Count of Aldol Reactions in Polyhydroxy Synthesis

To determine the number of aldol addition reactions occurring in the given chemical transformation, let us analyze the reaction mechanism step-by-step:

1. Identification of Reactants

• Acetaldehyde ($\text{CH}_3\text{CHO}$): Contains three acidic $\alpha$-hydrogens.
• Formaldehyde ($\text{HCHO}$): Contains no $\alpha$-hydrogens and acts exclusively as an electrophilic carbonyl compound.

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2. Step-by-Step Reaction Mechanism

Step 1: First Aldol Addition

In the presence of the base ($\text{NaOH}$), an $\alpha$-hydrogen of acetaldehyde is abstracted to form an enolate ion, which undergoes an aldol addition with the first molecule of formaldehyde ($\text{HCHO}$): $\text{CH}_3\text{CHO} + \text{HCHO} \xrightarrow{\text{OH}^-} \text{HO}-\text{CH}_2-\text{CH}_2\text{CHO}$

Step 2: Second Aldol Addition

The product, $3$-hydroxypropanal, still contains two $\alpha$-hydrogens. The base abstracts a second $\alpha$-hydrogen to form an enolate ion, which attacks a second molecule of formaldehyde ($\text{HCHO}$): $\text{HO}-\text{CH}_2-\text{CH}_2\text{CHO} + \text{HCHO} \xrightarrow{\text{OH}^-} (\text{HO}-\text{CH}_2)_2\text{CHCHO}$

Step 3: Third Aldol Addition

The product, $3$-hydroxy-$2$-(hydroxymethyl)propanal, still contains one remaining $\alpha$-hydrogen. A third abstraction occurs, and the enolate attacks a third molecule of formaldehyde ($\text{HCHO}$): $(\text{HO}-\text{CH}_2)_2\text{CHCHO} + \text{HCHO} \xrightarrow{\text{OH}^-} (\text{HO}-\text{CH}_2)_3\text{CCHO}$

At this point, all three $\alpha$-hydrogens of the original acetaldehyde molecule have been replaced by hydroxymethyl ($-\text{CH}_2\text{OH}$) groups via $3$ sequential Aldol addition reactions.

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Step 4: Cross-Cannizzaro Reaction

The intermediate $(\text{HO}-\text{CH}_2)_3\text{CCHO}$ (tris(hydroxymethyl)acetaldehyde) has no remaining $\alpha$-hydrogens. Therefore, it undergoes a Crossed Cannizzaro reaction with the fourth molecule of formaldehyde ($\text{HCHO}$) in the presence of concentrated aqueous $\text{NaOH}$: $(\text{HO}-\text{CH}_2)_3\text{CCHO} + \text{HCHO} + \text{NaOH} \longrightarrow \underset{\text{Pentaerythritol}}{\text{C}(\text{CH}_2\text{OH})_4} + \text{HCOONa}$

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Conclusion

• Number of Aldol addition reactions: $3$
• Number of Cannizzaro reactions: $1$

Thus, the total number of aldol reactions that take place is 3.

Correct Option: C