1. Identify the reaction between acetaldehyde and excess formaldehyde in concentrated NaOH

This is a crossed aldol / hydroxymethylation followed by Cannizzaro-type behavior typical of acetaldehyde with excess formaldehyde in strong base.

Acetaldehyde has \alpha-hydrogens, so it undergoes successive hydroxymethylation with formaldehyde.

The well-known overall product pair is:

• P = pentaerythritol, $\mathrm{C(CH_2OH)_4}$
• Q = sodium formate, $\mathrm{HCOONa}$

On acidification, $\mathrm{HCOONa}$ gives formic acid: $\mathrm{HCOONa + H^+ \to HCOOH + Na^+}$ Formic acid gives positive Tollens' test, so this matches the statement about $Q$.

Also, pentaerythritol is a polyhydric alcohol and does not give Tollens' test, so this matches $P$.

2. Write the structure of $P$

Pentaerythritol: $\mathrm{P = C(CH_2OH)_4}$

It has four hydroxymethyl groups attached to a central carbon.

3. Reaction of $P$ with excess cyclohexanone in presence of catalytic PTSA

A ketone reacts with alcohol groups under acid catalysis to form ketals.

Each pair of neighboring $-\mathrm{CH_2OH}$ groups of pentaerythritol can react with one molecule of cyclohexanone to form a spiro ketal ring.

Since pentaerythritol has four $-\mathrm{CH_2OH}$ groups, it can use them in two pairs, so with excess cyclohexanone it forms the di-ketal.

Thus $R$ is the bis(spiroketal) formed from:

• 1 molecule of pentaerythritol
• 2 molecules of cyclohexanone
• elimination of 2 molecules of water

Each cyclohexanone contributes one oxygen atom into the ketal; the original alcohol oxygens are retained in the ketal linkage. So total oxygen atoms in $R$ are: $4$

(Equivalently: pentaerythritol starts with 4 oxygens, and ketal formation removes water but does not change total oxygen count in the organic product from those alcohol oxygens.)

4. Count the number of methylene groups in $R$

From pentaerythritol part

Pentaerythritol has four $-\mathrm{CH_2}-$ groups: $4\text{ methylene groups}$

From cyclohexanone part

One cyclohexanone ring has structure with:

• one carbonyl carbon
• five $\mathrm{CH_2}$ groups

After ketal formation, the carbonyl carbon becomes acetal carbon, but the ring still contains five methylene groups.

Since 2 cyclohexanone molecules are used: $2 \times 5 = 10\text{ methylene groups}$

Total methylene groups in $R$

$4 + 10 = 14$

5. Add methylene groups and oxygen atoms

• Number of $-\mathrm{CH_2}-$ groups in $R$ = $14$
• Number of oxygen atoms in $R$ = $4$

Therefore, $14 + 4 = 18$

6. Final answer

$\boxed{18}$

The derived answer matches the stored correct answer.