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(C) The reaction of $CH_3-CH_2-CHO$ with excess $HCHO$ in the presence of an alkali involves two steps: $1$. Multiple cross-aldol condensation: The $\

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$CH_3-C(CH_2OH)_2-CHO$

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(C) The reaction of $CH_3-CH_2-CHO$ with excess $HCHO$ in the presence of an alkali involves two steps: $1$. Multiple cross-aldol condensation: The $\alpha$-hydrogens of propanal are replaced by hydroxymethyl groups $(CH_2OH)$ through reaction with $HCHO$. Since propanal has two $\alpha$-hydrogens,both are replaced to form $CH_3-C(CH_2OH)_2-CHO$. $2$. Cross-Cannizzaro reaction: The remaining aldehyde group $(-CHO)$ is then reduced to a primary alcohol $(-CH_2OH)$ by the excess $HCHO$,which itself gets oxidized to formate. The final product is $CH_3-C(CH_2OH)_3$.

The major product of the following reaction is :
$CH_3-CH_2-CHO \xrightarrow[\text{reflux}]{\text{excess } HCHO / \text{alkali}} ?$

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The major product of the following reaction is :$CH_3-CH_2-CHO \xrightarrow[\text{reflux}]{\text{excess } HCHO / \text{alkali}} ?$