Which one of the following reactions will not form acetaldehyde?

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${CH}_{3} {CH}_{2} OH \to_{}^{{CrO}_{3} - H_{2} {SO}_{4}}$

${CH}_{3} CN \to_{ii) H_{2} O}^{i) DIBAL - H}$

${CH}_{3} {CH}_{2} OH \to_{573 K}^{Cu}$

${CH}_{2} = {CH}_{2} + O_{2} \to_{H_{2} O}^{Pd (II) / Cu (II)}$

answer is C.

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Detailed Solution

$C H_{3} - C H_{2} - O H \to_{s t r o n g o x i d i \sin g a g e n t}^{C r O_{3} . H_{2} S O_{4}} C H_{3} - \overset{\overset{O}{| |}}{C} - O H$

$C H_{3} C H_{2} O H \to_{573 K}^{C u} C H_{3} - \overset{\overset{O}{| |}}{C} - H$

$C H_{3} C N \to_{(2) 510 K}^{(i) D i B A L - H} C H_{3} - \overset{\overset{O}{| |}}{C} - H$

$C H_{2} = C H_{2} + O_{2} \to_{H_{2} O}^{P d (I I) - C u (I I)} C H_{3} - \overset{\overset{O}{| |}}{C} - H$

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