/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 77 Identify \(\mathrm{X}\) and \(\m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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) \(\mathrm{X}=\mathrm{CH}_{3} \mathrm{NH}_{2}, \mathrm{Y}=\mathrm{HNO}_{2}\)

Short Answer

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

Step by step solution

01

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 \(\mathrm{Y}\).
02

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.
03

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.
04

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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An insecticide SEVIN is made by the reaction of 1 - naphthol with methyl isocyanate. A union carbide plant in Bhopal, India, were using this process to make SEVIN for use as an agricultural insecticide. On December 3,1984 , either by accident or by sabotage, a value was opened that admitted water to a large tank of methyl isocyanate. The pressure and temperature within the tank rose very high and the pressure relief value opened to prevent the tank from bursting. A large quantity of methyl isocyanate rushed through the pressure relief valves and the vapours flowed with the breeze into populated areas, killing about 2500 people and injuring many more. The temperature rise in the tank was attributed to (a) Dissolution of \(\mathrm{CH}_{3}-\mathrm{N}=\mathrm{C}=\mathrm{O}\) in water (b) Dissolution of \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) in water. (c) Dissolution of \(\mathrm{NH}_{3}\) in water. (d) Dissolution of \(\mathrm{CO}_{2}\) in water.

In an organic compound of molar mass is \(108 \mathrm{~g} \mathrm{~mol}^{-1}\). \(\mathrm{C}, \mathrm{H}\) and \(\mathrm{N}\) atoms are present in \(9: 1: 35\) by weight in it. Molecular formula can be (a) \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{~N}_{2}\) (b) \(\mathrm{C}_{7} \mathrm{H}_{10} \mathrm{~N}\) (c) \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{~N}_{3}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{18} \mathrm{~N}_{3}\)

Hinsberg reagent is (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}_{2} \mathrm{Cl}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{2} \mathrm{Cl}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{3} \mathrm{H}\)

Benzene diazonium chloride is obtained by the diazotization of aniline It gives two types of reactions in which nitrogen atoms are replaced and reactions in which nitrogen atoms are retained. For the above reaction, the reactivity of the diazonium ion will be maximum when \(\mathrm{G}\) is (a) \(-\mathrm{HSO}_{3}\) (b) \(-\mathrm{Cl}\) (c) \(-\mathrm{OCH}_{3}\) (d) \(-\mathrm{CH}_{3}\)

Among the following, the strongest base is (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) (b) \(\mathrm{p}-\mathrm{NO}_{2}-\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{NH}_{2}\) (c) \(\mathrm{m}-\mathrm{NO}_{2}-\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{NH}_{2}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.