Chapter 10 Haloalkanes and Haloarenes

1. Explain the difference between nucleophilic substitution and elimination reactions in haloalkanes. How does the nature of the halide influence the reaction pathway?

Answer: Nucleophilic substitution and elimination are two major types of reactions that haloalkanes undergo. These reactions are influenced by the nature of the halide (R-X) and the reaction conditions.

• Nucleophilic Substitution: In this reaction, the halogen atom (X) is replaced by a nucleophile (Nu⁻). The process can occur in two ways:

  • SN1 mechanism: This occurs in two steps. The first step is the dissociation of the halide to form a carbocation. The second step involves the nucleophile attacking the carbocation. This mechanism is favored by tertiary haloalkanes (R-CH₃), because the tertiary carbocation is more stable due to electron-donating groups.
  • SN2 mechanism: This occurs in a single step, where the nucleophile attacks the carbon from the opposite side of the leaving group (X), displacing it. This is a concerted mechanism, meaning the bond-breaking and bond-making occur simultaneously. The SN2 mechanism is favored by primary and secondary haloalkanes, as these have less steric hindrance, allowing the nucleophile to approach easily.

• Elimination Reaction: In an elimination reaction, a molecule of the halogen (X) and a hydrogen atom are removed, forming an alkene. There are two main types of elimination reactions:

  • E1 mechanism: This involves the formation of a carbocation intermediate after the halide ion (X⁻) leaves. The proton (H⁺) is then abstracted by a base. The E1 mechanism is favored in tertiary haloalkanes, as the carbocation intermediate is more stable.
  • E2 mechanism: This is a concerted mechanism, where the base removes a proton (H⁺) simultaneously with the departure of the halide ion (X⁻). The E2 mechanism is favored by strong bases and is more common in secondary and primary haloalkanes.

The nature of the halide affects the rate of reaction:

• Strong leaving groups (like iodine) favor nucleophilic substitution, as they can easily leave.
• Sterically hindered halides (like tertiary halides) tend to favor elimination reactions due to the difficulty in the nucleophile attacking the carbon center.

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2. Why is nucleophilic substitution reaction in alkyl halides favored by polar aprotic solvents and elimination reactions favored by polar protic solvents? Explain with examples.

Answer: The choice of solvent plays a crucial role in determining whether a nucleophilic substitution or elimination reaction will occur in alkyl halides.

• Polar Aprotic Solvents (e.g., DMSO, acetone): These solvents do not have hydrogen atoms that can form hydrogen bonds with the nucleophile. In polar aprotic solvents, the nucleophile is more "free" because it is not solvated by the solvent, meaning the nucleophile is more reactive. This promotes the SN2 mechanism, where the nucleophile attacks the carbon center and displaces the halide ion. In these solvents, the leaving group can easily depart, leading to a nucleophilic substitution reaction.

  • Example: In the reaction of CH₃Cl with NaOH in acetone, the hydroxide ion (OH⁻) can attack the carbon in methyl chloride, resulting in the formation of methanol (CH₃OH).

• Polar Protic Solvents (e.g., water, alcohols): These solvents can hydrogen bond with the nucleophile, thus stabilizing it. In a polar protic solvent, the nucleophile is somewhat "shielded" because it forms hydrogen bonds with the solvent, making it less reactive. However, these solvents can stabilize the leaving group (halide ion) better. This promotes elimination (E2) reactions, especially in situations where a strong base is used.

  • Example: In the reaction of C₂H₅Cl with NaOH in water, the hydroxide ion (OH⁻) can abstract a proton from the β-carbon, leading to the formation of ethylene (C₂H₄) via an E2 elimination reaction.

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3. Discuss the reactivity and structure of allylic and benzylic halides in nucleophilic substitution reactions. How do their structures affect the rate of reaction?

Answer: Allylic and benzylic halides are unique because of the resonance stabilization of their carbocation intermediates, which significantly enhances their reactivity in nucleophilic substitution reactions.

• Allylic Halides: These halides contain a halogen atom attached to a carbon that is adjacent to a double bond. The allyl carbocation (R-CH₂-CH=CH₂) is stabilized through resonance because the positive charge can be delocalized onto the adjacent carbon atoms. This resonance stabilization increases the stability of the carbocation intermediate formed during nucleophilic substitution, making the reaction faster.

