Classification

Organic compounds are formed when one or more hydrogen atoms in hydrocarbons (aliphatic or aromatic) are replaced by halogen atom(s). When the replacement occurs in an aliphatic hydrocarbon (like alkane, alkene, alkyne), the resulting compound is an alkyl halide or haloalkane. If the replacement is in an aromatic hydrocarbon (like benzene), the product is an aryl halide or haloarene.

In haloalkanes, the halogen atom is attached to a carbon atom that is sp$^3$ hybridised and part of an alkyl group. In contrast, haloarenes have the halogen atom bonded directly to a carbon atom that is sp$^2$ hybridised and is part of an aromatic ring structure.

Many naturally occurring organic compounds contain halogens and some possess significant medicinal properties. Synthetic halogen-containing compounds are also widely used as solvents, in industrial processes, and for the synthesis of numerous other organic compounds. For example, chloramphenicol is a chlorine-containing antibiotic effective against typhoid, thyroxine is an iodine-containing hormone produced by the body, chloroquine is used to treat malaria, and halothane has been used as an anaesthetic. Certain fluorinated compounds are even explored as blood substitutes.

On The Basis Of Number Of Halogen Atoms

Haloalkanes and haloarenes can be classified according to the number of halogen atoms present in their structure:

• Monohalogen compounds: Contain one halogen atom.

  Example: CH$_3$Cl (Chloromethane), C$_6$H$_5$Br (Bromobenzene)

• Dihalogen compounds: Contain two halogen atoms.

  Example: CH$_2$Cl$_2$ (Dichloromethane), C$_6$H$_4$Cl$_2$ (Dichlorobenzene)

• Polyhalogen compounds: Contain three or more halogen atoms (e.g., tri-, tetra-halogen compounds).

  Example: CHCl$_3$ (Trichloromethane or Chloroform), CCl$_4$ (Tetrachloromethane or Carbon tetrachloride)

Compounds Containing Sp3 C—X Bond (X= F, Cl, Br, I)

Monohalogen compounds are further categorised based on the hybridisation of the carbon atom directly bonded to the halogen. This class includes compounds where the halogen atom is attached to an sp$^3$-hybridised carbon atom.

• Alkyl halides or Haloalkanes (R—X):

  The halogen atom is bonded to an sp$^3$-hybridised carbon within an alkyl group. Alkyl halides form a homologous series with the general formula C$_n$H$_{2n+1}$X.

  They are further classified based on the degree of substitution of the carbon atom bearing the halogen:

  • Primary (1°) alkyl halide: Halogen is attached to a primary carbon atom (a carbon bonded to only one other carbon atom). Example: CH$_3$CH$_2$Cl (Chloroethane).
  • Secondary (2°) alkyl halide: Halogen is attached to a secondary carbon atom (a carbon bonded to two other carbon atoms). Example: CH$_3$CH(Cl)CH$_3$ (2-Chloropropane).
  • Tertiary (3°) alkyl halide: Halogen is attached to a tertiary carbon atom (a carbon bonded to three other carbon atoms). Example: (CH$_3$)$_3$CCl (2-Chloro-2-methylpropane).
• Allylic halides:

  These compounds have the halogen atom bonded to an sp$^3$-hybridised carbon atom which is located immediately adjacent to a carbon-carbon double bond (C=C). This sp$^3$ carbon is called an allylic carbon.

  (The image would show a general structure like H$_2$C=CH—CH$_2$—X, with the CH$_2$ attached to X labelled as the allylic carbon).

• Benzylic halides:

  In these compounds, the halogen atom is bonded to an sp$^3$-hybridised carbon atom that is directly attached to an aromatic ring (typically a benzene ring). This sp$^3$ carbon is called a benzylic carbon.

  (The image would show a general structure like C$_6$H$_5$—CH$_2$—X, with the CH$_2$ attached to X labelled as the benzylic carbon).

Compounds Containing Sp2 C—X Bond

This category includes monohalogen compounds where the halogen atom is directly bonded to an sp$^2$-hybridised carbon atom.

• Vinylic halides:

  These compounds have the halogen atom bonded to an sp$^2$-hybridised carbon atom that is part of a carbon-carbon double bond (C=C). The sp$^2$ carbon is a vinylic carbon.

  (The image would show a general structure like H$_2$C=CH—X, with the CH attached to X labelled as vinylic carbon).

• Aryl halides:

  In these compounds, the halogen atom is directly bonded to an sp$^2$-hybridised carbon atom that is part of an aromatic ring system.

  (The image would show a halogen atom directly attached to a benzene ring).

Nomenclature

Both common names and systematic IUPAC names are used for halogen-containing organic compounds.

For alkyl halides (haloalkanes), the common name is derived by naming the alkyl group followed by the name of the halide (e.g., methyl chloride). In the IUPAC system, they are named as halogen-substituted hydrocarbons (e.g., chloromethane). The position of the halogen and any alkyl substituents is indicated by numbers, ensuring the lowest possible numbers for substituents.

For mono-halogen substituted arenes (aryl halides), the common name and IUPAC name are generally the same (e.g., chlorobenzene). For di-halogen substituted arenes, common names use prefixes like ortho- (o- for 1,2 substitution), meta- (m- for 1,3 substitution), and para- (p- for 1,4 substitution). IUPAC names use numerical locants (1,2; 1,3; 1,4) to indicate the positions of the halogens.

Dihaloalkanes with the same type of halogen atoms have specific common names. If both halogen atoms are on the same carbon atom, they are called geminal dihalides or gem-dihalides, commonly named as alkylidene halides (e.g., ethylidene chloride for CH$_3$CHCl$_2$). If the halogen atoms are on adjacent carbon atoms, they are called vicinal dihalides or vic-dihalides, commonly named as alkylene dihalides (e.g., ethylene dichloride for CH$_2$ClCH$_2$Cl). In the IUPAC system, both gem- and vic-dihalides are systematically named as dihaloalkanes, using numerical locants to indicate the position of each halogen atom (e.g., 1,1-dichloroethane for CH$_3$CHCl$_2$, and 1,2-dichloroethane for CH$_2$ClCH$_2$Cl).

Structure	Common name	IUPAC name
CH$_3$CH$_2$CH(Cl)CH$_3$	sec-Butyl chloride	2-Chlorobutane
(CH$_3$)$_3$CCH$_2$Br	neo-Pentyl bromide	1-Bromo-2,2-dimethylpropane
(CH$_3$)$_3$CBr	tert-Butyl bromide	2-Bromo-2-methylpropane
CH$_2$ = CHCl	Vinyl chloride	Chloroethene
CH$_2$ = CHCH$_2$Br	Allyl bromide	3-Bromopropene
CH$_2$Cl$_2$	Methylene chloride	Dichloromethane
CHCl$_3$	Chloroform	Trichloromethane
CHBr$_3$	Bromoform	Tribromomethane
CCl$_4$	Carbon tetrachloride	Tetrachloromethane
CH$_3$CH$_2$CH$_2$F	n-Propyl fluoride	1-Fluoropropane
	o-Chlorotoluene	1-Chloro-2-methylbenzene or 2-Chlorotoluene
	Benzyl chloride	Chlorophenylmethane

(The images in the table rows would show the corresponding chemical structures).

Answer:

Answer:

Nature Of C-X Bond

The bond between a carbon atom and a halogen atom (C—X) is a polar covalent bond. This is because halogen atoms are generally more electronegative than carbon atoms.

In a C—X bond, the electron density is pulled more towards the halogen atom. This gives the halogen atom a partial negative charge ($\delta^-$) and the carbon atom bonded to it a partial positive charge ($\delta^+$):

$\overset{\delta+}{\text{C}}—\overset{\delta-}{\text{X}}$

Moving down Group 17 of the periodic table, the size of the halogen atom increases significantly (F < Cl < Br < I). Consequently, the length of the C—X bond also increases in the order C—F < C—Cl < C—Br < C—I. The increasing bond length generally correlates with decreasing bond strength or bond enthalpy.

The polarity of the C—X bond results in a dipole moment for alkyl halides. Although electronegativity decreases from F to I, the bond length increases. The dipole moment is proportional to the product of charge separation and bond length. For CH$_3$Cl, the dipole moment is slightly higher than CH$_3$F despite Fluorine being more electronegative. This is because the C—Cl bond is significantly longer than the C—F bond, compensating for the smaller charge separation.

