Question

Consider the reaction of isopropyl iodide with various nucleophiles. For each pair, predict which will give the larger substitution/elimination ratio. (a) I or \(\mathrm{Cl}\) (b) \(\mathrm{N}\left(\mathrm{CH}_{3}\right)_{3}\) or \(\mathrm{P}\left(\mathrm{CH}_{3}\right)_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{~S}^{-}\) or \(\mathrm{CH}_{3} \mathrm{O}^{-}\)

Short answer

(a) Cl will give the larger substitution/elimination ratio, as it is a stronger nucleophile than I. (b) \(\mathrm{N}\left(\mathrm{CH}_{3}\right)_{3}\) will give the larger substitution/elimination ratio, as it is a stronger nucleophile and base compared to \(\mathrm{P}\left(\mathrm{CH}_{3}\right)_{3}\). (c) \(\mathrm{CH}_{3} \mathrm{~S}^{-}\) will give the larger substitution/elimination ratio, as it is a better nucleophile but a weaker base than \(\mathrm{CH}_{3} \mathrm{O}^{-}\).

Step-by-step solution

(a) Compare I and Cl

(For this pair, we are comparing two halide ions: iodide (I) and chloride (Cl). In terms of nucleophilicity, iodide is a weaker nucleophile than chloride due to its larger size and lower charge density. This means that iodide will favor elimination over substitution more than chloride. Therefore, Cl will give the larger substitution/elimination ratio.)

(b) Compare \(\mathrm{N}\left(\mathrm{CH}_{3}\right)_{3}\) and \(\mathrm{P}\left(\mathrm{CH}_{3}\right)_{3}\)

(In this pair, we are comparing two similar molecules containing nitrogen (N) and phosphorus (P) with three methyl (CH3) groups attached. The key difference between these two molecules is that nitrogen is more electronegative and smaller than phosphorus. As a result, \(\mathrm{N}\left(\mathrm{CH}_{3}\right)_{3}\) is both a stronger nucleophile and a stronger base compared to \(\mathrm{P}\left(\mathrm{CH}_{3}\right)_{3}\). This makes \(\mathrm{N}\left(\mathrm{CH}_{3}\right)_{3}\) more likely to favor substitution over elimination. Therefore, \(\mathrm{N}\left(\mathrm{CH}_{3}\right)_{3}\) will give the larger substitution/elimination ratio.)

(c) Compare \(\mathrm{CH}_{3} \mathrm{~S}^{-}\) and \(\mathrm{CH}_{3} \mathrm{O}^{-}\)

(Here we are comparing two ions: the thiolate ion (\(\mathrm{CH}_{3} \mathrm{~S}^{-}\)) and the alkoxide ion (\(\mathrm{CH}_{3} \mathrm{O}^{-}\)). The sulfur in the thiolate ion is larger and less electronegative than the oxygen in the alkoxide ion. This difference makes the thiolate ion a better nucleophile but a weaker base than the alkoxide ion. Since a stronger nucleophile favors substitution over elimination, \(\mathrm{CH}_{3} \mathrm{~S}^{-}\) will give the larger substitution/elimination ratio.)

Key concepts

Nucleophilicity

Nucleophilicity is a crucial concept in organic chemistry, especially when discussing nucleophilic substitution reactions. A nucleophile is a species rich in electrons that seeks out positively charged or electron-deficient areas in molecules to form new chemical bonds. This is essential in nucleophilic substitution reactions, where a nucleophile replaces a leaving group in a molecule.
The strength of a nucleophile can greatly influence reaction pathways. It is mainly determined by:

• Charge: Negatively charged nucleophiles are generally stronger than their neutral counterparts.
• Electronegativity: Less electronegative atoms are more willing to share their electron pairs, making them stronger nucleophiles.
• Size and Polarizability: Larger atoms or groups are more polarizable and can easily displace a leaving group, often being better nucleophiles despite potentially weaker bonding.

Taking a closer look at the exercise, comparing iodide (I⁻) with chloride (Cl⁻) reveals that chloride is the stronger nucleophile due to its smaller size and higher charge density. This allows it to participate more effectively in substitution, compared to the larger, less effective iodide ion.

Substitution vs Elimination

In organic chemistry, understanding when a reaction undergoes substitution versus elimination is fundamental. Both reactions involve nucleophiles and can occur under similar conditions, but result in very different products.
Substitution reactions are classified into:

• SN1 (Substitution Nucleophilic Unimolecular): Occurs in two steps and is favored when the leaving group departs on its own to form a carbocation, followed by nucleophile attack.
• SN2 (Substitution Nucleophilic Bimolecular): Occurs in one concerted step where the nucleophile attacks the substrate, displacing the leaving group simultaneously.

Elimination reactions involve the removal of elements from a molecule, forming double bonds.
The decision between substitution and elimination is influenced by properties such as the strength of the base and the nucleophile, the steric hindrance present in the substrate, and the solvent conditions. Generally:

• Stronger nucleophiles favor substitution.
• Stronger bases, especially with sterically hindered substrates, drive elimination.

Continuing with our exercise, trimethylamine (\(\mathrm{N}(\mathrm{CH}_{3})_{3}) \) is a stronger nucleophile compared to trimethylphosphine (\(\mathrm{P}(\mathrm{CH}_{3})_3)\), promoting substitution over elimination.

Halide Comparison

Halides play a pivotal role as leaving groups in nucleophilic substitution and as agents in elimination reactions. Understanding their properties can help predict reaction outcomes.
Comparing different halide ions like iodide and chloride reveals important insights:

• Size and Polarizability: Larger halides like iodide are more polarizable. This means they can disperse charges more easily, which tends to make them better leaving groups, but weaker nucleophiles.
• Bond Strength: Halides with weaker bonds should theoretically be easier to remove during nucleophilic substitution, but their ability as nucleophiles also depends on their other chemical properties.

Considering chlorides and iodides, while chloride generally forms stronger bonds to carbon due to its smaller size and greater electronegativity, this also makes it a stronger nucleophile.
In the context of organic reactions involving these ions, chloride tends to engage more readily in substitution due to its nucleophilic prowess, whereas iodide might lean more towards elimination due to its larger size.