  • Example: The nucleophilic substitution of allyl chloride (CH₂=CH-CH₂Cl) with a nucleophile (e.g., NaOH) leads to the formation of an alcohol through the SN1 mechanism in a polar protic solvent, as the carbocation formed is stabilized by resonance.

• Benzylic Halides: Benzylic halides contain a halogen attached to a carbon adjacent to a benzene ring. The benzylic carbocation is even more stabilized than the allylic carbocation due to the delocalization of the positive charge onto the aromatic ring. This stabilization makes the benzylic carbocation highly stable, which enhances the reactivity of benzylic halides in nucleophilic substitution reactions, particularly the SN1 mechanism.

  • Example: Benzyl chloride (C₆H₅CH₂Cl) undergoes nucleophilic substitution with NaOH, forming benzyl alcohol (C₆H₅CH₂OH). The reaction proceeds through the SN1 mechanism due to the resonance stabilization of the benzylic carbocation.

Both allylic and benzylic halides undergo SN1 substitution more readily than alkyl halides due to the stability of their carbocation intermediates.

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4. Explain the concept of Walden inversion with an example. How does the stereochemistry of a chiral center change during an SN2 reaction?

Answer: Walden Inversion refers to the phenomenon where the stereochemistry of a chiral center is inverted during a nucleophilic substitution reaction. This is particularly significant in the SN2 mechanism, where the nucleophile attacks the carbon atom from the opposite side of the leaving group, leading to an inversion of configuration at the carbon center.

• Stereochemistry of SN2 Reaction: In an SN2 reaction, the nucleophile approaches the electrophilic carbon from the opposite side of the leaving group (backside attack). This leads to the simultaneous breaking of the C-X bond and the formation of the new bond between the carbon and the nucleophile. As a result, the stereochemistry of the carbon center is reversed (inverted) compared to the original configuration.

  • Example: In the reaction of (R)-2-bromobutane with NaOH, the hydroxide ion (OH⁻) attacks the carbon bearing the bromine atom from the opposite side, resulting in the formation of (S)-2-butanol. This inversion of configuration is the hallmark of Walden inversion.

• Walden Inversion and Chirality: If the substrate is chiral, the SN2 mechanism leads to an inversion of the stereochemical configuration at the chiral center. The reaction occurs through a single transition state, and the configuration of the product is the opposite of the starting compound.

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5. Why is the reactivity of alkyl halides in nucleophilic substitution reactions influenced by the size of the halogen atom? Explain with examples.

Answer: The reactivity of alkyl halides in nucleophilic substitution reactions is influenced by the size and polarizability of the halogen atom. The halogen affects the leaving group ability, which plays a crucial role in determining the rate of substitution.

• Smaller Halogens (e.g., Chlorine): Chlorine (Cl) is smaller and less polarizable compared to iodine. While chlorine is a good leaving group, it is less readily displaced by a nucleophile in comparison to larger halogens, because it holds onto the carbon more tightly.

  • Example: In the reaction of ethyl chloride (C₂H₅Cl) with a nucleophile (e.g., NaOH), the reaction occurs more slowly compared to an iodide compound, as chlorine is a poorer leaving group.

• Larger Halogens (e.g., Iodine): Iodine (I) is much larger and more polarizable, making it a better leaving group. The bond between carbon and iodine is weaker, allowing the nucleophile to displace the iodine more easily.

  • Example: Methyl iodide (CH₃I) reacts more quickly with a nucleophile (e.g., NaOH) than methyl chloride (CH₃Cl) due to the superior leaving group ability of iodine.

As the size of the halogen increases, the leaving group becomes more capable of leaving, enhancing the rate of nucleophilic substitution reactions.

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6. Why are alkyl halides more reactive in nucleophilic substitution reactions than aryl halides?

Answer: The increased reactivity of alkyl halides compared to aryl halides in nucleophilic substitution reactions is due to the following factors:

• Bond Strength: In alkyl halides, the bond between the carbon and the halogen is relatively weak, making it easier for the halogen to leave during nucleophilic substitution.

  • Example: The bond dissociation energy of C-I is lower than that of C-Cl or C-Br, which means that alkyl iodides react faster than alkyl chlorides or bromides.

