Problem 29 - A Complete Resource Book in Chemistry for JEE Main

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) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\)

Step-by-Step Solution

Answer

The compound (c) ( ( ext{CH}_3)_3 ext{S}^+ ext{I}^- ) will most readily react with NaOH to form methanol.

Step 1: Identify Nucleophilic Substitution

The reaction with NaOH involves a nucleophilic substitution process, where NaOH as a nucleophile will attack a compound to displace a leaving group, potentially forming methanol ( ext{CH}_3 ext{OH} ). Our task is to identify which compound has the best or most suitable leaving group that allows the formation of methanol.

Step 2: Analyze Leaving Groups in Each Compound

Analyze the options: (a) ( ( ext{CH}_3)_4 ext{N}^+ ext{I}^- ): The ( ext{I}^- ) can be displaced by ( ext{OH}^- ), forming ( ext{CH}_3 ext{OH} ) and leaving ( ( ext{CH}_3)_3 ext{N} ). (b) ( ext{CH}_3 ext{OCH}_3 ): No obvious leaving group without breaking the ( ext{C-O} ) bonds directly, which doesn't easily lead to methanol. (c) ( ( ext{CH}_3)_3 ext{S}^+ ext{I}^- ): The ( ext{I}^- ) can be displaced by ( ext{OH}^- ), forming ( ext{CH}_3 ext{OH} ) and leaving ( ( ext{CH}_3)_2 ext{S} ). (d) ( ( ext{CH}_3)_3 ext{CCl} ): The ( ext{Cl}^- ) is a leaving group, but forms carbocations that are less stable for forming methanol.

Step 3: Determine the Best Leaving Group

Among the compounds, ( ext{I}^- ) in option (a) ( ( ext{CH}_3)_4 ext{N}^+ ext{I}^- ) and (c) ( ( ext{CH}_3)_3 ext{S}^+ ext{I}^- ) serves as a good leaving group. However, in (c), the positively charged sulfur will stabilize the transition state better, making it more likely to react with ( ext{OH}^- ).

Step 4: Conclude Which Compound Forms Methanol

Considering both the leaving groups and the stability of the intermediates after the reaction, (c) ( ( ext{CH}_3)_3 ext{S}^+ ext{I}^- ) is most likely to react with ( ext{NaOH} ) to form methanol. The nucleophilic attack displaces ( ext{I}^- ) and results in the formation of ( ext{CH}_3 ext{OH} ).

Key Concepts

Leaving Group

In organic chemistry, a leaving group is an atom or group of atoms that can be readily displaced or removed during a chemical reaction, particularly during nucleophilic substitutions. For a good leaving group:

• It should be able to accept an electron pair.
• It is often more stable once it has left the molecule.
• Halides like iodide \( (\mathrm{I}^-) \) are common leaving groups due to their ability to stabilize the extra electrons.

In the exercise, the compounds \( \mathrm{CH}_3 \) and \( \mathrm{I}^- \) can act as leaving groups, but iodide is a more favorable option. This is due to its size and ability to stabilize, making it an excellent candidate for leaving group substitution when attacked by a nucleophile like \( \mathrm{OH}^- \).

Methanol Formation

Methanol (\( \mathrm{CH}_3\mathrm{OH} \)) is a simple alcohol that is often formed through nucleophilic substitution reactions. In the exercise, methanol forms when \( \mathrm{NaOH} \) acts as a nucleophile and replaces a leaving group in the reactant.During the substitution process:

• \( \mathrm{OH}^- \) attacks the carbon atom bonded to the leaving group, resulting in the detachment of that group.
• The result is the formation of a new bond between the \( \mathrm{OH}^- \) and the carbon, producing methanol.
• For example, in the compound \( \left(\mathrm{CH}_3\right)_3\mathrm{S}^+ \mathrm{I}^- \), replacing \( \mathrm{I}^- \) leads to methanol.

This process highlights how important it is for the leaving group to be stable; without this stability, methanol may not form efficiently.

Organic Chemistry Reactions

Organic chemistry reactions involve the transformation of organic compounds through different mechanisms. One primary type of reaction is nucleophilic substitution, where a nucleophile replaces a leaving group within a molecule.Key features of these reactions include:

• A nucleophile, which is electron-rich and seeks an electron-deficient site to bond with.
• A substrate containing a suitable leaving group that can be displaced.
• The stability of the transition state, which can affect the rate and favorability of the reaction.

In the context of the provided exercise, the nucleophilic substitution results in the formation of methanol. This occurs as \( \mathrm{NaOH} \) displaces an iodide ion \( \mathrm{I}^- \). Such reactions are fundamental in organic chemistry, forming the basis for more complex transformations and synthesis in the laboratory and industrial processes.