Physical & Chemical Properties of Haloalkanes

It means the class of organic compounds plays a great role not only in industrial applications but also in everyday life. Haloalkanes belong to the large classes of organic compounds and are characterized by the presence of one or more atoms of halogen, either fluorine, chlorine, bromine, or iodine, which are directly bonded to an alkyl group. Their chemical properties make haloalkanes quite useful in very different issues about pharmaceuticals and agriculture. In abundance, it becomes a part of countless everyday applications, haloalkanes turn out to be industrial refrigerants, solvents, and even anesthetics.

This Story also Contains

1. Haloalkanes and Their Properties
2. Haloalkanes Reactions
3. SN2 Mechanism
4. SN1 Reaction
5. Elimination-Addition Mechanism (I)
6. Importance and Applications of Haloalkanes
7. Some Solved Examples
8. Summary

Perhaps the most popular anesthesia in wide use is halothane; a haloalkane represents it. So haloalkanes have been so much useful because, with the high reactivity inside the organic character, they can undergo easily a number of chemical reactions that render them absolutely necessary intermediates in organic synthesis. This means haloalkanes along with the variation it shows in physical and chemical properties become attractive not only to chemists but also to engineers who are seeking better materials and processes.

Haloalkanes and Their Properties

Haloalkanes are a class of organic compounds mainly derived from alkanes by replacement of one or more numbers of hydrogen atoms with halide atoms like fluorine, chlorine, bromine, or iodine. Generally, haloalkanes can be represented by the general formula,$\mathrm{C}_n \mathrm{H}_{2_{n-1}} \mathrm{X}$. This indicates that, owing to the presence of polar carbon-halogen bonds, boiling and melting point and solubility characteristics, among others, are considerably varied for the mentioned compounds. The mentioned physical properties of haloalkanes are considerably influenced by the two following factors: the bond strength and the molecular weight. Chemically, several types of reactions can be carried out on the haloalkanes, most of which end up in nucleophilic substitution and elimination. Nucleophilic substitution occurs in two mechanisms: SN₁ and SN₂. It is the distinction between these mechanisms that explains ways in which haloalkanes react with nucleophiles resulting in a variety of different products. These theories thus allow the chemist to predict what product will be obtained from a set of starting materials, hence designing synthetic pathways for the target compounds. The most common bases that we have used are alc.KOH and aqueous NaOH. In alc.KOH, we have$\mathrm{EtO}^{-}\left(\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{O}^{-}\right)$as the base while in aqueous NaOH, the base is OH^-. The strengths of these two bases are given below:$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{O}^{-}>\mathrm{OH}^{-}$
Because $\mathrm{CH}_3-\mathrm{CH}_2$ is the releasing group and hence it unstabilizes the O^-, thus it reacts faster and hence it is a stronger base. The reactions occur as follows: For alc.KOH:$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}+\mathrm{KOH}(\mathrm{alc}) \longrightarrow \mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2+\mathrm{KBr}+\mathrm{H}_2 \mathrm{C}$

For aq.NaOH:

Haloalkanes Reactions

The important reaction of haloalkanes is the nucleophilic substitution, which in turn has been divided into two types of SN₁ and SN₂ mechanisms.

SN₂ Mechanism

In SN₂ mechanism, it is a single step in which a nucleophile attacks the electrophilic carbon atom of haloalkane and the product is formed with the displacement of leaving group that is, the halogen, in a single step.

This reaction is thus characterized with its second-order kinetics since the reaction rate depends on the haloalkane and the nucleophile equally. The best examples that can be considered in terms of an SN2 reaction include the conversion of a bromobutane when treated with hydroxide ions. This reaction will, therefore, tend to favor primary haloalkanes where steric hindrance is at its lowest, thus allowing effective nucleophilic attack.

The general reaction occurs as follows:
$\mathrm{R}-\mathrm{CH}_2-\mathrm{Cl}+\mathrm{OH}^{-} \rightarrow \mathrm{R}-\mathrm{CH}_2-\mathrm{OH}$

Mechanism

• It is a bimolecular nucleophilic substitution (SN₂) reaction.
• The rate of reaction follows second-order kinetics and depends upon the concentration of both the nucleophile as well as the substrate.

