Problem 77 - A Complete Resource Book in Chemistry for JEE Main

Question

Identify \(\mathrm{X}\) and \(\mathrm{Y}\) in the following sequence. \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Br} \stackrel{\mathrm{X}}{\longrightarrow}\) product \(\stackrel{\mathrm{Y}}{\longrightarrow} \mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2}\) (a) \(\mathrm{X}=\mathrm{KCN}, \mathrm{Y}=\mathrm{LiAlH}_{4}\) (b) \(\mathrm{X}=\mathrm{KCN}, \mathrm{Y}=\mathrm{H}_{3} \mathrm{O}^{+}\) (c) \(\mathrm{X}=\mathrm{CH}_{3} \mathrm{Cl}, \mathrm{Y}=\mathrm{AlCl}_{3} / \mathrm{HCl}\) (d) \(\mathrm{X}=\mathrm{CH}_{3} \mathrm{NH}_{2}, \mathrm{Y}=\mathrm{HNO}_{2}\)

Step-by-Step Solution

Answer

Option (a): \(\mathrm{X} = \mathrm{KCN}\), \(\mathrm{Y} = \mathrm{LiAlH}_{4}\).

Step 1: Analyze the Initial Reaction

The sequence begins with \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Br}\), which is bromoethane. The reagent \(\mathrm{X}\) needs to perform a reaction that will introduce a new functional group or extend the carbon chain to eventually form a product that will be further reacted by \(\mathrm{Y}\).

Step 2: Investigate the Potential Effect of Reagent X

Bromoethane \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Br}\) can undergo a nucleophilic substitution to form a carbon-carbon bond if reacted with \(\mathrm{KCN}\). \(\mathrm{KCN}\) introduces a cyano group \(-\mathrm{CN}\), resulting in \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CN}\), a nitrile.

Step 3: Analyze Reagent Y's Action

The second step involves converting the nitrile group into an amine to form \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2}\), which is propylamine. Reagent \(\mathrm{LiAlH}_{4}\) is known to reduce nitriles to primary amines.

Step 4: Match the Sequence with Given Options

Given this knowledge, reaction \(\mathrm{X}\) should be \(\mathrm{KCN}\) since it transforms bromoethane into a nitrile. The second reagent \(\mathrm{Y}\) should be \(\mathrm{LiAlH}_{4}\), as it reduces the nitrile to the amine. Reviewing the options, option (a) matches these requirements: \(\mathrm{X} = \mathrm{KCN}\), \(\mathrm{Y} = \mathrm{LiAlH}_{4}\).

Key Concepts

Nucleophilic Substitution

In organic chemistry, nucleophilic substitution is a fundamental type of reaction that is both common and important. This process involves the replacement of a leaving group, such as a halide or another electronegative group, with a nucleophile. A nucleophile is a chemical species that donates an electron pair to form a chemical bond. This type of reaction is essential for various synthetic applications.

In the given exercise, the reagent X, potassium cyanide ( KCN ), facilitates a nucleophilic substitution reaction. The initial compound, bromoethane ( C_{2}H_{3}Br ), contains a bromine atom, which acts as the leaving group. When KCN is added, the cyano group ( -CN ) from KCN substitutes the bromine. This transformation is key in extending the carbon chain to form a nitrile, C_{2}H_{5}CN .

• The leaving group, bromine, is detached from the ethyl group, making the carbon more reactive.
• The nucleophile, the cyano group, effectively bonds with the carbon, forming a new carbon-carbon linkage.

This process is critical for forming more complex molecules, as it introduces new functional groups capable of undergoing further chemical reactions.

Carbon-Carbon Bond Formation

The formation of carbon-carbon bonds is a central theme in organic chemistry because these bonds are the backbone of organic molecules. Establishing new carbon-carbon links is crucial for building larger, more complex structures from simpler ones. This type of reaction is particularly important in the synthesis of pharmaceuticals and other advanced materials.

In the exercise mentioned here, the carbon-carbon bond formation is achieved through the nucleophilic substitution reaction involving KCN , as described in the nucleophilic substitution section.

• When bromoethane reacts with KCN , the cyano group ( -CN ) attaches to the carbon of the ethyl group.
• This reaction results in the formation of a nitrile, specifically C_{2}H_{5}CN , which contains a new C-C bond.

Creating these connections is fundamental in organic synthesis because it allows for the development of molecules that are more complex and functionally diverse. This particular reaction is exemplary of how chemists construct new molecules by forging new bonds.

Reduction of Nitriles

Reduction of nitriles is an important process in organic chemistry since it transforms nitriles into primary amines. This conversion is significant because amines are vital building blocks for numerous chemicals, including pharmaceuticals, agrochemicals, and dyes.

In the exercise, the second reagent, LiAlH4 (lithium aluminium hydride), is used to reduce the nitrile, C_{2}H_{5}CN , that was formed in the previous step. Here's what happens:

• LiAlH_{4} , a powerful reducing agent, supplies the necessary electrons and hydrogen to break the triple bond of the nitrile group.
• This reduction process converts the cyano group ( -CN ) into a primary amine group ( -NH_{2} ).

The product of this reaction is propylamine, C_{3}H_{7}NH_{2} , which is a primary amine. Reduction reactions such as this one are crucial because they enable the transformation of functional groups, thereby allowing for further chemical functionalization and the synthesis of a wide array of chemical products.