• Resonance Effect: In aryl halides, the halogen is attached to a carbon that is part of an aromatic ring. The lone pairs of the halogen can delocalize into the ring, creating a partial bond character between the halogen and the carbon. This resonance effect makes the C-Hal bond in aryl halides stronger and less prone to breaking, making nucleophilic substitution more difficult.

  • Example: Bromobenzene (C₆H₅Br) does not readily undergo nucleophilic substitution because the electron density from the ring stabilizes the carbon-halogen bond, preventing the departure of the halogen.

Thus, alkyl halides are generally more reactive in nucleophilic substitution reactions compared to aryl halides.

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7. What is the role of the leaving group in the mechanism of nucleophilic substitution reactions in haloalkanes?

Answer: The leaving group plays a crucial role in the mechanism of nucleophilic substitution reactions in haloalkanes. The quality of the leaving group determines the ease with which it can depart, allowing the nucleophile to take its place. A good leaving group is one that can stabilize the negative charge formed when it departs from the substrate.

• Characteristics of a Good Leaving Group: A good leaving group is one that is weakly basic, polarizable, and capable of stabilizing the negative charge effectively after departure. Halogens like iodine (I⁻), bromine (Br⁻), and chlorine (Cl⁻) are good leaving groups, with iodine being the best, followed by bromine and chlorine.
  • Example: In an SN1 or SN2 reaction, the halogen (e.g., I⁻, Br⁻) departs with its negative charge, allowing the nucleophile to take its place. The ability of the leaving group to stabilize the departing charge affects the rate of the reaction.

A bad leaving group, like a hydroxide ion (OH⁻) or an alkoxide ion (RO⁻), is less likely to leave easily, slowing down the nucleophilic substitution reaction.

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8. How do the electronic effects (inductive and resonance effects) influence the reactivity of haloalkanes in nucleophilic substitution reactions?

Answer: The electronic effects—inductive and resonance effects—play a significant role in determining the reactivity of haloalkanes in nucleophilic substitution reactions.

• Inductive Effect: The presence of electron-withdrawing groups (EWGs) like halogens attached to the carbon chain increases the positive character of the carbon atom, making it more electrophilic and more prone to nucleophilic attack. This increases the reactivity of the haloalkane in the SN1 or SN2 reaction.

  • Example: Chloroethane (C₂H₅Cl) reacts more readily with nucleophiles than ethane (C₂H₅H) due to the inductive withdrawal of electron density from the carbon atom by the chlorine atom.

• Resonance Effect: The ability of halogens to participate in resonance with the substrate also affects the reactivity. Halogens have lone pairs that can delocalize into the carbon-halogen bond, which can affect the rate of the nucleophilic substitution. If the halogen has a strong resonance interaction, it will stabilize the transition state in the reaction and make it more likely to undergo substitution.

  • Example: In the case of benzyl chloride (C₆H₅CH₂Cl), the chlorine atom’s lone pairs participate in resonance with the benzene ring, stabilizing the carbocation intermediate formed during SN1 reactions.

9. Explain the mechanism of nucleophilic substitution in haloalkanes.

Answer: Nucleophilic substitution in haloalkanes occurs through two main mechanisms: SN1 (unimolecular nucleophilic substitution) and SN2 (bimolecular nucleophilic substitution). The choice of mechanism depends on factors like the structure of the haloalkane, the nature of the nucleophile, and the solvent.

SN2 Mechanism:

• Bimolecular: The rate of reaction depends on the concentration of both the haloalkane and the nucleophile.
• One-step process: The nucleophile attacks the electrophilic carbon (the carbon attached to the halogen) from the opposite side of the leaving group, while the halogen leaves.
• Backside attack: This leads to an inversion of configuration at the carbon center (stereochemical inversion).
• Example: In the reaction of methyl iodide (CH₃I) with a nucleophile like hydroxide (OH⁻), the nucleophile attacks the carbon from the opposite side of the iodine, displacing the iodine ion in one step.

