Question

Compare the rate of following reaction with same conditions: (a) \(\mathrm{CH}_{3 \mathrm{M}}-\mathrm{Br}+\underset{1 \mathrm{M}}{\mathrm{NaOH}} \longrightarrow\) (b) \(\mathrm{CH}_{3}-\mathrm{I}+\underset{1 \mathrm{M}}{\mathrm{NaOH}} \longrightarrow\) (c) \(\mathrm{CH}_{3}-\mathrm{I}+\underset{1 \mathrm{M}}{\mathrm{NaSH}} \longrightarrow\) (A) \(r_{\mathrm{a}}>r_{\mathrm{b}}>r_{\mathrm{c}}\) (B) \(r_{\mathrm{b}}>r_{\mathrm{a}}>r_{\mathrm{c}}\) (C) \(r_{\mathrm{c}}>r_{\mathrm{b}}>r_{\mathrm{a}}\) (D) \(r_{\mathrm{a}}=r_{\mathrm{b}}=r_{\mathrm{c}}\)

Short answer

The reaction rate order is \(r_c > r_b > r_a\), which corresponds to answer choice (C).

Step-by-step solution

Identify the main factors affecting the reaction rates

In the given reactions, we can identify two main factors that can affect the reaction rates: the nucleophilicity of the reacting species and the leaving group ability of the halogen atoms. We'll analyze each factor separately and then combine our observations to determine the overall reaction rates.

Compare the nucleophilicity of the reacting species

In reactions (a) and (b), NaOH is the reacting species, while in reaction (c), NaSH is the reacting species. Both NaOH and NaSH are nucleophiles, which means they can donate an electron pair to an electrophilic atom. The stronger the nucleophile, the faster the reaction will proceed. In both reactions (a) and (b), the nucleophile is OH⁻, while in reaction (c), the nucleophile is SH⁻. As an anion, SH⁻ is larger than OH⁻ and has a weaker O-H/S-H bond. This means that SH⁻ is a better nucleophile compared to OH⁻. Therefore, reaction (c) has the best nucleophile.

Compare the leaving group ability of the halogen atoms

In reactions (a), the leaving group is Br⁻, while in reactions (b) and (c), the leaving group is I⁻. The effectiveness of a leaving group depends on its stability once it has left. In general, the larger the halogen atom, the better the leaving group, as it can stabilize the negative charge more effectively. This means that I⁻ is a better leaving group than Br⁻. Therefore, reactions (b) and (c) have better leaving groups compared to reaction (a).

Combine the observations to determine the overall reaction rates

Based on our analysis, we can make the following observations: 1. Reaction (c) has the best nucleophile (SH⁻) and a good leaving group (I⁻). 2. Reactions (b) and (a) have the same nucleophile (OH⁻), but reaction (b) has a better leaving group (I⁻) than reaction (a) (Br⁻). Based on these observations, the reaction rate order should be \(r_c > r_b > r_a\) as reaction (c) has the best nucleophile and a good leaving group, while reaction (a) has the worst leaving group. This corresponds to answer choice (C).

Key concepts

Nucleophilicity

Understanding nucleophilicity is essential when comparing organic reaction rates. Nucleophilicity refers to the ability of a species to donate an electron pair to an electrophile, thus forming a chemical bond. It's often mistaken with basicity, which measures the ability to donate an electron pair to a proton specifically; however, in the context of reactions, we focus on the tendency to attack a carbon atom.

There are several factors that impact nucleophilicity, including charge, solvent, electronegativity, and steric hindrance. Generally, a negative charge increases nucleophilicity, polar solvents can enhance or diminish nucleophilicity, electronegative atoms tend to be less nucleophilic, and less steric hindrance allows for easier approach to the electrophile.

In the comparison of reaction rates, one key factor is assessing the nucleophile involved. For example, in the given exercise, hydroxide ion (OH⁻) is a strong nucleophile due to its negative charge and smaller size, but thiolate ion (SH⁻), being larger, is even more nucleophilic because it's less hindered and can spread its negative charge over a larger volume, making it more reactive towards electrophiles.

Leaving Group Ability

Leaving group ability is another critical concept in determining the rate of organic reactions. A leaving group is an atom or group that can depart with an electron pair from the substrate molecule, leading to the formation of a new product. The better a leaving group can stabilize the negative charge it takes with it as it leaves, the more readily it will depart.

Key factors that influence leaving group ability include the group's electronegativity, the ability to delocalize or stabilize negative charge, and the strength of the bond with the atom to which it was attached. Generally, halogens like iodide (I⁻) make excellent leaving groups because they are large, less electronegative than oxygen or nitrogen, and can effectively stabilize the negative charge.

In the exercise's scenario, iodide (reaction b and c) is a superior leaving group compared to bromide (reaction a). This superiority is due to iodide's larger size and lower electronegativity, allowing it to handle the negative charge better when it leaves the molecule, thereby speeding up the reaction.

Organic Reaction Mechanisms

Organic reaction mechanisms provide a step-by-step description of the process through which organic reactants transform into products. Understanding these mechanisms is key to predicting reaction outcomes and comparing reaction rates. Mechanisms include bond formations, bond breakages, electron pair movements, and the order in which these events occur.

Reactions can proceed via various mechanisms, including substitutions, eliminations, additions, and rearrangements. To deduce the rate comparison in the given exercise, one must think about the substitution mechanism proceeding in each reaction - whether it follows an SN1 or SN2 pathway, with SN2 involving a direct displacement and requiring a strong nucleophile and a good leaving group.

When comparing the reaction rates in the exercise, it is clear that the quality of both the nucleophile and the leaving group, as explained in the other sections, jointly influence the speed of the SN2 reaction. With the superior nucleophilicity of SH⁻ and the excellent leaving group ability of I⁻, reaction (c) takes precedence in rate over the others, demonstrating the holistic application of organic reaction mechanisms in analyzing reaction rates.