When CH_3Br and C_2H_5Br are subjected to Wur | vedclass

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(C) The Wurtz reaction involves the coupling of alkyl halides in the presence of sodium metal and dry ether to form higher alkanes.When a mixture of $

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$3$

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(C) The Wurtz reaction involves the coupling of alkyl halides in the presence of sodium metal and dry ether to form higher alkanes.When a mixture of $CH_3Br$ and $C_2H_5Br$ is used,the following free radicals are generated: $CH_3^{\bullet}$ and $C_2H_5^{\bullet}$.These radicals combine in all possible ways:$1$. $CH_3^{\bullet} + CH_3^{\bullet} \rightarrow CH_3-CH_3$ (Ethane)$2$. $C_2H_5^{\bullet} + C_2H_5^{\bullet} \rightarrow C_2H_5-C_2H_5$ ($n$-Butane)$3$. $CH_3^{\bullet} + C_2H_5^{\bullet} \rightarrow CH_3-C_2H_5$ (Propane)Thus,a total of $3$ different alkane products are formed.

Column $I$	Column $II$
$A$. $(CH_3)_3CH + O_2 \xrightarrow{KMnO_4} (CH_3)_3COH$	$i$. $CH_3COOH + H_2O$
$B$. $2CH_4 + O_2 \xrightarrow[100 \ atm]{Cu/523 \ K} 2CH_3OH$
$C$. $CH_4 + O_2 \xrightarrow[\Delta]{Mo_2O_3} HCHO + H_2O$	$iii$. $2CH_3OH$
$D$. $CH_3-CH_3 + \frac{3}{2}O_2 \xrightarrow{(CH_3COO)_2Mn} CH_3COOH + H_2O$	$iv$. $HCHO + H_2O$

When $CH_3Br$ and $C_2H_5Br$ are subjected to Wurtz reaction,the maximum number of possible alkane product$(s)$ formed is/are

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