When an alcoholic solution of ethylene dibromide $(Br-CH_2-CH_2-Br)$ is heated with granulated zinc,the compound formed is:
- A$Ethylene$
- B$Ethyne$
- C$Cyclobutane$
- D$Butane$
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$CH_3-C(CH_3)=CH-CH_3 \xrightarrow{Hg(OAc)_2/EtOH} (A) (90\%)$. The product $(A)$ of the above reaction is:
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View SolutionThe reaction is given as:
$(CH_3)_2C=C(CH_3)_2 + (CH_3)_3N^{+}-O^{-} + H_2O \xrightarrow{OsO_4 (10^{-4} \text{ mole})} A + (CH_3)_3N$
Product $(A)$ is:
$(CH_3)_2C=C(CH_3)_2 + (CH_3)_3N^{+}-O^{-} + H_2O \xrightarrow{OsO_4 (10^{-4} \text{ mole})} A + (CH_3)_3N$
Product $(A)$ is:
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