Methods Of Preparation Of Haloalkanes

Alkyl halides can be synthesised using several methods, starting from different classes of organic compounds.

From Alcohols

Alcohols (R—OH) are readily available and serve as a common starting material for preparing alkyl halides. The hydroxyl group (-OH) of an alcohol can be replaced by a halogen atom (X) through reaction with concentrated halogen acids (HX), phosphorus halides (PX$_3$ or PX$_5$), or thionyl chloride (SOCl$_2$).

• Using Hydrogen Halides (HX):

  Alcohols react with concentrated hydrochloric acid (HCl), hydrobromic acid (HBr), or hydroiodic acid (HI) to form alkyl chlorides, bromides, or iodides, respectively.

  R—OH + HX $\rightarrow$ R—X + H$_2$O

  The reaction requires a catalyst, typically anhydrous zinc chloride (ZnCl$_2$), for primary (1°) and secondary (2°) alcohols. Tertiary (3°) alcohols are more reactive and react simply by shaking with concentrated HCl at room temperature, without a catalyst.

  Alkyl bromides (R—Br) are commonly prepared by refluxing alcohols with constant boiling HBr (48%). Alkyl iodides (R—I) are often obtained by heating alcohols with sodium iodide (NaI) or potassium iodide (KI) in 95% orthophosphoric acid (H$_3$PO$_4$).

  The order of reactivity of alcohols with a given hydrogen halide is observed to be: Tertiary alcohol > Secondary alcohol > Primary alcohol.

• Using Phosphorus Halides (PX$_3$ or PX$_5$):

  Alcohols react with phosphorus trichloride (PCl$_3$), phosphorus pentachloride (PCl$_5$), phosphorus tribromide (PBr$_3$), or phosphorus triiodide (PI$_3$).

  R—OH + PCl$_5 \rightarrow$ R—Cl + POCl$_3$ + HCl

  3 R—OH + PCl$_3 \rightarrow$ 3 R—Cl + H$_3$PO$_3$

  For preparing alkyl bromides and iodides using phosphorus halides, phosphorus tribromide (PBr$_3$) and triiodide (PI$_3$) are often generated *in situ* (within the reaction mixture) by reacting red phosphorus with bromine or iodine, respectively, in the presence of the alcohol.

  3 R—OH + P + Br$_2 \rightarrow$ 3 R—Br + H$_3$PO$_3$

  3 R—OH + P + I$_2 \rightarrow$ 3 R—I + H$_3$PO$_3$

• Using Thionyl Chloride (SOCl$_2$):

  Alcohols react with thionyl chloride to produce alkyl chlorides, sulfur dioxide gas, and hydrogen chloride gas.

  R—OH + SOCl$_2 \rightarrow$ R—Cl + SO$_2$ $\uparrow$ + HCl $\uparrow$

  This method is particularly advantageous for preparing alkyl chlorides because the by-products, SO$_2$ and HCl, are gases that easily escape from the reaction mixture. This leaves behind a relatively pure alkyl chloride, making purification simpler.

Note that these methods are generally not suitable for preparing aryl halides from phenols. The carbon-oxygen bond in phenols possesses partial double bond character due to resonance between the oxygen's lone pair and the aromatic ring. This partial double bond is stronger and more difficult to cleave than a single bond in alcohols.

From Hydrocarbons

Haloalkanes can also be prepared by reactions starting from hydrocarbons.

• From Alkanes by Free Radical Halogenation:

  Alkanes react with halogens (Cl$_2$ or Br$_2$) in the presence of ultraviolet light or heat via a free radical mechanism (as studied in Unit 13, Class XI). This reaction substitutes one or more hydrogen atoms with halogen atoms.

  R—H + X$_2 \xrightarrow{\text{Light or Heat}}$ R—X + HX

  However, if the alkane has different types of hydrogen atoms (primary, secondary, or tertiary), free radical halogenation usually produces a complex mixture of isomeric monohaloalkanes, as well as polyhaloalkanes (where multiple hydrogen atoms are substituted). Separating pure individual compounds from this mixture is often difficult, leading to low yields of any specific product.

Answer:

      CH$_3$
      |
CH$_3$—CH—CH$_2$—CH$_3$
            

    CH$_3$(a)
    |
CH$_3$(b)—CH(c)—CH$_2$(d)—CH$_3$(e)
            

• From Alkenes:

  Alkenes can be converted into alkyl halides through addition reactions across the carbon-carbon double bond.

  • Addition of Hydrogen Halides (HX):

    Alkenes react with hydrogen halides (HCl, HBr, HI) to form alkyl halides. The hydrogen atom adds to one carbon of the double bond, and the halogen atom adds to the other.

    R—CH=CH$_2$ + HX $\rightarrow$ R—CHX—CH$_3$ (Markovnikov's Addition)

    When the alkene is unsymmetrical, the addition of HX follows Markovnikov's rule, which states that the hydrogen atom adds to the carbon atom of the double bond that has the greater number of hydrogen atoms, and the halogen atom adds to the carbon atom with fewer hydrogen atoms. This results in the more substituted alkyl halide as the predominant product.

    R—CH=CH$_2$ + HBr $\rightarrow$ R—CHBr—CH$_3$ (Major product, via Markovnikov)

    However, in the presence of peroxides, the addition of HBr to unsymmetrical alkenes follows the anti-Markovnikov rule, where the bromine adds to the less substituted carbon. This peroxide effect is specific to HBr.

  • Addition of Halogens (X$_2$):

    Alkenes react with halogens (Cl$_2$, Br$_2$) by adding across the double bond to form vicinal dihalides (halogens on adjacent carbons).

    R—CH=CH$_2$ + X$_2 \rightarrow$ R—CHX—CH$_2$X

    The addition of bromine (Br$_2$) in an inert solvent like carbon tetrachloride (CCl$_4$) is a common laboratory test for the presence of unsaturation (double or triple bonds). The reddish-brown colour of bromine solution is rapidly discharged as it reacts with the alkene to form a colourless vic-dibromide.

Halogen Exchange

Alkyl iodides and fluorides are often prepared by exchanging a less reactive halogen (like Cl or Br) in an alkyl halide with iodine or fluorine, respectively. These reactions are useful because direct synthesis of alkyl iodides (especially) or fluorides using other methods can be less effective or more hazardous.

• Finkelstein reaction:

  This reaction is used to prepare alkyl iodides (R—I) from alkyl chlorides (R—Cl) or bromides (R—Br). The alkyl halide is heated with sodium iodide (NaI) in a dry organic solvent, typically acetone.

  R—X + NaI $\xrightarrow{\text{dry acetone}}$ R—I + NaX (X = Cl, Br)

  The key to this reaction's effectiveness is that sodium chloride (NaCl) or sodium bromide (NaBr) formed as a by-product are insoluble in dry acetone and precipitate out of solution. According to Le Chatelier's principle, the removal of the product NaX shifts the equilibrium of the reaction to the right, favouring the formation of the alkyl iodide.

• Swarts reaction:

  This reaction is the best method for synthesising alkyl fluorides (R—F) from alkyl chlorides (R—Cl) or bromides (R—Br). The alkyl chloride or bromide is heated with a metallic fluoride, such as silver fluoride (AgF), mercurous fluoride (Hg$_2$F$_2$), cobalt(II) fluoride (CoF$_2$), or antimony(III) fluoride (SbF$_3$).

  R—X + AgF $\rightarrow$ R—F + AgX (X = Cl, Br)

  (Other metallic fluorides react similarly).

Preparation Of Haloarenes

Aryl halides can be prepared from aromatic hydrocarbons or aromatic amines.

From Hydrocarbons By Electrophilic Substitution

Aryl chlorides and bromides can be prepared by the direct electrophilic substitution of hydrogen atoms on the aromatic ring by chlorine or bromine, respectively. This reaction requires the presence of a Lewis acid catalyst, such as iron (Fe) or iron(III) halide (FeCl$_3$, FeBr$_3$). The catalyst helps to generate the electrophile (e.g., Cl$^+$ or Br$^+$).

(The image would show the reaction: Benzene + X$_2$ $\xrightarrow{\text{Lewis Acid}}$ Halobenzene + HX).

