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Some alkanes may be prepared by adding an alkyl iodide dropwise to sodium in a suitable inert solvent

e.g:$ 2CH_3I +2Na\rightarrow C_2H_6+2NaI$

if a mixture of iodomethane and 2-iodopropane is used as a starting material the product will contain

$\begin{array}{1 1}a)\; 2-methyl\; propane\\b)\; Methane, propene\; and\; propane\\c)\; Ethane\; and \;2.3\; dimethylbutane\\d)\; all\; of\; the \;above\end{array}$

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$2CH_3I+Na\rightarrow CH_3-CH_3$
Hence (c) is the correct answer.
answered Dec 4, 2013 by sreemathi.v

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