Which of the following pairs of compounds give the same product on treatment with hot $K M n O_{4}$ ?

|, C H_{3} C H = C H C H_{3}

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1,3 butadiene and 1,4 butadience

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C H_{3} C \equiv C C H_{3}, C H_{3} C H = C H C H_{3}

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C H_{2} = C H C H_{3} = C H_{2}, C H_{3} C H = C H C H_{3}

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Solution

The correct option is C $C H_{3} C \equiv C C H_{3}, C H_{3} C H = C H C H_{3}$
$C H_{3} C \equiv C C H_{3}$ and $C H_{3} C H = C H C H_{3}$ give the same product on treatment with hot $K M n O_{4}$ .
In both cases the product is acetic acid.
$C H_{3} C \equiv C C H_{3} h o t - ---- \to K M n O_{4} 2 C H_{3} C O O H$
and
$C H_{3} C H = C H C H_{3} h o t - ---- \to K M n O_{4} 2 C H_{3} C O O H$

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Similar questions

C H_{3} C H = C H_{2}

C H_{3} C H = C H C H_{2} C H_{3}

C H_{3} C H = C H C H_{3}

C H_{3} C H = C H C H_{3}

C H_{3} C H_{2} C H = C H_{2}

C H_{3} C \equiv C C H_{3} \to C H_{3} C H = C H C H_{3}

C H_{3} C H B r C H_{2} C H_{3} a l c . K O H ⟶ C H_{3} C H = C H C H_{3} (major product) + C H_{2} = C H C H_{2} C H_{3} (minor product)

{C H}_{3} C H = C H {C H}_{3} + {B r}_{2} \to

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