For substituted benzenes, this reaction can lead to the formation of isomeric products (ortho-, meta-, para-). For example, toluene undergoes electrophilic chlorination or bromination to give a mixture of ortho- and para-halotoluene.

(The image would show the reaction: Toluene + X$_2$ $\xrightarrow{\text{Lewis Acid}}$ Mixture of o-halotoluene and p-halotoluene).

The ortho- and para-isomers generally have significantly different melting points, which facilitates their separation by techniques like fractional crystallisation.

Direct iodination of arenes (with I$_2$) is possible, but it is a reversible reaction because the hydrogen iodide (HI) formed is a strong reducing agent and can reduce the aryl iodide back to the arene. To drive the reaction forward and obtain a good yield of the aryl iodide, an oxidising agent (like concentrated nitric acid HNO$_3$ or iodic acid HIO$_4$) must be added to oxidise the HI as it is formed.

Direct fluorination of arenes (with F$_2$) is usually not a practical method for preparing aryl fluorides due to the extremely high reactivity of fluorine, which can lead to uncontrolled reactions and often ring destruction.

From Amines By Sandmeyer’S Reaction

Aryl halides, particularly aryl chlorides and bromides, can be prepared from primary aromatic amines (like aniline) using the Sandmeyer's reaction. This method provides a controlled way to introduce a halogen onto the aromatic ring.

The reaction proceeds in two main steps:

1. Diazotisation: A primary aromatic amine (Ar—NH$_2$) is treated with sodium nitrite (NaNO$_2$) and a cold aqueous mineral acid (like HCl or HBr) at low temperature (typically 0-5 °C). This converts the amino group into a diazonium group (—N$_2^+$), forming an arenediazonium salt (Ar—N$_2^+$ X$^-$).

(The image would show the reaction: Ar—NH$_2$ + NaNO$_2$ + HX $\xrightarrow{0-5 \text{°C}}$ Ar—N$_2^+$ X$^-$ + NaCl + 2H$_2$O).

2. Replacement of the diazonium group: The freshly prepared solution of the arenediazonium salt is then mixed with a solution of the corresponding cuprous halide (Copper(I) chloride, CuCl, for chloroarenes; Copper(I) bromide, CuBr, for bromoarenes). This causes the diazonium group (—N$_2^+$) to be replaced by the halogen atom (—Cl or —Br), with the release of nitrogen gas.

(The image would show the reaction: Ar—N$_2^+$ X$^-$ $\xrightarrow{\text{CuX}}$ Ar—X + N$_2$ $\uparrow$).

For preparing aryl iodides (Ar—I) from arenediazonium salts, the Sandmeyer reaction conditions are not strictly required. The diazonium group can be replaced by iodine simply by shaking the arenediazonium salt solution with potassium iodide (KI).

Ar—N$_2^+$ X$^-$ + KI $\rightarrow$ Ar—I + KX + N$_2$ $\uparrow$

Answer:

(i) This is the reaction of a primary alcohol (propan-1-ol) with thionyl chloride. This reaction replaces the -OH group with -Cl, producing an alkyl chloride, SO$_2$ gas, and HCl gas.

CH$_3$CH$_2$CH$_2$OH + SOCl$_2 \rightarrow$ CH$_3$CH$_2$CH$_2$Cl + SO$_2$ + HCl

(Product: 1-Chloropropane)

(ii) This is the reaction of a primary alcohol (ethanol) with hydrobromic acid. This reaction replaces the -OH group with -Br, producing an alkyl bromide and water.

CH$_3$CH$_2$OH + HBr $\rightarrow$ CH$_3$CH$_2$Br + H$_2$O

(Product: Bromoethane)

(iii) This is the reaction of a tertiary alcohol (2-methylpropan-2-ol) with hydrochloric acid. Tertiary alcohols react readily with concentrated HCl without a catalyst at room temperature.

$\text{CH}_3\text{—}\underset{\text{CH}_3}{\overset{\text{CH}_3}{\text{C}}}\text{—OH} + \text{HCl} \rightarrow$ $\text{CH}_3\text{—}\underset{\text{CH}_3}{\overset{\text{CH}_3}{\text{C}}}\text{—Cl} + \text{H}_2\text{O}$

(Product: 2-Chloro-2-methylpropane or tert-Butyl chloride)

Physical Properties

The physical properties of haloalkanes and haloarenes, such as melting points, boiling points, solubility, and density, are influenced by their molecular structure, polarity, and intermolecular forces.

When pure, alkyl halides are generally colourless. However, alkyl bromides and iodides tend to develop colour upon exposure to light, likely due to decomposition and the liberation of free halogen.

Many volatile halogen-containing organic compounds have a characteristic sweet odour.

Melting And Boiling Points

Simple alkyl halides like methyl chloride (CH$_3$Cl), methyl bromide (CH$_3$Br), ethyl chloride (CH$_3$CH$_2$Cl), and some chlorofluoromethanes are gases at room temperature. Higher members of the homologous series are liquids or solids.

Compared to their parent hydrocarbons of similar molecular mass, haloalkanes have significantly higher boiling points. This is because molecules of organic halogen compounds are generally polar, possessing a dipole due to the polar C—X bond. This results in stronger dipole-dipole interactions between molecules. Additionally, they have higher molecular masses and more electrons than hydrocarbons, leading to stronger van der Waals forces (specifically London dispersion forces).

The strength of these intermolecular forces increases with increasing size and mass of the halogen atom. Therefore, for a given alkyl group, the boiling points of alkyl halides follow the order: R—I > R—Br > R—Cl > R—F.

(The image would show a graph plotting boiling points for CH$_3$X, CH$_3$CH$_2$X, CH$_3$CH$_2$CH$_2$X, etc., for different halogens X, illustrating the increase with molecular weight and the order RI > RBr > RCl > RF).

For isomeric haloalkanes (compounds with the same formula but different structures), the boiling point decreases with increasing branching in the alkyl chain. Increased branching leads to a more compact, spherical shape and reduced surface area, which weakens the van der Waals forces between molecules. For example, among the isomeric bromopentanes (C$_5$H$_{11}$Br), 2-bromo-2-methylpropane (a tertiary bromide) has the lowest boiling point.

Boiling points of isomeric dihalobenzenes (e.g., o-, m-, p-dichlorobenzene) are very similar. However, the melting points of the para-isomers are usually significantly higher than those of the ortho- and meta-isomers. This is because the symmetrical structure of the para-isomer allows its molecules to fit more closely and efficiently into the crystal lattice in the solid state. This tighter packing requires more energy to break, resulting in a higher melting point.

Solubility

Despite their polarity, haloalkanes are only very slightly soluble in water. To dissolve a haloalkane in water, energy is needed to overcome the intermolecular forces holding the haloalkane molecules together and to break the hydrogen bonds between water molecules. When new interactions form between haloalkane and water molecules, the energy released is less than the energy required to break the strong hydrogen bonds in water. As a result, the overall process is energetically unfavourable, leading to low solubility in water.

However, haloalkanes generally dissolve readily in organic solvents. This is because the intermolecular forces established between haloalkane molecules and organic solvent molecules are of comparable strength to the forces being broken within the separate haloalkane and solvent substances. The energy changes are balanced, favouring dissolution.

Density

The density of haloalkanes and haloarenes is generally higher than that of the corresponding hydrocarbons.

Bromo, iodo, and poly-chloro derivatives of hydrocarbons are typically heavier than water (density > 1 g/mL).

The density of these compounds increases with:

• Increasing number of carbon atoms in the alkyl chain.
• Increasing number of halogen atoms in the molecule.
• Increasing atomic mass of the halogen atom (density increases in the order RCl < RBr < RI for a given alkyl group).

Chemical Reactions

The chemical reactivity of haloalkanes and haloarenes is largely determined by the nature of the polar C—X bond and the electronic environment around it. Alkyl halides are generally more reactive than aryl halides towards nucleophilic substitution.

Reactions Of Haloalkanes

Alkyl halides (R—X) undergo three main types of reactions:

1. Nucleophilic substitution reactions
2. Elimination reactions
3. Reactions with metals

(1) Nucleophilic substitution reactions:

Nucleophiles are chemical species that are 'electron-rich', meaning they have a lone pair of electrons or a negative charge and are attracted to positively charged or electron-deficient centers. In haloalkanes, the carbon atom bonded to the halogen carries a partial positive charge ($\delta^+$) due to the polarity of the C—X bond. This makes the carbon atom an 'electrophilic center', susceptible to attack by nucleophiles.

