Pyridinium chloro chromate selectively oxidises a primary alcohol to an aldehyde e.g. \[RC{{H}_{2}}OH\xrightarrow[C{{H}_{2}}C{{l}_{2}}]{PCC}RCHO\]

No explanation is given for this question. Let's discuss the answer together.

\[\underset{A}{\mathop{{{C}_{3}}{{H}_{8}}O\xrightarrow{{{K}_{2}}C{{r}_{2}}{{O}_{7}}/{{H}^{+}}}}}\,\underset{B\left( -CHO \right)}{\mathop{{{C}_{3}}{{H}_{6}}O\xrightarrow{{{H}_{2}}NCONHN{{H}_{2}}}C}}\,\]
Since $$B$$  reduces Tollen’s reagent, it indicates that it has an $$-CHO$$   group, so it must be \[C{{H}_{3}}C{{H}_{2}}CHO.\]
Hence
\[\underset{\left[ A \right]}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}OH}}\,\to \underset{\left[ B \right]}{\mathop{C{{H}_{3}}C{{H}_{2}}CHO}}\,\]        \[\xrightarrow{{{H}_{2}}NNHCON{{H}_{2}}}\underset{\left[ C \right]}{\mathop{C{{H}_{3}}C{{H}_{2}}CH}}\,=\]       \[NNHCON{{H}_{2}}\]

4-Oxopentanal : \[C{{H}_{3}}\overset{\begin{smallmatrix} O \\ \parallel \end{smallmatrix}}{\mathop{-C-}}\,C{{H}_{2}}-C{{H}_{2}}-CHO\]

\[C{{H}_{3}}CHC{{l}_{2}}\xrightarrow{aq.\,\,KOH}\]     \[C{{H}_{3}}CH{{\left( OH \right)}_{2}}\xrightarrow{-{{H}_{2}}O}C{{H}_{3}}CHO\]

Aldehyde will be more reactive than ketone.

Base stronger than cyanide is used as a catalyst.

\[R-COOH\xrightarrow{-{{H}^{+}}}R-CO{{O}^{-}}\]
is a resonance hybrid of resonating structures.