Problem 21 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Which of the following is the correct method for synthesising methylt-butyl ether and why? (i) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}+\mathrm{NaOMe} \rightarrow\) (ii) \(\mathrm{CH}_{3} \mathrm{Br}+\mathrm{NaO}-\mathrm{t}-\mathrm{Bu} \rightarrow\)

Step-by-Step Solution

Answer

The correct method is (ii) because it leads to an \(S_N2\) reaction with methyl bromide and tert-butoxide.

Step 1: Analyze Reaction (i)

The reaction provided is \((\mathrm{CH}_{3})_{3} \mathrm{CBr}+\mathrm{NaOMe} \rightarrow\). This involves using \((\mathrm{CH}_{3})_{3} \mathrm{CBr}\), which has a bulky tertiary carbon, and sodium methoxide \(\mathrm{NaOMe}\). Tertiary bromides often do not undergo \(S_N2\) reactions due to steric hindrance. Instead, elimination reactions (\(E2\)) take place as the methoxide acts as a base and removes a proton, resulting in the formation of an alkene rather than an ether.

Step 2: Analyze Reaction (ii)

The reaction provided is \(\mathrm{CH}_{3} \mathrm{Br}+\mathrm{NaO}-\mathrm{t}-\mathrm{Bu} \rightarrow\). Here, methyl bromide \(\mathrm{CH}_{3} \mathrm{Br}\) is a small, primary substrate conducive for \(S_N2\) reactions. The tert-butoxide \(\mathrm{NaO}-\mathrm{t}-\mathrm{Bu}\), a strong nucleophile, can attack the primary carbon and displace the bromide ion, successfully forming \(\mathrm{CH}_{3} \mathrm{O}-\mathrm{t}-\mathrm{Bu}\), known as methyl tert-butyl ether.

Step 3: Evaluate and Conclude

By considering the two reactions, reaction (i) is unlikely to produce methyl tert-butyl ether due to steric hindrance leading to elimination rather than substitution. Reaction (ii), however, effectively utilizes a primary halide and a strong nucleophile for an \(S_N2\) substitution that results in the formation of methyl tert-butyl ether. Therefore, reaction (ii) is the correct method.

Key Concepts

Steric Hindrance

Steric hindrance occurs when bulky groups around a reactive site interfere with the approach of a nucleophile in a chemical reaction. In organic chemistry, it often impacts reactions like nucleophilic substitutions, where large groups prevent an effective attack by a nucleophile.

For instance, in reaction (i) involving \((\text{CH}_3)_3 \text{CBr}\), the bulky tertiary carbon is surrounded by three methyl groups. This setup creates significant steric hindrance, limiting the access of nucleophiles. As a result, tertiary substrates like \((\text{CH}_3)_3 \text{CBr}\) are poor candidates for \(S_N2\) reactions because the nucleophile struggles to approach and displace the leaving group.

The implication of steric hindrance is that it often leads to alternative reaction pathways. In the case of tertiary substrates, rather than substitution, elimination reactions may occur where a base removes a proton, leading to the formation of an alkene.

Nucleophilic Substitution

Nucleophilic substitution reactions are a cornerstone in organic synthesis whereby a nucleophile replaces a leaving group on a carbon atom. These reactions are generally classified into two types: \(S_N1\) and \(S_N2\).

The \(S_N2\) mechanism is a one-step reaction where the nucleophile attacks the electrophile simultaneously as the leaving group exits. This direct and concerted mechanism prefers primary and less hindered carbon atoms.

In reaction (ii) of the exercise, we see an example of an \(S_N2\) mechanism. The small and unhindered methyl bromide (\( \text{CH}_3 \text{Br} \) allows the nucleophile, \( \text{NaO}-\text{t}-\text{Bu}\), to effectively attack and displace the bromide ion. This results in the synthesis of methyl-tert-butyl ether. The efficiency of \(S_N2\) reactions is highly influenced by substrate and nucleophile characteristics, such as steric factors and nucleophilicity.

Elimination Reaction

Elimination reactions involve the removal of elements from a molecule to form a double bond, often competing with substitution reactions. These reactions can occur via \(E1\) or \(E2\) mechanisms.

For tertiary substrates like \( (\text{CH}_3)_3 \text{CBr} \) in reaction (i), elimination is favored over substitution due to increased steric hindrance. In such cases, bases such as \( \text{NaOMe}\) extract a proton, leading to the formation of an alkene and the loss of a leaving group, rather than forming an ether.

The \(E2\) mechanism, which is relevant here, involves a concerted one-step process. Both the proton removal and the leaving group elimination occur simultaneously. Also, strong bases tend to promote \(E2\) reactions, especially in sterically hindered substrates, where nucleophilic attack is challenging. Understanding the preference of a reaction towards elimination or substitution helps in predicting the outcome and planning effective synthetic routes.