In the reaction sequence $$2CH_3CHO \xrightarrow{OH^-} A \xrightarrow{\Delta} B$$; the product B is:

We start with two molecules of acetaldehyde, whose condensed formula is $$CH_3CHO$$. In the presence of a dilute hydroxide ion, an enolate ion is first generated from one of the aldehyde molecules.

The general enolate formation in basic medium is stated as:

$$RCH_2CHO + OH^- \rightarrow RCHCHO^- + H_2O$$

For acetaldehyde specifically we have:

$$CH_3CHO + OH^- \rightarrow CH_2CHO^- + H_2O$$

This enolate now acts as a nucleophile and attacks the carbonyl carbon of a second, non-enolised acetaldehyde molecule. Writing that nucleophilic addition step explicitly:

$$CH_2CHO^- + CH_3CHO \rightarrow CH_3CH(OH)CH_2CHO$$

The product obtained just after this step is a β-hydroxy aldehyde. It is commonly called an “aldol”; here the systematic name is 3-hydroxybutanal. We label this compound as $$A$$:

$$A = CH_3CH(OH)CH_2CHO$$

Now the question tells us that on heating (symbolised by $$\Delta$$) compound $$A$$ transforms into compound $$B$$. Aldol products eliminate water on heating to give an α,β-unsaturated aldehyde or ketone. The general dehydration step of an aldol is:

$$\text{β-hydroxy carbonyl} \; \xrightarrow{\Delta} \; \text{α,β-unsaturated carbonyl} + H_2O$$

Applying this to 3-hydroxybutanal, we remove the elements of water from the -OH group at C-3 and a hydrogen at C-2, generating a double bond between C-2 and C-3. Writing the dehydration step in detail:

$$CH_3CH(OH)CH_2CHO \; \xrightarrow{\Delta} \; CH_3CH = CHCHO + H_2O$$

The product $$B$$ obtained is therefore crotonaldehyde, whose condensed structural formula is

$$B = CH_3 - CH = CH - CHO$$

Looking at the options, this corresponds exactly to Option A.

Hence, the correct answer is Option A.