SN1 Mechanism:

• Unimolecular: The rate of reaction depends only on the concentration of the haloalkane.
• Two-step process: First, the bond between the carbon and the halogen breaks to form a carbocation intermediate. Then, the nucleophile attacks the carbocation.
• Carbocation stability: The more stable the carbocation (e.g., tertiary carbocations are more stable than primary ones), the more likely the reaction will proceed via the SN1 mechanism.
• Example: In tertiary butyl chloride (C₄H₉Cl), the chloride ion leaves, forming a tertiary carbocation, which is then attacked by a nucleophile like water.

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10. Why does the reactivity of haloalkanes increase with the size of the halogen (I, Br, Cl)?

Answer: The reactivity of haloalkanes increases with the size of the halogen due to the leaving group ability of the halogen. The larger halogens (iodine, bromine, chlorine) are better leaving groups because they can stabilize the negative charge better after departure.

• Iodine (I) is the largest and most polarizable halogen, which allows it to stabilize the negative charge effectively after leaving. This makes alkyl iodides react faster in nucleophilic substitution than alkyl bromides or chlorides.
• Bromine (Br) is smaller than iodine but still has good leaving group ability, making alkyl bromides also reactive in nucleophilic substitution.
• Chlorine (Cl) is smaller and less polarizable than iodine and bromine, making it a poorer leaving group, and hence alkyl chlorides react slower than alkyl iodides or bromides.

Thus, as the halogen size increases, the bond between the carbon and the halogen weakens, making the halogen a better leaving group and increasing the reactivity of haloalkanes in nucleophilic substitution reactions.

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11. What is the difference between nucleophilic substitution in haloalkanes and haloarenes?

Answer: The key differences in nucleophilic substitution between haloalkanes and haloarenes arise due to the structural differences between alkyl groups and aromatic rings.

• Haloalkanes (Alkyl Halides):

  • In haloalkanes, the carbon attached to the halogen is sp³ hybridized and has a relatively high electron density. This makes it more susceptible to nucleophilic attack.
  • Nucleophilic substitution can occur through either SN1 or SN2 mechanisms, depending on factors like the structure of the substrate, the nature of the nucleophile, and the solvent.

• Haloarenes (Aryl Halides):

  • In haloarenes, the carbon attached to the halogen is part of an aromatic ring. The halogen's lone pairs can engage in resonance with the ring, making the carbon-halogen bond stronger and more resistant to breaking.
  • As a result, nucleophilic substitution in haloarenes is typically less favorable. The reaction may require harsher conditions, such as the use of a strong nucleophile or high temperature.
  • The mechanism for nucleophilic substitution in haloarenes is usually unimolecular (SN1) if it occurs, but it is far less common than in alkyl halides.

Thus, haloalkanes are much more reactive than haloarenes in nucleophilic substitution reactions due to the stability of the aromatic ring and the resonance effect in haloarenes.

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12. What is the role of a solvent in nucleophilic substitution reactions of haloalkanes?

Answer: The solvent plays a significant role in influencing the rate and mechanism of nucleophilic substitution reactions in haloalkanes. Solvents are broadly classified into polar protic and polar aprotic solvents, and they affect the reaction in different ways:

• Polar Protic Solvents: These solvents can form hydrogen bonds with the nucleophile and stabilize the leaving group (e.g., water, alcohols).

  • Effect on SN1 reactions: Polar protic solvents stabilize the carbocation intermediate by solvation, thereby favoring the SN1 mechanism.
  • Effect on SN2 reactions: These solvents can solvate the nucleophile, reducing its nucleophilicity and slowing down the SN2 mechanism. However, they stabilize the leaving group, facilitating the departure.

• Polar Aprotic Solvents: These solvents cannot form hydrogen bonds with the nucleophile (e.g., DMSO, acetone, acetonitrile).

  • Effect on SN1 reactions: Polar aprotic solvents do not stabilize the carbocation, thus SN1 reactions are less favored in these solvents.
  • Effect on SN2 reactions: Polar aprotic solvents do not solvate the nucleophile as effectively as polar protic solvents, leading to a faster SN2 reaction, since the nucleophile remains highly reactive.

In summary, polar protic solvents favor SN1 reactions, while polar aprotic solvents favor SN2 reactions.

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13. Why are alkyl iodides more reactive than alkyl chlorides in nucleophilic substitution reactions?

Answer: Alkyl iodides are more reactive than alkyl chlorides in nucleophilic substitution reactions due to the following reasons:

1. Bond Strength: The carbon-iodine bond is weaker than the carbon-chlorine bond, making it easier for the iodine atom to leave.

