1. Tardigrade
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  3. Chemistry
  4. Me - C ≡ Me xrightarrow [E tOH ,-33 ° C]Na/NH3(liq) X xrightarrow textdil .alkaline KMnO4product (s) The product (s) from the above reaction will be

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Q. $Me - C \equiv Me \xrightarrow [E tOH ,-33 ^{\circ }C]{Na/NH_3(liq)} X \xrightarrow{\text{dil .alkaline} KMnO_4}$product (s)
The product (s) from the above reaction will be

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Solution:

$Na$ in liq. $NH _{3}$ reduces alkynes to trans alkenes. Cold alkaline $KMnO _{4}$ does syn dihydroxylation on alkene to produce vicinol diols.
Here trans alkenes give pair of enantiomers while cis alkenes give meso compound.
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