Which of the following alkyl halides will undergo $S_N2$ reaction most readily?

1-Bromobutane

The correct answer is Option 3. 1-Bromobutane.

In an \(S_N2\) (Substitution Nucleophilic Bimolecular) reaction, the nucleophile attacks the carbon atom that is bonded to the leaving group (in this case, bromine) from the opposite side. This results in the simultaneous formation of a new bond between the nucleophile and the carbon atom and the breaking of the bond between the carbon and the leaving group.

Key Points about \(S_N2\) Reactions:

Single Step: The reaction occurs in a single concerted step.

Backside Attack: The nucleophile attacks from the side opposite to the leaving group.

Inversion of Configuration: The reaction results in the inversion of the stereochemistry at the carbon center (if the center is chiral).

Rate of \(S_N2\) Reaction:

The rate of an SN2 reaction depends on:

Steric Hindrance: The nucleophile needs to access the carbon atom that is bonded to the leaving group. More steric hindrance around this carbon makes it harder for the nucleophile to approach, thereby slowing down the reaction.

Nature of Leaving Group: A good leaving group (like bromine) can facilitate the reaction. All the given alkyl halides have bromine as the leaving group, so this factor is constant among the options.

Analysis of Each Alkyl Halide:

1. 1-Bromo-2,2-Dimethylpropane:

The carbon attached to bromine is tertiary (the carbon is connected to three other carbons), and it has two bulky methyl groups attached to it. This significant steric hindrance around the carbon center makes it very difficult for the nucleophile to approach and attack the carbon atom. Due to high steric hindrance, this compound is the least reactive in an SN2 reaction.

2. 1-Bromo-2-Methylbutane:

The carbon attached to bromine is primary (the carbon is bonded to only one other carbon). It has a methyl group on the adjacent carbon, which adds some steric hindrance, but it is relatively minimal compared to more hindered structures. The reactivity in an \(S_N2\) reaction is higher than that of 1-Bromo-2,2-dimethylpropane due to lower steric hindrance.

3. 1-Bromobutane:

The carbon attached to bromine is primary (the carbon is bonded to only two hydrogens and no additional alkyl groups). This carbon center is relatively accessible to the nucleophile with minimal steric hindrance. The low steric hindrance allows the nucleophile to easily attack the carbon, making it highly reactive in \(S_N2\) reactions.

4. 1-Bromo-3-Methylbutane:

The carbon attached to bromine is primary, but there is a methyl group on the third carbon which introduces some steric hindrance. However, this is still less than the hindrance in 1-Bromo-2,2-dimethylpropane. Although it has less steric hindrance than 1-Bromo-2,2-dimethylpropane, it is not as favorable as 1-Bromobutane for \(S_N2\) reactions.

Conclusion

1-Bromobutane has the least steric hindrance around the carbon attached to the bromine, making it the most readily accessible for nucleophilic attack. In \(S_N2\) reactions, less steric hindrance leads to faster and more efficient reactions.

Therefore, the alkyl halide that will undergo an SN2 reaction most readily is 1-Bromobutane.