C omplete 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 begins with the complete reaction of acetaldehyde (${\mathrm{CH}}_3\mathrm{CHO}$) with excess formaldehyde ($\mathrm{HCHO}$) in the presence of a concentrated base ($\mathrm{NaOH}$).

Step 1: Aldol Condensation
Acetaldehyde has three $\alpha$-hydrogen atoms. In the presence of base, it undergoes successive aldol condensation steps with three molecules of formaldehyde.
${\mathrm{CH}}_3\mathrm{CHO}+3\mathrm{HCHO}\to C{\left({\mathrm{CH}}_2\mathrm{OH}\right)}_3\mathrm{CHO}$

Step 2: Cross-Cannizzaro Reaction
The intermediate tri-hydroxymethyl acetaldehyde, $C{\left({\mathrm{CH}}_2\mathrm{OH}\right)}_3\mathrm{CHO}$, has no remaining $\alpha$-hydrogens. In the presence of concentrated $\mathrm{NaOH}$ and excess formaldehyde, it undergoes a cross-Cannizzaro reaction:
$C{\left({\mathrm{CH}}_2\mathrm{OH}\right)}_3\mathrm{CHO}+\mathrm{HCHO}+{\mathrm{OH}}^{-}\to C{\left({\mathrm{CH}}_2\mathrm{OH}\right)}_4+{\mathrm{HCOO}}^{-}$
Thus, the products of this reaction are pentaerythritol, $C{\left({\mathrm{CH}}_2\mathrm{OH}\right)}_4$, and sodium formate, $\mathrm{HCOONa}$.

Step 3: Identification of P and Q
Pentaerythritol, $C{\left({\mathrm{CH}}_2\mathrm{OH}\right)}_4$ (an alcohol), does not give a Tollens' test. Thus, P is pentaerythritol.
Sodium formate, $\mathrm{HCOONa}$, on acidification gives formic acid ($\mathrm{HCOOH}$), which contains an aldehyde-like hydrogen and therefore gives a positive Tollens' test. Thus, Q is $\mathrm{HCOONa}$.

Step 4: Reaction of P with excess Cyclohexanone
Pentaerythritol P has four hydroxyl (-OH) groups. When treated with excess cyclohexanone in the presence of an acid catalyst (PTSA), it forms a di-acetal (specifically, a di-ketal) R by reacting with two molecules of cyclohexanone.
Each cyclohexanone molecule reacts with two adjacent hydroxyl groups of pentaerythritol to form a 1,3-dioxane ring. The structure of R contains a central carbon atom bonded to two spiro-linked 1,5-dioxaspiro[5.5]undecane systems:
Specifically, R is 3,9-bis(spiro[cyclohexane-1,2'-[1,3]dioxane]).

Step 5: Counting Methylene (-CH₂-) Groups and Oxygen Atoms in R
Let us determine the formula or count the groups directly:
1. Oxygen Atoms:
The product R is a di-ketal containing two 1,3-dioxane rings. Each ring contains 2 oxygen atoms.
Total number of oxygen atoms in R = 4.
2. Methylene (-CH₂-) Groups:
- From the pentaerythritol core: The central carbon is attached to four $-{\mathrm{CH}}_2-$ groups that are part of the dioxane rings. This gives 4 methylene groups.
- From the two cyclohexyl rings: Each cyclohexyl ring has five methylene ($-{\mathrm{CH}}_2-$) groups (since the carbon bonded to the two oxygens is a ketal carbon, $-C-$, not a methylene carbon).
For two cyclohexyl rings, the number of methylene groups = $2\times 5=10$ methylene groups.
Total number of methylene groups in R = $4+10=14$.

Conclusion:
Sum of the number of methylene groups and oxygen atoms in R = $14\left(\text{methylene groups}\right)+4\left(\text{oxygen atoms}\right)=18$.