In this kind of questions when aldehyde or ketone is given in presence of base then
aldol condensation or cannizzaro reaction either of the one takes place but for aldol presence of sp3 hybridization on carbonyls alpha position containing alpha hydrogen is a necessary condition but in formaldehyde there is absence of alpha hydrogen so cannizzaro reaction takes place. In cannizzaro 2 moles of product In presence of base give alcohol and carboxylic acid that is one mole gives alcohol and another gives carboxylic acid as a product so in this case products are Formic acid and methane alcohol I.E. methanol so answer is option (c)

When formaldehyde (CH2O) and KOH (potassium hydroxide) are heated, the reaction leads to the formation of methyl alcohol (CH3OH). Let's understand the reaction and its mechanism in detail:

Reaction Overview:
CH2O + KOH → CH3OH + KCHO

Mechanism:
1. The reaction starts with the deprotonation of formaldehyde (CH2O) by KOH. The hydroxide ion (OH-) from KOH acts as a base and removes a proton from the carbon atom in formaldehyde. This results in the formation of a carbanion (CH2O-) and water (H2O).

CH2O + OH- → CH2O- + H2O

2. The carbanion (CH2O-) then undergoes a nucleophilic attack on another formaldehyde molecule. The lone pair of electrons on the carbon atom of the carbanion attacks the carbon atom in another formaldehyde molecule, resulting in the formation of a C-C bond.

CH2O- + CH2O → CH3OHCH2O

3. The intermediate formed in the previous step rearranges to form methanol (CH3OH) and potassium formate (KCHO). This rearrangement involves the migration of a hydrogen atom from one carbon atom to another, leading to the formation of a methyl alcohol group and a formyl group.

CH3OHCH2O → CH3OH + KCHO

Overall, the reaction between formaldehyde and KOH results in the formation of methyl alcohol (CH3OH) and potassium formate (KCHO). This reaction is known as the Cannizzaro reaction, which involves the self-disproportionation of aldehydes or ketones in the presence of a strong base like KOH.

It's important to note that the other options (acetylene, methane, and ethyl formate) are not formed in this particular reaction. The formation of methyl alcohol is the primary outcome when formaldehyde and KOH are heated together.