A) catalytic hydrogenation
B) treatment with \[LiAl{{H}_{4}}\]
C) treatment with pyridinium chlorochromate
D) treatment with \[KMn{{O}_{4}}\]
Correct Answer: C
Solution :
| Ethanal \[\left( C{{H}_{3}}CHO \right)\] is an oxidized product of ethanol. Pyridinium chlorochromate \[\left( {{C}_{5}}{{H}_{5}}\overset{+}{\mathop{N}}\,\,HCl\,\overset{-}{\mathop{C}}\,\,r{{O}_{3}} \right)\] oxidises primary alcohols to aldehydes. Strong oxidising agents such as \[KMn{{O}_{4}}\] are used for getting-carboxylic acid from alcohols. |
| The oxidation process can be stopped at the aldehyde stage, if \[Cr\left( VI \right)\] reagents such as Collin's reagent\[\left( Cr{{O}_{3}}.2{{C}_{5}}{{H}_{5}}N \right)\] Corey's reagent or pyridinium chlorochromate are used PCG and pyridinium dichromate\[\left[ \left( {{C}_{5}}{{H}_{5}}NH \right)_{2}^{2+}C{{r}_{2}}O_{7}^{2-} \right]\] in anhydrous medium are used as the oxidising agent. So, the correct option is [c] |
| \[\underset{Ethanol}{\mathop{C{{H}_{3}}C{{H}_{2}}OH}}\,\underrightarrow{PCC}\underset{Ethanol}{\mathop{C{{H}_{3}}CHO}}\,\] |
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