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

The compound \\( (\mathrm{CH}_3)_3\mathrm{S}^+\mathrm{I}^- \\) is potentially related but doesn't readily give methanol directly.

Step-by-step solution

Understand the reaction condition

The exercise asks for the compound that reacts most readily with NaOH to form methanol. Methanol \(\text{(CH}_3\text{OH)}\) is an alcohol, and its formation suggests a substitution reaction mechanism.

Analyze each option

To determine which compounds can form methanol, consider the potential for leaving group presence and likelihood of nucleophilic substitution. - \((\text{CH}_3)_4N^+I^-\) does not have a suitable leaving group to form methanol.- \((\text{CH}_3)_3S^+I^-\) can undergo a reaction where NaOH could react with the methyl groups, but forms dimethyl sulfide, not methanol.- \((\text{CH}_3)_3CCl\) would typically form t-butanol upon reaction with NaOH due to SN1 mechanism.- \(\text{CH}_3OCH_3\) does not have any good leaving group for nucleophilic substitution with NaOH to readily form methanol.

Find the best candidate

Among the provided options, none seems to directly provide methanol via a ready substitution. However, based on substitution reaction mechanism contexts, \( (\text{CH}_3)_3S^+I^- \) can provide sulfur-bound leaving potential, but will form different compounds like dimethyl sulfide primarily and not methanol.

Make a conclusion

Given the contextual reasoning and typical organic chemistry reactivity with NaOH, the closest compound that undergoes a known related substitution which can make methanol indirectly in some reactions typically is \( (\text{CH}_3)_3S^+I^- \) due to it having three methyl groups, but it is not the ideal case here.

Key concepts

Methanol Formation

Methanol is a simple alcohol with the chemical formula \(\text{CH}_3\text{OH}\). It is commonly produced in organic chemistry by reactions involving substitution mechanisms.
In the context of this exercise, methanol forms when a methyl group \((\text{CH}_3^+)\) combines with a hydroxide ion \((\text{OH}^-)\). This is often achieved through nucleophilic substitution reactions where a suitable reactant and a strong base or nucleophile like \(\text{NaOH}\) are involved.

Understanding how methanol can be produced in a reaction requires knowledge of potential reactants and their structures. Typically, methanol formation asks for compounds with good leaving groups that can be displaced by a nucleophile, enabling the insertion of the hydroxyl group to create the alcohol. Unfortunately, in our given exercise, none of the reactants convert readily to methanol, reflecting real-world complexity in selecting the right chemical pathways.

Leaving Groups in Organic Chemistry

Leaving groups play a crucial role in nucleophilic substitution reactions. A leaving group is an atom or group that departs the molecule, making way for the incoming nucleophile to bind with the substrate.
In our exercise, a good leaving group would be an element like iodine (\(\text{I}^-\)) or chloride (\(\text{Cl}^-\)), which are both able to accept electrons and leave the reactant molecule.

• In an SN1 reaction, the leaving group departs on its own to form a carbocation, which then reacts with a nucleophile.
• In an SN2 reaction, the nucleophile attacks at the same time as the leaving group departs, leading to a direct substitution.

Effective leaving groups are often the conjugate bases of strong acids, meaning they are stable ions once they have left the starting compound. In our problem setup, although certain compounds had potential leaving groups, the specific structure and reactions did not readily yield methanol.

SN1 and SN2 Mechanisms

Nucleophilic substitution reactions are either SN1 or SN2 processes, each with distinct characteristics. These mechanisms explain how a nucleophile replaces a leaving group on a substrate.

**SN1 Mechanism**:

• Occurs in two steps: leaving group removal to form a carbocation, followed by nucleophile attack.
• Favors substrates with stable carbocations, like tertiary carbon centers.
• Example: \((\text{CH}_3)_3\text{CCl}\) would usually undergo SN1 to form t-butanol rather than methanol.

**SN2 Mechanism**:

• Single concerted step with a simultaneous attack by a nucleophile and departure of the leaving group.
• Favors less hindered substrates, typically primary or secondary carbons.
• Requires a strong nucleophile and a good leaving group for effective reactivity.

In our exercise, the specified compounds did not readily fit into either the SN1 or SN2 mechanisms for methanol synthesis, illustrating the importance of understanding which reactions are viable based on substitution type and molecular structure.