   • The bond dissociation energy of C-I is lower than C-Cl, so iodine leaves more easily, facilitating the nucleophilic substitution.

2. Leaving Group Ability: Iodine is a much better leaving group than chlorine because of its larger size and ability to stabilize the negative charge after departure. The larger atomic size of iodine allows it to stabilize the charge better than chlorine.

Thus, alkyl iodides undergo nucleophilic substitution reactions faster than alkyl chlorides due to the weaker C-I bond and the better leaving group ability of iodine.

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14. How does the steric hindrance of the substrate affect nucleophilic substitution in haloalkanes?

Answer: Steric hindrance significantly affects the rate of nucleophilic substitution reactions in haloalkanes, especially in SN2 mechanisms.

• Primary Haloalkanes: These compounds have less steric hindrance around the carbon attached to the halogen, making it easier for the nucleophile to attack the carbon. As a result, primary haloalkanes typically undergo nucleophilic substitution via the SN2 mechanism.

  • Example: Methyl iodide (CH₃I) reacts very quickly with a nucleophile like OH⁻ because of minimal steric hindrance.

• Secondary Haloalkanes: These compounds have more steric hindrance compared to primary halides but still undergo nucleophilic substitution, although at a slower rate than primary halides. They may undergo either SN1 or SN2 mechanisms, depending on conditions.

• Tertiary Haloalkanes: Tertiary haloalkanes experience significant steric hindrance around the carbon attached to the halogen, making it difficult for the nucleophile to attack. As a result, SN2 reactions are largely unfavorable for tertiary halides. Instead, these compounds undergo nucleophilic substitution via the SN1 mechanism, where the first step is the formation of a stable carbocation.

In conclusion, steric hindrance influences whether the SN2 or SN1 mechanism will dominate. SN2 reactions are favored with less steric hindrance, while SN1 is preferred with more steric hindrance.

15. Explain the preparation of haloalkanes from alcohols.

Answer: Haloalkanes can be prepared from alcohols through halogenation reactions, where the -OH group of alcohols is replaced by a halogen atom (Cl, Br, or I). There are different methods for preparing haloalkanes from alcohols, including reaction with halogenating agents, reduction with halogenating agents, and reaction with thionyl chloride (SOCl₂).

1. Reaction with Hydrogen Halides (HX):

• Alcohols react with hydrogen halides (HCl, HBr, HI) to form haloalkanes.

• Mechanism: The alcohol undergoes protonation to form a better leaving group (water), and then the halide ion attacks the carbocation (in the case of a secondary or tertiary alcohol).

  Example:

  $$R-OH + HX \rightarrow R-X + H_2O$$
  • Primary alcohols usually follow an SN2 mechanism.
  • Secondary and tertiary alcohols undergo SN1 mechanisms.

2. Reaction with Thionyl Chloride (SOCl₂):

• Thionyl chloride is commonly used to convert alcohols into alkyl chlorides.

• It is a chlorinating agent that reacts with alcohols to form haloalkanes. This reaction is preferred for primary and secondary alcohols.

  Example:

  $$R-OH + SOCl_2 \rightarrow R-Cl + SO_2 + HCl$$
  • This reaction proceeds through an SN2 mechanism, where the alcohol is converted directly into the halide.

3. Reaction with Phosphorus Trichloride (PCl₃):

• Alcohols react with PCl₃ to form alkyl chlorides. The reaction is commonly used for primary and secondary alcohols.

  Example:

  $$R-OH + PCl_3 \rightarrow R-Cl + H_3PO_3$$

4. Reaction with PBr₃ or PI₃:

• Similarly, PBr₃ or PI₃ can be used to prepare alkyl bromides or alkyl iodides from alcohols.

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16. What is the mechanism of electrophilic aromatic substitution in haloarenes?

Answer: Electrophilic aromatic substitution (EAS) in haloarenes involves the substitution of a hydrogen atom on the aromatic ring with an electrophile. The presence of the halogen (Cl, Br, I) on the ring influences the mechanism due to the halogen's ability to donate electrons through resonance.

