Problem 86 - General Chemistry: Principles and Modern Applications

Question

When \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{CBr}\) is added to \(\mathrm{CH}_{3} \mathrm{OH}\) at room temperature, the major product is \(\mathrm{CH}_{3} \mathrm{O}\left(\mathrm{CH}_{2} \mathrm{CH}_{3}\right)_{3}\) and a minor product is \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{C}\left(\mathrm{CH}_{2} \mathrm{CH}_{3}\right)_{2} .\) Write out the mechanisms for the reactions leading to these products and use curved arrows to show the movement of electrons.

Step-by-Step Solution

Answer

The major product results from a SN2 reaction where \(\mathrm{CH}_{3} \mathrm{OH}\) acts as a nucleophile, attacking the carbon in \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{CBr}\), resulting \(\mathrm{CH}_{3} \mathrm{O}\left(\mathrm{CH}_{2} \mathrm{CH}_{3}\right)_{3}\). The minor product comes from an elimination mechanism, where a $\beta -$hydrogen is removed along with the leaving group to form a double bond, forming $\mathrm{CH}_{3}\mathrm{CH}=\mathrm{C}\left(\mathrm{CH}_{2} \mathrm{CH}_{3}\right)_{2} .$ Curved arrow notation shows the flow of electrons.

Step 1: Mechanism for the Major Product

The major product is due to a Nucleophilic Substitution reaction. \(\mathrm{CH}_{3} \mathrm{OH}\) acts as a nucleophile and attacks the \(\mathrm{CBr}\), of \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{CBr}\) while \(\mathrm{Br}^{-}\) leaves, resulting in the product \(\mathrm{CH}_{3} \mathrm{O}\left(\mathrm{CH}_{2} \mathrm{CH}_{3}\right)_{3}\).

Step 2: Mechanism for the Minor Product

The minor product is due to an Elimination reaction. In this reaction, a \(\beta-\)hydrogen is removed along with \(\mathrm{Br}^{-}\) to form a double bond in $\mathrm{CH}_{3}\mathrm{CH}=\mathrm{C}\left(\mathrm{CH}_{2} \mathrm{CH}_{3}\right)_{2} .$ Despite being less predominant due to room temperature conditions, this reaction does result in a minor product.

Step 3: Showing Movement of Electrons

Curved arrow notation is used to show the movement of electrons during these reaction mechanisms. An arrow from \(\mathrm{CH}_{3} \mathrm{OH}\) to the Carbon atom of \(\mathrm{CBr}\) will depict the nucleophilic attack for step 1. In step 2, an arrow from the \(\beta-\)hydrogen to the adjacent Carbon followed by an arrow from the Carbon-Br bond to \(\mathrm{Br}\) will show the elimination process.

Key Concepts

Nucleophilic Substitution Mechanism

In a nucleophilic substitution reaction, a nucleophile replaces a leaving group in a compound. This process involves a couple of key components: the nucleophile, the substrate, and the leaving group.

• Nucleophile: In this reaction, \( \text{CH}_3\text{OH} \) acts as a nucleophile because it donates a pair of electrons to form a new bond. Nucleophiles are typically electron-rich species.
• Substrate: The substrate here is \( \left(\text{CH}_3\text{CH}_2\right)_3\text{CBr} \). It contains the carbon atom that gets attacked by the nucleophile.
• Leaving Group: The leaving group is \( \text{Br}^- \), a species that can easily depart with a pair of electrons.

During the reaction, \( \text{CH}_3\text{OH} \) attacks the carbon atom that is bonded to \( \text{Br} \), forming a new bond between the oxygen of methanol and the carbon atom. As this new bond forms, the carbon-bromine bond breaks, allowing \( \text{Br}^- \) to leave the molecule. This creates the major product \( \text{CH}_3\text{O}\left(\text{CH}_2\text{CH}_3\right)_3 \).
Curved arrows in the mechanism indicate the movement of electrons from the nucleophile to the substrate.

Elimination Reaction

Elimination reactions involve the loss of elements from a molecule to form a double bond. For the minor product, we see this type of reaction resulting in the formation of an alkene.
Here’s how it works:

• Formation of a Double Bond: The \( \beta \)-hydrogen, which is a hydrogen on the carbon next to the one bonded to \( \text{Br} \), plays a critical role. The removal of \( \text{Br}^- \) and the \( \beta \)-hydrogen creates a double bond between two carbon atoms.
• Leaving Group and Base: While \( \text{Br}^- \) acts as a leaving group, another base (like solvent molecules) could help remove the \( \beta \)-hydrogen.

The resulting double bond forms the alkene \( \text{CH}_3\text{CH} = \text{C}\left(\text{CH}_2\text{CH}_3\right)_2 \).

In the mechanism, curved arrows show the movement of electrons from the \( \beta \)-hydrogen to the adjacent carbon, and from the carbon-bromine bond to deliberately eject \( \text{Br}^- \).This reaction offers a secondary pathway, creating the minor product.

Electron Movement

Understanding electron movement is key to visualizing chemical reactions. Curved arrows are used in mechanisms to depict the flow of electrons for both substitution and elimination reactions.

• Curved Arrows: These arrows start where electrons originate and point to where they move to. For instance, if an electron pair moves from an oxygen atom to a carbon atom, the arrow starts from the oxygen.
• Depicting Bond Formation and Breaking: A single arrow can show both bond formations and bond breaks. In substitution, an arrow shows nucleophilic attack while another represents the leaving group’s departure.

During elimination, electrons from the \( \beta \)-hydrogen move to form a new bond, and another arrow indicates the breaking of the bond with the leaving group.
This visual tool helps track how electrons reorganize, allowing us to predict the structures of the resulting molecules in these reactions.
Always remember: electrons are the main players in chemical reactions, and understanding their movement is crucial for mastering organic chemistry mechanisms.