Reaction of Acetaldehyde and Formaldehyde Followed by Ketal Formation

To find the required sum, we analyze the reaction sequence step-by-step:

Step 1: Reaction of Acetaldehyde with excess Formaldehyde

When acetaldehyde ($\text{CH}_3\text{CHO}$) is treated with excess formaldehyde ($\text{HCHO}$) in the presence of concentrated $\text{NaOH}$ and heated:

1. Cross-Aldol Condensation: Acetaldehyde possesses 3 $\alpha$-hydrogens, so it undergoes cross-aldol condensation thrice with three molecules of formaldehyde to form an intermediate tri-hydroxy aldehyde: $\text{CH}_3\text{CHO} + 3\,\text{HCHO} \xrightarrow{\text{NaOH}} (\text{HOCH}_2)_3\text{C}-\text{CHO}$

2. Cross-Cannizzaro Reaction: Since the resulting aldehyde $(\text{HOCH}_2)_3\text{C}-\text{CHO}$ has no remaining $\alpha$-hydrogens, it undergoes a cross-Cannizzaro reaction with a fourth molecule of formaldehyde in the presence of concentrated $\text{NaOH}$: $(\text{HOCH}_2)_3\text{C}-\text{CHO} + \text{HCHO} + \text{NaOH} \rightarrow \underbrace{\text{C}(\text{CH}_2\text{OH})_4}_{\mathbf{P}} + \underbrace{\text{HCOONa}}_{\mathbf{Q}}$

• Compound $\mathbf{P}$ is Pentaerythritol [$\text{C}(\text{CH}_2\text{OH})_4$], which lacks an aldehyde group and therefore does not give Tollens' test.
• Compound $\mathbf{Q}$ is Sodium formate ($\text{HCOONa}$), which upon acidification yields formic acid ($\text{HCOOH}$). Formic acid contains a formyl group ($\text{H}-\text{C}(=\text{O})-\text{OH}$) and gives a positive Tollens' test.

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Step 2: Ketal Formation with Cyclohexanone

Treatment of pentaerythritol ($\mathbf{P}$) with excess cyclohexanone in the presence of a catalytic amount of $p$-toluenesulfonic acid ($\text{PTSA}$) leads to the condensation of its 1,3-diol units with cyclohexanone to form a bicyclic/spiro bis-ketal derivative, compound $\mathbf{R}$:

$\text{C}(\text{CH}_2\text{OH})_4 + 2\text{ Cyclohexanone} \xrightarrow{\text{PTSA}} \text{Product } \mathbf{R} + 2\,\text{H}_2\text{O}$

Molecular Structure of Product $\mathbf{R}$:

• Pentaerythritol unit: Consists of 1 central quaternary carbon and 4 methylene ($-\text{CH}_2-$) groups attached to oxygen atoms.
• Two Cyclohexyl rings: Each cyclohexanone ring contributes 1 ketal carbon (bonded to two oxygen atoms) and 5 methylene ($-\text{CH}_2-$) groups.

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Step 3: Counting Methylene Groups and Oxygen Atoms in $\mathbf{R}$

1. Number of Methylene ($-\text{CH}_2-$) groups:

   • From the pentaerythritol fragment: $4$
   • From the two cyclohexylidene rings: $2 \times 5 = 10$ $\text{Total number of } -\text{CH}_2- \text{ groups} = 4 + 10 = 14$

2. Number of Oxygen atoms:

   • From the two ketal rings (4 ether-like oxygens): $\text{Total number of oxygen atoms} = 4$

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Calculation:

$\text{Sum} = (\text{Number of } -\text{CH}_2- \text{ groups}) + (\text{Number of oxygen atoms}) = 14 + 4 = 18$

18