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 ...

The correct numerical answer is 18. Click to view the complete step-by-step calculation.

JEE Advanced 2024 Paper 1 Chemistry Question 10 Solution

Acetaldehyde ( $CH_3CHO$ ) reacts with excess formaldehyde ( $HCHO$ ) and conc. $NaOH$ via a combination of Aldol and crossed Cannizzaro reactions. $CH_3CHO + 3HCHO \xrightarrow{OH^-} (HOCH_2)_3C-CHO$ $(HOCH_2)_3C-CHO + HCHO \xrightarrow{OH^-} C(CH_2OH)_4 + HCOONa$
$P$ is Pentaerythritol ( $C(CH_2OH)_4$ ), which does not give Tollens' test. $Q$ is sodium formate ( $HCOONa$ ), which on acidification gives formic acid ( $HCOOH$ ), showing a positive Tollens' test.
Pentaerythritol ( $P$ ) reacts with 2 molecules of cyclohexanone in the presence of $H^+$ (PTSA) to form a spiro-cyclic diketal $R$ .
Structure of $R$ : The central carbon of pentaerythritol is connected to four $-CH_2-O-$ groups. These form two 6-membered rings with the carbonyl carbons of two cyclohexanone molecules. Each cyclohexanone ring contributes 5 methylene ( $-CH_2-$ ) groups.
Counting in $R$ :