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Which of the following yield both alkane and alkene?

$\begin{array}{1 1}(a)\;\text{Kolbe's reaction}\\(b)\; \text{Williamson's synthesis}\\(c)\;\text{Wurtz reaction}\\(d)\;\text{Sandmeyer reaction}\end{array}$

1 Answer

Kolbe's method:Electrolysis of a concentrated aqueous solution of either sodium or pottassium salts of saturated monocarboxylic acids yield higher alkanes at anode
$CH_3COONa \leftrightharpoons CH_3COO^-+Na^+$
$2Na^+\quad \underrightarrow{+2e} \quad2Na$
$2CH_3COO^-\quad\underrightarrow{-2e}\quad CH_3-CH_3+2CO_2$
$2Na+2H_2O\rightarrow 2NaOH+H_2$
However when sodium or pottassium salts of saturated dicarboxylic acids of succinic acid type are subjected to electrolysis in aqueous solution,alkenes are formed.
Hence (a) is the correct answer.
answered Mar 19, 2014 by sreemathi.v

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