Exhaustive Guide: Elimination & Haloarenes | Class 12 Chemistry | ChemCA

Elimination, Organometallics & Haloarenes

Module 3 | CBSE Class 12 Chemistry | Organic Chemistry

1. Elimination Reactions (β-Elimination)

When a haloalkane with a β-hydrogen atom is heated with an alcoholic solution of potassium hydroxide (alc. KOH), there is an elimination of a hydrogen atom from the β-carbon and a halogen atom from the α-carbon atom.

This results in the formation of an alkene. Because the β-hydrogen atom is involved, this reaction is often called β-elimination or dehydrohalogenation.

CH₃-CH₂-Br + KOH (alc.) →(heat) CH₂=CH₂ + KBr + H₂O

1.1 Saytzeff (Zaitsev) Rule

If an alkyl halide has more than one type of β-hydrogen, a mixture of alkenes is formed. The Russian chemist Alexander Zaitsev formulated a rule to predict the major product.

Saytzeff Rule: "In dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms." (The more highly substituted alkene is the major product).

Example: Dehydrohalogenation of 2-Bromobutane:
CH₃-CH₂-CH(Br)-CH₃ → (alc. KOH, heat)
1. CH₃-CH=CH-CH₃ (But-2-ene) - Major Product (81%) - Disubstituted alkene.
2. CH₃-CH₂-CH=CH₂ (But-1-ene) - Minor Product (19%) - Monosubstituted alkene.

1.2 Substitution vs. Elimination (Highly Tested)

A chemical reaction is often the result of a competition. Whether an alkyl halide undergoes substitution (S_N1/S_N2) or elimination depends on the nature of the nucleophile/base and the reaction conditions.

Aqueous KOH vs. Alcoholic KOH:
- Aqueous KOH (aq. KOH): KOH dissolves in water to give OH^- ions. OH^- is a strong nucleophile and easily replaces the halogen to give Alcohols (Substitution).
- Alcoholic KOH (alc. KOH): KOH dissolves in alcohol to produce alkoxide ions (e.g., C₂H₅O^-). Alkoxide ions are much stronger bases than OH^-. They powerfully abstract a proton (H⁺) from the β-carbon, leading to Alkenes (Elimination).

2. Reaction with Metals

2.1 Organometallic Compounds: Grignard Reagents

Most organic chlorides, bromides, and iodides react with certain metals to give compounds containing carbon-metal bonds. Such compounds are known as organometallic compounds.

Victor Grignard discovered an extremely important class of these compounds: Alkyl magnesium halides, commonly known as Grignard Reagents (R-Mg-X).

CH₃CH₂Br + Mg →(Dry ether) CH₃CH₂MgBr

The C-Mg bond is covalent but highly polar (Carbon is electronegative compared to Magnesium), while the Mg-X bond is essentially ionic.

Why must Grignard Reagents be prepared under strictly anhydrous conditions?
Grignard reagents are highly reactive. They react with any source of protons (like water, alcohols, amines) to form corresponding hydrocarbons. Even a trace amount of moisture will destroy the reagent by converting it into an alkane.
Reaction: R-Mg-X + H₂O → R-H (Alkane) + Mg(OH)X

2.2 Wurtz Reaction

Alkyl halides react with sodium in dry ether to give higher alkanes containing double the number of carbon atoms present in the original halide.

2R-X + 2Na →(Dry ether) R-R + 2NaX

3. Reactions of Haloarenes (Nucleophilic Substitution)

3.1 Why are Haloarenes Extremely Unreactive? (Board Favorite)

Aryl halides are extremely less reactive towards nucleophilic substitution reactions than alkyl halides due to four main reasons:

Resonance Effect: The lone pairs on the halogen atom are in conjugation with the π-electrons of the benzene ring. This imparts a partial double bond character to the C-X bond, making it shorter and much stronger (difficult to break) than the C-X single bond in haloalkanes.
Difference in Hybridization: In haloalkanes, the carbon is sp³ hybridized. In haloarenes, the carbon is sp² hybridized. sp² hybridized carbon has more s-character, making it more electronegative. Thus, it holds the electron pair of the C-X bond more tightly, resisting cleavage.
Instability of Phenyl Cation: In the S_N1 mechanism, cleavage would result in a phenyl cation. The phenyl cation is highly unstable because the positive charge is on an sp² hybridized carbon and cannot be stabilized by resonance. Hence, the S_N1 mechanism is ruled out.
Electronic Repulsions: It is difficult for an electron-rich nucleophile to approach the electron-rich arene ring due to repulsion.

3.2 Replacement by Hydroxyl Group (Dow's Process)

Because of their unreactivity, replacing the halogen requires extreme conditions. Chlorobenzene can be converted to phenol by heating it with aqueous NaOH at 623 K and 300 atm pressure.

Effect of Electron Withdrawing Groups (EWG): The presence of an electron withdrawing group (like -NO₂) at ortho and/or para positions vastly increases the reactivity of haloarenes. It stabilizes the intermediate carbanion through resonance.
Note: The presence of an EWG at the meta position has practically no effect on reactivity because the negative charge in the resonance hybrid does not appear on the meta carbon.

