Haloalkanes and Haloarenes: Class 12 Chemistry NCERT Guide

Classification and Nomenclature of Haloalkanes and Haloarenes

Understanding how to classify and name these compounds is the first step to mastering them.

Classification of Haloalkanes and Haloarenes:
We can classify them in two main ways:

1. Based on the number of halogen atoms:

• Mono, Di, or Polyhalogen Compounds: Depending on whether they contain one, two, or more halogen atoms. For example, C₂H₅Cl is a monohaloethane, while CH₂Cl₂ is a dihalomethane.
• Dihaloalkanes can be further classified as geminal (halogen atoms on the same carbon, e.g., 1,1-dichloroethane) or vicinal (halogen atoms on adjacent carbons, e.g., 1,2-dichloroethane).

1. Based on the hybridization of the carbon atom attached to the halogen (C-X bond):

• Compounds containing sp³ C-X bond:
• Alkyl Halides (Haloalkanes): The halogen is bonded to an sp³-hybridized carbon atom of an alkyl group. They are classified as primary (1°), secondary (2°), or tertiary (3°) based on the nature of this carbon.
• Allylic Halides: The halogen is bonded to an sp³-hybridized carbon atom next to a carbon-carbon double bond (C=C). Ex: CH₂=CH-CH₂Cl.
• Benzylic Halides: The halogen is bonded to an sp³-hybridized carbon atom attached to an aromatic ring. Ex: C₆H₅CH₂Cl.
• Compounds containing sp² C-X bond:
• Vinylic Halides: The halogen is bonded to an sp²-hybridized carbon of a C=C double bond. Ex: CH₂=CHCl.
• Aryl Halides (Haloarenes): The halogen is directly bonded to an sp²-hybridized carbon atom of an aromatic ring. Ex: C₆H₅Cl (Chlorobenzene).

Key Definitions in Haloalkanes and Haloarenes

Haloalkane (Alkyl Halide)

An organic compound derived from an alkane by replacing at least one hydrogen atom with a halogen atom (F, Cl, Br, I). The halogen is bonded to an sp³-hybridized carbon.

Haloarene (Aryl Halide)

An organic compound where a halogen atom is directly attached to an sp²-hybridized carbon atom of an aromatic ring.

Nucleophilic Substitution Reaction

A reaction in which a nucleophile (an electron-rich species) replaces a leaving group (like a halogen) on an atom. The main reactions for haloalkanes are SN1 and SN2.

Chirality

A geometric property of some molecules. A chiral molecule is one that is non-superimposable on its mirror image. The presence of an asymmetric carbon atom (a carbon bonded to four different groups) is a common cause of chirality.

Grignard Reagent

An organometallic compound with the general formula R-Mg-X, where R is an alkyl or aryl group and X is a halogen. They are highly reactive and are used to form new carbon-carbon bonds.

