JEE Advanced 2024 Paper 1 · Q10 · Cannizzaro / Acetal Protection

Question

Complete reaction of acetaldehyde with excess formaldehyde, upon heating with conc. NaOH solution, gives $\mathbf{P}$ and $\mathbf{Q}$. Compound $\mathbf{P}$ does not give Tollens' test, whereas $\mathbf{Q}$ on acidification gives positive Tollens' test. Treatment of $\mathbf{P}$ with excess cyclohexanone in the presence of catalytic amount of $p$-toluenesulfonic acid (PTSA) gives product $\mathbf{R}$. Sum of the number of methylene groups (-$CH_2$-) and oxygen atoms in $\mathbf{R}$ is ______.

SolveKar AI · Accepted Answer

Crossed aldol then crossed-Cannizzaro: $CH_3CHO$ + 4 $HCHO$ + NaOH → $C(CH_2OH)_4$ (pentaerythritol = $\mathbf{P}$) + $HCOO^-Na^+$ ($\mathbf{Q}$, sodium formate). Pentaerythritol has no –CHO so no Tollens'; $HCOO^-$ on acidification gives $HCOOH$ which gives positive Tollens'.
$\mathbf{P}$ (4 -OH) + 2 cyclohexanones / PTSA forms a spiro bis-cyclic acetal $\mathbf{R}$: each pair of -CH2OH groups condenses with one cyclohexanone C=O to form a 1,3-dioxane ring. $\mathbf{R}$ = 3,3,9,9-bis(spirocyclohexyl)-1,5,7,11-tetraoxa-spiro compound.
Count in $\mathbf{R}$: –CH2– groups = 4 (from C(CH2O)4) + 5×2 (from two cyclohexyl rings, each contributes 5 CH2 since the former C=O carbon is now sp³ quaternary) = 14. Oxygens = 4 (two acetal rings × 2 O). Sum = 14 + 4 = $\mathbf{18}$.

JEE Advanced 2024 Paper 1 · Q10 · Cannizzaro / Acetal Protection

Solution