How would #"ethane"# be prepared from #"propanoic acid"#?

2 Answers
Aug 24, 2017

Not straightforward.....you have to reduce, then oxidize, then reduce again........

Explanation:

A possible (if unlikely sequence) follows......

#H_3C-CH_2CO_2Hstackrel(LiAlH_4)rarrH_3C-CH_2CH_2OH#

Better methods of reduction may be available, i.e. #BMe_3# and #Me_2S#.... the given reduction would require prolonged reflux, and much time and trouble.

THe alcohol is eliminated to give propylene.....

#H_3C-CH_2CH_2OHstackrel(H_3PO_4)rarrH_2C=CH-CH_3#

The olefin is fiercely oxidized.....

#H_2C=CH-CH_3 stackrel(Cr_2O_7^(2-)"/"Delta)rarrH_3C-CO_2H#

The terminal methylene group is probably lost as carbon dioxide. And now we repeat the reduction process again; it would not be simple.

#H_3C-CO_2Hstackrel(LiAlH_4"/"Delta)rarrH_3C-CH_2OH#

#H_3C-CH_2OHstackrel(H_3PO_4"/"Delta)rarrH_2C=CH_2#

And finally reduction......

#H_2C=CH_2 stackrel(H_2, Pd)rarrH_3C-CH_3#

Note that this is an intellectual exercise ONLY. If you want ethane simply dig a hole in the ground and tap underground reserves.

May 3, 2018

How about this?

Explanation:

Step 1. Neutralization to sodium propanoate

#underbrace("CH"_3"CH"_2"COOH")_color(red)("propanoic acid") + "NaOH" → underbrace("CH"_3"CH"_2"COONa")_color(red)("sodium propanoate") + "H"_2"O"#

Step 2. Decarboxylation with soda lime

Soda lime is a mixture consisting mostly of calcium hydroxide with small amounts of sodium hydroxide and potassium hydroxide.

#"CH"_3"CH"_2"COONa + Ca(OH)"_2 stackrelcolor(blue)(Δcolor(white)(mm))(→) underbrace("CH"_3"CH"_3)_color(red)("ethane") + "CaCO"_3 + "NaOH"#