A nucleophilic substitution reaction involves a nucleophile (Nu$^-$ or Nu) attacking the carbon atom of the haloalkane. The incoming nucleophile forms a new bond with the carbon, and simultaneously, the halogen atom (which acts as a 'leaving group') departs as a halide ion (X$^-$).

Nu$^-$ + R—X $\rightarrow$ R—Nu + X$^-$

Since a nucleophile initiates this substitution, it is called a nucleophilic substitution reaction. These reactions are very common and useful for converting alkyl halides into a variety of other functional groups.

The leaving group ability of halide ions correlates with their basicity; weaker bases are better leaving groups. The order of leaving group ability is I$^-$ > Br$^-$ > Cl$^-$ >> F$^-$. This means alkyl iodides are generally the most reactive alkyl halides in substitution reactions, while alkyl fluorides are the least reactive.

Some nucleophiles are called ambident nucleophiles because they have two different atoms that can potentially act as the donor site and form a bond with the electrophilic carbon. Examples include the cyanide ion (CN$^-$) and the nitrite ion (NO$_2^-$).

• Cyanide ion (CN$^-$): It is a resonance hybrid [:C$\equiv$N $\leftrightarrow$ C=N:$^-$]. It can attack through the carbon atom (forming R—CN, an alkyl cyanide or nitrile) or through the nitrogen atom (forming R—NC, an alkyl isocyanide or isonitrile).
• Nitrite ion (NO$_2^-$): It is a resonance hybrid [O=N—O$^-$ $\leftrightarrow$ O$^-$—N=O]. It can attack through an oxygen atom (forming R—O—N=O, an alkyl nitrite) or through the nitrogen atom (forming R—NO$_2$, a nitroalkane).

The product formed depends on the nature of the reagent providing the ambident nucleophile (ionic vs. covalent). Ionic reagents (like KCN, KNO$_2$) tend to favour reaction through the more electronegative atom (N in CN$^-$, O in NO$_2^-$ resonance structures) or the atom that forms the stronger bond to carbon (C in CN$^-$). Covalent reagents (like AgCN, AgNO$_2$) tend to favour reaction through the less electronegative atom (N in CN$^-$ and NO$_2^-$) because the bonding is more directional.

Answer:

Mechanism of Nucleophilic Substitution: Nucleophilic substitution reactions generally follow two different mechanistic pathways: unimolecular (S$_\text{N}1$) and bimolecular (S$_\text{N}2$).

(a) Substitution Nucleophilic Bimolecular (S$_\text{N}2$):

This mechanism involves a single step where bond breaking (C—X) and bond formation (C—Nu) occur simultaneously. It is a concerted reaction with no intermediate formed.

The rate of the reaction depends on the concentration of both the alkyl halide and the nucleophile. Thus, it follows second-order kinetics: Rate = k [Alkyl halide] [Nucleophile].

Consider the reaction of methyl chloride with hydroxide ion:

CH$_3$Cl + OH$^-$ $\rightarrow$ CH$_3$OH + Cl$^-$

Mechanism:

(The image would show a diagram illustrating the concerted step: the nucleophile (e.g., OH$^-$) approaching the carbon from the back side (opposite to the leaving group X), formation of a transition state where carbon is partially bonded to both Nu and X, and the leaving group departing from the front side, resulting in the product).

As the nucleophile approaches the carbon atom bearing the leaving group, it attacks from the side opposite to the leaving group. This is necessary because the leaving group already occupies space on the "front" side, and the attacking nucleophile is also electron-rich and would be repelled by the leaving group's electron density. The back-side attack is favoured sterically and electronically.

During the transition state, the carbon atom is simultaneously bonded to the incoming nucleophile and the outgoing leaving group. This transition state is unstable and cannot be isolated. The carbon atom is momentarily bonded to five atoms in a trigonal bipyramidal arrangement, with the three original substituents in a plane and the incoming and leaving groups along an axis perpendicular to this plane.

As the nucleophile bonds to carbon and the leaving group departs, the spatial arrangement of the other three groups attached to the carbon atom flips from one side to the other. This is analogous to an umbrella being turned inside out by the wind. This flipping of configuration at the reacting carbon center is called inversion of configuration.

Steric effects are very important in S$_\text{N}2$ reactions. The approach of the nucleophile to the back side of the carbon is hindered by bulky substituents attached to that carbon or nearby carbons. The less hindered the carbon, the faster the S$_\text{N}2$ reaction.

Thus, the reactivity of alkyl halides towards S$_\text{N}2$ reactions decreases with increasing steric hindrance around the $\alpha$-carbon (the carbon bonded to the halogen):

Methyl halide > Primary halide > Secondary halide >> Tertiary halide

Methyl halides (like CH$_3$X) are the most reactive because they have only small hydrogen atoms attached to the carbon, providing minimal steric hindrance for the attacking nucleophile.

Tertiary halides (R$_3$CX) are the least reactive (or virtually unreactive) in S$_\text{N}2$ reactions because the three bulky alkyl groups severely block the approach of the nucleophile to the back side of the $\alpha$-carbon.

(The image would show diagrams of methyl, primary, secondary, and tertiary halides, illustrating the increasing bulk around the carbon attached to the halogen and the relative ease or difficulty of back-side attack by a nucleophile).

(b) Substitution Nucleophilic Unimolecular (S$_\text{N}1$):

This mechanism typically occurs in polar protic solvents (solvents that can form hydrogen bonds, like water, alcohols, acetic acid) and generally involves tertiary alkyl halides, as well as allylic and benzylic halides, which can form relatively stable carbocations.

The reaction proceeds in two steps, and the rate of the reaction depends only on the concentration of the alkyl halide. It follows first-order kinetics: Rate = k [Alkyl halide].

Consider the reaction of tert-butyl bromide with hydroxide ion:

(CH$_3$)$_3$CBr + OH$^-$ $\rightarrow$ (CH$_3$)$_3$COH + Br$^-$

Mechanism:

Step 1: Formation of Carbocation (Slow, Rate-determining step)

The alkyl halide undergoes slow, spontaneous ionisation in the polar protic solvent. The C—X bond breaks heterolytically, with the leaving group (halide ion) taking both bonding electrons. This forms a carbocation intermediate and a halide ion.

R—X $\xrightarrow{\text{slow, ionisation}}$ R$^+$ (Carbocation) + X$^-$ (Halide ion)

The energy required to break the C—X bond is offset by the solvation of the resulting ions (carbocation and halide ion) by the polar solvent. The stability of the carbocation is a major factor influencing the rate of this step. More stable carbocations are formed more easily and quickly.

Step 2: Attack by Nucleophile (Fast)

The nucleophile rapidly attacks the electron-deficient carbocation intermediate. Since the carbocation is planar (sp$^2$ hybridised), the nucleophile can attack from either side (front or back).

R$^+$ + Nu$^-$ $\xrightarrow{\text{fast}}$ R—Nu (Product)

The first step, the formation of the carbocation, is the slowest step in the mechanism and thus determines the overall rate of the reaction. Since this step involves only the alkyl halide, the rate is proportional only to the concentration of the alkyl halide.

The stability of the carbocation intermediate is key to the S$_\text{N}1$ mechanism. More stable carbocations are formed more easily, leading to faster S$_\text{N}1$ reactions. The order of carbocation stability is: Tertiary > Secondary > Primary > Methyl.

Therefore, the order of reactivity of alkyl halides towards S$_\text{N}1$ reactions is: Tertiary halide > Secondary halide > Primary halide >> Methyl halide.

Allylic (—CH$_2$—CH=CH$_2$) and benzylic (—CH$_2$—C$_6$H$_5$) halides show high reactivity towards S$_\text{N}1$ reactions because the carbocations formed from them are stabilised by resonance.

(The image would show resonance structures for an allylic carbocation (H$_2$C=CH—CH$_2^+ \leftrightarrow$ H$_2$C$^+$—CH=CH$_2$) and a benzylic carbocation (C$_6$H$_5$—CH$_2^+ \leftrightarrow$ resonance forms with positive charge delocalised onto the ring)).