Rate $\propto[\mathrm{R}-\mathrm{X}]\left[\mathrm{Nu}^{-}\right]$

• The rate-determining step depends on how fast the transition state is formed and also the stability of the transition state
• A stronger nucleophile is required as it has to attack and make the leaving group leave
• Polar aprotic solvents favor SN₂ reaction as they do not facilitate the formation of ions
• The reaction occurs in a concerted mechanism and inversion of configuration (Walden Inversion) takes place if the leaving group and the nucleophile have the same priority
• Steric hindrance in the substrate decreases the reactivity of the substrate toward SN₂ reaction

SN1 Reaction

On the other hand, the SN₁ mechanism involves a two-step reaction: first, a leaving group tendency by the halogen, followed by a carbocation intermediate; then, a reaction with a nucleophile forms the final product.SN₁ is very typical of a reaction in tertiary haloalkanes due to the stability of the carbocation formed. A typical example of an SN1 reaction is also the conversion of tert-butyl bromide into tert-butanol in water. This is because the formation of a carbocation would be the rate-determining step hence, it would follow first-order behavior in the reaction kinetics.

The general reaction occurs as follows:$\mathrm{R}-\mathrm{CH}_2-\mathrm{Cl}+\mathrm{OH}^{-} \rightarrow \mathrm{R}-\mathrm{CH}_2-\mathrm{OH}$

• The mechanism occurs as follows:
  
• SN₁ reactions are nucleophilic substitution reactions, involving a nucleophile replacing a leaving group.
• SN₁ reactions are unimolecular. The rate of this reaction depends only on the concentration of one reactant and does not depend upon the strength of the nucleophile

Rate $\propto[\mathrm{R}-\mathrm{X}]$

• The rate determining step depends on the stability of the intermediate carbocation which is obtained during the reaction

Rate $\propto$ stability of carbocation

• Since the mechanism involves the attack of nucleophiles on an already-formed carbocation, the strength of nucleophiles is unimportant for the rate of the reaction
• The rate of formation of the intermediate is independent of the concentration of nucleophiles and depends only on the concentration of reactants.
• Good ionizing solvents (polar protic solvents) are required to carry out the SN₁ reaction as there has to be a formation of ions
• Configuration of the product may be the same or inverted and in cases where the leaving group departs from a chiral center, racemization occurs.
  • If Nu^- attacks on the same side from where X^- leaves, then it is called 'Retention'.
  • If Nu^- attacks from the opposite side from where X^- leaves, then it is called 'Inversion'.
  • A racemic mixture is obtained when an equal amount of retention and inversion products are formed in the reaction.
  • Generally, partial racemization is seen in the reactions as it both SN₁ and SN₂ are competing

Elimination-Addition Mechanism (I)

Other than that, haloalkanes are capable of an elimination-addition mechanism in which HX, the abstracted hydrogen halide ends up producing alkenes. The above process is so important in organic chemistry when it is capable of leading to the production of double bonds, which can, in return be further tested for other reactions. The major building block of organic chemistry is the most important. That can be synthesized through dehydrohalogenation in Haloalkanes.

A very strong base such as sodium or potassium amide reacts with aryl halide, even those without electron-withdrawing substituents to give products corresponding to nucleophilic substitution of halide by the base.

Mechanism

1. Elimination stage: Amide ion is a very strong base and brings about the dehydrohalogenation of chlorobenzene by abstracting a proton from the carbon adjacent to the one that bears the leaving group. The product of this step is an unstable intermediate called benzyne.
   
2. 
   
3. Beginning of addition phase: Amide ion acts as a nucleophile and adds to one of the carbons of the triple bond. The product of this step is a carbanion.
4. Completion of addition phase: The aryl anion abstracts a proton from the ammonia used as the solvent in the reaction.
5. 

Importance and Applications of Haloalkanes

The significance of haloalkanes is hence not limited to their chemical properties but also in many real-life applications and industries. Haloalkanes act as the intermediate for the synthesis of many pharmaceutical industries. This can be proved by chloromethyl derivatives of haloalkanes applied in the synthesis of antitumor agents hence very important in medicinal chemistry.

