Question

In the preparation of alkanes; a concentrated aqueous solution of sodium or potassium salts of saturated carboxylic acid are subjected to .

• A. Electrolysis
• B. Hydrogenation
• C. Hydrolysis

Answer 1

The correct option is A Electrolysis

Do you remember Kolbe's electrolysis?

Electrolysis of an aqueous solution of sodium or potassium salt of a fatty acid gives higher alkane, and electrolysis of an aqueous solution of sodium or potassium salt of dibasic acid gives alkene, whereas the electrolysis of aqueous solution of sodium or ptassium salt of dibasic unsaturated fatty acid gives alkyne.

$2RCOONa\stackrel{Electrolysis}{\to }R-R+2C{O}_2+2\stackrel{-}{O}H+H_2$

Therefore,

Mechanism (free radical)

$2RCOONa$ $\to$ $2RCO{O}^{-}$ + $2N{a}^{+}$

At anode (oxidation)

$2RCO{O}^{-}\to 2\stackrel{-}{e}+2RCO\stackrel{\cdot }{O}$
$2\stackrel{-}{e}+2RCO\stackrel{\cdot }{O}\to 2\stackrel{\cdot }{R}+2C{O}_2\uparrow$
$2\stackrel{\cdot }{R}\to R-R$

At cathode (reduction): reduction of $H_{2}O$ takes place since reduction potential of is greater than $N{a}^{⨁}$

$2H_{2}O+2e^{-}\to 2O{H}^{-}+H_2$

Methane cannot be prepared by this method

$2C{H}_{3}COOK+2H_{2}O\stackrel{Electrolysis}{\to }C{H}_3-C{H}_3+2C{O}_2+2KOH+H_2$