Question

Write a note on Wurtz reaction and mention any 2 uses of iodoform.

Answer 1

Hint: The Wurtz reaction is a coupling type of reaction and one can elaborate it in detail. The iodoform is used for medical purposes which can be elaborated in answer.

Complete step by step answer:
1) The Wurtz reaction is an organic reaction which is a type of coupling reaction. In this reaction, sodium metal is used and it is reacted with two alkyl halides in the conditions where a solution of dry ether is used to yield a higher alkane and the compounds containing sodium and the halogen as a side product.

2) Wurtz reaction general representation can be written as follows:
$2R - X + 2Na \to R - R + 2N{a^ + }{X^ - }$
In the above reaction, two alkyl groups are joined to form a higher alkane and sodium and halogen ionic side product.

3) The detailed reaction mechanism for the Wurtz reaction is as follows:
One electron from sodium metal is transferred to a halide atom in the alkyl halide which gives metal halide and alkyl radical. The schematic representation of this step is as follows.
$R - X + M \to {R^ \bullet } + {M^ + }{X^ - }$
The alkyl radical formed then receives an electron from sodium metal and then forms alkyl anion.
${R^ \bullet } + M \to {R^ - }{M^ + }$
In the next step, the alkyl anion formed will attack the initial reactant alkyl halide and form a higher alkane. This step in the reaction is a type of nucleophilic substitution reaction. This reaction follows the $SN_2$ reaction which is the type of nucleophilic substitution reaction.
${R^ - }{M^ + } + R - X \to R - R + {M^ + }{X^ - }$
Therefore, the higher alkane is formed as the reaction product.

4) Uses of iodoform are as follows:
i) The iodoform is used as a disinfectant.
ii) It is also used as an antiseptic for the dressing of wounds.
iii) It is also used as a sterilizing agent on wounds and fire burns.

Note:
When the alkyl halide bond breaks and the formation of radical happens because halide gets an electron from sodium metal forming an ionic bond with sodium. By adding a single electron to the alkyl radial will form the anionic alkyl which has a negative charge on it. In a nucleophilic substitution reaction, the nucleophile is replaced with the strong nucleophile which attacks the reactant.