4. Electrophilic Substitution Reactions

Haloarenes undergo the usual electrophilic substitution reactions of the benzene ring: Halogenation, Nitration, Sulfonation, and Friedel-Crafts reactions.

4.1 Directive Influence of Halogens (The Clash of Effects)

Halogen atoms are highly electronegative. Therefore, they have a strong -I effect (inductive effect), which withdraws electrons from the benzene ring, deactivating it. Consequently, electrophilic substitution in haloarenes occurs slowly and requires drastic conditions compared to benzene.

However, halogens also possess lone pairs that can be donated to the ring via the +R effect (resonance effect). Resonance structures show that electron density increases specifically at the ortho and para positions.

Conclusion: Halogens are deactivating groups (due to -I > +R), but they are ortho-para directing (because +R effect stabilizes the intermediate carbocation specifically at ortho/para positions).

5. Reaction of Haloarenes with Metals

A. Wurtz-Fittig Reaction:
A mixture of an alkyl halide and an aryl halide reacts with sodium in dry ether to give an alkylarene.

C₆H₅X + R-X + 2Na →(Ether) C₆H₅-R (Alkylbenzene) + 2NaX

B. Fittig Reaction:
Aryl halides alone react with sodium in dry ether to form analogous compounds where two aryl groups are joined together.

2 C₆H₅X + 2Na →(Ether) C₆H₅-C₆H₅ (Biphenyl) + 2NaX

6. NCERT Solved Examples (Step-by-Step)

NCERT Example 10.9: Although chlorine is an electron withdrawing group, yet it is ortho-, para- directing in electrophilic aromatic substitution reactions. Why?

Solution:
Chlorine shows two opposing effects:
1. -I Effect (Inductive): It withdraws electrons due to high electronegativity, thereby deactivating the ring overall making electrophilic attack slower than in benzene.
2. +R Effect (Resonance): It donates its lone pair to the benzene ring. When an electrophile attacks at the ortho or para position, the intermediate carbocation (sigma complex) formed is resonance-stabilized by the lone pair of the chlorine atom. This specific stabilization does not occur if the attack is at the meta position.
Since the +R effect dictates the orientation by stabilizing the intermediate, chlorine directs the incoming electrophile to the ortho and para positions.

7. Previous Year Questions (PYQs) & Exhaustive Question Bank

Part A: Conceptual (1-2 Marks)

[CBSE 2018, 2020]

Q1. Why is Grignard reagent prepared under strictly anhydrous conditions?

Answer: Grignard reagents (R-Mg-X) are highly reactive and act as strong bases. If even a trace amount of moisture (H₂O) is present, the Grignard reagent will immediately react with the acidic hydrogen of water to form an alkane, thereby destroying the reagent.
Reaction: RMgX + H₂O → R-H + Mg(OH)X

[CBSE 2017, 2019]

Q2. Account for the following: Aryl halides are extremely less reactive towards nucleophilic substitution reactions than alkyl halides.

Answer: This is mainly due to resonance. The lone pair of electrons on the halogen atom is in conjugation with the pi-electrons of the benzene ring. This imparts a partial double bond character to the C-X bond, making it shorter, stronger, and much more difficult to cleave than the purely single C-X bond in alkyl halides. Additionally, the sp² hybridized carbon of the ring holds the electron pair more tightly.

Part B: Assertion-Reason Type (1 Mark)

[CBSE Sample Paper 2023]

Q3. Assertion (A): Treatment of chloroethane with aqueous KOH gives ethanol, while with alcoholic KOH it gives ethene.
Reason (R): Aqueous KOH acts as a strong nucleophile, whereas alcoholic KOH acts as a strong base.

Answer: Both Assertion and Reason are correct, and Reason is the correct explanation for Assertion. In aqueous solution, KOH provides OH^- which substitutes the halogen (S_N2). In an alcoholic solution, alkoxide ions (RO^-) are formed which are much stronger bases and abstract a β-proton, leading to an elimination reaction (dehydrohalogenation).

Part C: Conversions & Applications (3 Marks)

[CBSE 2016, 2022]

Q4. Write the major products in the following reactions:
(a) 2-Bromopentane is treated with alcoholic KOH.
(b) Chlorobenzene is treated with Sodium in dry ether.
(c) Bromobenzene is treated with CH₃Br and Na in dry ether.

Answer:
(a) According to Saytzeff's rule, the more highly substituted alkene is the major product. Dehydrohalogenation of 2-Bromopentane yields Pent-2-ene as the major product.
(b) This is the Fittig reaction. Two molecules of chlorobenzene couple in the presence of sodium to form Biphenyl.
(c) This is the Wurtz-Fittig reaction. Bromobenzene and bromomethane couple to form an alkylarene, specifically Toluene (Methylbenzene).

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Exhaustive Guide: Elimination & Haloarenes | Class 12 Chemistry