The correct order of reactivity of following compounds in ${S_{N}}^{2}$ reaction is:

Q: The correct order of reactivity of following compounds in ${S_{N}}^{2}$ reaction is:

The correct order of reactivity of following compounds in ${S_{N}}^{2}$ reaction is:

Question

Accepted Answer

In an ${S_{N}}^{2}$ reaction, the reactivity of alkyl halides is influenced by steric hindrance. The ${S_{N}}^{2}$ mechanism involves a backside attack by the nucleophile, which is more efficient when there is less steric hindrance around the carbon atom bonded to the leaving group (in this case, bromine).

Let's analyze the given compounds in Option 3:

1. **$CH_{3}-CH_{2}-CH_{2}-CH_{2}-Br$ (1-bromobutane):** This is a primary alkyl halide with minimal steric hindrance, making it highly reactive in ${S_{N}}^{2}$ reactions.

2. **$(CH_{3})_{2}CH-CH_{2}Br$ (2-bromo-2-methylpropane):** This is a secondary alkyl halide. Although it has more steric hindrance than a primary halide, it is still relatively reactive in ${S_{N}}^{2}$ reactions compared to tertiary halides.

3. **$CH_{3}-CH_{2}-CH(Br)-CH_{3}$ (2-bromobutane):** This is also a secondary alkyl halide, similar to the previous compound, but with slightly more steric hindrance due to the presence of an additional ethyl group.

4. **$(CH_{3})_{3}C-Br$ (tert-butyl bromide):** This is a tertiary alkyl halide with significant steric hindrance, making it the least reactive in ${S_{N}}^{2}$ reactions.

Therefore, the correct order of reactivity in ${S_{N}}^{2}$ reactions, from most to least reactive, is: $CH_{3}-CH_{2}-CH_{2}-CH_{2}-Br > (CH_{3})_{2}CH-CH_{2}Br > CH_{3}-CH_{2}-CH(Br)-CH_{3} > (CH_{3})_{3}C-Br$. This matches Option 3.

Now, let's explain why the other options are incorrect:

1. **Option 1:** This option incorrectly places the tertiary alkyl halide, $(CH_{3})_{3}C-Br$, as more reactive than the secondary and primary halides, which contradicts the principle of steric hindrance affecting ${S_{N}}^{2}$ reactivity.

2. **Option 2:** This option incorrectly suggests that the tertiary alkyl halide is the least reactive, which is correct, but it places the primary alkyl halide, $CH_{3}-CH_{2}-CH_{2}-CH_{2}Br$, as less reactive than the secondary halides, which is incorrect.

4. **Option 4:** This option incorrectly suggests that the secondary alkyl halide, $CH_{3}-CH_{2}-CH(Br)-CH_{3}$, is more reactive than the primary alkyl halide, $CH_{3}-CH_{2}-CH_{2}-CH_{2}-Br$, which contradicts the ${S_{N}}^{2}$ mechanism's preference for less steric hindrance.

In conclusion, Option 3 correctly reflects the order of reactivity for the given compounds in an ${S_{N}}^{2}$ reaction, based on the degree of steric hindrance.