Question

Which compound in each of the following pairs will react faster in an \(\mathrm{S}_{\mathrm{N}}^{2}\) reaction with \(\mathrm{OH}^{-} ?\) (a) \(\mathrm{CH}_{3} \mathrm{Br}\) or \(\mathrm{CH}_{3} \mathrm{I}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I}\) in ethanol or in dimethyl sulfoxide (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) or \(\mathrm{CH}_{3} \mathrm{Cl}\) (d) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CHBr}\) or \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCH}_{2} \mathrm{Br}\)

Short answer

(a) \( \mathrm{CH}_{3} \mathrm{I} \); (b) in DMSO; (c) \( \mathrm{CH}_{3} \mathrm{Cl} \); (d) \( \mathrm{H}_{2} \mathrm{C} = \mathrm{CHCH}_{2} \mathrm{Br} \).

Step-by-step solution

Analyzing the Leaving Group

In an \( \mathrm{S}_{\mathrm{N}}^{2} \) reaction, the speed of the reaction depends significantly on the leaving group. A better leaving group facilitates a faster reaction. Iodide (I\(^-\)) is a better leaving group than bromide (Br\(^-\)) due to its larger size and weaker bond with carbon. Therefore, \( \mathrm{CH}_{3} \mathrm{I} \) will react faster than \( \mathrm{CH}_{3} \mathrm{Br} \) with \( \mathrm{OH}^{-} \).

Solvent Effect Analysis

The choice of solvent impacts \( \mathrm{S}_{\mathrm{N}}^{2} \) reactions significantly. Polar aprotic solvents such as dimethyl sulfoxide (DMSO) are preferred because they help stabilize the transition state and do not solvate the nucleophile. Ethanol is a polar protic solvent and tends to solvate nucleophiles, slowing down the reaction. Hence, \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I} \) will react faster in DMSO than in ethanol.

Size and Sterics of the Substrate

In \( \mathrm{S}_{\mathrm{N}}^{2} \) reactions, less steric hindrance allows for faster reaction rates as the nucleophile can approach the carbon more easily. Tertiary alkyl halides like \( (\mathrm{CH}_{3})_{3} \mathrm{CCl} \) are more hindered compared to primary alkyl halides like \( \mathrm{CH}_{3} \mathrm{Cl} \). Therefore, \( \mathrm{CH}_{3}\mathrm{Cl} \) will react faster than \( (\mathrm{CH}_{3})_{3} \mathrm{CCl} \) with \( \mathrm{OH}^{-} \).

Double Bond Effect

The presence of a double bond adjacent to the leaving group affects the reactivity in \( \mathrm{S}_{\mathrm{N}}^{2} \) reactions. A double bond in \( \mathrm{H}_{2} \mathrm{C} = \mathrm{CHBr} \) causes resonance stabilization, reducing the partial positive charge on the carbon and making it less susceptible to nucleophilic attack. Thus, \( \mathrm{H}_{2} \mathrm{C} = \mathrm{CHCH}_{2}\mathrm{Br} \), without such stabilization by resonance, will react faster with \( \mathrm{OH}^{-} \).

Key concepts

Leaving Group

In an \( \mathrm{S}_{\mathrm{N}}^{2} \) reaction, the leaving group plays a crucial role in determining the reaction rate. A leaving group is the atom or group that detaches from the reaction center, making way for the nucleophile to attack. Generally, the better the leaving group, the faster the reaction.

Iodide (I\(^-\)) is considered a superior leaving group compared to bromide (Br\(^-\)) because it is larger and forms a weaker bond with the carbon atom. This makes the bond easier to break, facilitating a quicker reaction. Hence, in a comparison between \( \mathrm{CH}_{3} \mathrm{I} \) and \( \mathrm{CH}_{3} \mathrm{Br} \), \( \mathrm{CH}_{3} \mathrm{I} \) reacts faster in an \( \mathrm{S}_{\mathrm{N}}^{2} \) process with \( \mathrm{OH}^- \).

In summary, always look for larger atoms like iodide as better leaving groups when predicting \( \mathrm{S}_{\mathrm{N}}^{2} \) reaction rates.

Solvent Effects

The choice of solvent has significant implications for \( \mathrm{S}_{\mathrm{N}}^{2} \) reactions. Solvents can either speed up or slow down these reactions markedly.

Especially, polar aprotic solvents, such as dimethyl sulfoxide (DMSO), are preferred in \( \mathrm{S}_{\mathrm{N}}^{2} \) reactions. These solvents can stabilize the transition state effectively without solvating the nucleophile. This leads to a faster reaction rate.

On the other hand, polar protic solvents, like ethanol, have a tendency to solvate nucleophiles due to their hydrogen bonding capacity. This solvation reduces the nucleophile's reactivity, thus slowing the reaction rate. Consequently, \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I} \) reacts more rapidly in DMSO than in ethanol.

Steric Hindrance

Steric hindrance refers to the physical obstruction that bulky groups provide around the reactive site, impeding the nucleophile's approach.

In an \( \mathrm{S}_{\mathrm{N}}^{2} \) reaction, where the nucleophile needs to access the reactive center efficiently, less steric hindrance is favorable for a fast reaction. Primary alkyl halides, like \( \mathrm{CH}_{3} \mathrm{Cl} \), have less steric hindrance due to fewer bulky groups around the carbon atom.

In contrast, tertiary alkyl halides, such as \( (\mathrm{CH}_{3})_{3} \mathrm{CCl} \), contain more substituents that block the nucleophile's path, thereby slowing the reaction. Therefore, \( \mathrm{CH}_{3} \mathrm{Cl} \) reacts more swiftly in an \( \mathrm{S}_{\mathrm{N}}^{2} \) scenario compared to more crowded counterparts like \((\mathrm{CH}_{3})_{3} \mathrm{CCl} \).

Understanding steric effects helps in predicting the efficiency and speed of \( \mathrm{S}_{\mathrm{N}}^{2} \) reactions.

Double Bond Effect

The presence of double bonds near the reactive center can greatly influence \( \mathrm{S}_{\mathrm{N}}^{2} \) reaction rates. This is known as the double bond effect.

In molecules like \( \mathrm{H}_{2} \mathrm{C} = \mathrm{CHBr} \), resonance effects from the double bond stabilize the molecule, which can stabilize the carbon undergoing nucleophilic attack. This resonance stabilization reduces the partial positive charge on the carbon, making it less reactive to a nucleophile.

Conversely, in molecules like \( \mathrm{H}_{2} \mathrm{C} = \mathrm{CHCH}_{2} \mathrm{Br} \), where the leaving group is not directly bonded to an alkene, the lack of resonance stabilization means that the carbon center remains more reactive to nucleophilic attack. Thus, \( \mathrm{H}_{2} \mathrm{C} = \mathrm{CHCH}_{2} \mathrm{Br} \) reacts faster in \( \mathrm{S}_{\mathrm{N}}^{2} \) reactions.