Question

In the preparation of \(\mathrm{CH}_{3}-\mathrm{O}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\), two methods, namely (i) and (ii) are suggested. Which method is preferable and why? (i) \(\mathrm{CH}_{3} \mathrm{ONa}+\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Br} \rightarrow \mathrm{CH}_{3}-\mathrm{O}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\) (ii) \(\mathrm{CH}_{3} \mathrm{Br}+\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{ONa} \rightarrow \mathrm{CH}_{3}-\mathrm{O}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\)

Short answer

Answer: The preferable method for the synthesis of CH3-O-C(CH3)3 is method (i) because the methoxide ion is a strong nucleophile and can easily replace the bromine atom in the SN1 reaction of the tertiary alkyl bromide. Method (ii) is not preferred because the stronger base would rather eliminate a proton, forming an alkene via the E2 mechanism.

Step-by-step solution

Step 1. Identify the two given methods

Compare the two methods (i) and (ii) provided for the synthesis of \(\mathrm{CH}_{3}-\mathrm{O}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\): (i) \(\mathrm{CH}_{3} \mathrm{ONa}+\left(\mathrm{CH}_{3}\right)_{3}\mathrm{C}-\mathrm{Br} \rightarrow \mathrm{CH}_{3}-\mathrm{O}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\) (ii) \(\mathrm{CH}_{3} \mathrm{Br}+\left(\mathrm{CH}_{3}\right)_{3}\mathrm{C}-\mathrm{ONa} \rightarrow \mathrm{CH}_{3}-\mathrm{O}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\) Both reactions involve the replacement of a halogen (bromine) with a methoxy group (\(\mathrm{CH}_{3}\mathrm{O}^{-}\)).

Step 2. Examine the nature of the reactants

Notice that the reactants in both methods are different: (i) In method (i), the nucleophile is a methoxide ion (\(\mathrm{CH}_{3}\mathrm{O}^{-}\)) and the electrophile is tertiary alkyl bromide: \(\left(\mathrm{CH}_{3}\right)_{3}\mathrm{C}-\mathrm{Br}\). (ii) In method (ii), the nucleophile is a bromide ion (\(\mathrm{Br}^{-}\)) and the electrophile is tertiary alkoxide: \(\left(\mathrm{CH}_{3}\right)_{3}\mathrm{C}-\mathrm{O}\mathrm{Na}\).

Step 3. Evaluate the feasibility of nucleophilic substitution reactions

Tertiary alkyl halides follow the SN1 mechanism for nucleophilic substitution, which involves the formation of a tertiary carbocation intermediate. Method (i) is likely to be feasible because the methoxide ion is a strong nucleophile, and it can readily attack the electrophilic carbon center in the alkyl halide. On the other hand, method (ii) involves the replacement of a strong base (\(\mathrm{CH}_{3}\mathrm{O}^{-}\)) with a weaker one (\(\mathrm{Br}^{-}\)). This is not favored because the stronger base (\(\mathrm{CH}_{3}\mathrm{O}^{-}\)) would rather eliminate a proton from the tertiary alkyl halide, leading to alkene formation via the E2 mechanism instead of the desired product.

Determine the preferable method and explain why

Based on the analysis in Step 3, the preferable method to synthesize \(\mathrm{CH}_{3}-\mathrm{O}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\) is method (i): \(\mathrm{CH}_{3} \mathrm{ONa}+\left(\mathrm{CH}_{3}\right)_{3}\mathrm{C}-\mathrm{Br} \rightarrow \mathrm{CH}_{3}-\mathrm{O}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\) This is because the methoxide ion is a strong nucleophile and can easily replace the bromine atom in the SN1 reaction of the tertiary alkyl bromide. In contrast, method (ii) is not preferred because the stronger base would rather eliminate a proton, forming an alkene via the E2 mechanism.

Key concepts

SN1 Mechanism

The SN1 mechanism is a type of nucleophilic substitution reaction characterized by a two-step process involving a carbocation intermediate. In the context of organic chemistry, ‘SN’ stands for substitution nucleophilic, and ‘1’ denotes the unimolecular nature of the rate-determining step.

In the first step, the leaving group (often a halogen) departs from the substrate, resulting in the formation of a highly reactive carbocation. The rate of this step is independent of the nucleophile's concentration because it involves only the substrate. Consequently, the reaction rate is determined solely by this first step, which is why it's labeled unimolecular.

During the second step, the nucleophile attacks the carbocation, usually leading to the formation of a new covalent bond. Due to the planar structure of a carbocation, nucleophilic attack can occur from either side, often resulting in a racemic mixture of enantiomers when the carbon center is chiral.

E2 Elimination Mechanism

The E2 mechanism stands for bimolecular elimination and is another important reaction in organic chemistry. This mechanism is a one-step process where the base (the nucleophile in this context) removes a proton, which in turn facilitates the expulsion of a leaving group, like a halogen. This concerted process leads to the formation of a double bond, resulting in an alkene.

One key feature of the E2 mechanism is the need for the base and the leaving group to be in a particular spatial arrangement, called the antiperiplanar configuration. This means that the hydrogen atom being abstracted and the leaving group should be on opposite sides of the molecule but in the same plane for the reaction to proceed most effectively.

The E2 mechanism typically occurs when a strong base is present and tends to be favored in systems with steric hindrance that make SN2 reactions less likely. Since the base directly removes a proton in a single concerted step, the rate of an E2 reaction depends on both the concentration of the substrate and the base, hence the term bimolecular.

Tertiary Alkyl Halides

Tertiary alkyl halides are organic compounds containing a carbon atom bonded to three other carbon atoms and a halogen. These are prone to nucleophilic substitution through the SN1 mechanism due to the stable tertiary carbocations they can form upon loss of the leaving group.

As the alkyl group branches increase, so does the ability of carbocations to disperse their positive charge, making them more stable and lowering the energy barrier for their formation. This stability is enhanced by hyperconjugation and the inductive effect of the surrounding alkyl groups.

The nature of tertiary alkyl halides dictates the type of reactions they're likely to undergo. Although prone to SN1, these molecules are often reactive in elimination reactions like E2, especially in the presence of strong bases. Deciding which pathway a reaction will take can hinge on the specific conditions, including the solvent, the temperature, and the strength of the nucleophile or base.

Strong Nucleophile

A strong nucleophile is a species with a heightened tendency to donate an electron pair, resulting in the formation of a new chemical bond. In SN1 and SN2 reactions, the strength of a nucleophile is a critical factor for the reaction’s efficiency and outcome.

Strong nucleophiles are characterized by their negative charge or lone pairs of electrons that are not stabilized by resonance or other electronic effects. Factors that influence nucleophilicity include charge, electronegativity, steric bulk, and the solvent. Ionizable substances in polar protic solvents, such as water or alcohol, tend to form strong nucleophiles.

The presence of a strong nucleophile in a reaction mixture often pushes the pathway toward an SN2 mechanism, especially with primary and secondary substrates. With tertiary substrates, even strong nucleophiles may end up promoting an E2 elimination instead, unless the conditions are specifically geared to favor SN1 substitution — for instance, by using a polar protic solvent to stabilize the carbocation intermediate.