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.