Mechanism:

1. Generation of the Electrophile: The electrophile (E⁺) is generated. For example, in the case of bromination, the electrophile is the bromonium ion (Br⁺), which is generated by reacting bromine with a Lewis acid like FeBr₃.
2. Attack by the Electrophile: The electron-rich aromatic ring attacks the electrophile, forming a positively charged intermediate called the sigma complex (also known as the arenium ion).
3. Restoration of Aromaticity: The positive charge on the sigma complex is delocalized, and the hydrogen atom is removed by a base, restoring the aromaticity of the ring.

Effect of Halogens:

• Halogens are electron-withdrawing through induction but electron-donating through resonance. This dual effect makes them ortho-para directing in EAS reactions, meaning the substitution occurs at the ortho or para positions relative to the halogen.

Example:

Bromination of chlorobenzene:

$$C_6H_5Cl + Br_2 \xrightarrow{FeBr_3} o\text{-}C_6H_4Br + p\text{-}C_6H_4Br$$

In this reaction, the halogen (Cl) directs the substitution to the ortho and para positions due to its resonance-donating nature.

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17. Explain the preparation of haloarenes by the reaction of arenes with halogens in the presence of a catalyst.

Answer: Haloarenes can be prepared by reacting arenes (aromatic compounds) with halogens (Cl₂, Br₂, I₂) in the presence of a Lewis acid catalyst such as FeCl₃, FeBr₃, or AlCl₃. This process is known as halogenation of aromatics and is an example of an electrophilic aromatic substitution reaction.

Mechanism:

1. Generation of the Electrophile: The halogen molecule (e.g., Cl₂ or Br₂) is activated by the catalyst (FeCl₃ or FeBr₃), which helps in the generation of the halonium ion (Cl⁺ or Br⁺). The catalyst is necessary because the halogen molecule is not electrophilic enough on its own to attack the aromatic ring.
2. Attack by the Electrophile: The electron-rich benzene ring attacks the halonium ion, forming a positively charged sigma complex (arenium ion).
3. Restoration of Aromaticity: The sigma complex loses a proton (H⁺) in the presence of the catalyst, and the aromaticity of the ring is restored, leading to the formation of haloarenes.

Example:

Bromination of benzene in the presence of FeBr₃:

$$C_6H_6 + Br_2 \xrightarrow{FeBr_3} C_6H_5Br + HBr$$

This reaction substitutes a hydrogen atom of the benzene ring with a bromine atom, forming bromobenzene.

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18. How can haloalkanes be converted into alcohols?

Answer: Haloalkanes can be converted into alcohols by nucleophilic substitution reactions, where the halogen atom is replaced by a hydroxyl group (-OH). This reaction typically occurs in the presence of a strong nucleophile such as hydroxide ions (OH⁻).

Reaction with Aqueous NaOH (Sodium Hydroxide):

• When haloalkanes react with an aqueous solution of sodium hydroxide (NaOH), the hydroxyl ion (OH⁻) displaces the halogen atom, forming an alcohol.

  Example (SN2 mechanism for primary haloalkane):

  $$R-CH_2Cl + OH^- \rightarrow R-CH_2OH + Cl^-$$

  This reaction proceeds through a backside attack in the SN2 mechanism.

• For secondary and tertiary haloalkanes, the reaction may proceed via the SN1 mechanism due to the formation of a more stable carbocation.

  Example (SN1 mechanism for tertiary haloalkane):

  $$(CH_3)_3CCl + OH^- \rightarrow (CH_3)_3COH + Cl^-$$

  In this case, the first step involves the formation of a tertiary carbocation.

Other Methods:

• Reduction with Sodium in Dry Ether: In the case of alkyl halides, a reduction reaction with sodium in dry ether can convert haloalkanes into alkanes, and further oxidation can yield alcohols.

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19. What are the uses of haloalkanes in industry?

Answer: Haloalkanes have a wide range of industrial applications due to their chemical properties. Some important uses include:

• Solvents: Many haloalkanes, such as chloroform (CHCl₃), carbon tetrachloride (CCl₄), and dichloromethane (CH₂Cl₂), are used as solvents in laboratories and industries because they can dissolve a wide variety of organic and inorganic compounds.