Worked Examples: SN1 and SN2 Mechanisms

• Example 1: The SN1 Mechanism (Hydrolysis of tert-Butyl Bromide) Let's analyze the reaction of tert-butyl bromide with a weak nucleophile like water: (CH₃)₃C-Br + H₂O → (CH₃)₃C-OH + HBr. This reaction follows the SN1 (Substitution Nucleophilic Unimolecular) mechanism, which occurs in two steps. Step 1: Formation of a Carbocation (Rate-Determining Step) The C-Br bond breaks heterolytically, with the bromine atom taking both electrons. This step is slow and reversible. It forms a stable tertiary (3°) carbocation. (CH₃)₃C-Br (slow) ⇌ (CH₃)₃C⁺ + Br⁻ Tutor Tip: This step is the slowest because it requires energy to break a covalent bond. The stability of the carbocation is key; 3° carbocations are most stable due to hyperconjugation and inductive effects, which is why SN1 is favored for tertiary halides. Step 2: Attack of the Nucleophile The nucleophile (H₂O) is weak, but it can now quickly attack the positively charged carbocation from either side. (CH₃)₃C⁺ + H₂O (fast) → (CH₃)₃C-OH₂⁺ Step 3: Deprotonation A final, fast step where a water molecule removes a proton to give the final product, tert-butanol. (CH₃)₃C-OH₂⁺ + H₂O → (CH₃)₃C-OH + H₃O⁺ Final Answer: The reaction proceeds via an SN1 mechanism, forming a stable tertiary carbocation intermediate.
• Example 2: The SN2 Mechanism (Hydrolysis of Methyl Bromide) Now, consider the reaction of a primary haloalkane like methyl bromide with a strong nucleophile like hydroxide ion: CH₃-Br + OH⁻ → CH₃-OH + Br⁻. This follows the SN2 (Substitution Nucleophilic Bimolecular) mechanism, which is a single-step process. Mechanism: A Concerted, Single Step Both the bond-breaking (C-Br) and bond-making (C-OH) occur simultaneously. The nucleophile (OH⁻) attacks the carbon atom from the side opposite to the leaving group (Br⁻). This is called backside attack. [HO⁻] + CH₃-Br → [HO---CH₃---Br]⁻ (Transition State) → HO-CH₃ + [Br⁻] Tutor Tip: The structure in the middle is a transition state, not an intermediate. It's a high-energy, unstable state where the carbon is momentarily bonded to five groups. Because the attack happens from the back, the configuration of the molecule gets inverted, like an umbrella flipping inside out in a strong wind. This is known as Walden Inversion. Why SN2 for primary halides? Primary halides like methyl bromide have very little steric hindrance. This means there's plenty of space for the nucleophile to approach and attack the carbon atom. Tertiary halides are too bulky for this backside attack, which is why they prefer the SN1 pathway. Final Answer: The reaction proceeds via a concerted SN2 mechanism involving a single transition state and results in an inversion of configuration.

Exam Traps: SN1 vs SN2 and Elimination Reactions

One of the most common areas of confusion for students is deciding whether a reaction will be SN1, SN2, E1, or E2. Here’s how to think like an expert:

• Substrate is King: The structure of the haloalkane is the most important factor.
• Primary (1°): Strongly favors SN2. Never SN1 (primary carbocation is too unstable).
• Tertiary (3°): Strongly favors SN1 (forms stable carbocation). Cannot undergo SN2 due to steric hindrance.
• Secondary (2°): This is the battleground. Both SN1 and SN2 are possible. You must look at other factors.

• Nucleophile Strength & Concentration:
• Strong Nucleophiles (like OH⁻, RO⁻, CN⁻) favor SN2.
• Weak Nucleophiles (like H₂O, ROH) favor SN1.

• The Solvent Matters:
• Polar Protic Solvents (like water, alcohol) can stabilize both the carbocation and the leaving group, favoring SN1.
• Polar Aprotic Solvents (like acetone, DMSO) favor SN2 because they don't solvate the nucleophile as strongly, leaving it 'naked' and more reactive.

• Substitution vs. Elimination: Remember that elimination reactions are always in competition with substitution. High temperatures and the use of strong, bulky bases (like potassium tert-butoxide) strongly favor elimination over substitution.

