Question

Reaction \[{C_{2}}{H_5}I + {C_{5}}{H_{11}}I + 2Na \to {C_2}{H_5} - {C_5}{H_{11}} + 2NaI\] is called:
A. Hoffmann’s Reaction
B. Dow’s Reaction
C. Wurtz’s Reaction
D. Riemer-Tiemann reaction

Answer 1

Hint: The reaction that is given here consists of two alkyl halides reacting in the presence of sodium metal to give a higher alkane. This reaction involves the formation of carbon-carbon bonds in a nucleophilic substitution step.

Complete Step by Step Solution:
In this reaction, we have two alkyl halides reacting in the presence of sodium metal to give an alkane. The alkane is formed out of the two alkyl groups of the alkyl halides being bonded to each other via carbon-carbon bonds. This type of coupling reaction is called the Wurtz Reaction named after Charles Adolphe Wurtz. Its general scheme is given below:
\[{R^1} - X + {R^2} - X \xrightarrow{2Na} {R^1} - {R^2} + 2NaX\] where \[{R^1}\]and\[{R^2}\]are two different alkyl groups.

The Wurtz Reaction between ethyl iodide (\[{C_2}{H_5}I\]) and pentyl iodide (\[{C_5}{H_{11}}I\]) to produce heptane (given in the question) proceeds in the following way.
In the first step, one electron from the metallic sodium is transferred to \[{C_2}{H_5}I\] producing an ethyl radical as shown below.

Image: Formation of ethyl radical

The ethyl radical then accepts an electron from another sodium atom to form an ethyl carbanion as shown here.
 
Image: Formation of ethyl carbanion

The nucleophilic ethyl carbanion then reacts with pentyl iodide in an \[{S_N}2\] reaction to form heptane as shown below.

Image: Formation of heptane by \[{S_N}2\] attack.
Thus, the correct option is C.

Note: Some students might question why the first step of the reaction involved the formation of ethyl radical from ethyl iodide and not the formation of pentyl radical from pentyl iodide. This is because the ethyl radical is much more stable than the pentyl radical. The ethyl radical has 3 \[\alpha \]-hydrogens whereas the pentyl radical has 2.

Image: Ethyl and Pentyl radical

This means that there is a higher hyperconjugation effect in the ethyl radical compared to the pentyl radical and the higher the hyperconjugation, the more stable the radical will be. Hyperconjugation is not the only factor governing radical stability. It is advised that students learn about all the factors that affect the stability of reactive intermediates.