how can be conversion, ethyl iodide to propyl iodide
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You can do this using a classic halogen exchange reaction called a Finkelstein reaction.
CH33CH22Br +NaI ⇌ CH33CH22I + NaBr
Just add ethyl bromide (a.k.a. bromoethane) to anhydrous acetone (a.k.a. propan-2-one) and Sodium Iodide.
The reaction proceeds via an SN2 mechanism and shifts to the right due to the precipitation of NaBr which is almost insoluble in acetone in contrast to sodium iodide which is soluble
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CH33CH22Br +NaI ⇌ CH33CH22I + NaBr
Just add ethyl bromide (a.k.a. bromoethane) to anhydrous acetone (a.k.a. propan-2-one) and Sodium Iodide.
The reaction proceeds via an SN2 mechanism and shifts to the right due to the precipitation of NaBr which is almost insoluble in acetone in contrast to sodium iodide which is soluble
Mark me as a Brainliest.
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Ch3ch2I+ KCN(alc) ➡️ ch3ch2CN
ch3ch2CN + LiAlH4 ➡️ ch3ch2ch2nh2
Ch3ch2ch2nh2 + KI ➡️ ch3ch2ch2I + KNH2.
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