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

Question

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 (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

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. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/93ec3c733d2c8cc8d6c2949a43bf669a-.png" />

Q. $M e - C \equiv M e N a / N H_{3} (l i q) EtO H, - 3 3^{\circ} C X dil .alkaline K M n O_{4}$ product (s)
The product (s) from the above reaction will be

$N a$ in liq. $N H_{3}$ reduces alkynes to trans alkenes. Cold alkaline $K M n O_{4}$ does syn dihydroxylation on alkene to produce vicinol diols.
Here trans alkenes give pair of enantiomers while cis alkenes give meso compound.

Q. Me−C≡MeNa/NH3​(liq)EtOH,−33∘C​Xdil .alkalineKMnO4​​product (s) The product (s) from the above reaction will be

Na in liq. NH3​ reduces alkynes to trans alkenes. Cold alkaline KMnO4​ does syn dihydroxylation on alkene to produce vicinol diols. Here trans alkenes give pair of enantiomers while cis alkenes give meso compound.

Q. $M e - C \equiv M e N a / N H_{3} (l i q) EtO H, - 3 3^{\circ} C X dil .alkaline K M n O_{4}$ product (s)
The product (s) from the above reaction will be

$N a$ in liq. $N H_{3}$ reduces alkynes to trans alkenes. Cold alkaline $K M n O_{4}$ does syn dihydroxylation on alkene to produce vicinol diols.
Here trans alkenes give pair of enantiomers while cis alkenes give meso compound.