Practice Questions with Solutions

• Q: Give the IUPAC name of the following compound: CH₃-CH(Br)-CH₂-CH(Cl)-CH₃ A: Step 1: Identify the longest carbon chain. The longest chain has 5 carbon atoms, so the parent alkane is pentane. Step 2: Number the chain to give the substituents the lowest possible locants. Numbering from the left gives substituents at positions 2 (Bromo) and 4 (Chloro). Numbering from the right gives 2 (Chloro) and 4 (Bromo). The sum of locants is 2+4=6 in both cases. Step 3: When locants are a tie, give the lower number to the group that comes first alphabetically. 'Bromo' comes before 'Chloro'. Therefore, numbering starts from the left. Final answer: The correct IUPAC name is 2-Bromo-4-chloropentane.
• Q: Predict the major product when 2-bromopentane is treated with alcoholic KOH. Name the rule that governs the product formation. A: Step 1: Identify the reactants and reaction type. 2-bromopentane is a secondary haloalkane. Alcoholic KOH is a strong base, and the use of alcohol as a solvent along with heating favors elimination reaction (dehydrohalogenation) over substitution. Step 2: Identify possible products. The β-hydrogens (hydrogens on carbons adjacent to the carbon with bromine) can be eliminated. There are two types of β-hydrogens: on C1 and on C3. - Elimination of H from C1 gives Pent-1-ene. - Elimination of H from C3 gives Pent-2-ene. Step 3: Apply the relevant rule. According to Zaitsev's (or Saytzeff's) rule, in dehydrohalogenation reactions, the preferred product is the alkene that has the greater number of alkyl groups attached to the doubly bonded carbon atoms (i.e., the more substituted alkene is more stable and is the major product). Step 4: Determine the major product. Pent-2-ene is a more substituted alkene than Pent-1-ene. Final answer: The major product is Pent-2-ene. The reaction is governed by Zaitsev's rule.
• Q: Why is the boiling point of bromobenzene higher than that of toluene, even though they have nearly the same molecular mass? A: Step 1: Identify the compounds and their molecular masses. Bromobenzene (C₆H₅Br) has a mass of approx. 157 g/mol. Toluene (C₆H₅CH₃) has a mass of approx. 92 g/mol. The question's premise is slightly off, as their masses aren't that close, but the concept is about intermolecular forces. Let's compare bromobenzene (157 u) and, for a better comparison, fluorobenzene (96 u) which is closer to toluene (92 u). Step 2: Analyze the intermolecular forces in each molecule. Toluene is a nonpolar molecule (or very weakly polar), and the primary intermolecular forces are weak van der Waals forces (dispersion forces). Step 3: Analyze the intermolecular forces in bromobenzene. The C-Br bond is polar due to the difference in electronegativity between carbon and bromine. This creates a permanent dipole in the molecule. Therefore, in addition to van der Waals forces, bromobenzene molecules experience stronger dipole-dipole interactions. Step 4: Relate intermolecular forces to boiling point. Stronger intermolecular forces require more energy (higher temperature) to overcome, leading to a higher boiling point. The dipole-dipole interactions in bromobenzene are stronger than the weak van der Waals forces in toluene. Final answer: Bromobenzene is a polar molecule and exhibits stronger dipole-dipole interactions compared to the weak van der Waals forces in the nearly nonpolar toluene. Stronger intermolecular forces lead to a higher boiling point.
• Q: How would you distinguish between Chloroform (CHCl₃) and Carbon Tetrachloride (CCl₄)? A: Step 1: Analyze the structure of both compounds. Chloroform has a C-H bond, while Carbon Tetrachloride does not. Step 2: Choose a chemical test that can differentiate them based on their structure. The carbylamine test (or isocyanide test) is a specific test for primary amines, but its first step involves reacting chloroform with a primary amine and alcoholic KOH. A key reaction of chloroform is its hydrolysis with aqueous KOH, which produces formate salts, but a more distinctive test is the Carbylamine reaction itself, where chloroform is a reactant. Step 3: A better distinguishing test is the reaction of Chloroform with a primary amine (like aniline) in the presence of alcoholic KOH. Chloroform gives a foul-smelling isocyanide (carbylamine). CHCl₃ + R-NH₂ + 3KOH → R-NC (foul smell) + 3KCl + 3H₂O Step 4: Test the second compound. Carbon tetrachloride (CCl₄) does not have the necessary structure to undergo the carbylamine reaction. It will not produce a foul-smelling product under the same conditions. Final answer: Chloroform can be distinguished from carbon tetrachloride by the carbylamine test. When warmed with aniline and alcoholic KOH, chloroform produces phenyl isocyanide, which has a very unpleasant odour. Carbon tetrachloride does not give this test.

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