Do you think the arrangement of atoms in space affects how they react with other atoms to form compounds? Yes! Stereochemistry plays a very important role in determining the formation of the product in chemical reactions of haloalkanes. But what exactly is stereochemistry? Let us study how stereochemistry affects the formation of the product. In this article, we will learn about the other two types of chemical reactions of haloalkanes- elimination reactions and reaction with metals and how stereochemistry affects these reactions. Let’s begin.
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Stereochemistry and Reaction of Haloalkanes
Stereochemistry refers to the study of the arrangement of atoms in space. Stereoisomers are compounds having the same molecular formula but they differ in the spatial arrangement(i.e. arrangement in space). Stereoisomers fall into two main categories:
- Geometric Isomers
- Optical Isomers
It is important to understand optical isomers and their involvement in these chemical reactions.
Browse more Topics under Haloalkanes And Haloarenes
- Introduction and Classification
- Nomenclature and Nature of C-X bond
- Physical properties
- Methods of Preparation
- Chemical Reactions – Nucleophilic Substitution Reactions
- Polyhalogen compounds
- Reactions of haloarenes
Optical Isomers
Optical isomers are compounds having the same molecular formula but they differ in the manner they rotate a plane polarized light. Plane polarized light formation takes place when an ordinary light passes through a Nicol prism thereby causing diffraction and resulting in the formation of a single wavelength light. This light of the single wavelength is plane polarized light.
Set-Up of Polarimeter
The instrument that helps in the generation of plane polarized light from an ordinary light is a polarimeter.
Asymmetry of a Molecule
Asymmetry of a molecule also determines whether the molecule will be chiral or achiral. There are objects or compounds that are non-superimposable. These types of objects are known as chiral objects or chiral compound. However, there are many compounds/objects, such as a sphere, that is identical with its mirror image and it is superimposable as well. This type of compounds/objects is called achiral object. Therefore, asymmetry of a carbon atom in a particular molecule will determine the chirality of that molecule.
Why are we studying the chirality? This is because it will help us to study about enantiomers. Enantiomers are nothing but the stereoisomers that cannot form superimposable mirror images of each other. Enantiomers have same physical properties such as boiling point, index of refraction etc. The compounds vary in the manner of rotation of a plane polarized light (dextro or laevo).
However, if the particular composition of a mixture contains an equal amount of both the types of enantiomers, the net optical rotation becomes zero because dextro and laevo cancel out each other. This type of mixture is a racemic mixture and the process is racemization.
Stereochemistry – Configuration
In an organic compound, there are different types of spatial arrangement. In this section, we will study about the spatial arrangements of an organic compound with respect to stereochemistry.
Retention of Configuration
Retention of configuration is the ability of an asymmetric carbon to retain the spatial arrangement of the substituents/groups during a chemical reaction. This can only happen when the bond of carbon stereocenter remain intact. Therefore, a substituent will be replaced by other without disturbing the structure. Thus, the relative arrangement of the reactant and the product can be correlated.
For example, the reaction of 2 − methylbutan − 1 – ol upon heating with conc. HCl.
Inversion of Configuration
If an asymmetric carbon atom is not able to retain its spatial arrangement during a chemical reaction is known as inversion. An inversion occurs during SN2 reactions. This is because the nucleophile attacks the C-X bond of an alkyl halide by the backside method. As a result of which, the configuration of the molecule gets inverted. Therefore, optically active alkyl halides in an SN2 reaction leads to the inversion of configuration.
Racemization
Considering the two-step process of SN1 reactions, we will note that in the first step planar carbocation generation and in the second step attack of a nucleophile occur. The carbocation is planar which allow the nucleophile in the second step to attack the carbon atom either from below or above. This, in turn, results in both configurations- retention, and inversion.
There is also a possibility that 50 percent of the mixture will have the inversion of configuration and 50 percent will have retention of configuration. In the below example, both inversion and retention of configuration have taken place with respect to the actual reactant. Retention ((−)Butan − 2 − ol) Inversion ((+)Butan − 2 − ol).
Thus a reaction of an asymmetric carbon can lead to the formation of three different configurations- Retention of Configuration, Inversion of Configuration, and Racemization.
Chemical Reactions of Haloalkanes- Elimination Reactions
The primary condition for a haloalkane to undergo elimination reaction is the presence of β −hydrogen atom.
Dehydrohalogenation or β – elimination
Elimination reaction occurs when an alkyl halide reacts with alcoholic KOH solution under heating conditions. The elimination takes place of hydrogen from β –carbon and the elimination of halogen member occur from α –carbon. The reaction will result in the formation of alkene as the final product.