• Refrigerants: Haloalkanes such as Freon (CCl₂F₂), CFCs, and HFCs have been used as refrigerants in air conditioning and refrigeration systems. However, due to environmental concerns over their role in ozone depletion, their use is being phased out in favor of more environmentally friendly alternatives.

• Pesticides: Some haloalkanes are used as pesticides and herbicides, such as DDT (Dichlorodiphenyltrichloroethane), which was widely used in agriculture before being banned due to its environmental impact.

• Pharmaceuticals: Chlorhexidine (used as an antiseptic) and other haloalkane-based compounds are important in medical applications, particularly as disinfectants and antiseptics.

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20. Discuss the environmental impact of haloalkanes.

Answer: Haloalkanes, particularly chlorofluorocarbons (CFCs), hydrochlorofluorocarbons (HCFCs), and hydrofluorocarbons (HFCs), have significant environmental impacts, especially on the ozone layer and the global climate.

1. Ozone Depletion:

   • CFCs and HCFCs contain chlorine and bromine atoms that, when released into the atmosphere, can reach the stratosphere. There, they break down the ozone layer, which protects the Earth from harmful ultraviolet (UV) radiation.
   • Chlorine and bromine atoms catalyze the breakdown of ozone (O₃) into oxygen (O₂) molecules, leading to thinning of the ozone layer, commonly referred to as the ozone hole.

2. Global Warming Potential:

   • Some haloalkanes, especially HFCs, are potent greenhouse gases and contribute to global warming by trapping heat in the atmosphere. While HFCs do not deplete the ozone layer, their high global warming potential makes them significant contributors to climate change.

3. Regulations:

   • Due to their environmental effects, the use of CFCs and HCFCs is being phased out under the Montreal Protocol. Many countries are shifting toward using HFCs and other alternatives, though these still pose a threat to climate change.

21. Explain the reactivity of haloalkanes with sodium metal.

Answer: Haloalkanes can react with sodium metal in the presence of dry ether to form alkanes. This reaction is called the Wurtz reaction, and it involves the coupling of two alkyl radicals or ions. This reaction is mainly used to prepare alkanes from haloalkanes and is particularly useful in organic synthesis for the formation of carbon-carbon bonds.

Mechanism:

1. Formation of Alkyl Radicals: Sodium metal donates electrons to the haloalkane, resulting in the formation of alkyl radicals or carbanions.

2. Coupling Reaction: Two alkyl radicals or carbanions combine to form a new alkane, with the displacement of the halide ion (X⁻).

   Example:

   $$2R-X + 2Na \rightarrow R-R + 2NaX$$

   In this reaction, sodium metal reduces the haloalkane (R-X) by removing the halogen, forming the corresponding alkane (R-R).

   • Example for ethyl chloride: $$2C_2H_5Cl + 2Na \rightarrow C_2H_5-C_2H_5 + 2NaCl$$ This results in the formation of ethane (C₂H₆).

Limitations:

• The reaction is more efficient with primary haloalkanes, as secondary and tertiary haloalkanes may lead to side reactions or may not undergo the reaction effectively.

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22. Discuss the industrial preparation of chlorobenzene from benzene.

Answer: Chlorobenzene is an important industrial chemical, and it is primarily prepared by the chlorination of benzene using chlorine (Cl₂) in the presence of a catalyst like iron(III) chloride (FeCl₃) or aluminum chloride (AlCl₃). This process is an example of electrophilic aromatic substitution.

Mechanism:

1. Generation of the Electrophile: Chlorine reacts with the Lewis acid catalyst (e.g., FeCl₃) to form the chloronium ion (Cl⁺), which is highly electrophilic.
2. Attack by the Electrophile: The electron-rich benzene ring attacks the chloronium ion, forming a sigma complex (arenium ion), where the positive charge is delocalized.
3. Restoration of Aromaticity: The complex loses a proton (H⁺), restoring the aromaticity of the ring and forming chlorobenzene.

Example:

$$C_6H_6 + Cl_2 \xrightarrow{FeCl_3} C_6H_5Cl + HCl$$

In this reaction, benzene (C₆H₆) reacts with chlorine gas in the presence of FeCl₃ to form chlorobenzene (C₆H₅Cl) and hydrogen chloride (HCl) as a byproduct.

