What would be the product of a reaction between methanal and pyrrolidine in acid?

I know that normally a reaction between a carbonyl and a secondary amine would form an enamine. Here I am confused because there are no hydrogens that can be removed to form the enamine as this is methanal. I'm guessing that the nitrogen will attack the C=O double bond and then water will be formed as a leaving group, the nitrogen lone pair will come down to form a double bond and push this off. I'm not sure where to go from here as there will be a positive charge on the N.

To determine the product of the reaction between methanal (formaldehyde) and pyrrolidine in an acidic medium, let's consider the proposed mechanism you mentioned where the nitrogen attacks the carbon-oxygen double bond.

1. The nitrogen lone pair attacks the carbon-oxygen double bond, forming a temporary intermediate known as a tetrahedral intermediate.

H₂C=O + H₃C-CHNH → H₂C-C(H₃)-CHNH⁺ (Tetrahedral intermediate)

2. The oxygen atom loses a proton (H⁺) to the acid and forms a bond with a hydrogen atom from the acid. This leads to the formation of a protonated alcohol intermediate.

H₂C-C(H₃)-CHNH⁺ → H₂C-C(H₃)-CHOH⁺ (Protonated alcohol intermediate)

3. In the presence of acid, a water molecule acts as a nucleophile and attacks the protonated alcohol intermediate, displacing the proton from the oxygen atom.

H₂C-C(H₃)-CHOH⁺ + H₂O → H₂C-C(H₃)-CHOH + H₃O⁺ (Alcohol intermediate)

4. Finally, the alcohol intermediate loses a proton to the acid, resulting in the formation of the final product.

H₂C-C(H₃)-CHOH → H₂C=C(H₃)-CH₂OH (Final product)

Therefore, the product of the reaction between methanal and pyrrolidine in an acidic medium is 2-methyl-2-propanol (also known as tert-butanol). The nitrogen atom of pyrrolidine attacks the methanal molecule, leading to the formation of the alcohol product.