Summary of relative reactivities in S$_\text{N}1$ and S$_\text{N}2$ reactions (considering the structure of the alkyl group):

S$_\text{N}1$ Reactivity: Tertiary > Secondary > Primary > Methyl (Stability of carbocation)

S$_\text{N}2$ Reactivity: Methyl > Primary > Secondary >> Tertiary (Steric hindrance)

For a given alkyl group, the reactivity of the alkyl halide (R—X) depends on the leaving group ability of the halide ion. Since I$^-$ is the best leaving group (weakest base), followed by Br$^-$ and Cl$^-$, and F$^-$ is a very poor leaving group, the reactivity order for R—X with a fixed R group is consistent for both S$_\text{N}1$ and S$_\text{N}2$: R—I > R—Br > R—Cl >> R—F.

Answer:

Answer:

(c) Stereochemical aspects of nucleophilic substitution reactions:

To fully understand how S$_\text{N}1$ and S$_\text{N}2$ reactions proceed, particularly when they occur at a carbon atom with four different substituents (a stereocenter), we need to consider their stereochemical outcomes. This involves concepts like optical activity, chirality, retention, inversion, and racemisation.

(i) Optical activity: Certain compounds have the ability to rotate the plane of plane-polarised light when it passes through their solution. Such compounds are called optically active. Plane-polarised light is created by passing ordinary light through a polariser (like a Nicol prism).

The amount and direction of rotation are measured using an instrument called a polarimeter. If a compound rotates the plane of polarised light to the right (clockwise), it is called dextrorotatory (or d-form), indicated by a plus sign (+) before the observed angle of rotation (e.g., (+)-butan-2-ol). If it rotates the light to the left (counter-clockwise), it is called levorotatory (or l-form), indicated by a minus sign (-) (e.g., (-)-butan-2-ol).

Compounds that exist as (+) and (-) forms are called optical isomers or enantiomers, and the phenomenon is called optical isomerism.

(ii) Molecular asymmetry, chirality, and enantiomers:

The ability of some molecules to exist as optical isomers is related to their molecular symmetry. A molecule is considered chiral if it is non-superimposable on its mirror image. This property is called chirality or handedness, like a left and right hand. Chiral molecules are typically optically active.

Conversely, a molecule that is superimposable on its mirror image is called achiral. Achiral molecules are generally optically inactive.

A common structural feature that leads to chirality in organic molecules is the presence of an asymmetric carbon atom (also known as a chiral carbon or stereocenter). This is a carbon atom bonded to four different atoms or groups.

Consider propan-2-ol (CH$_3$CHOHCH$_3$). The carbon bonded to -OH is also bonded to a hydrogen, a methyl group, and another methyl group. Since two groups are identical (methyl), this carbon is not asymmetric. Propan-2-ol is achiral and optically inactive.

(The image would show a 3D representation of propan-2-ol (A) and its mirror image (B). Rotating B by 180 degrees gives C, which is identical to A, showing superimposability).

Now consider butan-2-ol (CH$_3$CHOHCH$_2$CH$_3$). The carbon bonded to -OH is also bonded to a hydrogen, a methyl group (CH$_3$), an ethyl group (CH$_2$CH$_3$), and a hydroxyl group (-OH). These four groups are all different. This carbon is an asymmetric carbon atom. Butan-2-ol is chiral and exists as two optical isomers, which are non-superimposable mirror images.

(The image would show 3D representations of the two enantiomers of butan-2-ol (D and E) as mirror images. Rotating E to get F shows that F cannot be superimposed on D).

Other examples of chiral molecules with a single asymmetric carbon include 2-chlorobutane, 2-bromopropanoic acid, etc.

The stereoisomers that are non-superimposable mirror images of each other are called enantiomers. The two enantiomers of a chiral compound have identical physical properties (melting point, boiling point, density, refractive index, solubility in achiral solvents, etc.). Their only difference lies in their interaction with plane-polarised light (rotating it in opposite directions) and their reactivity with other chiral molecules.

A mixture containing equal molar amounts of two enantiomers (one d-form and one l-form) is called a racemic mixture or racemic modification. Racemic mixtures are optically inactive because the rotation of plane-polarised light caused by one enantiomer is exactly cancelled out by the equal and opposite rotation caused by the other. A racemic mixture is often designated by prefixing 'dl-' or '($\pm$)' before the compound's name, e.g., ($\pm$)-butan-2-ol.

The process of converting an enantiomer or a non-racemic mixture into a racemic mixture is called racemisation.

(iii) Retention of configuration:

Retention of configuration refers to a chemical reaction or transformation where the spatial arrangement of groups around a stereocenter remains unchanged relative to the starting material. This typically happens when bonds to the stereocenter are not broken during the reaction.

For example, if a reaction occurs at a position other than the asymmetric carbon, or if a group attached to the asymmetric carbon is replaced in such a way that the relative positions of the other groups are maintained, the configuration at the stereocenter is retained. However, the sign of optical rotation might change if the product is a different compound with a different magnitude or direction of rotation, even if the configuration at the stereocenter is the same.

Consider the reaction of (-)-2-methylbutan-1-ol with concentrated HCl:

(The image would show the reaction of (-)-2-methylbutan-1-ol with HCl to form (+)-1-chloro-2-methylbutane. Structures would clearly show that the bond broken is O-H, not the bond to the stereocenter (marked), and the configuration at the stereocenter is preserved).

In this reaction, the chiral center is C2, which is not involved in the bond-breaking and bond-forming steps that occur at C1. Thus, the spatial arrangement around C2 is unchanged, and the configuration is retained. Even though the product (+)-1-chloro-2-methylbutane rotates light in the opposite direction compared to the reactant (-)-2-methylbutan-1-ol, this is consistent with retention because they are different compounds. The change in sign doesn't automatically mean inversion; one must look at the relative spatial arrangement around the stereocenter.

(iv) Inversion, retention, and racemisation in S$_\text{N}1$ and S$_\text{N}2$ reactions:

When a nucleophilic substitution reaction occurs at a carbon atom that is a stereocenter (i.e., an asymmetric carbon), the outcome depends on the reaction mechanism (S$_\text{N}1$ or S$_\text{N}2$).

• S$_\text{N}2$ Reaction at a Stereocenter:

  As discussed in the S$_\text{N}2$ mechanism, the nucleophile attacks the carbon atom from the back side, opposite to the leaving group. This simultaneous bond formation and bond breaking process causes the tetrahedral geometry around the carbon to flip. This leads to a complete inversion of configuration at the stereocenter.

  If the reactant is an optically active alkyl halide (pure enantiomer), the S$_\text{N}2$ reaction will result in the formation of the product with the inverted configuration. The product will be the other enantiomer of the corresponding alcohol (or other functional group if the nucleophile is different).

  For example, the reaction of (-)-2-bromooctane (an optically active alkyl halide) with sodium hydroxide (a nucleophile) produces (+)-octan-2-ol. The product has the inverted configuration relative to the reactant.

  (The image would show the S$_\text{N}2$ reaction of (-)-2-bromooctane with OH$^-$ to form (+)-octan-2-ol, using 3D structures that clearly illustrate the inversion of the stereocenter configuration).

  Thus, S$_\text{N}2$ reactions of optically active alkyl halides proceed with 100% inversion of configuration.

• S$_\text{N}1$ Reaction at a Stereocenter:

  In the S$_\text{N}1$ mechanism, the slow step is the formation of a carbocation intermediate. When the reaction occurs at a stereocenter, the leaving group departs from the asymmetric carbon, producing a carbocation. This carbocation is sp$^2$ hybridised and has a trigonal planar geometry.

  The second step involves the attack of the nucleophile on this planar carbocation. Since the carbocation is planar (achiral), the nucleophile can attack from either side of the plane with equal probability.

  Attack from the "front" side (where the leaving group was located) results in a product with the same configuration as the original alkyl halide.

  Attack from the "back" side (opposite to where the leaving group was) results in a product with the inverted configuration relative to the original alkyl halide.

  If the starting material is a pure enantiomer of an alkyl halide, and the nucleophile attacks the planar carbocation equally from both sides, a mixture of the two enantiomers of the product will be formed. If the attack from both sides is exactly equally probable, a 50:50 mixture of the two enantiomers will result. This 50:50 mixture is a racemic mixture, and it is optically inactive.

