CH_3-CH(OH)-CH_2-CHO 3HCHO + A xrightarrow[40 | vedclass

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(C) The reaction of acetaldehyde $(CH_3CHO)$ with formaldehyde $(HCHO)$ in the presence of a base $(Na_2CO_3)$ is a crossed aldol condensation followe

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$(HOCH_2)_3C-CHO$

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(C) The reaction of acetaldehyde $(CH_3CHO)$ with formaldehyde $(HCHO)$ in the presence of a base $(Na_2CO_3)$ is a crossed aldol condensation followed by Cannizzaro reaction or simply a multi-step aldol addition. Acetaldehyde has three $\alpha$-hydrogens. In the presence of $3$ moles of formaldehyde,all three $\alpha$-hydrogens are replaced by hydroxymethyl $(-CH_2OH)$ groups. The reaction proceeds as: $CH_3CHO + 3HCHO \xrightarrow{Na_2CO_3} (HOCH_2)_3C-CHO$ (Pentaerythritol precursor). Thus,the product $(B)$ is $(HOCH_2)_3C-CHO$.

$CH_3-CH(OH)-CH_2-CHO$
$3HCHO + A \xrightarrow[40^o C]{Na_2CO_3} (B) (82\%)$
Product $(B)$ of the above reaction is

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$CH_3-CH(OH)-CH_2-CHO$ $3HCHO + A \xrightarrow[40^o C]{Na_2CO_3} (B) (82\%)$ Product $(B)$ of the above reaction is