MCQ
Methyl alcohol on oxidation with acidified ${K_2}C{r_2}{O_7}$ gives
- A$C{H_3}COC{H_3}$
- B$C{H_3}CHO$
- ✓$HCOOH$
- D$C{H_3}COOH$
✓
Answer
Correct option: C.
$HCOOH$
c
(c)$C{{H}_{3}}OH\underset{{{H}_{2}}S{{O}_{4}}}{\mathop{\xrightarrow{{{K}_{2}}C{{r}_{2}}{{O}_{7}}}}}\,HCOOH$
(c)$C{{H}_{3}}OH\underset{{{H}_{2}}S{{O}_{4}}}{\mathop{\xrightarrow{{{K}_{2}}C{{r}_{2}}{{O}_{7}}}}}\,HCOOH$
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