Question

Identify the nucleophile, electrophile, and leaving group in each of the following substitution reactions. Predict whether equilibrium favors the reactants or products: (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{ONa}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{I} \rightleftharpoons\) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{3}+\mathrm{NaI}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{3}^{+}+\mathrm{KI} \rightleftharpoons\) \(\mathrm{NH}_{3}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I}+\mathrm{K}^{+}\)

Short answer

For the first reaction, the nucleophile is \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{ONa}\), the electrophile is \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{I}\), and the leaving group is \(I^{-}\). The reaction favors the products. For the second reaction, the nucleophile is \(KI\), the electrophile is \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{3}^{+}\), and the leaving group is \(\mathrm{NH}_{3}\). The reaction predominantly favors the reactants.

Step-by-step solution

Identifying reaction components for first reaction

First, let's focus on the first reaction. The formula is \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{ONa}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{I} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{3}+\mathrm{NaI}\). \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{ONa}\) acts as the nucleophile, as the negatively charged oxygen can donate a pair of electrons. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{I}\) is the electrophile, because it has a slightly positive carbon due to the electronegativity of iodine, making it susceptible to an attack from a nucleophile. Iodine (\(I^{-}\)) is the leaving group as it gets displaced in the course of reaction.

Predicting equilibrium for first reaction

The reaction favors the products because the leaving group is a stable iodide ion (\(I^{-}\)), and thus a good leaving group. Furthermore, formation of sodium iodide is driven by ionic bond formation which is a strong interaction.

Identifying reaction components for second reaction

Now, let's look at the second reaction. The formula is \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{3}^{+}+\mathrm{KI} \rightleftharpoons \mathrm{NH}_{3}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I}+\mathrm{K}^{+}\). Potassium iodide (\(KI\)) acts as the nucleophile because iodine (\(I^{-}\)) can donate a pair of electrons. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{3}^{+}\) is the electrophile as it has a slightly positive carbon. The leaving group here is \(\mathrm{NH}_{3}\).

Predicting equilibrium for second reaction

The reaction predominantly favors the reactants as \(\mathrm{NH}_{3}\), the leaving group, is a poor leaving group. It is a strong base and has difficulty maintaining a positive charge after leaving, making it less likely for the reaction to proceed in the direction of the products.

Key concepts

Nucleophile and Electrophile

Understanding the actors in chemical reactions is crucial for mastering organic chemistry. In the world of substitution reactions, nucleophiles and electrophiles play starring roles.

A nucleophile is like a chemical 'Robin Hood'—it's generous, willing to donate its pair of electrons to form a bond. Typically, nucleophiles are negative ions or have a partial negative charge. They seek out positively charged or electron-deficient areas in other molecules. In our exercise, \(\text{CH}_3\text{CH}_2\text{ONa})\) and \(\text{KI}\) are examples of nucleophiles.

On the opposite end, we have electrophiles. These molecules are like the 'Sheriff of Nottingham' in our story—they are electron-deficient and are on the quest to accept a pair of electrons. Often, they are positive ions or have a partial positive charge. They're the target for the noble nucleophiles. In the given reactions, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{I})\) and \(\text{CH}_3\text{CH}_2\text{NH}_3^+}\) are the electrophiles.

Equilibrium Prediction

Imagine a tug-of-war between reactants and products in a chemical reaction. Equilibrium is the referee that decides which side has the winning force. It's a state where the rate at which the reactants convert to products is equal to the rate at which products revert back to reactants. Predicting where the equilibrium will favor is a bit like predicting the weather—there are signs and rules.

In substitution reactions, the nature of the leaving group and the strength of the new bond formed are crucial clues. A strong, stable leaving group will favor products, as seen in our first reaction with the iodide ion (\(I^-\)), where products are favored. Conversely, if the leaving group is weak and unstable, the reactants are more likely not to 'let go,' hence favoring the reactants side, as demonstrated in the second reaction with \(NH_3\) as the leaving group. Other factors like temperature, solvent, and the concentration of reactants/products also sway the equilibrium's position.

Leaving Group Identification

The concept of a 'leaving group' is essential in substitution reactions. It's the part of the molecule that leaves - taking the exit door during the chemical 'party.' Identifying a good leaving group is akin to spotting someone at a party who's had enough and is ready to leave quietly without causing a fuss.

Good leaving groups are usually weak bases that can stabilize the negative charge they carry after departure. They are stable on their own, just like iodide (\(I^-\)) in the first reaction. Conversely, a poor leaving group, such as \(NH_3\) in the second reaction, is a strong base and is like a partygoer who does not want to leave — making the reaction less likely to proceed. When identifying leaving groups, consider their stability after they've left because that will greatly affect the reaction's outcome.