${C H}_{3} C H = C H C H O$ is oxidised to ${C H}_{3} - C H = C H C O O H$ , using oxidising agent as :

alkaline

K M n O_{4}

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K_{2} {C r}_{2} O_{7} / c o n c . H_{2} {S O}_{4}

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ammoniacal

A g {N O}_{3}

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dilute4

H {N O}_{3}

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Solution

The correct option is C ammoniacal $A g {N O}_{3}$
$C H_{3} C H = C H C H O + [O] \to C H_{3} - C H = C H C O O H$

on analyzing above reaction we find that we have to use that oxidizing agent which retain the double bond and oxidized $- C H O$ into $- C O O H .$

Now,

we have Option A,B,and D are strong oxidizing agent which can oxidised $- C H O a s w e l l a s d o u b l e$

Thus, we will use mild oxidising agent $a m m o n i c a l A g N O_{3} (T o l l e n^{'} s r e a g e n t)$ to carry out above reaction.

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Similar questions

K_{2} {C r}_{2} O_{7}

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K_{2} {C r}_{2} O_{7} + c o n c . H_{2} {S O}_{4} + H_{2} O_{2} + e t h e r ⟶

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K_{2} {C r}_{2} O_{7} + K I + H_{2} {S O}_{4} \to K_{2} {S O}_{4} + {C r}_{2} {({S O}_{4})}_{3} + I_{2} + H_{2} O

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K M {n O}_{4} + H_{2} C_{2} O_{4} + H_{2} {S O}_{4} \to K_{2} {S O}_{4} + {M n S O}_{4} + {C O}_{2} + H_{2} O

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