The above reaction mechanism demonstrates a haloalkane. The α –carbon contains a halogen group whereas the β –carbon contains the hydrogen. When the base reacts with the hydrogen of the β –carbon, bond shifting occurs. Ond shifting results in the elimination of halogen group present in the reactant thereby forming an alkene as the major product and a corresponding by-product.
However, if there is more than 1β −hydrogen atom then multiple eliminations will occur thereby forming multiple alkenes. In such cases, the major product will be the most stable alkenes out of all the alkenes. Generally, the alkene which undergoes most substitution is generally the most stable. The major product is also known as Zaitsev′s product.
Zaitsev′s Product
The most substituted alkene refers to the alkene which has the greater number of alkyl group joined with the double bond. Hence, the more the number of alkyl group present with the double bond, more is the stability of the alkene. This happens due to the process of hyperconjugation. Example: Dehydrohalogenation of 2-Bromobutane
When alkyl halides react with nucleophiles there is a continuous competition whether the elimination or substitution reaction will occur. This mainly depends upon a certain factor which will determine what type of reaction will take place.
Factors Responsible for Favouring Elimination/Substitution Reactions
- Nature of the substrate present in the reaction- Reaction will depend upon the type of haloalkane present (primary, secondary or tertiary).
- Size and strength of the base/nucleophile- If the nucleophile is highly reactive there is a possibility that the reaction taking place will be a substitution reaction (SN2). However, if the nucleophile is bulky then due to steric hindrance substitution reaction is difficult. In such cases, elimination reaction takes place.
- Conditions involved in reaction (solvent used in the reaction)
Summary of Chemical Reactions
| Type of Haloalkane | Elimination Reaction | Substitution Reaction & its Types | |
| Primary | Not possible | Substitution Reaction Possible (SN2) | |
| Secondary | Possible- On the basis of base strength | Substitution Reaction Possible
(SN1 or SN2)- On the basis of nucleophile strength |
|
| Tertiary | Possible-On the basis of alkene stability | Possible- SN1
On the basis of carbocation stability |
Chemical Reactions of Haloalkanes- Reactions with Metals
Haloalkanes on reaction with metals form a special class of compounds – organometallic compounds. Alkyl Halides or R-X reacts with certain metals to produce compounds containing C-metal bonds.
Grignard Reagent
In the year 1910, a renowned scientist “Victor Grignard” discovered a special class of organometallic compounds/reagents known as a Grignard Reagent. The Grignard reagent is Alkyl magnesium halide (R-MgX). This reagent formation is possible when a haloalkane (in this case ethyl bromide) react with Magnesium in the presence of dry ether. This will result in the formation of R-MgX.
C-Mg bond is polar and covalent in nature. This is due to the difference in electronegativity between the carbon and magnesium atom. Thus, the carbon atom will attract the electron towards itself resulting in the formation of partial positive charge (δ +) on magnesium atom and partial negative charge (δ -) on the carbon atom. However, Mg-X bond in the same compound is ionic in nature.
Therefore, the Grignard reagent is highly reactive in nature. It reacts with an acidic proton bearing compound to produce hydrocarbon. Thus substances such as H2O, R − OH, and R − NH2 are enough acidic and can react with Grignard reagents to form hydrocarbon (RH).
For example, if ethyl magnesium bromide reacts with an alcohol to form the respective hydrocarbon. Anhydrous condition (dry ether) is used in this reaction in order to avoid traces of water.
Wurtz Reaction
In this reaction, haloalkane reacts with sodium in presence of dry ether to form hydrocarbons with twice the number of carbon atoms originally present in the reactant.
Solved Example for You
Q. Reaction of alkyl chloride with KOH (aqueous) produces alcohols. However, the same reaction with alcoholic KOH produces alkene as the reaction’s major product. Justify
Solution: The aqueous solution of KOH breaks down into K+ and OH− due to ionization. OH− act as the nucleophile and undergo nucleophilic substitution reaction with alkyl chlorides thereby forming alcohols. Additionally, the highly hydrating property of aqueous solution decreases the basic property of OH−. Thus the reaction is unable to form an alkene.
However, the alcoholic KOH solution has alkoxide ions (RO−). They demonstrate highly basic property in comparison to OH−ions. Therefore, alcoholic KOH solution is able to preferentially remove a molecule from HCl from the substrate (alkyl chloride) and form an alkene.
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