  The formation of a racemic mixture from an optically active reactant is called racemisation.

  For example, the hydrolysis of optically active 2-bromobutane via an S$_\text{N}1$ mechanism produces ($\pm$)-butan-2-ol (a racemic mixture).

  (The image would show the S$_\text{N}1$ reaction of an enantiomer of 2-bromobutane. Step 1 shows the formation of the planar carbocation. Step 2 shows the attack of the nucleophile (H$_2$O) from both sides of the planar carbocation, leading to the formation of both enantiomers of butan-2-ol in equal amounts).

  Thus, S$_\text{N}1$ reactions of optically active alkyl halides are typically accompanied by racemisation. However, complete 100% racemisation may not always be observed, sometimes a slight preference for inversion occurs if the leaving group hasn't fully diffused away from the carbocation before nucleophilic attack (ion pair formation).

2. Elimination reactions:

Alkyl halides containing at least one hydrogen atom on a carbon atom adjacent to the carbon bearing the halogen ($\alpha$-carbon) can undergo elimination reactions. These adjacent carbon atoms are called $\beta$-carbons, and the hydrogen atoms on them are called $\beta$-hydrogen atoms.

When a haloalkane with a $\beta$-hydrogen is heated with a strong base (like an alcoholic solution of potassium hydroxide), the base removes a $\beta$-hydrogen atom, and simultaneously, the halogen atom is removed from the $\alpha$-carbon. This process involves the removal of a hydrogen and a halogen from adjacent carbons, resulting in the formation of a carbon-carbon double bond (an alkene) and a molecule of HX (which reacts with the base).

(The image would show the general reaction scheme: R—CH(X)—CH$_2$—H + Base $\rightarrow$ R—CH=CH$_2$ + H—Base + X$^-$).

Since a $\beta$-hydrogen is involved in the elimination, this type of reaction is specifically called $\beta$-elimination or dehydrohalogenation (removal of hydrogen and halogen).

If an alkyl halide has $\beta$-hydrogen atoms on more than one different $\beta$-carbon, or if removing a hydrogen from a specific $\beta$-carbon can lead to isomeric alkenes (e.g., due to the position of the double bond), a mixture of alkenes may be formed. In such cases, one alkene usually predominates as the major product.

The regioselectivity (which alkene isomer is formed) in dehydrohalogenation reactions is often governed by Zaitsev's rule (Saytzeff's rule). Formulated by Russian chemist Alexander Zaitsev, this rule states that in dehydrohalogenation reactions, the major alkene product is the one that is most substituted, meaning it has the greater number of alkyl groups attached to the doubly bonded carbon atoms.

Example: Dehydrohalogenation of 2-bromopentane with alcoholic KOH:

(The image would show the reaction: CH$_3$CH$_2$CH$_2$CH(Br)CH$_3$ + Alc. KOH $\rightarrow$ CH$_3$CH$_2$CH=CHCH$_3$ (pent-2-ene, major) + CH$_3$CH$_2$CH$_2$CH=CH$_2$ (pent-1-ene, minor)). It would indicate that pent-2-ene is more substituted (two alkyl groups, ethyl and methyl, attached to the double bond carbons) than pent-1-ene (one alkyl group, propyl, attached to the double bond carbons), thus making it the major product according to Zaitsev's rule.

Competition between Substitution and Elimination:

Alkyl halides, particularly those with $\beta$-hydrogens, can undergo both nucleophilic substitution (S$_\text{N}1$ or S$_\text{N}2$) and elimination (E1 or E2) reactions when treated with a base or nucleophile. These reactions are often competing pathways.

The favoured pathway (substitution or elimination) depends on several factors:

• Structure of the alkyl halide: Primary halides tend to favour S$_\text{N}2$. Tertiary halides favour S$_\text{N}1$ (in protic solvents) or E1/E2. Secondary halides can undergo S$_\text{N}2$, S$_\text{N}1$, or elimination, depending on conditions.
• Strength and size of the base/nucleophile: Strong bases/nucleophiles (like OH$^-$ or OR$^-$) favour S$_\text{N}2$ (with primary halides) or E2. Weak bases/nucleophiles (like H$_2$O or ROH) favour S$_\text{N}1$/E1 (with tertiary/secondary halides). Bulky bases/nucleophiles (like tert-butoxide, (CH$_3$)$_3$CO$^-$) are sterically hindered and prefer to abstract a proton (act as a base) leading to elimination (E2) rather than attacking the carbon center (substitution).
• Reaction conditions: High temperature generally favours elimination reactions over substitution reactions. The solvent also plays a role (e.g., polar protic solvents favour S$_\text{N}1$/E1, polar aprotic solvents favour S$_\text{N}2$).

Understanding the interplay of these factors is necessary to predict the major product of a reaction involving an alkyl halide and a basic or nucleophilic reagent.

3. Reaction with metals:

Alkyl halides react with certain metals to form compounds that contain a carbon-metal bond. These are known as organometallic compounds.

• Formation of Grignard Reagents:

  One of the most important classes of organometallic compounds is alkyl magnesium halides, R—Mg—X, commonly known as Grignard reagents. These are prepared by reacting an alkyl halide (R—X, typically chloride, bromide, or iodide) with magnesium metal in dry ether (like diethyl ether or tetrahydrofuran, THF).

  R—X + Mg $\xrightarrow{\text{dry ether}}$ R—Mg—X (X = Cl, Br, I)

  In a Grignard reagent (R—Mg—X), the bond between the carbon atom and the magnesium atom (C—Mg) is highly polar covalent, with a partial negative charge on the carbon atom (due to magnesium being electropositive) and a partial positive charge on magnesium. The magnesium-halogen bond (Mg—X) is essentially ionic.

  $\overset{\delta-}{\text{R}}—\overset{\delta+}{\text{Mg}}—\text{X}$

  Grignard reagents are extremely reactive due to the highly polarised C—Mg bond and the carbanionic nature of the carbon atom. They react readily with any source of protons (even weak acids) to form hydrocarbons. This includes reaction with water, alcohols, amines, carboxylic acids, terminal alkynes, etc.

  R—Mg—X + H—Z $\rightarrow$ R—H + Mg(Z)X (where H—Z is a proton source like H$_2$O, ROH, RNH$_2$, RCOOH, RC$\equiv$CH)

  Because they react vigorously with moisture, Grignard reagents must be prepared and handled under strictly anhydrous conditions in an inert atmosphere (like nitrogen or argon). The reaction with water provides a method for converting an alkyl halide to a hydrocarbon.

• Wurtz reaction:

  Alkyl halides react with sodium metal in dry ether to produce alkanes containing twice the number of carbon atoms as the original alkyl halide. This coupling reaction is known as the Wurtz reaction (discussed in Unit 13, Class XI).

  2 R—X + 2 Na $\xrightarrow{\text{dry ether}}$ R—R + 2 NaX

  This reaction is useful for preparing symmetrical alkanes (R—R) from primary alkyl halides. A mixture of two different alkyl halides in a Wurtz reaction can lead to a mixture of three possible alkane products (R—R, R—R', R'—R'), making it less useful for preparing unsymmetrical alkanes (R—R').

Reactions Of Haloarenes

Aryl halides are significantly less reactive towards nucleophilic substitution reactions compared to alkyl halides. This difference in reactivity is attributed to several factors related to the presence of the halogen atom on the aromatic ring.

Reasons for the low reactivity of aryl halides towards nucleophilic substitution:

1. Resonance effect: In haloarenes, the lone pairs of electrons on the halogen atom are in conjugation with the $\pi$-electron system of the aromatic ring. This leads to resonance structures where a negative charge is delocalised onto the ortho and para positions of the ring, and the carbon-halogen bond acquires partial double bond character.

   (The image would show the resonance structures of a halobenzene, indicating partial double bond character in the C-X bond and negative charge on the ring ortho and para carbons).

   Due to this partial double bond character, the C—X bond in haloarenes is stronger and shorter than a typical C—X single bond in alkyl halides. Breaking this bond is more difficult, making nucleophilic substitution less favourable.

