Problem 31 - A Complete Resource Book in Chemistry for JEE Main

Question

The major produet in the reaction $$ \begin{aligned} &\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{Br} \stackrel{\mathrm{AgCN}}{\longrightarrow} ? \text { is } \\ &\text { (a) } \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CN} \\ &\text { (b) } \mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{Ag} \end{aligned} $$ (c) \(\mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{NC}\) (d) none of these

Step-by-Step Solution

Answer

The major product is \( \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{NC} \) (option c).

Step 1: Understanding the Reactants

The given reactant is ethyl bromide ( \( \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{Br} \)). The reagent used for the reaction is silver cyanide ( \( \mathrm{AgCN} \)). We need to determine the major product when ethyl bromide reacts with silver cyanide.

Step 2: Reactivity of Silver Cyanide

Silver cyanide is primarily an ionic compound, and its reactivity is such that it helps in the formation of isocyanides. Consequently, when \( \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{Br} \) reacts with \( \mathrm{AgCN} \), the nucleophile \( \mathrm{CN}^- \) attacks an electron-deficient carbon atom of ethyl bromide to form isocyanides.

Step 3: Identifying the Major Product

In this reaction, the nucleophile preferentially attaches itself in a way that forms an isocyanide rather than a cyanide. Therefore, the major product is \( \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{NC} \), which corresponds to the formation of ethyl isocyanide.

Step 4: Selecting the Correct Option

From the given options, option (c) \( \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{NC} \) correctly represents the major product of the reaction between \( \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{Br} \) and \( \mathrm{AgCN} \).

Key Concepts

Nucleophilic Substitution

Nucleophilic substitution is a fundamental concept in organic chemistry, involving the replacement of a leaving group by a nucleophile. A nucleophile is an electron-rich species that typically carries a negative charge or has lone pairs that can be donated. In the context of our exercise, ethyl bromide, which is the starting material, contains a bromine atom that acts as the leaving group.

When a nucleophile, such as the cyanide ion (CN⁻), attacks the positively charged or partially positive carbon of ethyl bromide, it results in the nucleophile replacing the bromine atom. This forms a new chemical bond with carbon. The effectiveness of the nucleophilic substitution depends on factors such as the strength of the leaving group, the nature of the nucleophile, the solvent used, and the structure of the substrate.

In many substitution reactions, such as the one described, we often encounter a competition between isocyanide and cyanide formation, which can be dictated by the type of nucleophile and reaction conditions.

Reaction Mechanism

Understanding the reaction mechanism is essential to know how reactants transform into products at the molecular level. For the reaction between ethyl bromide and silver cyanide, the key part of the mechanism involves the attack of the carbon atom by the cyanide nucleophile.

Silver cyanide adds a unique twist to this nucleophilic substitution. This compound is not just a straightforward source of CN⁻ ions, due to its ionic and covalent bonding nature which tends to favor the formation of isocyanides rather than cyanides. The nucleophile generated from silver cyanide is not simple CN⁻ but interacts in a manner that favors binding to carbon through nitrogen, thereby forming isocyanides.

During the process, the negatively charged CN⁻ acts as the nucleophile and seeks out the electron-deficient carbon atom in the ethyl bromide. Once the nucleophile attacks, the bromine, as a good leaving group, is expelled from the carbon chain, resulting in the formation of ethyl isocyanide rather than ethyl cyanide.

Isocyanide Formation

The formation of isocyanides is a notable outcome when specific reagents like silver cyanide are used in substitution reactions. They differ structurally from regular cyanides by the way they bond to the carbon chain.

Isocyanides are represented with the -NC functional group, where the nitrogen atom joins to the carbon chain, transforming ethyl bromide into ethyl isocyanide. The reason silver cyanide tends to produce isocyanides under nucleophilic substitution is because of its unique property of promoting nitrogen bonding with carbon.

This phenomenon is partly due to the partial covalent nature of the bond in AgCN, where communication between the components favors nitrogen linkage over carbon linkage. It is interesting to note that isocyanides are generally less common in organic reactions compared to cyanides but hold significant importance in synthetic chemistry due to their role in multistep synthesis and complex molecule formations.

In summary, the reaction mechanism and the type of nucleophile used play a vital role in determining whether a cyanide or isocyanide is produced, and this exercise provides a clear example of how different conditions can lead to diverse chemical products.