Industrial Use:

Chlorobenzene is used as a solvent and an intermediate in the production of other chemicals like phenol, anisole, and dyes. It is also used in the manufacture of pesticides, herbicides, and various pharmaceutical products.

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23. How does the presence of halogens influence the physical properties of haloalkanes?

Answer: The presence of halogens (such as chlorine, bromine, or iodine) in haloalkanes significantly influences their physical properties, including their boiling point, melting point, density, and solubility.

1. Boiling and Melting Points:

• Boiling and melting points of haloalkanes increase with the increase in the atomic size and polarizability of the halogen.
  • For example, the boiling point of chloromethane (CH₃Cl) is -24.2°C, whereas bromomethane (CH₃Br) has a boiling point of 3.6°C, and iodomethane (CH₃I) has a boiling point of 42.1°C.
  • This increase in boiling point can be attributed to the increasing London dispersion forces due to the larger halogen atoms.

2. Solubility:

• Haloalkanes are less soluble in water compared to alcohols and ethers, primarily because they are less polar than alcohols. However, they are soluble in nonpolar solvents such as benzene or ether.
• The solubility decreases as the size of the halogen increases (i.e., HI is less soluble in water than HCl).

3. Density:

• The density of haloalkanes is typically higher than that of water, and it increases with the increase in the atomic size of the halogen. This is because heavier halogens contribute more mass to the molecule.
• For example, iodoalkanes are denser than chloroalkanes.

4. Polarity:

• The polarity of the haloalkane also increases with the electronegativity of the halogen. Fluorine is highly electronegative, and haloalkanes containing fluorine (e.g., CF₄) are more polar compared to those containing chlorine or iodine.

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24. Describe the industrial preparation of iodoform from methyl ketones.

Answer: Iodoform (CHI₃) is an important organic iodine compound that can be prepared industrially through the reaction of methyl ketones (acetone or other similar compounds) with iodine in the presence of a base like sodium hydroxide (NaOH). This reaction is known as the haloform reaction.

Mechanism:

1. Halogenation: The methyl ketone is first treated with iodine (I₂) and a strong base (NaOH). The base deprotonates the methyl group (CH₃) adjacent to the carbonyl group, forming an enolate ion.
2. Substitution of Hydrogen: The enolate ion reacts with iodine to form a triiodinated intermediate.
3. Cleavage: The triiodinated compound undergoes cleavage, releasing iodoform (CHI₃) and leaving behind a carboxylate ion.
4. Final Product: The iodoform is then isolated as a yellow precipitate.

Example:

$$CH_3COCH_3 + 3I_2 + 4NaOH \rightarrow CHI_3 + CH_3COONa + 3NaI + 3H_2O$$

In this reaction, acetone (CH₃COCH₃) reacts with iodine in the presence of sodium hydroxide to produce iodoform (CHI₃) along with sodium acetate (CH₃COONa), sodium iodide (NaI), and water.

Industrial Use:

Iodoform is used as a disinfectant, antiseptic, and in the synthesis of other chemicals. It also has applications in the pharmaceutical and chemical industries.

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25. Explain the reaction of haloalkanes with alcoholic KOH and its significance.

Answer: The reaction of haloalkanes with alcoholic potassium hydroxide (KOH) is an important reaction that results in the formation of alkenes through elimination. This reaction is known as the dehydrohalogenation reaction.

Mechanism:

• The hydroxide ion (OH⁻) from the alcoholic KOH attacks the carbon bonded to the halogen in the haloalkane, leading to the elimination of a hydrogen atom from the adjacent carbon (β-hydrogen) and the halide ion (X⁻).
• The result is the formation of an alkene (double bond between the carbons).

Example (For Ethyl Chloride):

$$C_2H_5Cl + KOH \xrightarrow{\text{alcoholic}} C_2H_4 + KCl + H_2O$$

In this reaction, ethyl chloride (C₂H₅Cl) reacts with alcoholic KOH to form ethene (C₂H₄), potassium chloride (KCl), and water.

Significance:

1. Alkene Formation: This reaction is important for the synthesis of alkenes from haloalkanes, which are key intermediates in the production of plastics, pharmaceuticals, and dyes.
2. Elimination Reaction (E2): This reaction follows the E2 mechanism (a type of elimination reaction), which is a concerted process where the bond-breaking and bond-forming steps occur simultaneously.