Question

Which of the following reactions is an example of nucleophilic substitution reaction?
A. $RX + Mg \to RMgX$
B. $RX + KOH \to ROH + KX$
C. $2RX + 2Na \to R-R + 2NaX$
D. $RX + {H_2} \to RH + HX$

Answer 1

Nucleophilic substitution reactions involve the replacement of one or more atoms in a compound with a nucleophile (species with a negative charge). Thus, by closely examining the mechanisms of each of these reactions, we can figure out whether the intermediates or attacking species involved are nucleophiles. The correct option would be the substitution reaction with a nucleophile as the predominant attacking species.

Complete step by step answer:
let us first look at the first reaction given:
As we know, this is the formation of a Grignard reagent. This is done through the single electron transfer mechanism, involving free radicals. The alkyl halide first splits up into an alkyl free radical and a halide free radical. The halide free radical then abstracts one electron from the magnesium atom (which has two valence electrons in total) to form the magnesium halide free radical:
$R - X \to \mathop R\limits^ \bullet + \mathop X\limits^ \bullet$
$\mathop X\limits^ \bullet + \mathop {Mg}\limits_ \bullet ^ \bullet \to {X^ - } - \mathop {Mg}\limits_ \oplus ^ \bullet$
This magnesium halide free radical then combines with the alkyl free radical produced in the first step to give the Grignard reagent:
${X^ - } - \mathop {Mg}\limits_ \oplus ^ \bullet + \mathop R\limits^ \bullet \to R - MgX$
Hence, this is not a nucleophilic substitution reaction.
Now let us look at the second reaction:
In this, the $KOH$ splits up into its ions in solution, thus producing the nucleophile $O{H^ - }$. This nucleophile then attacks the alkyl halide and replaces the halogen atom:
$KOH \rightleftharpoons {K^ + } + O{H^ - }$
$R - X + O{H^ - } \to R - OH + {X^ - }$
Since a nucleophile is replacing another atom, this reaction is a nucleophilic substitution reaction.
The third reaction happens as follows:
One electron is transferred from the sodium atom to the halogen to produce a sodium halide and an alkyl radical:
$R-X + Na \to R \bullet + N{a^ + }{X^ - }$
The alkyl radical then accepts an electron from another sodium atom to form an alkyl anion:
$R \bullet + Na \to {R^ - }N{a^ + }$
The carbon atom of the alkyl anion, which is nucleophilic in nature, then displaces the halide to produce the carbon-carbon covalent bond:
${R^ - }N{a^ + } + R-X \to R-R + N{a^ + }{X^ - }$
Although the last step in this reaction involves a nucleophile, predominantly it is a free radical mechanism., and hence, cannot be counted as a strictly nucleophilic substitution reaction.
The last reaction is also a free radical mechanism, and is almost identical to the third reaction mechanism given above:
The first step involves the splitting of the alkyl halide and the hydrogen molecule into free radicals:
$R - X \to \mathop R\limits^ \bullet + \mathop X\limits^ \bullet$
${H_2} \to \mathop H\limits^ \bullet + \mathop H\limits^ \bullet$
One hydrogen free radical combines with the alkyl free radical, while the other combines with the halogen free radical:
$\mathop H\limits^ \bullet + \mathop R\limits^ \bullet \to R - H$
$\mathop H\limits^ \bullet + \mathop X\limits^ \bullet \to H - X$
Thus, even this is not a nucleophilic substitution reaction.
Hence, as we have seen, only the second reaction is a nucleophilic substitution reaction.

So, the correct answer is Option B .

Note: Nucleophilic substitution reactions are mainly of two types: $S{N^1}$ (first order) and $S{N^2}$ (second order). $S{N^1}$ is mainly given by bulky molecules and doesn’t change the optical activity of molecules. However, $S{N^2}$ reaction leads to racemisation, that is, produces optical inversion.