Question

The compound that will react most readily with \(\mathrm{NaOH}\) to form methanol is (a) \(\left(\mathrm{CH}_{3}\right)_{4} \mathrm{~N}^{+} \mathrm{I}\) (b) \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~S}^{+} \mathrm{I}^{-}\) (d) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\)

Short answer

Compound (c) \( (\mathrm{CH}_{3})_{3} \mathrm{~S}^{+} \mathrm{I}^{-} \) reacts most readily with \( \mathrm{NaOH} \) to form methanol.

Step-by-step solution

Identify Potential Reactions

We will assess each compound to determine its potential to produce methanol upon reaction with \( \mathrm{NaOH} \). Methanol formation typically involves the production of \( \mathrm{CH}_3 \mathrm{OH} \) from a methyl source with removal of a leaving group.

Analyze Compound (a)

Compound (a) is \( \left(\mathrm{CH}_{3}\right)_{4} \mathrm{~N}^{+} \mathrm{I} \). This is a quaternary ammonium salt that is stable and doesn't easily release \( \mathrm{CH}_3^+ \), necessary for forming methanol. Thus, it doesn't readily react with \( \mathrm{NaOH} \) to produce methanol.

Analyze Compound (b)

Compound (b) is \( \mathrm{CH}_{3} \mathrm{OCH}_{3} \), dimethyl ether, which is stable and does not have a good leaving group to facilitate substitution with hydroxide to form methanol.

Analyze Compound (c)

Compound (c) is \( \left(\mathrm{CH}_{3}\right)_{3} \mathrm{~S}^{+} \mathrm{I}^{-} \), a sulfonium salt. This compound can easily undergo nucleophilic substitution by \( \mathrm{NaOH} \), where the \( \mathrm{I}^- \) leaves and methanol \( \mathrm{CH}_3 \mathrm{OH} \) is formed from one of the \( \mathrm{CH}_3 \) groups.

Analyze Compound (d)

Compound (d) is \( \left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl} \). This is a tertiary alkyl chloride, which can undergo \( \mathrm{E2} \) elimination reactions rather than forming methanol directly when reacting with \( \mathrm{NaOH} \), as \( \mathrm{Cl}^- \) is a good leaving group, but a beta-elimination is more favorable.

Determine Most Readily Reacting Compound

Based on the analysis, compound (c) \( \left(\mathrm{CH}_{3}\right)_{3} \mathrm{~S}^{+} \mathrm{I}^{-} \) will react most readily with \( \mathrm{NaOH} \) to form methanol, as it can undergo a substitution reaction, easily forming \( \mathrm{CH}_3\mathrm{OH} \).

Key concepts

Nucleophilic substitution

In the world of organic chemistry, nucleophilic substitution reactions play a crucial role. These reactions involve an electron-rich nucleophile attacking a carbon atom that is bonded to a leaving group. The leaving group is displaced, allowing the nucleophile to form a new bond with the carbon atom. The key players in such reactions are:

• Nucleophiles: Species that have a pair of electrons ready to form a bond. In the context of our exercise, the hydroxide ion (\( \mathrm{OH}^- \)) from \( \mathrm{NaOH} \) serves as the nucleophile.
• Substrate: The molecule that contains the carbon atom to be attacked by the nucleophile. In a potential reaction with \( \mathrm{NaOH} \), this could be a molecule like a sulfonium salt with a leaving group, such as \( \mathrm{I}^- \).
• Leaving Group: An atom or group of atoms that can easily depart with a pair of electrons. Effective leaving groups in nucleophilic substitution are usually weak bases, such as iodide ions (\( \mathrm{I}^- \)).

To predict which reactions will occur, consider the type of substrate and the strength of the nucleophile and leaving group. The better the leaving group, the more likely a nucleophilic substitution reaction will happen. Ensure the nucleophile can easily reach the carbon center for an efficient attack.

Sulfonium salts

Sulfonium salts are fascinating compounds in organic chemistry. They involve a sulfur atom bonded to three alkyl or aryl groups and carry a positive charge. A common type of sulfonium salt, like \( (\mathrm{CH}_3)_3 \mathrm{S}^+ \), is involved directly in our exercise. These salts are characterized by:

• Structure: A sulfur atom connected to three methyl groups (tri-methyl sulfonium) and associated with a negative counterion, typically an iodide (\( \mathrm{I}^- \)), forming a neutral compound overall.
• Reactivity: The positive charge on the sulfur makes the adjacent methyl groups more susceptible to attack by nucleophiles. This predisposes sulfonium salts to engage in nucleophilic substitution reactions.
• Application: In reactions with strong nucleophiles, like \( \mathrm{NaOH} \), sulfonium salts can effectively transfer methyl groups, as the counterion (\( \mathrm{I}^- \)) acts as a facile leaving group.

Through nucleophilic substitution in these salts, one can transform certain methyl groups into methanol, which is especially useful for this type of reactive chemistry.

Methanol formation

The formation of methanol during organic reactions arises through the replacement of a group attached to a methyl group with \( \mathrm{OH}^- \). Methanol, known scientifically as \( \mathrm{CH}_3 \mathrm{OH} \), is a vital alcohol with many practical applications. Here's how it forms:

• Reaction Mechanism: In our discussed exercise, methanol is formed when \( \mathrm{NaOH} \) provides \( \mathrm{OH}^- \) as a nucleophile. This hydroxide ion attacks a susceptible methyl group in the sulfonium salt \( (\mathrm{CH}_3)_3 \mathrm{S}^+ \), displacing the iodide ion \( \mathrm{I}^- \) and forming methanol.
• Substitution Process: Here, the sulfur methyl bond in the sulfonium salt is broken as the iodide leaves, allowing \( \mathrm{OH}^- \) to form a new bond with the methyl carbon, resulting in methanol.
• Significance: Methanol is not only a fundamental solvent with industrial importance, but its formation from such reactions also highlights the dynamic nature of organic substitution reactions, showcasing the reactivity of simple chemical structures when subjected to nucleophiles.

Understanding methanol formation helps students and chemists alike grasp how simple molecular interactions can lead to the creation of widely-used and chemically significant compounds.