An equimolar mixture of sodium propanoate and sodium 2-methylpropanoate is electrolysed in concentrated aqueous solution (Kolbe conditions). Consider only radical coupling products (disproportionation products are ignored). Which hydrocarbons are formed?

A2-methylbutane
Bbutane
C2,3-dimethylbutane
D2-methylpentane

Answer: A, B, C

Decarboxylation gives

CH X_{3} CH X_{2} \cdot

and

(CH X_{3}) X_{2} CH \cdot

radicals. Coupling of

C X_{2} H X_{5}

with

C X_{2} H X_{5}

gives butane,

C X_{2} H X_{5}

with isopropyl gives 2-methylbutane, and isopropyl with isopropyl gives 2,3-dimethylbutane. 2-Methylpentane would need a propyl and an isopropyl radical.

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An equimolar mixture of sodium propanoate and sodium 2-methylpropanoate is electrolysed in concentrated aqueous solution (Kolbe conditions). Consider only radical coupling products (disproportionation products are ignored). Which hydrocarbons are formed?

Answer: A, B, C

More questions on Alkanes: preparation (Wurtz, Kolbe, decarboxylation) and reactions