2. Difference in hybridisation of carbon atom in C—X bond: In haloalkanes, the carbon atom bonded to the halogen is sp$^3$ hybridised. In haloarenes, the carbon atom bonded to the halogen is sp$^2$ hybridised. An sp$^2$ hybrid orbital has a greater s-character (33% s-character) compared to an sp$^3$ hybrid orbital (25% s-character).

   The greater s-character of the sp$^2$ hybrid orbital means that the electrons in this orbital are held more tightly by the nucleus. Consequently, the sp$^2$ hybridised carbon atom in haloarenes is more electronegative than the sp$^3$ hybridised carbon in alkyl halides. This more electronegative carbon holds the electron pair of the C—X bond more tightly. This results in a shorter and stronger C—X bond in haloarenes (e.g., C—Cl bond length is 169 pm in chlorobenzene vs. 177 pm in methyl chloride), making it more difficult to cleave.

3. Instability of phenyl cation: If haloarenes were to undergo an S$_\text{N}1$ type reaction, it would involve the ionisation of the C—X bond to form a phenyl carbocation (C$_6$H$_5^+$). This phenyl cation is highly unstable because the positive charge is on an sp$^2$ hybridised carbon of the ring and cannot be effectively stabilised by resonance with the $\pi$ electrons (as the vacant p orbital needed for resonance is perpendicular to the ring's $\pi$ system). The instability of the phenyl cation means that the S$_\text{N}1$ mechanism is very unlikely for aryl halides under normal conditions.
4. Repulsion between nucleophile and arene $\pi$ electron cloud: The aromatic ring in haloarenes is electron-rich due to the delocalised $\pi$ electrons. An attacking nucleophile is also electron-rich. There can be electrostatic repulsion between the incoming nucleophile and the electron-rich aromatic ring, making the approach of the nucleophile less favourable compared to the attack on the partially positive carbon of an alkyl halide.

Despite their low reactivity towards nucleophilic substitution under normal conditions, nucleophilic substitution of halogens on an aromatic ring can be achieved under specific, harsh conditions or when the aromatic ring is activated by the presence of electron-withdrawing groups.

• Replacement by Hydroxyl group (Formation of Phenols):

  Chlorobenzene can be converted into phenol by heating it with an aqueous solution of sodium hydroxide (NaOH) at high temperature (around 623 K) and high pressure (around 300 atm). This is known as Dow's process.

  (The image would show the reaction: C$_6$H$_5$Cl + NaOH(aq) $\xrightarrow{623 \text{ K}, 300 \text{ atm}}$ C$_6$H$_5$ONa + HCl, followed by C$_6$H$_5$ONa + H$^+$ $\rightarrow$ C$_6$H$_5$OH).

• Effect of Electron-Withdrawing Groups:

  The presence of electron-withdrawing groups (EWGs) like the nitro group (—NO$_2$) on the aromatic ring, particularly at the ortho (o-) and para (p-) positions relative to the halogen, significantly increases the reactivity of aryl halides towards nucleophilic substitution. The more EWGs present at these positions, the greater the reactivity.

  For example, the reactivity order towards nucleophilic substitution for chlorobenzene and nitrophenols is:

  Chlorobenzene < p-Chloronitrobenzene < 2,4-Dichloronitrobenzene < 2,4,6-Trichloronitrobenzene (Picryl chloride)

  Reactions with activated aryl halides:

  p-Chloronitrobenzene reacts with NaOH at 443 K:

  2,4-Dichloronitrobenzene reacts with NaOH at 368 K:

  2,4,6-Trichloronitrobenzene reacts with H$_2$O on warming:

  The effect of the nitro group is strongest at the ortho and para positions. An EWG at the meta position has little or no activating effect on nucleophilic substitution.

  The mechanism involves the attack of the nucleophile on the carbon bearing the halogen, forming a negatively charged intermediate (a carbanion). This carbanion is resonance stabilised. The negative charge is delocalised onto the ring, specifically appearing on the carbon atoms at the ortho and para positions relative to the carbon undergoing substitution. Electron-withdrawing groups (like -NO$_2$) at the ortho or para positions can effectively stabilise this negative charge through resonance and inductive effects, making the intermediate more stable and thus lowering the activation energy for the reaction.

  (The image would show the mechanism for the reaction of p-chloronitrobenzene with OH$^-$: Attack by OH$^-$ forming a carbanion, resonance structures showing the negative charge delocalised including onto the carbon bearing the NO$_2$ group, expulsion of Cl$^-$ to form the product).

  When an EWG is at the meta position, the negative charge in the intermediate does not appear on the carbon atom directly attached to the EWG in any significant resonance structure. Therefore, EWGs at the meta position do not effectively stabilise the carbanion intermediate via resonance, and their inductive effect alone is less effective at activating the ring towards nucleophilic substitution compared to ortho/para substitution.

2. Electrophilic substitution reactions:

Similar to benzene, haloarenes undergo characteristic electrophilic substitution reactions on the aromatic ring. These include reactions like halogenation, nitration, sulfonation, and Friedel-Crafts reactions (alkylation and acylation).

Halogen atoms attached to an aromatic ring have two opposing effects on electrophilic substitution:

• Inductive effect (-I effect): Halogens are highly electronegative and withdraw electron density from the aromatic ring through the sigma bond. This electron withdrawal deactivates the ring towards electrophilic attack by making it less electron-rich.
• Resonance effect (+R effect): Halogens have lone pairs of electrons that can be donated into the aromatic ring via resonance, increasing electron density within the ring, particularly at the ortho and para positions. This effect is activating towards o,p positions.

Overall, the inductive effect (deactivating) is stronger than the resonance effect (activating for o,p). Therefore, haloarenes are slightly deactivated towards electrophilic substitution compared to benzene, and reactions usually require more vigorous conditions (higher temperatures or stronger catalysts).

However, because the resonance effect selectively increases electron density at the ortho and para positions, the incoming electrophile will preferentially attack these positions. Thus, halogens are ortho, para-directing groups.

The mechanism for electrophilic substitution involves the formation of a carbocation intermediate (a $\sigma$-complex). This intermediate is stabilised by resonance. When the attack occurs at the ortho or para position, the positive charge in the intermediate is delocalised onto carbons that are adjacent or further removed from the halogen-bearing carbon. The halogen, by resonance, can donate electron density (lone pair) to these positively charged carbons in the ortho and para intermediates, providing additional stabilisation. This stabilisation is not possible for attack at the meta position because the positive charge does not appear on the carbon atoms adjacent to the halogen-bearing carbon in the meta intermediate.

(The image would show the sigma complexes (carbocation intermediates) for ortho, meta, and para attack on a halobenzene, illustrating how the halogen's lone pair can contribute resonance structures only for ortho and para attack, stabilising the positive charge).

So, the stronger inductive effect makes the ring less reactive overall (deactivation), while the resonance effect controls *where* the substitution occurs (o,p-direction) by offering greater stabilisation to the intermediates formed from attack at these positions.

Specific Electrophilic Substitution Reactions of Haloarenes:

• (i) Halogenation: Halobenzenes react with halogens (e.g., Cl$_2$, Br$_2$) in the presence of a Lewis acid catalyst (FeCl$_3$, FeBr$_3$) to form dihalobenzenes, primarily a mixture of ortho- and para-isomers.

  (The image would show the reaction: Chlorobenzene + Cl$_2$ $\xrightarrow{\text{FeCl}_3}$ o-dichlorobenzene + p-dichlorobenzene).

• (ii) Nitration: Halobenzenes react with a nitrating mixture (conc. HNO$_3$ + conc. H$_2$SO$_4$) to form nitrohalobenzenes, mainly the ortho- and para-isomers.

  (The image would show the reaction: Chlorobenzene + HNO$_3$ $\xrightarrow{\text{conc. H}_2\text{SO}_4}$ o-chloronitrobenzene + p-chloronitrobenzene).

• (iii) Sulphonation: Halobenzenes react with concentrated sulfuric acid (often heated) to form halobenzenesulfonic acids, predominantly the ortho- and para-isomers.

  (The image would show the reaction: Chlorobenzene + conc. H$_2$SO$_4$ $\xrightarrow{\Delta}$ o-chlorobenzenesulfonic acid + p-chlorobenzenesulfonic acid).

