A) nucleophilic attack, transfer of \[{{H}^{-}}\] and transfer of \[{{H}^{+}}\]
B) transfer of \[{{H}^{-}}\], transfer of \[{{H}^{+}}\] and nucleophiiic attack
C) transfer of\[{{H}^{+}}\], nucleophilic attack and transfer of\[{{H}^{-}}\]
D) electrophilic attack by \[O{{H}^{-}}\], transfer of \[{{H}^{+}}\] and transfer of \[{{H}^{-}}\]
Correct Answer: A
Solution :
The Cannizzaro reaction is as \[HCHO+HCHO\xrightarrow{KOH\,\,(con)}\underset{methyl\text{ }alcohol}{\mathop{C{{H}_{3}}OH}}\,\] \[\underset{acetic\text{ }acid}{\mathop{+HCOO{{K}^{+}}}}\,\] The mechanism of Cannizzaro reaction is as Step I Attack of nudeophile \[O{{H}^{-}}\] to the carbonyl carbon Step II The transfer of hydride ion from anion (I) to second molecule of aldehyde and finally rapid transfer of proton takes place. \[H-\overset{\begin{smallmatrix} O \\ | \end{smallmatrix}}{\mathop{\underset{\begin{smallmatrix} | \\ H \\ alcohol \end{smallmatrix}}{\mathop{C}}\,}}\,-H\xleftarrow{Proton\text{ }exchange\text{ }fast}H-\overset{\begin{smallmatrix} {{O}^{(-)}} \\ | \end{smallmatrix}}{\mathop{\underset{\begin{smallmatrix} | \\ H \\ alchol \end{smallmatrix}}{\mathop{C}}\,}}\,-{{H}^{+}}\] \[\underset{acid}{\mathop{+H-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-\overset{-}{\mathop{O}}\,\,{{K}^{+}}}}\,\]You need to login to perform this action.
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