Complete reaction of acetaldehyde with excess formaldehyde, upon heating with conc. NaOH solution, gives P and Q. Compound P does not give Tollens' test, whereas Q on acidification gives positive Tollens' test. Treatment of P with excess cyclohexanone in the presence of catalytic amount of p-toluenesulfonic acid (PTSA) gives product R.

Sum of the number of methylene groups ($$-CH_2-$$) and oxygen atoms in R is ______.

The reaction sequence is a standard industrial route to the poly-alcohol pentaerythritol and its ketal derivatives. We analyse every step.

Step 1 : Reaction of acetaldehyde with excess formaldehyde in conc. NaOH
• Formaldehyde has no $$\alpha$$-hydrogen, so in strong alkali it undergoes the Cannizzaro reaction: $$2\,HCHO + OH^- \;\longrightarrow\; HCOO^- + CH_3OH$$

• In the presence of acetaldehyde, a cross-Cannizzaro/Aldol sequence occurs.   1 mol acetaldehyde first gives an enolate which adds to two molecules of formaldehyde (cross-aldol).   The aldehydic group thus produced is then reduced by hydride transfer from a third molecule of formaldehyde (Cannizzaro step).
The overall stoichiometry is

$$CH_3CHO + 3\,HCHO + 3\,NaOH \; \xrightarrow{\;\;\text{heat}\;\;} C(CH_2OH)_4 \;+\; 3\,NaHCOO$$

• The poly-alcohol obtained is pentaerythritol, $$P = C(CH_2OH)_4$$.   It possesses no -CHO group, hence it does not respond to Tollens’ reagent.

• The salt $$Q = NaHCOO$$ on acidification furnishes formic acid, $$HCOOH$$, which does reduce Tollens’ reagent. Thus the statements regarding $$P$$ and $$Q$$ are consistent.

Step 2 : Ketal formation between $$P$$ and excess cyclohexanone in the presence of catalytic p-toluenesulfonic acid (PTSA)
Acid catalyses the reaction of a ketone with two alcohol functions to give a ketal. Each molecule of cyclohexanone requires two -OH groups:

$$\text{cyclohexanone}+2\,ROH \;\xrightarrow{\;\;H^+\;\;} \text{cyclohexanone ketal}\;(\;C=O \text{ is replaced by }C(OR)_2\;)$$

Pentaerythritol has four -OH groups, therefore it can furnish the necessary two alcohol units twice. Consequently, with excess cyclohexanone it forms a diketal:

$$R = \bigl[\;C(CH_2O{-})_4\bigr]$$ Each pair $$\;{-}OCH_2{-}$$ originates from one molecule of cyclohexanone, so $$R$$ contains two cyclohexanone rings attached through ketal linkages.

Counting -CH2- groups in $$R$$
• Central pentaerythritol carbon: four $$CH_2$$ groups → $$4$$
• Each cyclohexanone ring (after ketal formation) supplies five $$CH_2$$ groups   (the carbonyl carbon becomes quaternary, carries no hydrogens).
  Two rings ⇒ $$2 \times 5 = 10$$
Total methylene groups $$= 4 + 10 = 14$$

Counting oxygen atoms in $$R$$
Every ketal linkage contains the two original alcohol oxygens; thus the four -OH groups of pentaerythritol become four ether oxygens. (The carbonyl oxygens of the ketones are lost as water during ketalisation).
Total oxygen atoms $$= 4$$

Required sum
$$\text{(number of }{-}CH_2{-}\text{ groups)} + \text{(number of O atoms)} = 14 + 4 = 18$$

Therefore, the asked sum is 18.