1. Identify the substrate and reagent

The given aldehyde is propanal: $\mathrm{CH_3CH_2CHO}$

It is treated with excess formaldehyde in alkali under reflux.

2. Recall the reaction type

An aldehyde having at least one $\alpha$-hydrogen undergoes crossed aldol reaction with formaldehyde in base.

Since formaldehyde has no $\alpha$-hydrogen, it acts only as the electrophile. Propanal forms an enolate and adds to formaldehyde.

Repeated hydroxymethylation can occur at the $\alpha$-carbon as long as $\alpha$-hydrogens are available.

For propanal, the $\alpha$-carbon is the carbon next to $-CHO$: $\mathrm{CH_3-CH_2-CHO}$ It has 2 $\alpha$-hydrogens.

So, with excess HCHO, both $\alpha$-hydrogens are replaced by $\mathrm{-CH_2OH}$ groups: $\mathrm{CH_3-CH(CH_2OH)_2-CHO}$

3. What happens under strongly basic conditions with excess formaldehyde?

Now the product still contains an aldehyde group, but it has no $\alpha$-hydrogen left at the carbon adjacent to the aldehyde carbonyl (that carbon is now bonded to $\mathrm{CH_3}$, two $\mathrm{CH_2OH}$ groups, and $\mathrm{CHO}$).

Such aldehydes undergo Cannizzaro reaction in presence of formaldehyde/alkali.

In crossed Cannizzaro with formaldehyde:

• formaldehyde is oxidized to formate,
• the other aldehyde is reduced to alcohol.

Thus, $\mathrm{CH_3-CH(CH_2OH)_2-CHO} \longrightarrow \mathrm{CH_3-CH(CH_2OH)_2-CH_2OH}$

4. Final major product

Therefore the major product is: $\boxed{\mathrm{CH_3-CH(CH_2OH)-CH_2OH\; with\ another\ }CH_2OH\text{ on the same central carbon}}$ More clearly: $\boxed{\mathrm{CH_3-C(CH_2OH)_2-CH_2OH}}$

This is a triol formed by double hydroxymethylation followed by crossed Cannizzaro reduction.

5. Option matching

• Option B is $\mathrm{CH_3CH_2CH_2OH}$, which is just 1-propanol, so it is not correct.
• The correct structure is the triol $\mathrm{CH_3-C(CH_2OH)_2-CH_2OH}$.

Hence the correct option is C.

6. Comparison with stored answer

Stored correct answer: C

My derived answer: C

So they agree.