$C H_{3} - C \equiv C - C H_{3}$ can be converted into $C H_{3} - C O - C O - C H_{3}$ by the action of :

alkaline

K M n O_{4}

at room temperature

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selenium dioxide

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H g^{2 +}

H_{2} S O_{4}

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O_{3}

and

Z n

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Solution

The correct options are
A alkaline $K M n O_{4}$ at room temperature
B selenium dioxide
D $O_{3}$ and $Z n$
The reaction is as follows:
$C H_{3} - C \equiv C - C H_{3} \to C H_{3} - C O - C O - C H_{3}$
This conversion is affected by cold dilute $K M n O_{4}$ (Baeyer's reagent), $S e O_{2}$ or ozonolysis in the presence of zinc dust. $H g^{2}$ in the presence of $H_{2} S O_{4}$ converts alkyne into ketone only. The reaction is as follows:
$C H_{3} - C \equiv C - C H_{3} + H_{2} O H g^{2 +} - -- \to H^{+} C H_{3} - C (O H) = C H - C H_{3} \to C H_{3} - C O - C H_{2} - C H_{3}$

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Q. Which is the most suitable reagent among the following to distinguish compound (iii) from rest of the compounds?

(i) C H_{3} - C \equiv C - C H_{3}

(i i) C H_{3} - C H_{2} - C H_{2} - C H_{3}

(i i i) C H_{3} - C H_{2} - C \equiv C H

(i v) C H_{3} - C H = C H_{2}

C H_{3} - C H_{3} | C | C H_{3} - C H_{2} B r C_{2} H_{5} O H \to

products.
Products can be:

{C H}_{3} {C H}_{2} O H

can be converted into

{C H}_{3} C H O

by:

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CH3−C≡C−CH3 can be converted into CH3−CO−CO−CH3 by the action of :

$C H_{3} - C \equiv C - C H_{3}$ can be converted into $C H_{3} - C O - C O - C H_{3}$ by the action of :