Question

What is the minimum quantity (in grams) of methyl iodide required for preparing $1$ mole of ethane by Wurtz reaction? [At. Wt. of iodine $= 127$ ].

Answer 1

Hint : To calculate the minimum quantity of methyl iodide required for preparing $1$ mole ethane we have to write the chemical equation involved in the Wurtz reaction. Then with the help of stoichiometric coefficients, we can find the quantity of $C{H_3}I$ required by using the stoichiometric ratios.

Complete Step By Step Answer:
Wurtz’s reaction is an example of coupling reaction in which sodium metal is treated with two alkyl halides in the presence of a dry ether solution to obtain a higher alkane along with a compound having sodium and the halogen.
It is a very useful reaction for the production of alkanes and other than sodium metals, silver, zinc, indium, and iron can also be used in this process.
Wurtz reaction mechanism includes a free radical species represented by $\mathop R\limits^*$ which is a part of a halogen-metal exchange. The general form of this reaction is written as:
$2\,R - X + 2Na{ } \to { }R - R + \,\,2N{a^ + }{X^ - }$
In this reaction, when $2$ moles of alkyl halide is reacted with sodium metal they form $1$ mole of higher alkane. The reaction for this is represented as:
$\;2C{H_3}I + 2Na \to CH3\, - CH3\, + \,\,2NaI$
From this reaction, we can deduce that $1$ mole of ethane is produced from $2$ moles of methyl iodide. Now, the Molecular weight of methyl iodide $= 124g$
The total amount of methyl iodide required $= 2 \times \,124 = 248g$ .
Hence, we can say that $248g$ methyl iodide is required to produce $1$ mole of ethane.
Therefore, option (D) is correct.

Note :
In the Wurtz reaction $2$ moles of alkyl halides reacts with $2$ moles of sodium atom in the presence of dry ether to produce $1$ mole of alkane. It is a $3$ step radical mechanism process. Using Wurtz reaction you can only produce symmetrical alkanes. It cannot be used to produce unsymmetrical alkane.