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

  • A
    $C(CH_2OH)_4$
  • B
    $(HOCH_2)_2C(CHO)_2$
  • C
    $(HOCH_2)_3C-CHO$
  • D
    $(HOCH_2)_2C(CHO)-CH_2-CH_2OH$

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