JEE Main 2024 · Chemistry · Carbonyl Compounds, Carboxylic Acids and Amines · hard

Acetaldehyde is treated with excess formaldehyde in the presence of concentrated NaOH. The organic products formed are

Single correct option

A. $\mathrm{C(CH_2OH)_4}$ and $\mathrm{HCOONa}$
B. $\mathrm{CH_3CH(OH)CH_2CHO}$ only
C. $\mathrm{(HOCH_2)_3CCHO}$ and $\mathrm{CH_3OH}$
D. $\mathrm{HOCH_2CH_2CH_2OH}$ and $\mathrm{HCOONa}$

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Answer: (A) $\mathrm{C(CH_2OH)_4}$ and $\mathrm{HCOONa}$

Acetaldehyde has three

\alpha

-hydrogens and formaldehyde has none, so three successive crossed aldol additions with HCHO give

(\mathrm{HOCH_2})_3\mathrm{C{-}CHO}

. This aldehyde has no

\alpha

-H and, in the presence of excess HCHO and strong base, undergoes crossed Cannizzaro reaction: HCHO is oxidised to formate and the hindered aldehyde is reduced to

\mathrm{CH_2OH}

. The product is pentaerythritol,

\mathrm{C(CH_2OH)_4}

, with sodium formate.

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