Consider the compounds$$\left( I \right){\text{ }}R - I{\text{ }}\left( {II} \right){\text{ }}R - Br{\text{ }}\left( {III} \right){\text{ }}R - Cl{\text{ }}\left( {IV} \right){\text{ }}R - F$$. The rate of $$S{N_2}$$reaction is $$ A. I II III IV \\ B. II I III IV \\ C. I = II = III = IV \\ D.{\text{}}IV III II I \\ $$

Question

Consider the compounds$$\left( I \right){\text{ }}R - I{\text{ }}\left( {II} \right){\text{ }}R - Br{\text{ }}\left( {III} \right){\text{ }}R - Cl{\text{ }}\left( {IV} \right){\text{ }}R - F$$. The rate of $$S{N_2}$$reaction is $$  A. I > II > III > IV   \\  B. II > I > III > IV   \\  C. I = II = III = IV   \\  D.{\text{}}IV > III > II > I   \\  $$

Vedantu Content Team · Accepted Answer

Hint: First we must know that the $$S{N_2}$$ stands for Nucleophilic Substitution, Second Order reaction. Types of Alkyl halide, solvent, nucleophilicity, leaving group, etc are some factors that affect the rate of $$S{N_2}$$ reaction. Complete step by step answer:There are various factors that may affect the $$S{N_2}$$ reaction. Leaving the group is one the important factor that impacts the $$S{N_2}$$ reaction. The leaving group is almost always excluded with a full negative charge, so those leaving groups are best that can best stabilize an anion. $$^ - OTos{,^ - }OMs{,^ - }Br{,^ - }I{,^ - }Cl{,^ - }NH2{,^ - }OH{,^ - }F,{\text{ }}etc.$$are some examples of leaving group. In the $$S{N_2}$$ mechanism, the reactivity of halides for the same alkyl group increases in the order. This happens because as the size increases, the halide ion becomes a better leaving group.The reactivity of the halogen group decreases with an increase in the $$C - X$$ bond energy. Bond energy of $$C - F$$ is maximum and so fluorides are least reactive. On the other hand, the bond energy of $$C - I$$ is minimum and hence, iodides are most reactive. In other words, iodide is the best leaving group among the four halide ions, making alkyl iodide the most reactive.Thus, Alkyl halides follow the reactivity sequence, $$R - F{\text{ }} < {\text{ }}R - Cl{\text{ }} < {\text{ }}R - Br{\text{ }} < {\text{ }}R - I$$Hence, Option A is the correct option.Note:$$S{N_2}$$ reactions are particularly sensitive to crowding or steric factors. $$S{N_2}$$ reactions get greatly affected by steric hindrance at the site of reaction. So, the order of reactivity of alkyl halides in $$S{N_2}$$ reactions is as follows: Methyl > primary halide > secondary halidesThe tertiary alkyl halides have high steric factors or are so crowded that they do not generally react by an $$S{N_2}$$ mechanism.

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