MCQ
$CO + NaOH \to $
- ✓$HCOONa$
- B${C_2}{H_2}{O_4}$
- C$HCOOH$
- D$C{H_3}COOH$
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compound $'C'$ is
$C{H_3}C{H_2}C{H_2}OH \xrightarrow{PCl_3}A\xrightarrow{Alco.KOH}B$

Figure $\xrightarrow[{pyridine}]{{C{H_3}COCl}}$ $P$ $\xrightarrow[{C{H_3}COOH}]{{B{r_2}}}Q\,\xrightarrow{{{H_3}{O^ + }}}R$