MCQ
$Y$ is


- A

- B

- ✓

- D







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$3HCHO+A\xrightarrow[{{40}\,^{o}}C]{N{{a}_{2}}C{{O}_{3}}}\underset{(82\%)}{\mathop{(B)}}$
Product $(B)$ of the above reaction is

Compound $'C'$ is