A) \[HCHO\]
B) \[C{{H}_{3}}CHO\]
C) \[C{{H}_{3}}C{{H}_{2}}CHO\]
D) \[C{{H}_{3}}COC{{H}_{3}}\]
Correct Answer: A
Solution :
Key Idea: The given description is of Cannizaro reaction. This is a disproportionation reaction which is given by aldehydes which does not have a-hydrogen atom. During this reaction half of the molecules are oxidised and other half of the molecules are reduced. [a] \[H-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-H\] No \[\alpha \]- hydrogen atom [b] \[{{H}_{3}}\overset{\alpha }{\mathop{C}}\,-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-H\] Has a-hydrogen atom [c] \[C{{H}_{3}}-\overset{\alpha }{\mathop{C}}\,{{H}_{2}}-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-H\] Has a-hydrogen atom [d] \[\overset{\alpha }{\mathop{C}}\,{{H}_{3}}-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-C{{H}_{3}}\] Has a-hydrogen atom \[\therefore \] Only \[HCHO\] (formaldehyde) gives Cannizaro reaction. \[\therefore \] \[HCHO+NaOH\to \underset{formic\text{ }acid}{\mathop{HCOOH}}\,\,\,+\underset{methyl\text{ }alcohol}{\mathop{C{{H}_{3}}OH}}\,\]
You need to login to perform this action.
You will be redirected in
3 sec
