- Formaldehyde (HCHO)

### Step 2: Understand the Reaction Conditions
The reaction takes place in the presence of 50% aqueous KOH. This indicates that the reaction is likely to involve a condensation reaction, specifically an aldol condensation, where aldehydes can react in the presence of a base.

### Step 3: Reaction of 4-Methylbenzaldehyde with Formaldehyde
In the presence of KOH, formaldehyde (HCHO) can undergo a reaction to form a hydroxymethyl group. The 4-methylbenzaldehyde can also participate in the reaction. The aldehyde groups will react to form a β-hydroxy aldehyde.

1. **Formaldehyde (HCHO)** can be converted to a hydroxymethyl group (–CH2OH).
2. **4-Methylbenzaldehyde** will remain as it is but will also participate in the aldol reaction.

### Step 4: Formation of the Aldol Product
The aldol product formed from the reaction of 4-methylbenzaldehyde and formaldehyde will be a β-hydroxy aldehyde. The structure will look like this:

- The product will have a hydroxymethyl group attached to the benzaldehyde structure.

### Step 5: Acidification
After the aldol product is formed, the mixture is acidified. Acidification will lead to dehydration of the β-hydroxy aldehyde, resulting in the formation of an α,β-unsaturated aldehyde.

### Step 6: Identify the Final Products
The final products after acidification will be:
- 4-methylcinnamaldehyde (the dehydrated product of the aldol condensation)
- Formic acid (HCOOH) from the oxidation of formaldehyde.

### Final Answer
The products formed from the reaction of 50% aqueous KOH on an equimolar mixture of 4-methylbenzaldehyde and formaldehyde followed by acidification are:
- 4-methylcinnamaldehyde
- Formic acid (HCOOH)