The reagent used to accomplish the given transformation has to be identified. Concept Introduction: Elimination reactions are the one in which the groups are lost and the saturated bonds are converted to unsaturated bonds. Usually the substitution reaction compete with elimination reaction. In elimination reaction, the beta proton is removed together with the leaving group to form a double bond. E2 reaction proceeds through a single step without any formation of intermediates. The base abstracts a proton form the substrate and the loss of leaving group also happens resulting in the formation of double bond. E2 stands for bimolecular elimination. The proton can be abstracted by the base in two different ways leading to a regiochemical outcome. In the given reaction below, two alkenes are formed as products due to regiochemistry. The two alkenes that are formed can be given different names. If the alkene is more substituted means it is known as Zaitsev product, and if the alkene formed is less substituted means it is known as Hofmann product. Generally more substituted product is the major product (Zaitsev product). If the reaction is performed with a sterically hindered base, the major product will be less substituted alkene (Hofmann product). The regiochemical outcome can be decided by choosing the base. Some of the sterically hindered bases are, Addition reactions are the one in which the groups are added to an unsaturated double bond. The double bond is destroyed because the groups are added across the double bond. If the considered alkene is a symmetrical one, then there is no difference in the product obtained but if the alkene is an unsymmetrical one, then regiochemistry decides the products that are obtained. When two similar groups are added across a double bond, regiochemistry is irrelevant. When adding two different groups across the double bond of an unsymmetrical alkene, the regiochemistry becomes relevant. In simple words, it can be said that, when two different groups are added across a double bond, then regiochemistry becomes relevant. Markovnikov Addition In the vinylic bond, if the bulky group gets substituted in the carbon atom that is more substituted means, it is known as Markovnikov’s addition. anti-Markovnikov Addition In the vinylic bond, if the bulky group gets substituted in the carbon atom that is less substituted means, it is known as anti-Markovnikov’s addition. If hydrogen halide is used in presence of hydrogen peroxide, anti-Markovnikov addition takes place. This means the halide ends up on less substituted carbon atom. This is because when peroxide is used the mechanism that the reaction undergoes is radical mechanism. This involves a radical intermediate instead of ionic intermediate. The main difference between ionic mechanism and radical mechanism is that proton is installed first in ionic mechanism while in radical mechanism, halide is installed first. This is the reason why ionic mechanism gives Markovnikov product and radical mechanism gives anti-Markovnikov product.

Question

Want to see the full answer?

Answer 1

Question

Chapter 11.8C, Problem 11.90P

Interpretation Introduction

Interpretation:

The reagent used to accomplish the given transformation has to be identified.

Concept Introduction:

Elimination reactions are the one in which the groups are lost and the saturated bonds are converted to unsaturated bonds. Usually the substitution reaction compete with elimination reaction.

In elimination reaction, the beta proton is removed together with the leaving group to form a double bond.

E2 reaction proceeds through a single step without any formation of intermediates. The base abstracts a proton form the substrate and the loss of leaving group also happens resulting in the formation of double bond. E2 stands for bimolecular elimination.

The proton can be abstracted by the base in two different ways leading to a regiochemical outcome. In the given reaction below, two alkenes are formed as products due to regiochemistry.

The two alkenes that are formed can be given different names. If the alkene is more substituted means it is known as Zaitsev product, and if the alkene formed is less substituted means it is known as Hofmann product. Generally more substituted product is the major product (Zaitsev product).

If the reaction is performed with a sterically hindered base, the major product will be less substituted alkene (Hofmann product). The regiochemical outcome can be decided by choosing the base. Some of the sterically hindered bases are,

Addition reactions are the one in which the groups are added to an unsaturated double bond. The double bond is destroyed because the groups are added across the double bond. If the considered alkene is a symmetrical one, then there is no difference in the product obtained but if the alkene is an unsymmetrical one, then regiochemistry decides the products that are obtained.

When two similar groups are added across a double bond, regiochemistry is irrelevant. When adding two different groups across the double bond of an unsymmetrical alkene, the regiochemistry becomes relevant. In simple words, it can be said that, when two different groups are added across a double bond, then regiochemistry becomes relevant.

Markovnikov Addition

In the vinylic bond, if the bulky group gets substituted in the carbon atom that is more substituted means, it is known as Markovnikov’s addition.

anti-Markovnikov Addition

In the vinylic bond, if the bulky group gets substituted in the carbon atom that is less substituted means, it is known as anti-Markovnikov’s addition.

If hydrogen halide is used in presence of hydrogen peroxide, anti-Markovnikov addition takes place. This means the halide ends up on less substituted carbon atom. This is because when peroxide is used the mechanism that the reaction undergoes is radical mechanism. This involves a radical intermediate instead of ionic intermediate. The main difference between ionic mechanism and radical mechanism is that proton is installed first in ionic mechanism while in radical mechanism, halide is installed first. This is the reason why ionic mechanism gives Markovnikov product and radical mechanism gives anti-Markovnikov product.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

1. Provide a complete IUPAC name for each of the following compounds. a) b) c) OH OH OH a) b) c) 2. Provide a complete IUPAC name for each of the following compounds. a) b) a) OH b) он c) OB >=

c) 3. Provide a common name for each of the following alcohols. a) a) OH b) OH c) HO b) c) 4. Provide a common name for each of the following compounds. b) OH a) 5 a) Y OH c) OH

Using the critical constants for water (refer to the table in the lecture slides), calculate the second virial coefficient. Assume that the compression factor (Z) is expressed as an expansion series in terms of pressure.