Haloalkanes have been largely taken up in agriculture for formulating pesticides and herbicides in the agrochemical sector. The very selective way of interaction with the biological system can make these molecules important in developing efficient agricultural products. This helps bring about an increased yield with less impact on the environment.

Another area where the uses of haloalkanes have been applied is refrigerants. Halogenated hydrocarbons such as chlorofluorocarbons have been broadly used in refrigeration because of their stability and non-flammable nature. However, most of those compounds have currently been banned because of their effects on the environment, more specifically knocking up the rates of ozone depletion, and have been replaced by other more friendly compounds like hydrofluorocarbons.

Laboratory experiments for haloalkanes are important in any academic curriculum and in the understanding of the mechanism of reaction and organic synthesis. Giving students bench time in the execution of nucleophilic substitution and elimination reactions of haloalkanes practically enriches the theoretical part of this course to have a better understanding of the subject of organic chemistry.

Aryl halides are extremely less reactive towards nucleophilic substitution reactions due to the following reasons:

• Resonance effect: In haloarenes, the electron pairs on the halogen atom are in conjugation with π-electrons of the ring and the following resonating structures are possible.
  
  C—Cl bond acquires a partial double bond character due to resonance. As a result, the bond cleavage in haloarene is more difficult than in haloalkane and therefore, they are less reactive towards nucleophilic substitution reaction.
• The difference in hybridization of carbon atom in C—X bond: In haloalkane, the carbon atom attached to halogen is sp³ hybridized while in the case of haloarene, the carbon atom attached to halogen is sp²-hybridised. The sp² hybridized carbon with a greater s-character is more electronegative and can hold the electron pair of the C—X bond more tightly than sp³ -hybridized carbon in haloalkane with less s-character. Thus, the C—Cl bond length in haloalkane is 177pm while in haloarene is 169 pm. Since it is more difficult to break a shorter bond than a longer bond, therefore, haloarenes are less reactive than haloalkanes toward nucleophilic substitution reaction.
• Instability of phenyl cation: In the case of haloarenes, the phenyl cation formed as a result of self-ionization will not be stabilized by resonance and therefore, S_N1 mechanism is ruled out.
• Because of the possible repulsion, it is less likely for the electron-rich nucleophile to approach electron-rich arenes.

Chlorobenzene can be converted into phenol by heating in aqueous sodium hydroxide solution at a temperature of 623K and a pressure of 300 atmospheres.

The presence of an electron-withdrawing group (-NO₂ ) at ortho- and para-positions increases the reactivity of haloarenes.

Recommended topic video on(Physical & Chemical Properties of Haloalkanes)

Some Solved Examples

Example 1
Question:

The major product of the following reaction is:

1)

2)

3)

4) (correct)

Solution:

The given reaction sequence occurs as

Therefore option 4 is correct.

Example 2
Question:

The major product of the following reaction is:

1) (correct)

2)

3)

4)

Solution:

The given reaction sequence occurs as given below:

Hence, the correct answer is Option (1)

Example 3
Question:

The major product obtained in the following reaction is :

1)$
(+) \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}(\mathrm{OtBu}) \mathrm{CH}_2 \mathrm{C}_6 \mathrm{H}_5
$

2)$
(-) C_6 H_5 C H(O t B u) \mathrm{CH}_2 C_6 H_5
$

3)$
( \pm) C_6 H_5 C H(O t B u) C H_2 C_6 H
$

4) $
\mathrm{C}_6 H_5 \mathrm{CH}=\mathrm{CHC}_6 H_5
$

Solution:

As we have learned,

E₂ the reaction mechanism is favored with bulky reactants at higher temperatures. The reaction will lead to dehydrohalogenation and the major product obtained will be an alkene.

The reaction occurs as

Hence, the correct answer is Option (4)

Summary

Introduce haloalkanes or alkyl halides as organic compounds that bear distinct physical and chemical properties, making them quite essential in varied fields and applications. This essay will cover haloalkanes' unique physical properties and their chemical reactivity in regard to nucleophilic substitution mechanisms involving the SN₁ and SN₂ reactions and elimination-addition mechanisms.