Question

\(\mathrm{CH}_{3} \mathrm{Br}+\mathrm{Nu}^{-} \longrightarrow \mathrm{CH}_{3}-\mathrm{Nu}+\mathrm{Br}^{-}\) The decreasing order of the rate of the above reaction with nucleophiles \(\left(\mathrm{Nu}^{-}\right)\) a to \(\mathrm{d}\) is \(\left[\mathrm{Nu}^{-}=\begin{array}{lll}\text { (a) } \mathrm{PhO}^{-} & \text {(b) } \mathrm{AcO}^{-} & \text {(c) } \mathrm{HO}^{-} & \text {(d) } \mathrm{CH}_{3} \mathrm{O}^{-}\end{array}\right]\) (a) \(d>c>a>b\) (b) \(\mathrm{d}>\mathrm{c}>\mathrm{b}>\mathrm{a}\) (c) \(\mathrm{a}>\mathrm{b}>\mathrm{c}>\mathrm{d}\) (d) \(\mathrm{b}>\mathrm{d}>\mathrm{c}>\mathrm{a}\)

Short answer

The correct order is (a) \(d > c > a > b\).

Step-by-step solution

Identify the Reaction Type

The given reaction involves the substitution of the bromine atom in \(\mathrm{CH}_3 \mathrm{Br}\) by a nucleophile \(\mathrm{Nu}^{-}\). This is a nucleophilic substitution reaction, typically described by the S\(\mathrm{N}2\) mechanism, where the nucleophile attacks the carbon atom bonded to the leaving group.

Assess Nucleophilic Strength

Nucleophilic strength is affected by factors like basicity and steric hindrance. Generally, stronger bases are better nucleophiles in protic solvents, and less sterically hindered species are more effective nucleophiles.

Compare Basicity of Nucleophiles

Compare the basicity of the given nucleophiles: \(\mathrm{PhO}^{-}, \mathrm{AcO}^{-}, \mathrm{HO}^{-}, \mathrm{CH}_3 \mathrm{O}^{-}\). The order of basicity is: \(\mathrm{CH}_3 \mathrm{O}^{-} > \mathrm{HO}^{-} > \mathrm{PhO}^{-} > \mathrm{AcO}^{-}\). Methoxide (\(\mathrm{CH}_3 \mathrm{O}^{-}\)) and hydroxide (\(\mathrm{HO}^{-}\)) are strong bases due to the small, highly electronegative oxygen atom.

Evaluate Steric Hindrance

Among the nucleophiles, steric hindrance can also impact reactivity. Both \(\mathrm{CH}_3 \mathrm{O}^{-}\) and \(\mathrm{HO}^{-}\) are less hindered compared to \(\mathrm{PhO}^{-}\) and \(\mathrm{AcO}^{-}\), which makes them more reactive in S\(\mathrm{N}2\) reactions.

Rank the Nucleophiles

Using both basicity and steric hindrance, the order of reactivity is \(\mathrm{CH}_3 \mathrm{O}^{-} > \mathrm{HO}^{-} > \mathrm{PhO}^{-} > \mathrm{AcO}^{-}\). This ranking is based on their ability to donate electrons and bond with carbon, as well as minimal hindrance in the rate-limiting step.

Match the Order with Answer Choices

The calculated order corresponds to option (a) \(\mathrm{d} > \mathrm{c} > \mathrm{a} > \mathrm{b}\).

Key concepts

SN2 Mechanism

An SN2 mechanism refers to a nucleophilic substitution reaction that occurs in one step. The 'S' stands for substitution, the 'N' for nucleophilic, and the '2' indicates bimolecular—meaning that two molecules are involved in the rate-determining step. In the example given, the nucleophile \( \mathrm{Nu}^{-} \) attacks the carbon atom of \( \mathrm{CH}_3 \mathrm{Br} \), leading to the displacement of the bromine atom.

This type of reaction is characterized by a backside attack. The nucleophile approaches the carbon opposite to the leaving group, which in this case is bromide (\( \mathrm{Br}^{-} \)). Due to the single-step nature of the SN2 reaction, both the nucleophile and the substrate (\( \mathrm{CH}_3 \mathrm{Br} \)) affect the rate. Therefore, an SN2 mechanism is highly sensitive to the structure of both components. Ideally, less hindered substrates accelerate the reaction. This is because they allow the nucleophile better access to the electrophilic center.

• Occurs in one concerted step.
• Involves a backside attack from the nucleophile.
• Rate depends on both the nucleophile and the substrate.

Nucleophilic Strength

Nucleophilic strength measures the ability of a nucleophile to donate an electron pair to form a new bond. Several factors influence nucleophilic strength, including the nucleophile’s charge, electronegativity, the solvent, and the presence of bulky groups.

Charged nucleophiles, like \( \mathrm{CH}_3 \mathrm{O}^{-} \) or \( \mathrm{HO}^{-} \), are typically stronger nucleophiles because they carry a negative charge. This negative charge indicates an excess of electrons, making these molecules more eager to participate in reactions by donating their electrons to electrophiles.

However, the solvent can dramatically affect nucleophilic strength. In protic solvents, which have hydrogen-bonding capability (like water or alcohols), smaller nucleophiles like \( \mathrm{F}^{-} \) are heavily solvated and hence less nucleophilic, whereas larger anions like \( \mathrm{I}^{-} \) might show greater nucleophilicity in such environments.

• Negative charge often increases nucleophilic strength.
• Solvents affect how nucleophiles interact with reaction partners.
• The size and electronegativity can alter nucleophilicity.

Basicity

Basicity and nucleophilicity, though related, are not the same. Basicity describes how well a substance can accept a proton, while nucleophilicity is about electron pair donation. Despite these differences, a strong base often turns out to be a good nucleophile, particularly in protic solvents.

In our case, \( \mathrm{CH}_3 \mathrm{O}^{-} \) and \( \mathrm{HO}^{-} \) are strong bases. This is due to the high electronegativity of oxygen and the small atomic size, which allows these ions to hold onto their electrons tightly. Their basic nature corresponds to a high preference for participating in SN2 reactions where electron-donation to an electrophile is necessary. However, basicity is not an absolute determinant of nucleophilic strength because factors like steric hindrance can also limit the ability of a base to act as a nucleophile.

• Basicity is about proton acceptance.
• Strong bases often are stronger nucleophiles.
• Factors beyond basicity can affect reaction outcomes.

Steric Hindrance

Steric hindrance refers to the physical obstruction of a site by surrounding atoms or groups of atoms, making it less accessible to reactants. In an SN2 reaction, the nucleophile must approach the central carbon atom closely. Therefore, it's critical for the carbon to be as free from surrounding bulk as possible.

In our example, \( \mathrm{CH}_3 \mathrm{O}^{-} \) and \( \mathrm{HO}^{-} \) exhibit minimal steric hindrance compared to \( \mathrm{PhO}^{-} \) and \( \mathrm{AcO}^{-} \), thus making them more effective in SN2 reactions. Steric hindrance can slow down or even prevent a reaction if the nucleophile cannot adequately access the electrophilic center due to bulky groups around the carbon atom.

This factor is critical in determining the rate of SN2 reactions because no matter how strong the nucleophile might be, if it physically cannot interact with the substrate, the reaction will proceed very slowly or not at all.

• Steric hindrance affects reaction rates significantly.
• Less bulk around the reacting site means faster SN2 reactions.
• Steric hindrance is a crucial factor in practical chemical synthesis.