• (iv) Friedel-Crafts reactions (Alkylation and Acylation): Halobenzenes undergo Friedel-Crafts alkylation with alkyl halides (R—X) and Friedel-Crafts acylation with acyl halides (RCO—X) or acid anhydrides, in the presence of a Lewis acid catalyst like anhydrous AlCl$_3$. A mixture of ortho- and para-substituted products is obtained.

  Friedel-Crafts Alkylation:

  (The image would show the reaction: Chlorobenzene + CH$_3$Cl $\xrightarrow{\text{anhydrous AlCl}_3}$ o-chlorotoluene + p-chlorotoluene).

  Friedel-Crafts Acylation:

  (The image would show the reaction: Chlorobenzene + CH$_3$COCl $\xrightarrow{\text{anhydrous AlCl}_3}$ o-chloroacetophenone + p-chloroacetophenone).

Answer:

3. Reaction with metals:

Like alkyl halides, aryl halides can also react with certain metals to form organometallic compounds or undergo coupling reactions.

• Wurtz-Fittig reaction:

  This reaction is a variation of the Wurtz reaction and is used to prepare alkylarenes (an alkyl chain attached to an aromatic ring). It involves treating a mixture of an alkyl halide and an aryl halide with sodium metal in dry ether.

  Ar—X + R—X + 2 Na $\xrightarrow{\text{dry ether}}$ Ar—R + 2 NaX

• Fittig reaction:

  This reaction is analogous to the Wurtz reaction but involves only aryl halides. When two molecules of an aryl halide are treated with sodium metal in dry ether, they couple to form a biaryl (or diphenyl) compound, where two aromatic rings are joined by a single bond.

  2 Ar—X + 2 Na $\xrightarrow{\text{dry ether}}$ Ar—Ar + 2 NaX

Polyhalogen Compounds

Organic compounds containing more than one halogen atom are termed polyhalogen compounds. Several of these compounds are important in industrial processes, agriculture, and have various practical uses, although some pose significant environmental and health concerns.

Dichloromethane (Methylene Chloride)

Chemical formula: CH$_2$Cl$_2$.

Uses: Widely used as a solvent, particularly as a paint remover. It is also used as a propellant in some aerosol sprays, as a process solvent in the manufacturing of drugs, and for cleaning and finishing metal surfaces.

Health Effects: Methylene chloride affects the human central nervous system (CNS). Low-level exposure can cause impaired hearing and vision. Higher concentrations in the air can lead to dizziness, nausea, tingling and numbness in extremities. Direct contact with the skin can cause severe burning and redness. Eye contact can damage the cornea.

Trichloromethane (Chloroform)

Chemical formula: CHCl$_3$.

Uses: Traditionally used as a solvent for various organic substances like fats, alkaloids, and iodine. Its primary industrial use today is in the production of the refrigerant Freon-22 (CHClF$_2$). It was historically used as a general anaesthetic in surgery but has been replaced due to safety concerns.

Health Effects: Inhaling chloroform vapours depresses the central nervous system, consistent with its former use as an anaesthetic. Exposure to even relatively low levels in air (e.g., 900 ppm) can cause dizziness, fatigue, and headache. Chronic exposure can damage the liver (as it is metabolised to the highly toxic phosgene) and kidneys. Skin exposure can cause sores.

Storage: Chloroform is sensitive to light and air. It slowly undergoes photo-oxidation in the presence of air and light to form phosgene (carbonyl chloride, COCl$_2$), a highly poisonous gas.

2 CHCl$_3$ + O$_2 \xrightarrow{\text{light}}$ 2 COCl$_2$ (phosgene) + 2 HCl

To prevent phosgene formation, chloroform is stored in dark-coloured, airtight bottles that are completely filled to exclude air.

Triiodomethane (Iodoform)

Chemical formula: CHI$_3$.

Uses: Previously used as an antiseptic. However, its antiseptic action is not due to the compound itself but results from the slow liberation of free iodine upon contact with skin or mucous membranes.

Note: Due to its strong, unpleasant odour, iodoform has been largely replaced by other iodine-containing antiseptic formulations.

Tetrachloromethane (Carbon Tetrachloride)

Chemical formula: CCl$_4$.

Uses: Used in large quantities for manufacturing refrigerants (like some freons) and propellants for aerosol cans. It serves as a feedstock in synthesising other chlorofluorocarbons and chemicals. It finds use in pharmaceutical manufacturing and as a general solvent. Until the mid-1960s, it was widely used as a cleaning solvent (degreasing agent) and in fire extinguishers, but these uses have declined due to health and environmental concerns.

Health Effects: Exposure to CCl$_4$ can cause liver cancer in humans and permanent damage to nerve cells. Common effects include dizziness, lightheadedness, nausea, and vomiting. Severe exposure can lead to stupor, coma, unconsciousness, or death. It can also cause irregular heartbeats or cardiac arrest. Skin and eye contact can cause irritation.

Environmental Impact: When released into the environment, CCl$_4$ is volatile and rises into the atmosphere. In the stratosphere, it contributes to the depletion of the ozone layer. Ozone depletion increases human exposure to harmful ultraviolet (UV) radiation, leading to higher risks of skin cancer, eye diseases (like cataracts), and potential harm to the immune system.

Freons

Freons are a group of chlorofluorocarbon compounds derived from methane (CH$_4$) and ethane (C$_2$H$_6$). They contain carbon, fluorine, and chlorine atoms (and sometimes hydrogen).

Properties: Freons are typically very stable, unreactive, non-toxic, non-corrosive, and easily liquefied gases. These properties made them attractive for various applications.

Example: Freon-12 (CCl$_2$F$_2$) is a common freon. It is manufactured from carbon tetrachloride by the Swarts reaction (halogen exchange with SbF$_3$ or other metallic fluorides).

CCl$_4$ + SbF$_3 \xrightarrow{\text{Heat}}$ CCl$_2$F$_2$ + SbCl$_3$ (unbalanced, simplified)

Uses: Historically, freons were widely used as refrigerants in refrigerators and air conditioners, and as propellants in aerosol cans. They were also used as solvents and in foam blowing.

Environmental Impact: Due to their stability and unreactivity, freons released into the atmosphere persist and eventually diffuse up into the stratosphere. In the stratosphere, UV radiation causes freons to break down, releasing chlorine free radicals ($\text{Cl}^{\cdot}$). These chlorine radicals catalyse the destruction of ozone molecules (O$_3$), disrupting the natural balance of the ozone layer. This depletion contributes to increased UV radiation reaching the Earth's surface.

P,P’-Dichlorodiphenyltrichloroethane( Ddt)

Chemical name: p,p'-Dichlorodiphenyltrichloroethane. Formula: (ClC$_6$H$_4$)$_2$CHCCl$_3$.

History: DDT was the first chlorinated organic insecticide. It was first synthesised in 1873, but its effectiveness as an insecticide was discovered much later in 1939 by Paul Muller. Muller was awarded the Nobel Prize in Medicine in 1948 for this discovery.

Uses: After World War II, DDT was used extensively worldwide, primarily for controlling insect-borne diseases like malaria (spread by mosquitoes) and typhus (spread by lice). Its effectiveness in public health made it a very important compound initially.

Problems and Environmental Impact: Problems associated with widespread DDT use emerged in the late 1940s.
1. Many insect species developed resistance to DDT over time.
2. It was found to have high toxicity towards fish and other aquatic life.
3. DDT is highly chemically stable and not easily broken down by microorganisms in the environment (persistence).
4. It is fat-soluble and not easily metabolised by animals. When animals ingest contaminated food, DDT accumulates in their fatty tissues (bioaccumulation) and can increase in concentration up the food chain (biomagnification). This can lead to harmful effects on wildlife (like thinning of bird eggshells) and potentially on humans.

Due to these concerns, the use of DDT has been banned in many countries, including the United States (since 1973), although it is still used in some parts of the world for controlling malaria.

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Question 10.5. Draw the structures of major monohalo products in each of the following reactions:

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Question 10.7. Which alkyl halide from the following pairs would you expect to react more rapidly by an $S_N2$ mechanism? Explain your answer.

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Question 10.8. In the following pairs of halogen compounds, which compound undergoes faster $S_N1$ reaction?

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Question 10.9. Identify A, B, C, D, E, R and R1 in the following:

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