Which of the following is the correct order of decreasing SN2 reactivity ?

To determine the correct order of decreasing SN2 reactivity, we need to consider the factors that influence the SN2 reaction mechanism. The SN2 reaction is a bimolecular nucleophilic substitution process where the nucleophile attacks the electrophile, leading to the displacement of a leaving group. The key factors affecting reactivity include steric hindrance, the nature of the leaving group, and the solvent used.

Factors Influencing SN2 Reactivity

1. **Steric Hindrance**: In SN2 reactions, steric hindrance plays a significant role. The more crowded the carbon atom is, the less accessible it becomes for the nucleophile. Therefore, primary alkyl halides are more reactive than secondary, and tertiary alkyl halides are the least reactive in SN2 reactions.

2. **Leaving Group Ability**: A good leaving group can stabilize the negative charge after it departs, making the reaction more favorable. Common good leaving groups include halides (like I^- and Br^-) and tosylate. The better the leaving group, the faster the reaction will proceed.

3. **Nucleophile Strength**: Stronger nucleophiles can attack the electrophile more effectively. For example, I^- is a stronger nucleophile than F^- due to its larger size and lower electronegativity, which makes it less hindered by steric factors.

Order of Reactivity

Taking these factors into account, the general order of decreasing SN2 reactivity can be summarized as follows:

• Primary alkyl halides (most reactive)
• Secondary alkyl halides
• Tertiary alkyl halides (least reactive)

For example, if we compare the reactivity of the following compounds:

• 1-bromopropane (primary)
• 2-bromobutane (secondary)
• 2-bromo-2-methylpropane (tertiary)

The order of reactivity would be:

• 1-bromopropane > 2-bromobutane > 2-bromo-2-methylpropane

Example of Reactivity in Action

Let’s say we have a reaction involving these three compounds with a strong nucleophile like sodium iodide (NaI). The primary alkyl halide (1-bromopropane) will react the fastest because it has the least steric hindrance, allowing the iodide ion to attack the carbon atom easily. The secondary alkyl halide (2-bromobutane) will react more slowly due to increased steric hindrance, and the tertiary alkyl halide (2-bromo-2-methylpropane) will not undergo an SN2 reaction at all, as it is too hindered and will instead favor an SN1 mechanism if conditions allow.

In summary, when considering the order of decreasing SN2 reactivity, remember that primary alkyl halides are the most reactive, followed by secondary, and tertiary alkyl halides are the least reactive due to steric hindrance and the nature of the leaving group. This understanding is crucial for predicting the outcomes of nucleophilic substitution reactions in organic chemistry.