convert ethylens oxide to n-propyl alcohol
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CH2=CH2(ehtylene) on treating with Br2 and CCl4 gives BrCH2-CH2Br(Ethylene dibromide).
On treating ethylene dibromide with NaNH2or Liq.NH3 at 196 K, you get HC=CH(acetylene). This process is called dehydrobromination reaction.
On treating this acetylene again with NaNH2 in liq. NH3 at 196 K, you get HC=C-Na+ (Sodium acetylide).
on treating this sodium acetylide with CH3I, NaI gets removed and you get HC=C-CH3 (Propyne).
now, he only have to convert triple bond into double bond . for this reduce propyne using H2-Pd/BaSO4 . You get H2C=CH-CH3
(Propylene) . All you have to now do is to do an anti-markovnikov addition using H20 in the presence of H2O2 to yield n-PROPYL ALCOHOL.
That's it.
My friend has written this answer.. and it's 100 percent correct.. so hope this helps you out!
On treating ethylene dibromide with NaNH2or Liq.NH3 at 196 K, you get HC=CH(acetylene). This process is called dehydrobromination reaction.
On treating this acetylene again with NaNH2 in liq. NH3 at 196 K, you get HC=C-Na+ (Sodium acetylide).
on treating this sodium acetylide with CH3I, NaI gets removed and you get HC=C-CH3 (Propyne).
now, he only have to convert triple bond into double bond . for this reduce propyne using H2-Pd/BaSO4 . You get H2C=CH-CH3
(Propylene) . All you have to now do is to do an anti-markovnikov addition using H20 in the presence of H2O2 to yield n-PROPYL ALCOHOL.
That's it.
My friend has written this answer.. and it's 100 percent correct.. so hope this helps you out!
mansigupta23:
thank you
formula of ethylene oxide is C2H4O
but in this you use the formula of ethylene
ok no problem
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